WEBVTT 00:00:00.000 --> 00:00:01.944 align:middle line:90% [SQUEAKING] 00:00:01.944 --> 00:00:02.916 align:middle line:90% [RUSTLING] 00:00:02.916 --> 00:00:04.860 align:middle line:90% [CLICKING] 00:00:04.860 --> 00:00:13.247 align:middle line:90% 00:00:13.247 --> 00:00:15.080 align:middle line:84% CASEY RODRIGUEZ: Last time, so we're talking 00:00:15.080 --> 00:00:17.930 align:middle line:90% about a set of real numbers. 00:00:17.930 --> 00:00:22.970 align:middle line:84% And last time, I stated the following theorem 00:00:22.970 --> 00:00:26.180 align:middle line:84% about the existence and properties of R 00:00:26.180 --> 00:00:28.110 align:middle line:90% that make it special. 00:00:28.110 --> 00:00:50.650 align:middle line:84% So the theorem is, there exists a unique ordered field 00:00:50.650 --> 00:01:02.110 align:middle line:84% with the least upper bound property containing 00:01:02.110 --> 00:01:03.110 align:middle line:90% the rational numbers. 00:01:03.110 --> 00:01:07.870 align:middle line:84% So remember, a field was a set that 00:01:07.870 --> 00:01:12.850 align:middle line:84% had operations plus and multiplication. 00:01:12.850 --> 00:01:17.080 align:middle line:84% An ordered field also this set has an order on it. 00:01:17.080 --> 00:01:23.770 align:middle line:84% And this order interacts with the operations 00:01:23.770 --> 00:01:26.590 align:middle line:84% of addition and multiplication in natural ways. 00:01:26.590 --> 00:01:28.120 align:middle line:84% And the least upper bound property 00:01:28.120 --> 00:01:32.560 align:middle line:84% means every non-empty subset which is bounded above 00:01:32.560 --> 00:01:37.090 align:middle line:90% has a supremum in the set. 00:01:37.090 --> 00:01:44.730 align:middle line:90% 00:01:44.730 --> 00:01:47.900 align:middle line:84% So least upper bound property, meaning the supremum 00:01:47.900 --> 00:01:51.420 align:middle line:84% belongs in R, not necessarily in the non-empty set, 00:01:51.420 --> 00:01:52.430 align:middle line:90% which is bounded above. 00:01:52.430 --> 00:01:54.980 align:middle line:90% 00:01:54.980 --> 00:01:57.680 align:middle line:84% And we saw last time that q, the rational numbers, 00:01:57.680 --> 00:01:59.300 align:middle line:90% does not have this property. 00:01:59.300 --> 00:02:03.950 align:middle line:84% By looking at that set, q positive rational q 00:02:03.950 --> 00:02:06.110 align:middle line:90% squared less than 2. 00:02:06.110 --> 00:02:08.030 align:middle line:84% That set was bounded above but didn't 00:02:08.030 --> 00:02:10.400 align:middle line:84% have a supremum in rational numbers, 00:02:10.400 --> 00:02:12.950 align:middle line:84% because the square root of 2 essentially 00:02:12.950 --> 00:02:14.950 align:middle line:90% is not a rational number. 00:02:14.950 --> 00:02:18.560 align:middle line:84% And the square root of 2 would be the supremum. 00:02:18.560 --> 00:02:22.340 align:middle line:84% So just based on this theorem which characterizes 00:02:22.340 --> 00:02:26.390 align:middle line:84% and constructs R, the set of real numbers, 00:02:26.390 --> 00:02:30.200 align:middle line:84% we're now going to prove facts about real numbers 00:02:30.200 --> 00:02:33.740 align:middle line:90% and soon turn to limits. 00:02:33.740 --> 00:02:36.290 align:middle line:84% So as I said in the beginning, limits 00:02:36.290 --> 00:02:41.480 align:middle line:84% is or are the central object of study in analysis. 00:02:41.480 --> 00:02:43.640 align:middle line:84% That's what analysis is, the study of limits. 00:02:43.640 --> 00:02:46.580 align:middle line:84% The real analysis part, or the real, 00:02:46.580 --> 00:02:48.410 align:middle line:84% and that is that we're doing this 00:02:48.410 --> 00:02:50.165 align:middle line:84% within the setting of the real numbers. 00:02:50.165 --> 00:02:56.480 align:middle line:90% 00:02:56.480 --> 00:03:00.920 align:middle line:84% So now, let me just point out something kind of simple 00:03:00.920 --> 00:03:03.330 align:middle line:90% about the real numbers. 00:03:03.330 --> 00:03:10.790 align:middle line:84% So I mean, extremely simple fact about the real numbers 00:03:10.790 --> 00:03:13.730 align:middle line:84% is it's not discrete like the integers are. 00:03:13.730 --> 00:03:17.170 align:middle line:84% So for the integers, if I take one integer and another, 00:03:17.170 --> 00:03:18.920 align:middle line:84% it's not necessarily the case that there's 00:03:18.920 --> 00:03:20.480 align:middle line:84% an integer strictly in between them. 00:03:20.480 --> 00:03:22.520 align:middle line:84% There's not an integer between 0 and 1. 00:03:22.520 --> 00:03:25.910 align:middle line:90% 00:03:25.910 --> 00:03:29.060 align:middle line:84% And but R does satisfy this property, 00:03:29.060 --> 00:03:31.670 align:middle line:84% essentially because it's a field. 00:03:31.670 --> 00:03:33.980 align:middle line:90% So and it's an ordered field. 00:03:33.980 --> 00:03:41.300 align:middle line:84% So what's the simple fact, if x, y are real numbers, 00:03:41.300 --> 00:03:47.390 align:middle line:84% and x is less than y, then there exists a real number 00:03:47.390 --> 00:03:55.670 align:middle line:84% r, a little r, so that x is less than r is less than y. 00:03:55.670 --> 00:04:03.350 align:middle line:84% And I can give you r explicitly. r is equal to x plus y over 2. 00:04:03.350 --> 00:04:05.890 align:middle line:90% 00:04:05.890 --> 00:04:08.950 align:middle line:84% All right, that's not so surprising. 00:04:08.950 --> 00:04:12.530 align:middle line:84% Now, this statement here is also true 00:04:12.530 --> 00:04:15.550 align:middle line:84% if I replace r with rational numbers. 00:04:15.550 --> 00:04:20.260 align:middle line:84% Namely if I had two rational numbers, 1 less than the other, 00:04:20.260 --> 00:04:23.590 align:middle line:84% then there exists a rational number in between that's 00:04:23.590 --> 00:04:25.580 align:middle line:90% between x and y. 00:04:25.580 --> 00:04:28.360 align:middle line:84% I just again, define r in this way. 00:04:28.360 --> 00:04:37.940 align:middle line:84% Now, a natural question is the following, 00:04:37.940 --> 00:04:47.700 align:middle line:84% is that if x and y are in R, and x is less than y, 00:04:47.700 --> 00:04:49.640 align:middle line:90% then does there-- 00:04:49.640 --> 00:04:51.710 align:middle line:84% I'll put a question mark over that-- does there 00:04:51.710 --> 00:04:59.360 align:middle line:84% exist an R in q such that x is less than R is less than y? 00:04:59.360 --> 00:05:03.540 align:middle line:84% I can't necessarily define R by this formula here 00:05:03.540 --> 00:05:07.950 align:middle line:84% and guaranteed that R will be rational. 00:05:07.950 --> 00:05:18.180 align:middle line:84% So for example, if x equals root 2, 00:05:18.180 --> 00:05:26.880 align:middle line:84% and let's say y equals 2 root 2, then x plus y over 2 00:05:26.880 --> 00:05:32.780 align:middle line:84% equals 3/2 root 2, which is not a rational number. 00:05:32.780 --> 00:05:34.530 align:middle line:84% Because if it were a rational number, then 00:05:34.530 --> 00:05:38.070 align:middle line:84% I can multiply 2 through by 2/3 and say square root of 2 00:05:38.070 --> 00:05:39.340 align:middle line:90% is a rational number. 00:05:39.340 --> 00:05:43.650 align:middle line:84% So all of that to say, that simply 00:05:43.650 --> 00:05:46.650 align:middle line:84% by doing this trick of just taking the average does not 00:05:46.650 --> 00:05:49.080 align:middle line:84% necessarily mean that if I take two real numbers, 00:05:49.080 --> 00:05:51.630 align:middle line:84% 1 less than the other, then there exists a rational 00:05:51.630 --> 00:05:53.400 align:middle line:90% in between them. 00:05:53.400 --> 00:05:57.580 align:middle line:90% That's not so clear to see. 00:05:57.580 --> 00:05:59.790 align:middle line:90% But this is one of the most-- 00:05:59.790 --> 00:06:03.000 align:middle line:84% this is one of the basic facts about R, 00:06:03.000 --> 00:06:06.840 align:middle line:84% is that the answer to this question is, yes. 00:06:06.840 --> 00:06:09.060 align:middle line:84% And that in some sense the rational numbers 00:06:09.060 --> 00:06:11.310 align:middle line:90% are dense in the real numbers. 00:06:11.310 --> 00:06:13.350 align:middle line:84% For any to real numbers, I can find a rational 00:06:13.350 --> 00:06:15.420 align:middle line:90% in between them. 00:06:15.420 --> 00:06:21.720 align:middle line:84% It's the first main property of the real numbers we'll prove. 00:06:21.720 --> 00:06:29.852 align:middle line:84% And it'll be a consequence of a second property we'll prove, 00:06:29.852 --> 00:06:31.560 align:middle line:84% which is called the Archimedean property. 00:06:31.560 --> 00:06:34.390 align:middle line:90% So this theorem has two parts. 00:06:34.390 --> 00:06:44.150 align:middle line:84% Its first is called Archimedean property of R. 00:06:44.150 --> 00:06:54.050 align:middle line:84% And the statement of this is, if x and y are in R, 00:06:54.050 --> 00:07:03.200 align:middle line:84% and x is positive, then there exists a natural number N, 00:07:03.200 --> 00:07:07.760 align:middle line:90% such that nx is bigger than y. 00:07:07.760 --> 00:07:16.520 align:middle line:90% 00:07:16.520 --> 00:07:19.310 align:middle line:84% And then the second part of this theorem 00:07:19.310 --> 00:07:22.880 align:middle line:84% is the statement that the rationales are dense. 00:07:22.880 --> 00:07:24.860 align:middle line:84% That the answer to this question is, yes. 00:07:24.860 --> 00:07:31.190 align:middle line:84% So this usually goes by the name of the density 00:07:31.190 --> 00:07:34.130 align:middle line:90% of the rationales. 00:07:34.130 --> 00:07:43.020 align:middle line:84% So it states that if x and y and in R, and x is less than y, 00:07:43.020 --> 00:07:50.300 align:middle line:84% then there exists a rational number little r, 00:07:50.300 --> 00:07:53.330 align:middle line:84% such that x is less than r is less than y. 00:07:53.330 --> 00:08:00.220 align:middle line:90% 00:08:00.220 --> 00:08:02.800 align:middle line:84% So let's prove the first theorem, 00:08:02.800 --> 00:08:05.840 align:middle line:84% the first part of this theorem, the Archimedean property. 00:08:05.840 --> 00:08:10.090 align:middle line:84% So what do we have to use to prove this theorem? 00:08:10.090 --> 00:08:12.640 align:middle line:84% Just what we know about the real numbers, the fact 00:08:12.640 --> 00:08:15.977 align:middle line:84% that it's an ordered field with the least upper bound property. 00:08:15.977 --> 00:08:18.310 align:middle line:84% And you'll see how this least upper bound property plays 00:08:18.310 --> 00:08:22.900 align:middle line:84% a major role in all of these elementary things we 00:08:22.900 --> 00:08:27.020 align:middle line:90% prove about R. 00:08:27.020 --> 00:08:33.470 align:middle line:90% So let's restate our hypothesis. 00:08:33.470 --> 00:08:38.360 align:middle line:90% So suppose x and y are in r. 00:08:38.360 --> 00:08:40.559 align:middle line:90% And x is less than-- 00:08:40.559 --> 00:08:42.215 align:middle line:90% and x is bigger than 0. 00:08:42.215 --> 00:08:47.450 align:middle line:90% 00:08:47.450 --> 00:08:48.935 align:middle line:90% So we wish to show-- 00:08:48.935 --> 00:08:54.850 align:middle line:90% 00:08:54.850 --> 00:08:56.920 align:middle line:84% so just restating this inequality here, 00:08:56.920 --> 00:08:57.876 align:middle line:90% we wish to show-- 00:08:57.876 --> 00:09:01.450 align:middle line:90% 00:09:01.450 --> 00:09:05.410 align:middle line:84% the word show, you should read as synonymous with prove-- 00:09:05.410 --> 00:09:10.600 align:middle line:84% that there exists a natural number N, 00:09:10.600 --> 00:09:14.425 align:middle line:84% such that n is bigger than y over x. 00:09:14.425 --> 00:09:17.030 align:middle line:90% 00:09:17.030 --> 00:09:18.570 align:middle line:90% That's just restating that. 00:09:18.570 --> 00:09:20.720 align:middle line:84% So we're going to prove this by contradiction. 00:09:20.720 --> 00:09:23.400 align:middle line:90% 00:09:23.400 --> 00:09:28.500 align:middle line:84% So the proof will go by contradiction. 00:09:28.500 --> 00:09:34.850 align:middle line:84% So this is, again, this is our hypothesis, 00:09:34.850 --> 00:09:36.700 align:middle line:84% which we will assume throughout the proof. 00:09:36.700 --> 00:09:39.400 align:middle line:84% The thing we're trying to prove is this statement here, 00:09:39.400 --> 00:09:40.690 align:middle line:90% the second sentence. 00:09:40.690 --> 00:09:43.450 align:middle line:84% So this is the thing that we will negate. 00:09:43.450 --> 00:09:45.850 align:middle line:84% We will not negate the hypothesis, 00:09:45.850 --> 00:09:48.970 align:middle line:84% we're negating what we want to prove in the end. 00:09:48.970 --> 00:09:52.220 align:middle line:84% And then arriving at a false statement. 00:09:52.220 --> 00:09:56.420 align:middle line:84% Therefore showing our conclusion is true. 00:09:56.420 --> 00:09:59.510 align:middle line:90% 00:09:59.510 --> 00:10:02.960 align:middle line:84% So we assume that the second statement is false. 00:10:02.960 --> 00:10:05.100 align:middle line:84% Again, we're assuming the first statement. 00:10:05.100 --> 00:10:09.700 align:middle line:90% 00:10:09.700 --> 00:10:13.915 align:middle line:90% So suppose not, i.e. 00:10:13.915 --> 00:10:18.730 align:middle line:90% 00:10:18.730 --> 00:10:25.750 align:middle line:84% for all N a natural number, n is less than or equal to y over x. 00:10:25.750 --> 00:10:30.250 align:middle line:84% That's the negation of the fact of there exists an integer 00:10:30.250 --> 00:10:31.750 align:middle line:90% that's bigger than y over x. 00:10:31.750 --> 00:10:39.000 align:middle line:90% 00:10:39.000 --> 00:10:46.090 align:middle line:84% So then that means then the natural numbers 00:10:46.090 --> 00:10:54.250 align:middle line:84% as a subset of the real numbers is bounded above. 00:10:54.250 --> 00:11:03.780 align:middle line:90% 00:11:03.780 --> 00:11:06.870 align:middle line:84% Now, the set of natural numbers is a non-empty subset 00:11:06.870 --> 00:11:08.100 align:middle line:90% of R, which is bounded above. 00:11:08.100 --> 00:11:11.400 align:middle line:90% Therefore, it has a supremum. 00:11:11.400 --> 00:11:25.370 align:middle line:84% So by the least upper bound property of R, 00:11:25.370 --> 00:11:39.460 align:middle line:84% N has a supremum, call it a in R. 00:11:39.460 --> 00:11:54.070 align:middle line:84% So since a is the supremum of N, anything smaller than a cannot 00:11:54.070 --> 00:11:56.020 align:middle line:84% be an upper bound for the natural numbers. 00:11:56.020 --> 00:11:58.840 align:middle line:90% 00:11:58.840 --> 00:12:02.050 align:middle line:84% Because remember, a is supposed to be the least upper bound. 00:12:02.050 --> 00:12:05.950 align:middle line:84% Anything smaller than that cannot be an upper bound 00:12:05.950 --> 00:12:08.770 align:middle line:90% for the natural numbers. 00:12:08.770 --> 00:12:20.120 align:middle line:90% Is not an upper bound for N. 00:12:20.120 --> 00:12:21.920 align:middle line:90% But what does that mean? 00:12:21.920 --> 00:12:24.220 align:middle line:84% So if a minus 1 is not an upper bound for N, 00:12:24.220 --> 00:12:28.540 align:middle line:84% that means there must exist some integer in N which is strictly 00:12:28.540 --> 00:12:32.245 align:middle line:90% bigger than a minus 1. 00:12:32.245 --> 00:12:34.180 align:middle line:84% There exists a natural number, I'm 00:12:34.180 --> 00:12:43.090 align:middle line:84% going to call it m, such that a minus 1 is less than m. 00:12:43.090 --> 00:12:48.490 align:middle line:84% But then, this implies that a is less than m 00:12:48.490 --> 00:12:56.070 align:middle line:84% plus 1, which implies that a is not 00:12:56.070 --> 00:13:03.620 align:middle line:84% an upper bound for the natural numbers. 00:13:03.620 --> 00:13:07.290 align:middle line:84% Because m plus 1-- m is an integer, a natural number. 00:13:07.290 --> 00:13:08.810 align:middle line:90% So m plus 1's a natural number. 00:13:08.810 --> 00:13:11.480 align:middle line:84% And I've just found a natural number bigger than a. 00:13:11.480 --> 00:13:14.780 align:middle line:84% That means a can't be an upper bound for the natural numbers. 00:13:14.780 --> 00:13:20.450 align:middle line:84% But and therefore a does not equal the sup of N. 00:13:20.450 --> 00:13:22.700 align:middle line:90% And this is a contradiction. 00:13:22.700 --> 00:13:33.440 align:middle line:84% Because A is defined as a sup of N. 00:13:33.440 --> 00:13:35.630 align:middle line:84% So from the assumption, just to recap, 00:13:35.630 --> 00:13:40.760 align:middle line:84% just from the assumption that for all natural numbers 00:13:40.760 --> 00:13:43.940 align:middle line:84% N, N is less than or equal to y over x. 00:13:43.940 --> 00:13:46.250 align:middle line:84% We concluded that it has a supremum that's 00:13:46.250 --> 00:13:48.020 align:middle line:90% not its supremum. 00:13:48.020 --> 00:13:49.370 align:middle line:90% So that's a false statement. 00:13:49.370 --> 00:13:52.490 align:middle line:84% And therefore, our initial assumption 00:13:52.490 --> 00:13:56.060 align:middle line:84% that for all natural numbers N, N 00:13:56.060 --> 00:13:58.740 align:middle line:84% is less than or equal to y over x, that must be false. 00:13:58.740 --> 00:14:01.010 align:middle line:84% And therefore, there exist an N, which 00:14:01.010 --> 00:14:04.580 align:middle line:84% is bigger than y over x, which is what we wanted to prove. 00:14:04.580 --> 00:14:07.750 align:middle line:90% 00:14:07.750 --> 00:14:13.810 align:middle line:84% So for the proof of the second theorem, the density 00:14:13.810 --> 00:14:17.320 align:middle line:90% of rational numbers, we'll do-- 00:14:17.320 --> 00:14:19.075 align:middle line:84% so we have three cases to consider. 00:14:19.075 --> 00:14:23.110 align:middle line:90% 00:14:23.110 --> 00:14:30.790 align:middle line:84% So both x and y are real numbers. 00:14:30.790 --> 00:14:31.885 align:middle line:90% x is less than y. 00:14:31.885 --> 00:14:37.560 align:middle line:90% 00:14:37.560 --> 00:14:47.570 align:middle line:84% There are three cases, call it A, 00:14:47.570 --> 00:14:50.750 align:middle line:90% x is less than 0 is less than y. 00:14:50.750 --> 00:14:56.240 align:middle line:84% B, 0 is bigger than or equal to x is less than y. 00:14:56.240 --> 00:15:02.480 align:middle line:84% And C, x is less than y is less than or equal to 0. 00:15:02.480 --> 00:15:07.490 align:middle line:84% So we want to find a rational number between x and y. 00:15:07.490 --> 00:15:09.930 align:middle line:84% So for this case, this is pretty simple. 00:15:09.930 --> 00:15:11.630 align:middle line:90% We just take R to be 0. 00:15:11.630 --> 00:15:16.340 align:middle line:84% So I'm not going to say anything about case A. I'm just 00:15:16.340 --> 00:15:28.980 align:middle line:90% going to move on to case B. 00:15:28.980 --> 00:15:41.800 align:middle line:84% So case B. So suppose x is bigger than or equal to 0 00:15:41.800 --> 00:15:45.080 align:middle line:90% and less than y. 00:15:45.080 --> 00:15:51.760 align:middle line:90% Then by AP I'll refer to-- 00:15:51.760 --> 00:15:55.690 align:middle line:84% So that'll be shorthand notation for the Archimedean property. 00:15:55.690 --> 00:15:59.170 align:middle line:84% So part one, which we've already proven is true. 00:15:59.170 --> 00:16:03.130 align:middle line:84% By the Archimedean property, there exists a natural number 00:16:03.130 --> 00:16:11.910 align:middle line:84% N, such that N times y minus x is bigger than 1. 00:16:11.910 --> 00:16:20.220 align:middle line:90% 00:16:20.220 --> 00:16:25.535 align:middle line:84% Now, we're going to use Archimedean property again. 00:16:25.535 --> 00:16:33.780 align:middle line:90% 00:16:33.780 --> 00:16:36.720 align:middle line:84% There exists an integer l and a natural number 00:16:36.720 --> 00:16:44.610 align:middle line:84% l, such that l is bigger than n times x. 00:16:44.610 --> 00:16:51.090 align:middle line:90% 00:16:51.090 --> 00:16:56.850 align:middle line:90% Thus, the set of all integer-- 00:16:56.850 --> 00:17:00.510 align:middle line:84% of natural numbers k, such that k 00:17:00.510 --> 00:17:06.780 align:middle line:90% is bigger than nx is non-empty. 00:17:06.780 --> 00:17:10.650 align:middle line:90% 00:17:10.650 --> 00:17:12.670 align:middle line:84% So this is a subset of the natural numbers. 00:17:12.670 --> 00:17:15.839 align:middle line:90% So it has a least element. 00:17:15.839 --> 00:17:24.270 align:middle line:84% By the well ordering property of the natural numbers, 00:17:24.270 --> 00:17:31.520 align:middle line:90% S has a least element m. 00:17:31.520 --> 00:17:35.990 align:middle line:90% 00:17:35.990 --> 00:17:38.570 align:middle line:90% Now what does this mean? 00:17:38.570 --> 00:17:40.640 align:middle line:90% What does m have to satisfy? 00:17:40.640 --> 00:17:44.060 align:middle line:84% What we're going to show is that m over n 00:17:44.060 --> 00:17:50.000 align:middle line:84% is in fact, a rational number that is between x and y. 00:17:50.000 --> 00:17:54.030 align:middle line:90% So the goal is, I guess-- 00:17:54.030 --> 00:17:59.010 align:middle line:84% so let me just make a comment here. 00:17:59.010 --> 00:18:01.460 align:middle line:90% So once we've made-- 00:18:01.460 --> 00:18:05.420 align:middle line:84% once we've found this natural number, little n, 00:18:05.420 --> 00:18:08.077 align:middle line:84% so that n times y minus x is bigger than 1, 00:18:08.077 --> 00:18:08.910 align:middle line:90% what does this mean? 00:18:08.910 --> 00:18:15.590 align:middle line:84% This means n times y is bigger than nx plus 1. 00:18:15.590 --> 00:18:18.170 align:middle line:90% So our goal is then to find-- 00:18:18.170 --> 00:18:22.040 align:middle line:90% 00:18:22.040 --> 00:18:25.310 align:middle line:84% so this stuff in brackets, you should not 00:18:25.310 --> 00:18:26.610 align:middle line:90% consider as part of the proof. 00:18:26.610 --> 00:18:31.850 align:middle line:84% This is me trying to explain to you where from here 00:18:31.850 --> 00:18:34.520 align:middle line:84% we would like to go to conclude the proof that there exists 00:18:34.520 --> 00:18:36.770 align:middle line:84% a rational number between y and x. 00:18:36.770 --> 00:18:40.220 align:middle line:84% So we have this inequality, the Archimedean property 00:18:40.220 --> 00:18:42.930 align:middle line:84% tells us there's a natural number that satisfies this. 00:18:42.930 --> 00:18:50.460 align:middle line:84% So what we would like to do is find another natural number. 00:18:50.460 --> 00:18:55.530 align:middle line:84% And I'm foreshadowing what's to come-- a natural number m-- 00:18:55.530 --> 00:18:56.850 align:middle line:90% well, let's not-- 00:18:56.850 --> 00:18:59.937 align:middle line:84% I don't want you to give you any false hopes. 00:18:59.937 --> 00:19:01.020 align:middle line:90% I mean, they're not false. 00:19:01.020 --> 00:19:02.062 align:middle line:90% It'll happen in a minute. 00:19:02.062 --> 00:19:06.530 align:middle line:90% But let's call this j. 00:19:06.530 --> 00:19:09.890 align:middle line:90% So that two things happen. 00:19:09.890 --> 00:19:13.280 align:middle line:90% n times x is less than j. 00:19:13.280 --> 00:19:25.770 align:middle line:84% And j is less than or equal to nx plus 1. 00:19:25.770 --> 00:19:30.486 align:middle line:84% So if we're able to find such an integer j satisfying this, 00:19:30.486 --> 00:19:34.680 align:middle line:84% these two inequalities, then from this inequality 00:19:34.680 --> 00:19:38.415 align:middle line:84% we would get that ny is bigger than-- 00:19:38.415 --> 00:19:40.970 align:middle line:90% 00:19:40.970 --> 00:19:44.720 align:middle line:84% or this is bigger than nx plus 1, 00:19:44.720 --> 00:19:50.180 align:middle line:84% which is bigger than or equal to j, which is bigger than nx. 00:19:50.180 --> 00:19:52.370 align:middle line:90% I.e. 00:19:52.370 --> 00:19:55.460 align:middle line:84% let me just rewrite this. nx is less than j, 00:19:55.460 --> 00:19:58.250 align:middle line:90% is less than n times y. 00:19:58.250 --> 00:20:03.170 align:middle line:84% Or in other words, x is less than j over m is less than y. 00:20:03.170 --> 00:20:06.215 align:middle line:84% And here's our rational number which we would choose. 00:20:06.215 --> 00:20:12.220 align:middle line:90% 00:20:12.220 --> 00:20:15.970 align:middle line:84% So that's the game plan, which maybe I 00:20:15.970 --> 00:20:19.690 align:middle line:84% should have said right after I came up 00:20:19.690 --> 00:20:21.970 align:middle line:90% with this integer little n. 00:20:21.970 --> 00:20:24.790 align:middle line:84% You may be asking why I come up with this integer little n. 00:20:24.790 --> 00:20:26.740 align:middle line:84% Well, we have to start somewhere. 00:20:26.740 --> 00:20:29.110 align:middle line:84% And the Archimedean property gives us this. 00:20:29.110 --> 00:20:32.620 align:middle line:84% And this somehow gives us a scale to at least a scale 00:20:32.620 --> 00:20:35.630 align:middle line:90% 1 over n to kind of work on. 00:20:35.630 --> 00:20:38.690 align:middle line:84% But anyway, so going back to the proof. 00:20:38.690 --> 00:20:40.760 align:middle line:84% So we have by the well ordering property 00:20:40.760 --> 00:20:42.860 align:middle line:84% of the natural numbers, this set S, 00:20:42.860 --> 00:20:44.820 align:middle line:84% which is a set of all natural numbers 00:20:44.820 --> 00:20:50.000 align:middle line:84% so that k is bigger than nx has a least element little m. 00:20:50.000 --> 00:20:55.190 align:middle line:90% 00:20:55.190 --> 00:21:02.390 align:middle line:84% Now m, so S has a least element m, means m is in S. 00:21:02.390 --> 00:21:09.590 align:middle line:84% So since m is in S, this means simply 00:21:09.590 --> 00:21:17.480 align:middle line:84% by the definition of what S is, nx is less than m. 00:21:17.480 --> 00:21:18.980 align:middle line:90% That's very good. 00:21:18.980 --> 00:21:20.720 align:middle line:84% That's one of the properties we wanted-- 00:21:20.720 --> 00:21:23.210 align:middle line:84% I wrote it as j, but we wanted an integer to satisfy. 00:21:23.210 --> 00:21:27.248 align:middle line:90% 00:21:27.248 --> 00:21:48.620 align:middle line:84% Since m is the least element of S, m minus 1 is not in S. 00:21:48.620 --> 00:21:55.760 align:middle line:84% So that implies that m minus 1 not in S means m minus 1 00:21:55.760 --> 00:21:59.580 align:middle line:84% is less than or equal to n times x, i.e. 00:21:59.580 --> 00:22:03.386 align:middle line:90% 00:22:03.386 --> 00:22:07.490 align:middle line:84% n is less than or equal to n times x plus 1. 00:22:07.490 --> 00:22:13.130 align:middle line:84% Now, we'll combine these two inequalities along 00:22:13.130 --> 00:22:17.470 align:middle line:84% with the first one involving n times y minus x. 00:22:17.470 --> 00:22:19.490 align:middle line:84% So I'm just basically rewriting what 00:22:19.490 --> 00:22:21.965 align:middle line:84% I told you our goal was here in this bracket. 00:22:21.965 --> 00:22:27.540 align:middle line:90% 00:22:27.540 --> 00:22:33.090 align:middle line:90% Then, n times x is less than m. 00:22:33.090 --> 00:22:36.030 align:middle line:84% And this is less than or equal to n times 00:22:36.030 --> 00:22:39.750 align:middle line:84% x plus 1, which by our definition of n 00:22:39.750 --> 00:22:43.110 align:middle line:84% in the first inequality we have up there involving y 00:22:43.110 --> 00:22:45.180 align:middle line:90% is less than n times y. 00:22:45.180 --> 00:22:48.547 align:middle line:84% So we have nx is less than m, is less than y. 00:22:48.547 --> 00:22:54.765 align:middle line:84% So that means x is less than m over n is this less than y. 00:22:54.765 --> 00:22:57.420 align:middle line:90% 00:22:57.420 --> 00:23:10.680 align:middle line:84% So r equals m over n, which is a proportion of natural numbers, 00:23:10.680 --> 00:23:16.508 align:middle line:90% is our choice. 00:23:16.508 --> 00:23:18.300 align:middle line:84% That's an awkward way to finish a sentence. 00:23:18.300 --> 00:23:20.490 align:middle line:90% But my mind went blank. 00:23:20.490 --> 00:23:25.890 align:middle line:84% Anyways, so that handles the case B, 00:23:25.890 --> 00:23:27.810 align:middle line:84% that x is bigger than or equal to 0. 00:23:27.810 --> 00:23:36.260 align:middle line:90% 00:23:36.260 --> 00:23:41.520 align:middle line:90% So one small thing that I-- 00:23:41.520 --> 00:23:42.960 align:middle line:90% just a very minor comment. 00:23:42.960 --> 00:23:45.730 align:middle line:84% If you've got all this, and you understand it, that's fine. 00:23:45.730 --> 00:23:48.270 align:middle line:84% But you may be asking since I never actually said it 00:23:48.270 --> 00:23:50.092 align:middle line:84% out loud, where did I use the fact that x 00:23:50.092 --> 00:23:51.300 align:middle line:90% is bigger than or equal to 0? 00:23:51.300 --> 00:24:28.330 align:middle line:90% 00:24:28.330 --> 00:24:32.590 align:middle line:84% So where I used that-- so let's just assume x is bigger than 0. 00:24:32.590 --> 00:24:47.650 align:middle line:84% So what where did I use is really right in this place 00:24:47.650 --> 00:24:48.150 align:middle line:90% here. 00:24:48.150 --> 00:24:50.250 align:middle line:84% So I don't want to spend too much time on that. 00:24:50.250 --> 00:25:11.850 align:middle line:84% But by being able to claim that m minus 1 is not in S. 00:25:11.850 --> 00:25:14.640 align:middle line:90% So what about case C? 00:25:14.640 --> 00:25:29.020 align:middle line:84% So case C, we'll just reduce to case B. Suppose 00:25:29.020 --> 00:25:34.040 align:middle line:84% x is less than y, which is less than or equal to 0. 00:25:34.040 --> 00:25:40.630 align:middle line:84% Then 0 is bigger than or equal to minus y, which 00:25:40.630 --> 00:25:42.910 align:middle line:90% is bigger than minus x. 00:25:42.910 --> 00:25:56.110 align:middle line:84% So by case B, there exists a rational number, R tilde, 00:25:56.110 --> 00:26:04.420 align:middle line:84% such that minus y is less than R tilde is less than minus x. 00:26:04.420 --> 00:26:12.510 align:middle line:90% 00:26:12.510 --> 00:26:22.820 align:middle line:84% Then this implies that x is less than minus R 00:26:22.820 --> 00:26:25.130 align:middle line:90% tilde is less than y. 00:26:25.130 --> 00:26:34.550 align:middle line:84% So R equal to minus R tilde does the job. 00:26:34.550 --> 00:26:36.110 align:middle line:90% So that proves the theorem. 00:26:36.110 --> 00:26:45.383 align:middle line:90% 00:26:45.383 --> 00:26:46.800 align:middle line:84% So we'll use this in just a minute 00:26:46.800 --> 00:26:55.710 align:middle line:84% to prove a simple statement about sups and infs. 00:26:55.710 --> 00:27:09.550 align:middle line:84% But before I use it, let me go on. 00:27:09.550 --> 00:27:14.680 align:middle line:84% And what I'd like to do is state a different way 00:27:14.680 --> 00:27:20.870 align:middle line:84% to verify that a number is a sup of a set or the inf of a set. 00:27:20.870 --> 00:27:24.340 align:middle line:84% So I'm just going to state it for sups. 00:27:24.340 --> 00:27:26.830 align:middle line:84% And there's an analogous statement for infs. 00:27:26.830 --> 00:27:30.820 align:middle line:90% So the theorem is this. 00:27:30.820 --> 00:27:32.860 align:middle line:84% And this will actually be on the assignment. 00:27:32.860 --> 00:27:52.040 align:middle line:84% So assume S is a subset of R is non-empty and bounded above. 00:27:52.040 --> 00:27:54.260 align:middle line:90% So S has a supremum. 00:27:54.260 --> 00:27:56.750 align:middle line:84% So I'm going to tell you what the supremum satisfies. 00:27:56.750 --> 00:27:59.090 align:middle line:84% And it's an if and only if statement. 00:27:59.090 --> 00:28:09.890 align:middle line:84% Then some number x equals a sup of S if and only if-- 00:28:09.890 --> 00:28:12.500 align:middle line:84% so either I'll do a double arrow, 00:28:12.500 --> 00:28:16.340 align:middle line:90% or I'll write if and only if-- 00:28:16.340 --> 00:28:18.870 align:middle line:90% two conditions are satisfied. 00:28:18.870 --> 00:28:31.660 align:middle line:84% The first is that x is an upper bound for S. 00:28:31.660 --> 00:28:36.350 align:middle line:84% And the second, so at one point, or in the very near future, 00:28:36.350 --> 00:28:38.320 align:middle line:84% we're going to be seeing epsilons and deltas. 00:28:38.320 --> 00:28:40.850 align:middle line:90% So I'll state it this way. 00:28:40.850 --> 00:28:44.380 align:middle line:84% So you should start seeing them now. 00:28:44.380 --> 00:28:50.500 align:middle line:84% For every epsilon positive, there exists an element y 00:28:50.500 --> 00:28:56.290 align:middle line:84% in S such that x minus epsilon is less 00:28:56.290 --> 00:28:59.095 align:middle line:84% than y is less than or equal to x. 00:28:59.095 --> 00:29:02.570 align:middle line:90% 00:29:02.570 --> 00:29:05.870 align:middle line:90% So why is this-- 00:29:05.870 --> 00:29:08.850 align:middle line:90% why should this be the case? 00:29:08.850 --> 00:29:16.070 align:middle line:84% Well, if S is all over here to the left of x, then 00:29:16.070 --> 00:29:18.680 align:middle line:84% and if I take something smaller, so I'm 00:29:18.680 --> 00:29:23.070 align:middle line:84% actually kind of giving you a picture proof of one direction. 00:29:23.070 --> 00:29:28.250 align:middle line:84% And so, if I take something smaller, x minus epsilon, 00:29:28.250 --> 00:29:31.010 align:middle line:84% then this cannot be an upper bound for S. 00:29:31.010 --> 00:29:34.940 align:middle line:84% So therefore, there must exist some y in S 00:29:34.940 --> 00:29:36.440 align:middle line:90% bigger than x minus epsilon. 00:29:36.440 --> 00:29:37.820 align:middle line:84% And it's less than or equal to x, 00:29:37.820 --> 00:29:39.890 align:middle line:84% because x is an upper bound for S. 00:29:39.890 --> 00:29:44.340 align:middle line:84% So in fact, this picture is a proof of this direction, 00:29:44.340 --> 00:29:47.270 align:middle line:84% namely if x is the supremum of S, 00:29:47.270 --> 00:29:51.150 align:middle line:84% then these two conditions are satisfied. 00:29:51.150 --> 00:29:52.715 align:middle line:84% And then, on the assignment I'll have 00:29:52.715 --> 00:29:56.990 align:middle line:84% you prove the other direction, basically, 00:29:56.990 --> 00:30:01.400 align:middle line:84% that these two conditions imply that a real number is 00:30:01.400 --> 00:30:08.710 align:middle line:90% the supremum of S. 00:30:08.710 --> 00:30:14.290 align:middle line:84% So let's use this theorem and Archimedean 00:30:14.290 --> 00:30:17.440 align:middle line:84% property to prove a simple statement 00:30:17.440 --> 00:30:21.370 align:middle line:90% about the sup of a simple set. 00:30:21.370 --> 00:30:23.950 align:middle line:84% So I should-- so let me also write here-- 00:30:23.950 --> 00:30:28.987 align:middle line:90% 00:30:28.987 --> 00:30:30.820 align:middle line:84% that I'm not going to state it as a theorem. 00:30:30.820 --> 00:30:34.990 align:middle line:84% But I'll just-- this is a remark meaning I'm being a little bit 00:30:34.990 --> 00:30:37.060 align:middle line:84% loose with what I'm writing down, but it's true-- 00:30:37.060 --> 00:30:39.910 align:middle line:90% 00:30:39.910 --> 00:30:41.920 align:middle line:90% x equals inf of S. 00:30:41.920 --> 00:30:47.080 align:middle line:84% So for S, a non-empty subset which is bounded below, 00:30:47.080 --> 00:30:49.900 align:middle line:84% there's an analogous characterization 00:30:49.900 --> 00:30:55.480 align:middle line:90% as x is a lower bound for s. 00:30:55.480 --> 00:31:02.410 align:middle line:84% And for all epsilon positive, there exists a y in S, 00:31:02.410 --> 00:31:05.920 align:middle line:84% so that x is less than or equal to y is less than x 00:31:05.920 --> 00:31:07.630 align:middle line:90% plus epsilon. 00:31:07.630 --> 00:31:10.720 align:middle line:84% So this is the analogous statement for the inf, 00:31:10.720 --> 00:31:12.086 align:middle line:90% for something to be the inf. 00:31:12.086 --> 00:31:18.740 align:middle line:90% 00:31:18.740 --> 00:31:20.810 align:middle line:84% So this is, I guess you could call it a theorem. 00:31:20.810 --> 00:31:22.850 align:middle line:84% It's not a very spectacular theorem. 00:31:22.850 --> 00:31:31.090 align:middle line:84% But if I look at the set 1 minus 1 over N, in a natural number, 00:31:31.090 --> 00:31:39.040 align:middle line:84% and I take its sup, this is equal to 1. 00:31:39.040 --> 00:31:42.490 align:middle line:84% So this shouldn't surprise you too much. 00:31:42.490 --> 00:31:43.690 align:middle line:90% Because what is this set? 00:31:43.690 --> 00:31:47.440 align:middle line:90% 00:31:47.440 --> 00:31:58.000 align:middle line:84% This is the set 0, 1/2, 2/3, 3/4 4/5, and so on. 00:31:58.000 --> 00:31:59.890 align:middle line:84% So you see 1 is always bigger than 00:31:59.890 --> 00:32:01.400 align:middle line:90% or equal to everything in here. 00:32:01.400 --> 00:32:03.040 align:middle line:84% So it's certainly an upper bound. 00:32:03.040 --> 00:32:05.950 align:middle line:84% And everything is kind of progressively getting closer 00:32:05.950 --> 00:32:06.730 align:middle line:90% to 1. 00:32:06.730 --> 00:32:09.190 align:middle line:84% So it should satisfy the second property. 00:32:09.190 --> 00:32:11.470 align:middle line:84% Namely, if I go a little bit to the left of 1, 00:32:11.470 --> 00:32:14.620 align:middle line:84% then I'll be able to find something less than 00:32:14.620 --> 00:32:18.580 align:middle line:84% or equal to 1 and bigger than that thing to the left of 1. 00:32:18.580 --> 00:32:23.170 align:middle line:84% And that'll be how we prove this theorem. 00:32:23.170 --> 00:32:26.260 align:middle line:90% 00:32:26.260 --> 00:32:38.460 align:middle line:84% So first off, since 1 minus 1 over n is less than 1, 00:32:38.460 --> 00:32:42.930 align:middle line:84% for all natural members N, this implies 00:32:42.930 --> 00:32:47.350 align:middle line:84% 1 is an upper bound for this set. 00:32:47.350 --> 00:32:55.300 align:middle line:90% 00:32:55.300 --> 00:32:58.590 align:middle line:84% Now we're going to verify that one satisfies 00:32:58.590 --> 00:33:00.090 align:middle line:84% that second property of the theorem, 00:33:00.090 --> 00:33:05.190 align:middle line:84% that for every epsilon positive, I can find a natural number 00:33:05.190 --> 00:33:07.650 align:middle line:84% little n, so that 1 minus 1 over n 00:33:07.650 --> 00:33:11.480 align:middle line:90% is bigger than 1 minus epsilon. 00:33:11.480 --> 00:33:13.480 align:middle line:84% And we'll use the Archimedean property for that. 00:33:13.480 --> 00:33:20.860 align:middle line:90% 00:33:20.860 --> 00:33:27.370 align:middle line:84% So if you're ever supposed to show 00:33:27.370 --> 00:33:29.600 align:middle line:84% something for every epsilon positive, 00:33:29.600 --> 00:33:35.680 align:middle line:84% then every proof should start off with epsilon be positive. 00:33:35.680 --> 00:33:38.450 align:middle line:84% In fact, I'll give you a couple of points on the exam 00:33:38.450 --> 00:33:44.470 align:middle line:84% if there's epsilon delta m's whatever proofs, which 00:33:44.470 --> 00:33:46.960 align:middle line:84% means you should prove something for all epsilon-- 00:33:46.960 --> 00:33:49.600 align:middle line:84% I'll give you a couple of points on the exam 00:33:49.600 --> 00:33:52.070 align:middle line:84% if you just at least state let epsilon be positive. 00:33:52.070 --> 00:33:54.370 align:middle line:90% So let epsilon be positive. 00:33:54.370 --> 00:34:00.200 align:middle line:90% 00:34:00.200 --> 00:34:06.710 align:middle line:90% So then, there exists-- 00:34:06.710 --> 00:34:16.600 align:middle line:84% so by the Archimedean property, there exist a natural number N, 00:34:16.600 --> 00:34:24.260 align:middle line:84% such that 1 over epsilon is less than N. This is taking, 00:34:24.260 --> 00:34:29.699 align:middle line:84% for example, in the statement of the Archimedean property, 00:34:29.699 --> 00:34:34.305 align:middle line:84% if you like, it's taking x to be 1 and y to be 1 over epsilon. 00:34:34.305 --> 00:34:42.110 align:middle line:90% 00:34:42.110 --> 00:34:47.550 align:middle line:90% Then 1 minus 1 over n-- 00:34:47.550 --> 00:34:52.210 align:middle line:90% 00:34:52.210 --> 00:34:56.739 align:middle line:90% so let's do it this way. 00:34:56.739 --> 00:34:59.820 align:middle line:90% Then 1 minus epsilon is-- 00:34:59.820 --> 00:35:03.420 align:middle line:90% so since epsilon is less than-- 00:35:03.420 --> 00:35:05.850 align:middle line:84% 1 over epsilon is less than n, this 00:35:05.850 --> 00:35:09.690 align:middle line:84% is equivalent to saying 1 over n is less than epsilon. 00:35:09.690 --> 00:35:12.300 align:middle line:90% 00:35:12.300 --> 00:35:18.260 align:middle line:84% And therefore, minus epsilon is less than minus 1 over n. 00:35:18.260 --> 00:35:21.200 align:middle line:84% So then, 1 minus epsilon is less than 1 minus 1 00:35:21.200 --> 00:35:25.460 align:middle line:84% over n, which is as we have up here, is less than 1. 00:35:25.460 --> 00:35:29.390 align:middle line:90% 00:35:29.390 --> 00:35:33.290 align:middle line:84% Thus, there exists a natural number N, so that-- 00:35:33.290 --> 00:35:43.830 align:middle line:90% 00:35:43.830 --> 00:35:46.810 align:middle line:90% so that went kind of quick. 00:35:46.810 --> 00:35:49.050 align:middle line:84% So what I did was, I found a natural number 00:35:49.050 --> 00:35:50.910 align:middle line:84% N, which is bigger than 1 over epsilon. 00:35:50.910 --> 00:35:53.910 align:middle line:84% And from now on, typically, I won't state it like this, 00:35:53.910 --> 00:35:56.680 align:middle line:90% I'll state it more like this. 00:35:56.680 --> 00:36:00.997 align:middle line:90% Which this follows from this. 00:36:00.997 --> 00:36:03.330 align:middle line:84% So when I make a statement like, choose a natural number 00:36:03.330 --> 00:36:07.240 align:middle line:84% so that 1 over n is less than epsilon, 00:36:07.240 --> 00:36:08.990 align:middle line:84% that follows from the Archimedean property 00:36:08.990 --> 00:36:11.630 align:middle line:84% just by taking 1 over this inequality. 00:36:11.630 --> 00:36:14.940 align:middle line:90% 00:36:14.940 --> 00:36:15.440 align:middle line:90% And so. 00:36:15.440 --> 00:36:17.900 align:middle line:84% From there we concluded that 1 minus epsilon 00:36:17.900 --> 00:36:21.620 align:middle line:84% is less than 1 minus 1 over n, which is still less than 1. 00:36:21.620 --> 00:36:25.640 align:middle line:84% And therefore, we have proven the second-- 00:36:25.640 --> 00:36:29.510 align:middle line:90% number two property of 1. 00:36:29.510 --> 00:36:32.060 align:middle line:84% And thus by the theorem, 1 is equal to the supremum 00:36:32.060 --> 00:36:32.840 align:middle line:90% of this set. 00:36:32.840 --> 00:36:51.030 align:middle line:90% 00:36:51.030 --> 00:36:53.910 align:middle line:84% So let's get a little more familiarity 00:36:53.910 --> 00:36:59.573 align:middle line:84% with using sups and infs, and in particular using 00:36:59.573 --> 00:37:00.990 align:middle line:84% that theorem, which I stated right 00:37:00.990 --> 00:37:04.050 align:middle line:84% there, about this characterization of the sup 00:37:04.050 --> 00:37:06.930 align:middle line:84% as being an upper bound and satisfying the second property 00:37:06.930 --> 00:37:09.012 align:middle line:84% that for every epsilon positive you 00:37:09.012 --> 00:37:10.470 align:middle line:84% can find something in the set which 00:37:10.470 --> 00:37:15.100 align:middle line:90% is bigger than x minus epsilon. 00:37:15.100 --> 00:37:17.925 align:middle line:84% So to do that, let's look at a couple of different sets-- 00:37:17.925 --> 00:37:24.870 align:middle line:90% 00:37:24.870 --> 00:37:26.530 align:middle line:84% types of subsets of real numbers. 00:37:26.530 --> 00:37:36.150 align:middle line:84% So for x a real number and a subset, 00:37:36.150 --> 00:37:40.230 align:middle line:90% we define two sets, x plus a-- 00:37:40.230 --> 00:37:42.600 align:middle line:90% this is just a shift of a by x. 00:37:42.600 --> 00:37:47.640 align:middle line:84% So this is a set of all elements of the form x plus little a, 00:37:47.640 --> 00:37:52.950 align:middle line:84% where little a is a capital A. x times a, 00:37:52.950 --> 00:37:59.838 align:middle line:84% this is a set of all elements of the form x times a, 00:37:59.838 --> 00:38:02.960 align:middle line:84% where a is in capital A. So from now 00:38:02.960 --> 00:38:04.970 align:middle line:84% on basically towards the end, until almost 00:38:04.970 --> 00:38:07.340 align:middle line:84% the end of the course, we're working as in subsets 00:38:07.340 --> 00:38:08.460 align:middle line:90% of the real numbers. 00:38:08.460 --> 00:38:12.110 align:middle line:84% So these things are meaningful, plus and multiplication. 00:38:12.110 --> 00:38:15.540 align:middle line:90% 00:38:15.540 --> 00:38:22.420 align:middle line:84% So we have the following theorem. 00:38:22.420 --> 00:38:25.840 align:middle line:84% So let's-- let me state the assumptions. 00:38:25.840 --> 00:38:29.200 align:middle line:84% And then we'll probably guess the conclusions. 00:38:29.200 --> 00:38:39.420 align:middle line:84% If x is a real number, and A is bounded above, 00:38:39.420 --> 00:38:49.360 align:middle line:84% then the conclusion is x plus a is bounded above. 00:38:49.360 --> 00:38:54.040 align:middle line:84% And the supremum of this set x plus A 00:38:54.040 --> 00:38:59.230 align:middle line:84% is equal to x plus the supremum of A. 00:38:59.230 --> 00:39:00.760 align:middle line:90% We should be able to guess this. 00:39:00.760 --> 00:39:08.380 align:middle line:84% Because let's assume A is this interval here. 00:39:08.380 --> 00:39:12.610 align:middle line:90% That would make the sup A-- 00:39:12.610 --> 00:39:15.310 align:middle line:84% this, let's say right endpoint of this interval, 00:39:15.310 --> 00:39:22.070 align:middle line:84% then when I shift everything by x, then this point 00:39:22.070 --> 00:39:26.540 align:middle line:84% sup A shifts to x plus sup A, which 00:39:26.540 --> 00:39:31.120 align:middle line:84% should be the supremum of x plus capital A. 00:39:31.120 --> 00:39:35.260 align:middle line:90% So this is not too surprising. 00:39:35.260 --> 00:39:37.300 align:middle line:84% There are surprising theorems in analysis. 00:39:37.300 --> 00:39:41.560 align:middle line:84% We already saw of one, at least, I thought it was surprising. 00:39:41.560 --> 00:39:44.380 align:middle line:84% Hopefully you found it surprising, 00:39:44.380 --> 00:39:46.330 align:middle line:84% about the cardinality of the power set 00:39:46.330 --> 00:39:50.500 align:middle line:84% compared to the cardinality of the original set. 00:39:50.500 --> 00:39:53.560 align:middle line:84% You come up with this strange-looking-- 00:39:53.560 --> 00:39:57.010 align:middle line:84% or the proof of that, remember, you came up with this 00:39:57.010 --> 00:40:01.000 align:middle line:84% strange-looking set which you had to-- 00:40:01.000 --> 00:40:07.590 align:middle line:84% which in the end basically referenced itself 00:40:07.590 --> 00:40:13.830 align:middle line:84% in its definition, which led to the conclusion we wanted. 00:40:13.830 --> 00:40:15.230 align:middle line:90% But this one is not so scary. 00:40:15.230 --> 00:40:16.980 align:middle line:84% This one you should be able-- this theorem 00:40:16.980 --> 00:40:19.470 align:middle line:84% you should be able to believe and maybe even prove 00:40:19.470 --> 00:40:22.440 align:middle line:90% without me telling you how to. 00:40:22.440 --> 00:40:25.320 align:middle line:90% If not, that's fine too. 00:40:25.320 --> 00:40:30.900 align:middle line:84% The other statement is that if x is positive 00:40:30.900 --> 00:40:42.690 align:middle line:84% and A is bounded above, then x plus A, or x times A, 00:40:42.690 --> 00:40:51.390 align:middle line:90% sorry, is bounded above. 00:40:51.390 --> 00:41:03.180 align:middle line:84% And sup of x times A equals x times the sup of A. 00:41:03.180 --> 00:41:05.980 align:middle line:84% And again, so if I were to draw a picture, and let's 00:41:05.980 --> 00:41:11.200 align:middle line:84% say A is kind of symmetric with respect to 0. 00:41:11.200 --> 00:41:17.260 align:middle line:84% So there's sup of A, and I multiply it by x. 00:41:17.260 --> 00:41:20.890 align:middle line:84% Then this either fattens the interval or makes it smaller. 00:41:20.890 --> 00:41:23.620 align:middle line:90% 00:41:23.620 --> 00:41:25.420 align:middle line:90% So let's say I made it smaller. 00:41:25.420 --> 00:41:32.720 align:middle line:84% Now to x times A, then x times sup of A 00:41:32.720 --> 00:41:34.340 align:middle line:84% would be the supremum of this set. 00:41:34.340 --> 00:41:37.220 align:middle line:90% 00:41:37.220 --> 00:41:38.510 align:middle line:90% Why x positive? 00:41:38.510 --> 00:41:42.710 align:middle line:84% Well, the reason is because if I multiply by x negative, 00:41:42.710 --> 00:41:45.630 align:middle line:84% this not only shrinks it but it flips things. 00:41:45.630 --> 00:41:47.780 align:middle line:84% So in fact, there's a statement in the book 00:41:47.780 --> 00:41:52.910 align:middle line:84% about if x is negative, then here you 00:41:52.910 --> 00:41:55.410 align:middle line:84% would need to assume A is bounded below. 00:41:55.410 --> 00:41:58.130 align:middle line:84% So the corresponding statement for x negative 00:41:58.130 --> 00:42:02.760 align:middle line:84% is, if x is less than 0 and A is bounded below, 00:42:02.760 --> 00:42:05.510 align:middle line:90% then x times A is bounded above. 00:42:05.510 --> 00:42:09.920 align:middle line:84% Because multiplying by something negative flips inequalities. 00:42:09.920 --> 00:42:14.780 align:middle line:84% And the sup of x times A is equal to minus 00:42:14.780 --> 00:42:20.410 align:middle line:84% x times the inf of A. Or there's no minus. 00:42:20.410 --> 00:42:22.300 align:middle line:84% This should be sup of x times A is 00:42:22.300 --> 00:42:29.460 align:middle line:90% equal to x times the inf of A. 00:42:29.460 --> 00:42:29.960 align:middle line:90% All right. 00:42:29.960 --> 00:42:38.330 align:middle line:84% So let's prove these two theorems 00:42:38.330 --> 00:42:45.560 align:middle line:84% using this previous theorem that I stated without proof, 00:42:45.560 --> 00:42:47.900 align:middle line:84% but you'll prove in the assignment about how 00:42:47.900 --> 00:42:50.990 align:middle line:84% to characterize sups of sets as upper bounds 00:42:50.990 --> 00:42:52.760 align:middle line:84% and satisfying this epsilon property. 00:42:52.760 --> 00:43:02.500 align:middle line:90% 00:43:02.500 --> 00:43:05.290 align:middle line:84% So let me just restate our assumptions. 00:43:05.290 --> 00:43:15.545 align:middle line:84% So suppose x is in R, and A is bounded above. 00:43:15.545 --> 00:43:22.990 align:middle line:90% 00:43:22.990 --> 00:43:32.095 align:middle line:84% Then the number sup A exists in R, 00:43:32.095 --> 00:43:33.970 align:middle line:84% because R has the least upper bound property. 00:43:33.970 --> 00:43:36.950 align:middle line:90% A is a non-empty subset. 00:43:36.950 --> 00:43:39.240 align:middle line:84% So I should have said that, so all of this 00:43:39.240 --> 00:43:43.120 align:middle line:84% is for a non-empty subset of A, so that I'm 00:43:43.120 --> 00:43:46.150 align:middle line:90% talking about something. 00:43:46.150 --> 00:43:48.310 align:middle line:84% So I have a non-empty subset, which 00:43:48.310 --> 00:43:50.800 align:middle line:84% is bounded above by the least upper bound property. 00:43:50.800 --> 00:43:57.660 align:middle line:84% The supremum of A exists in R. So now 00:43:57.660 --> 00:44:00.780 align:middle line:84% I'm going to show that x times the sup of A 00:44:00.780 --> 00:44:04.800 align:middle line:84% satisfies that it's an upper bound for x times A. 00:44:04.800 --> 00:44:09.360 align:middle line:84% Then for all little a in A, since sup 00:44:09.360 --> 00:44:12.900 align:middle line:84% is an upper bound for capital A, little a 00:44:12.900 --> 00:44:20.070 align:middle line:84% is less than or equal to sup A, which implies that for all a 00:44:20.070 --> 00:44:23.670 align:middle line:84% in capital A, if I multiply through by x, 00:44:23.670 --> 00:44:34.370 align:middle line:84% I mean, if I add x to both sides, x plus little a 00:44:34.370 --> 00:44:37.670 align:middle line:84% is less than or equal to x plus sup of A. 00:44:37.670 --> 00:44:43.400 align:middle line:84% Which implies that x plus sup of A 00:44:43.400 --> 00:44:52.840 align:middle line:84% is an upper bound for the set x plus A. 00:44:52.840 --> 00:44:54.430 align:middle line:84% So that's the first property we wanted 00:44:54.430 --> 00:44:58.700 align:middle line:84% to prove of x plus sup A. Now, we'll 00:44:58.700 --> 00:45:02.720 align:middle line:84% prove the second property, this epsilon property. 00:45:02.720 --> 00:45:06.410 align:middle line:90% Let epsilon be positive. 00:45:06.410 --> 00:45:19.080 align:middle line:84% Then by the previous theorem, there exist a y in A such 00:45:19.080 --> 00:45:26.760 align:middle line:84% that sup minus A is less than epsilon is less than y 00:45:26.760 --> 00:45:30.090 align:middle line:90% is less than or equal to sup A. 00:45:30.090 --> 00:45:33.900 align:middle line:84% This is just from the fact that the supremum of A 00:45:33.900 --> 00:45:38.310 align:middle line:84% satisfies those two conditions up there. 00:45:38.310 --> 00:45:42.990 align:middle line:84% Now I just add x through all of these inequalities, which 00:45:42.990 --> 00:45:53.490 align:middle line:84% implies there exists a y in A, such that x plus 00:45:53.490 --> 00:45:59.040 align:middle line:84% sup A minus epsilon is less than x 00:45:59.040 --> 00:46:08.850 align:middle line:84% plus y, which is less than or equal to x plus sup of A. 00:46:08.850 --> 00:46:11.408 align:middle line:84% And that proves the second property. 00:46:11.408 --> 00:46:12.450 align:middle line:90% Because what have I done? 00:46:12.450 --> 00:46:16.470 align:middle line:84% I found that for every epsilon positive, an element of the set 00:46:16.470 --> 00:46:18.300 align:middle line:90% x plus A-- 00:46:18.300 --> 00:46:20.820 align:middle line:84% so I'll even-- I'll restate this-- 00:46:20.820 --> 00:46:25.320 align:middle line:84% which implies there exists an element z in the set 00:46:25.320 --> 00:46:28.305 align:middle line:84% x plus capital A, namely x plus y. 00:46:28.305 --> 00:46:37.650 align:middle line:90% 00:46:37.650 --> 00:46:46.800 align:middle line:84% So that x plus sup A minus epsilon is less than z 00:46:46.800 --> 00:46:55.050 align:middle line:84% is less than or equal to x plus sup A. 00:46:55.050 --> 00:46:57.420 align:middle line:84% And that's the second property which 00:46:57.420 --> 00:47:00.030 align:middle line:84% we wanted to prove, this epsilon property. 00:47:00.030 --> 00:47:07.320 align:middle line:84% Now we've proven it for x plus sup A, for the set x plus A. 00:47:07.320 --> 00:47:19.550 align:middle line:84% Thus, sup of x plus A equals x plus sup A. 00:47:19.550 --> 00:47:27.190 align:middle line:84% So we proved that x plus sup A is the supremum of x plus A 00:47:27.190 --> 00:47:32.330 align:middle line:84% by showing it was an upper bound and it satisfied this epsilon 00:47:32.330 --> 00:47:32.830 align:middle line:90% property. 00:47:32.830 --> 00:47:41.010 align:middle line:90% 00:47:41.010 --> 00:47:45.090 align:middle line:84% I mean, it's essentially the same proof 00:47:45.090 --> 00:48:01.620 align:middle line:84% for x times capital A. You just replace pluses 00:48:01.620 --> 00:48:02.820 align:middle line:90% with multiplication. 00:48:02.820 --> 00:48:11.580 align:middle line:90% 00:48:11.580 --> 00:48:14.472 align:middle line:90% How much time do we have? 00:48:14.472 --> 00:48:17.740 align:middle line:84% So in fact, I'm running a little short on time. 00:48:17.740 --> 00:48:20.270 align:middle line:90% 00:48:20.270 --> 00:48:22.990 align:middle line:84% I guess a little slow-moving today. 00:48:22.990 --> 00:48:27.160 align:middle line:84% So I will not go through actually 00:48:27.160 --> 00:48:29.680 align:middle line:84% write the proof of the second part, 00:48:29.680 --> 00:48:34.870 align:middle line:84% simply because the same logic works. 00:48:34.870 --> 00:48:38.420 align:middle line:84% Only now, I replace everything by multiplication 00:48:38.420 --> 00:48:42.098 align:middle line:90% by x instead of addition by x. 00:48:42.098 --> 00:48:42.640 align:middle line:90% Well, almost. 00:48:42.640 --> 00:48:47.020 align:middle line:90% 00:48:47.020 --> 00:48:47.660 align:middle line:90% OK, why not. 00:48:47.660 --> 00:48:49.550 align:middle line:84% Let's go through the proof real quick. 00:48:49.550 --> 00:48:53.040 align:middle line:84% I just won't spend as much time explaining stuff. 00:48:53.040 --> 00:48:55.790 align:middle line:90% So now we want to do-- 00:48:55.790 --> 00:49:04.180 align:middle line:84% we want to show x times sup A is a supremum of x times A. 00:49:04.180 --> 00:49:13.945 align:middle line:84% So suppose x is positive and A is bounded above. 00:49:13.945 --> 00:49:17.110 align:middle line:90% 00:49:17.110 --> 00:49:23.260 align:middle line:84% Then sup A, again, exists in R, because A is bounded above. 00:49:23.260 --> 00:49:28.678 align:middle line:84% And because sup A is an upper bound for capital A, 00:49:28.678 --> 00:49:32.560 align:middle line:84% then for all a in capital A, a is less than 00:49:32.560 --> 00:49:41.440 align:middle line:84% or equal to sup A, which implies that for all a in capital A, 00:49:41.440 --> 00:49:45.430 align:middle line:84% x times little a is less than or equal to x 00:49:45.430 --> 00:50:00.480 align:middle line:84% times sup A. Which implies that x times sup A is an upper bound 00:50:00.480 --> 00:50:03.810 align:middle line:90% for x times A. 00:50:03.810 --> 00:50:08.610 align:middle line:84% So we've proven that x times sup A is an upper bound for the set 00:50:08.610 --> 00:50:12.840 align:middle line:84% x times A. We now want to verify this epsilon property for x 00:50:12.840 --> 00:50:19.030 align:middle line:84% times sup A with respect to the set x times A. 00:50:19.030 --> 00:50:23.218 align:middle line:90% So let epsilon be positive. 00:50:23.218 --> 00:50:25.010 align:middle line:84% And I'm going to put it in brackets, again, 00:50:25.010 --> 00:50:29.060 align:middle line:84% what we want to do, what we want to define. 00:50:29.060 --> 00:50:34.440 align:middle line:84% We want to show there exists z in x times A, 00:50:34.440 --> 00:50:44.670 align:middle line:84% so that x times sup A minus epsilon is less than z. 00:50:44.670 --> 00:50:47.860 align:middle line:84% z is always less than or equal to x times sup A, 00:50:47.860 --> 00:50:51.210 align:middle line:84% since we've proven x time sup A is an upper bound for this set. 00:50:51.210 --> 00:50:54.360 align:middle line:84% So I'm not going to keep writing the second inequality. 00:50:54.360 --> 00:50:59.320 align:middle line:90% 00:50:59.320 --> 00:51:01.180 align:middle line:84% So this is what we want to prove. 00:51:01.180 --> 00:51:02.740 align:middle line:90% We haven't proven it yet. 00:51:02.740 --> 00:51:04.360 align:middle line:90% So let epsilon be positive. 00:51:04.360 --> 00:51:10.250 align:middle line:84% Then just as we did in the plus case, 00:51:10.250 --> 00:51:18.820 align:middle line:84% then there exists an element y in A such that sup A minus-- 00:51:18.820 --> 00:51:24.640 align:middle line:84% now here, we're going to choose y not exactly for epsilon here. 00:51:24.640 --> 00:51:26.230 align:middle line:84% Remember, the statement for number two 00:51:26.230 --> 00:51:29.860 align:middle line:84% holds for sup A for every epsilon. 00:51:29.860 --> 00:51:33.640 align:middle line:84% In particular, I can choose anything I want here 00:51:33.640 --> 00:51:43.540 align:middle line:84% and find a y in between sup A minus whatever I want here, 00:51:43.540 --> 00:51:46.080 align:middle line:90% which is positive in sup A. 00:51:46.080 --> 00:51:47.970 align:middle line:84% So instead of putting epsilon here 00:51:47.970 --> 00:51:52.080 align:middle line:84% like I did before, I'm going to put epsilon over x. 00:51:52.080 --> 00:51:54.460 align:middle line:84% Which I can do, because x is positive, 00:51:54.460 --> 00:51:57.360 align:middle line:84% which means epsilon over x is some positive number. 00:51:57.360 --> 00:52:00.360 align:middle line:90% 00:52:00.360 --> 00:52:05.500 align:middle line:84% Now, why did I choose epsilon over x? 00:52:05.500 --> 00:52:06.550 align:middle line:90% Because magic happens. 00:52:06.550 --> 00:52:10.090 align:middle line:90% 00:52:10.090 --> 00:52:13.600 align:middle line:84% Then that means there exists a y in A, 00:52:13.600 --> 00:52:22.240 align:middle line:84% such that if I multiply through by x I get x times sup. 00:52:22.240 --> 00:52:27.040 align:middle line:84% A minus epsilon is less than x times y, is less than-- 00:52:27.040 --> 00:52:28.960 align:middle line:84% so I'm going to stop writing this inequality. 00:52:28.960 --> 00:52:32.050 align:middle line:84% Because this is always true, because x times sup A is bigger 00:52:32.050 --> 00:52:33.790 align:middle line:90% than or equal to x times y. 00:52:33.790 --> 00:52:37.450 align:middle line:90% 00:52:37.450 --> 00:52:43.770 align:middle line:84% Which implies there exists a z in x times A. Namely, 00:52:43.770 --> 00:52:48.510 align:middle line:84% z equals x times y, where y is from here, 00:52:48.510 --> 00:52:56.290 align:middle line:84% such that x times sup A minus epsilon is less than z. 00:52:56.290 --> 00:53:01.990 align:middle line:84% And therefore, x times sup A satisfies the second epsilon 00:53:01.990 --> 00:53:16.549 align:middle line:84% property with respect to S given by x times A. 00:53:16.549 --> 00:53:18.850 align:middle line:90% And that's the end. 00:53:18.850 --> 00:53:23.680 align:middle line:84% So you see that I wanted this in the end. 00:53:23.680 --> 00:53:28.270 align:middle line:84% So I chose y to give me this slightly different thing 00:53:28.270 --> 00:53:29.200 align:middle line:90% for epsilon over x. 00:53:29.200 --> 00:53:32.380 align:middle line:84% Because in the end, I would multiply through by x. 00:53:32.380 --> 00:53:34.860 align:middle line:90% And I wanted this. 00:53:34.860 --> 00:53:38.500 align:middle line:84% When we do proofs for limits, you'll 00:53:38.500 --> 00:53:41.210 align:middle line:84% see we're always trying to make something less than epsilon. 00:53:41.210 --> 00:53:46.700 align:middle line:84% So quite often, we'll have to choose something 00:53:46.700 --> 00:53:49.760 align:middle line:84% to be less than epsilon over 5, or epsilon 00:53:49.760 --> 00:53:53.600 align:middle line:84% over some number for everything to work out in the end, 00:53:53.600 --> 00:53:54.563 align:middle line:90% just like we did here. 00:53:54.563 --> 00:53:56.480 align:middle line:84% So that's kind of a preview of things to come. 00:53:56.480 --> 00:54:02.390 align:middle line:90% 00:54:02.390 --> 00:54:08.740 align:middle line:84% And one last simple theorem about sups and infs. 00:54:08.740 --> 00:54:11.770 align:middle line:84% This really doesn't use that theorem, but just basically 00:54:11.770 --> 00:54:17.230 align:middle line:84% what the definition of sups and infs are. 00:54:17.230 --> 00:54:26.500 align:middle line:84% Namely that if A and B are subsets of R, 00:54:26.500 --> 00:54:47.050 align:middle line:84% and for all, let's say, with A bounded above, B bounded below, 00:54:47.050 --> 00:54:59.660 align:middle line:84% and for all x, y, for all x in A, and for all y in B, 00:54:59.660 --> 00:55:01.430 align:middle line:90% x is less than or equal to y. 00:55:01.430 --> 00:55:05.950 align:middle line:90% 00:55:05.950 --> 00:55:13.870 align:middle line:84% Then sup of A is less than or equal to the inf of B. 00:55:13.870 --> 00:55:21.160 align:middle line:84% So picture-wise, here's A. Everything here 00:55:21.160 --> 00:55:25.760 align:middle line:84% sits below everything in B. So B has to be over here. 00:55:25.760 --> 00:55:28.210 align:middle line:84% And therefore, the sup of A, which is there, 00:55:28.210 --> 00:55:31.210 align:middle line:84% has to be less than or equal to the INF of B, which is there. 00:55:31.210 --> 00:55:34.513 align:middle line:84% That's the picture that goes along with this. 00:55:34.513 --> 00:55:35.930 align:middle line:84% But how do we actually prove this? 00:55:35.930 --> 00:55:39.070 align:middle line:84% I mean, we have to use the definitions. 00:55:39.070 --> 00:55:41.990 align:middle line:90% Pictures do not suffice. 00:55:41.990 --> 00:55:45.000 align:middle line:84% Although they inform us, they don't suffice. 00:55:45.000 --> 00:55:48.005 align:middle line:90% So this is quite simple. 00:55:48.005 --> 00:55:51.500 align:middle line:90% 00:55:51.500 --> 00:55:53.470 align:middle line:84% So I'm not going to rewrite the hypotheses now, 00:55:53.470 --> 00:55:55.840 align:middle line:90% because it'd take a little bit. 00:55:55.840 --> 00:56:08.820 align:middle line:90% But so if for all x in-- 00:56:08.820 --> 00:56:12.390 align:middle line:84% let me-- basically what we're going to do 00:56:12.390 --> 00:56:19.590 align:middle line:84% is, we're going to take a sup and then an inf. 00:56:19.590 --> 00:56:31.800 align:middle line:84% So let y be in B. Then for all x in A, 00:56:31.800 --> 00:56:34.350 align:middle line:84% x is less than or equal to y, which 00:56:34.350 --> 00:56:45.206 align:middle line:84% implies y is an upper bound for A. 00:56:45.206 --> 00:56:50.030 align:middle line:84% And therefore, the supremum of A, which 00:56:50.030 --> 00:56:53.915 align:middle line:84% is the least upper bound, has to be smaller than or equal to y. 00:56:53.915 --> 00:56:59.350 align:middle line:90% 00:56:59.350 --> 00:57:08.290 align:middle line:84% Thus, we've proven for all y in B, sup A is less than 00:57:08.290 --> 00:57:09.620 align:middle line:90% or equal to y. 00:57:09.620 --> 00:57:22.720 align:middle line:84% Which implies that sup A is a lower bound for B. 00:57:22.720 --> 00:57:25.010 align:middle line:84% And by that same logic a minute ago, 00:57:25.010 --> 00:57:29.570 align:middle line:84% so remember the infemum of B is the greatest lower bound. 00:57:29.570 --> 00:57:32.170 align:middle line:84% So if I take any lower bound of B, 00:57:32.170 --> 00:57:37.330 align:middle line:84% it has to be less than or equal to the inf of B. So 00:57:37.330 --> 00:58:14.160 align:middle line:84% which implies sup A is less than or equal to inf B. 00:58:14.160 --> 00:58:19.540 align:middle line:84% So we're starting to close out here our discussion of-- 00:58:19.540 --> 00:58:23.260 align:middle line:84% or at least our discussion of the elementary properties 00:58:23.260 --> 00:58:25.130 align:middle line:90% of the real numbers. 00:58:25.130 --> 00:58:26.920 align:middle line:84% So let me say just a couple of things 00:58:26.920 --> 00:58:28.030 align:middle line:90% about the absolute value. 00:58:28.030 --> 00:58:35.880 align:middle line:90% 00:58:35.880 --> 00:58:38.175 align:middle line:84% And let me recall how this is defined. 00:58:38.175 --> 00:58:40.168 align:middle line:90% 00:58:40.168 --> 00:58:42.210 align:middle line:84% At least, this is how it should have been defined 00:58:42.210 --> 00:58:45.450 align:middle line:90% from your calculus class. 00:58:45.450 --> 00:58:50.400 align:middle line:84% If x is in R, we define the absolute value of x. 00:58:50.400 --> 00:58:56.400 align:middle line:84% This is either x if x is bigger than or equal to 0, 00:58:56.400 --> 00:59:01.350 align:middle line:84% or minus x if x is less than or equal to 0. 00:59:01.350 --> 00:59:04.260 align:middle line:84% Note that these two things agree when x is 0. 00:59:04.260 --> 00:59:06.060 align:middle line:84% So that I'm not defining the absolute value 00:59:06.060 --> 00:59:09.645 align:middle line:84% of x to be two different things when x equals 0. 00:59:09.645 --> 00:59:12.710 align:middle line:90% 00:59:12.710 --> 00:59:24.680 align:middle line:84% So and what is this really meant to be? 00:59:24.680 --> 00:59:26.570 align:middle line:90% It's supposed to be-- 00:59:26.570 --> 00:59:30.380 align:middle line:84% it's supposed to measure the distance from-- 00:59:30.380 --> 00:59:32.540 align:middle line:90% or it's supposed to represent-- 00:59:32.540 --> 00:59:37.100 align:middle line:84% I shouldn't say, is the distance, 00:59:37.100 --> 00:59:40.640 align:middle line:84% because I haven't told you what a distance means. 00:59:40.640 --> 00:59:42.980 align:middle line:84% But what is it supposed to represent? 00:59:42.980 --> 00:59:54.160 align:middle line:84% It's supposed to represent the distance from x to 0, which is 00:59:54.160 --> 00:59:55.570 align:middle line:90% why it's always non-negative. 00:59:55.570 --> 00:59:59.240 align:middle line:84% I mean, if I have x was over here. 00:59:59.240 --> 01:00:02.610 align:middle line:84% And this distance is meant to be absolute value 01:00:02.610 --> 01:00:06.440 align:middle line:90% of x [INAUDIBLE] that y. 01:00:06.440 --> 01:00:11.720 align:middle line:90% 01:00:11.720 --> 01:00:14.810 align:middle line:84% So in fact, that's the first thing that [INAUDIBLE].. 01:00:14.810 --> 01:00:17.480 align:middle line:84% I'm just going to prove some very simple properties 01:00:17.480 --> 01:00:18.740 align:middle line:90% of the absolute value. 01:00:18.740 --> 01:00:21.290 align:middle line:84% I mean, again, these properties should not 01:00:21.290 --> 01:00:22.520 align:middle line:90% be surprising to you. 01:00:22.520 --> 01:00:25.880 align:middle line:90% You should know all of them. 01:00:25.880 --> 01:00:30.020 align:middle line:84% But we're trying to get familiar-- 01:00:30.020 --> 01:00:31.550 align:middle line:90% that's a tough word to say-- 01:00:31.550 --> 01:00:37.490 align:middle line:84% familiarities, or familiarity with proofs. 01:00:37.490 --> 01:00:41.370 align:middle line:84% So I'm going to do is many proofs as I can for you. 01:00:41.370 --> 01:00:47.570 align:middle line:84% So the first statement is, for all x in R, 01:00:47.570 --> 01:00:49.880 align:middle line:84% absolute value of x is bigger than 0-- 01:00:49.880 --> 01:00:51.650 align:middle line:90% bigger than or equal to 0. 01:00:51.650 --> 01:00:58.880 align:middle line:84% And the absolute value equals 0 if and only if x equals 0. 01:00:58.880 --> 01:01:02.780 align:middle line:90% 01:01:02.780 --> 01:01:07.370 align:middle line:84% The second property is that for all x in R, 01:01:07.370 --> 01:01:10.685 align:middle line:84% the absolute value of x equals the absolute value of minus x. 01:01:10.685 --> 01:01:15.120 align:middle line:90% 01:01:15.120 --> 01:01:22.750 align:middle line:84% For all xy in R, if I take the absolute value of the product, 01:01:22.750 --> 01:01:25.035 align:middle line:84% this is equal to the product of the absolute values. 01:01:25.035 --> 01:01:32.600 align:middle line:90% 01:01:32.600 --> 01:01:36.410 align:middle line:84% For all x in R, the absolute value of x squared 01:01:36.410 --> 01:01:40.720 align:middle line:84% equals the absolute value of x squared. 01:01:40.720 --> 01:01:50.134 align:middle line:90% 01:01:50.134 --> 01:01:54.910 align:middle line:84% The fifth property is, if x and y is in R, 01:01:54.910 --> 01:02:02.100 align:middle line:84% then x is less than or equal to y if and only 01:02:02.100 --> 01:02:06.210 align:middle line:84% if minus y is less than or equal to x is less than 01:02:06.210 --> 01:02:07.830 align:middle line:90% or equal to y. 01:02:07.830 --> 01:02:15.170 align:middle line:84% And the sixth property that we'll prove 01:02:15.170 --> 01:02:22.353 align:middle line:84% is that for all x in R, x is less than 01:02:22.353 --> 01:02:23.645 align:middle line:90% or equal to its absolute value. 01:02:23.645 --> 01:02:28.800 align:middle line:90% 01:02:28.800 --> 01:02:32.280 align:middle line:84% So again, these are not too surprising. 01:02:32.280 --> 01:02:35.900 align:middle line:84% But we'll go through the proofs, because that's 01:02:35.900 --> 01:02:36.900 align:middle line:90% the point of this class. 01:02:36.900 --> 01:02:40.060 align:middle line:90% 01:02:40.060 --> 01:02:43.980 align:middle line:84% Later on in life you'll come across some more interesting 01:02:43.980 --> 01:02:46.080 align:middle line:84% theorems than definitely this one. 01:02:46.080 --> 01:02:48.790 align:middle line:90% 01:02:48.790 --> 01:02:54.760 align:middle line:84% So if-- so we're going to prove this first statement, that if x 01:02:54.760 --> 01:02:59.320 align:middle line:84% is in R, the absolute value of x is bigger than or equal to 0. 01:02:59.320 --> 01:03:05.110 align:middle line:84% So if x is bigger than or equal to 0, 01:03:05.110 --> 01:03:07.810 align:middle line:84% then the absolute value of x is by definition 01:03:07.810 --> 01:03:10.555 align:middle line:84% equal to x, which is bigger than or equal to 0. 01:03:10.555 --> 01:03:13.670 align:middle line:90% 01:03:13.670 --> 01:03:18.560 align:middle line:84% If x is less than or equal to 0, then the absolute-- 01:03:18.560 --> 01:03:23.900 align:middle line:84% then minus x is bigger than or equal to 0. 01:03:23.900 --> 01:03:27.180 align:middle line:84% And by the definition of the absolute value, 01:03:27.180 --> 01:03:30.912 align:middle line:84% the absolute value of x is equal to minus x, which 01:03:30.912 --> 01:03:32.120 align:middle line:90% is bigger than or equal to 0. 01:03:32.120 --> 01:03:34.220 align:middle line:84% So that's proven that the absolute value of x 01:03:34.220 --> 01:03:36.005 align:middle line:84% is always bigger than or equal to 0. 01:03:36.005 --> 01:03:38.950 align:middle line:90% 01:03:38.950 --> 01:03:40.450 align:middle line:84% So now, let's prove this statement. 01:03:40.450 --> 01:03:43.360 align:middle line:84% Absolute value of x equals 0 if and only if x equals 0. 01:03:43.360 --> 01:03:46.510 align:middle line:90% 01:03:46.510 --> 01:03:50.030 align:middle line:84% So when you see an if and only if, or two arrows here, 01:03:50.030 --> 01:03:55.050 align:middle line:84% that means there's two statements you need to prove, 01:03:55.050 --> 01:03:58.900 align:middle line:84% that this implies this, and this implies this. 01:03:58.900 --> 01:04:03.900 align:middle line:84% So let's start with this direction. 01:04:03.900 --> 01:04:07.230 align:middle line:84% Usually in if and only if, there is an easy direction. 01:04:07.230 --> 01:04:12.450 align:middle line:84% If x equals 0, then this is clear simply 01:04:12.450 --> 01:04:14.370 align:middle line:90% from the definition. 01:04:14.370 --> 01:04:18.135 align:middle line:84% Then absolute value of x equals x equals 0. 01:04:18.135 --> 01:04:20.880 align:middle line:90% 01:04:20.880 --> 01:04:26.120 align:middle line:84% Let's go with the other direction. 01:04:26.120 --> 01:04:31.225 align:middle line:84% So suppose the absolute value of x equals 0. 01:04:31.225 --> 01:04:34.120 align:middle line:90% 01:04:34.120 --> 01:04:36.850 align:middle line:84% And if x is bigger than or equal to 0, 01:04:36.850 --> 01:04:40.240 align:middle line:84% we get that x is equal to its absolute value of x 01:04:40.240 --> 01:04:41.140 align:middle line:90% is equal to 0. 01:04:41.140 --> 01:04:44.700 align:middle line:90% 01:04:44.700 --> 01:04:52.320 align:middle line:84% If x is less than or equal to 0, then minus x 01:04:52.320 --> 01:04:56.370 align:middle line:84% is equal to its absolute value of x, which equals to 0, 01:04:56.370 --> 01:04:58.860 align:middle line:90% or x equals 0. 01:04:58.860 --> 01:05:01.890 align:middle line:84% So assuming the absolute value of x is 0, 01:05:01.890 --> 01:05:04.200 align:middle line:84% we've proven that x equals 0 in both cases. 01:05:04.200 --> 01:05:10.410 align:middle line:90% 01:05:10.410 --> 01:05:12.045 align:middle line:84% So that proves the first property. 01:05:12.045 --> 01:06:07.810 align:middle line:90% 01:06:07.810 --> 01:06:10.465 align:middle line:84% So now, we're on to proving number two. 01:06:10.465 --> 01:06:18.931 align:middle line:90% 01:06:18.931 --> 01:06:23.050 align:middle line:84% So number two, for all x in R, the absolute value of x 01:06:23.050 --> 01:06:24.380 align:middle line:90% equals minus x. 01:06:24.380 --> 01:06:28.240 align:middle line:84% So we'll do that by, again, we have to consider two cases. 01:06:28.240 --> 01:06:31.550 align:middle line:84% x is non-negative or x is non-positive. 01:06:31.550 --> 01:06:40.120 align:middle line:84% So if x is bigger than or equal to 0, 01:06:40.120 --> 01:06:44.230 align:middle line:84% then minus x is less than or equal to 0, 01:06:44.230 --> 01:06:47.800 align:middle line:84% which implies that the absolute value of minus x 01:06:47.800 --> 01:06:52.832 align:middle line:84% equals minus minus x, which equals x, which equals 01:06:52.832 --> 01:06:54.790 align:middle line:84% the absolute value of x since we're in the case 01:06:54.790 --> 01:06:56.370 align:middle line:90% that x is non-negative. 01:06:56.370 --> 01:07:00.510 align:middle line:90% 01:07:00.510 --> 01:07:04.020 align:middle line:84% If x is less than or equal to 0, n minus x 01:07:04.020 --> 01:07:05.760 align:middle line:84% is bigger than or equal to 0, which 01:07:05.760 --> 01:07:10.380 align:middle line:84% implies that the absolute value of minus x 01:07:10.380 --> 01:07:14.640 align:middle line:84% is equal to minus x, which is, since x is less than 01:07:14.640 --> 01:07:16.890 align:middle line:84% or equal to 0, and by the definition 01:07:16.890 --> 01:07:20.580 align:middle line:84% of the absolute value, equal to absolute value of x. 01:07:20.580 --> 01:07:21.750 align:middle line:90% So that proves too. 01:07:21.750 --> 01:07:33.560 align:middle line:90% 01:07:33.560 --> 01:07:39.440 align:middle line:84% So for all xy real number, the absolute value of x times y 01:07:39.440 --> 01:07:42.890 align:middle line:84% is equal to the absolute value of x times absolute value of y. 01:07:42.890 --> 01:07:44.790 align:middle line:84% So we need to consider two cases. 01:07:44.790 --> 01:07:46.220 align:middle line:90% One of them is-- 01:07:46.220 --> 01:07:50.630 align:middle line:84% both of them are non-negative, both of them are non-positive. 01:07:50.630 --> 01:07:54.920 align:middle line:90% And both of them are-- 01:07:54.920 --> 01:07:59.240 align:middle line:84% or one of them is positive, one of them is non-negative, 01:07:59.240 --> 01:08:04.950 align:middle line:84% or one of them is non-negative, one of them is non-positive. 01:08:04.950 --> 01:08:08.870 align:middle line:84% So if x is bigger than or equal to 0, 01:08:08.870 --> 01:08:13.250 align:middle line:84% and y is bigger than or equal to 0, then x times y 01:08:13.250 --> 01:08:14.900 align:middle line:84% is bigger than or equal to 0, which 01:08:14.900 --> 01:08:19.460 align:middle line:84% implies x, the absolute value of x times y 01:08:19.460 --> 01:08:21.680 align:middle line:90% is equal to x times y. 01:08:21.680 --> 01:08:23.540 align:middle line:84% And since x is non-negative, that's 01:08:23.540 --> 01:08:24.859 align:middle line:90% equal to its absolute value. 01:08:24.859 --> 01:08:27.830 align:middle line:84% Since y is non-negative, that's equal to its absolute value. 01:08:27.830 --> 01:08:33.580 align:middle line:90% 01:08:33.580 --> 01:08:37.859 align:middle line:84% If x is bigger than or equal to 0, 01:08:37.859 --> 01:08:41.590 align:middle line:84% and y is less than or equal to 0, 01:08:41.590 --> 01:08:52.090 align:middle line:84% then minus x times y is bigger than or equal to 0, 01:08:52.090 --> 01:08:55.840 align:middle line:84% which implies that the absolute value of x times y 01:08:55.840 --> 01:08:58.290 align:middle line:90% is equal to-- 01:08:58.290 --> 01:09:04.770 align:middle line:84% so I should say, let's write it this way. 01:09:04.770 --> 01:09:11.040 align:middle line:84% This is equal to minus xy, which is equal to x times minus y. 01:09:11.040 --> 01:09:15.000 align:middle line:84% And since y is negative, minus y is equal to its absolute value. 01:09:15.000 --> 01:09:18.700 align:middle line:84% And since x is non-negative, x is equal to its absolute value. 01:09:18.700 --> 01:09:21.990 align:middle line:90% So this is equal to that. 01:09:21.990 --> 01:09:25.740 align:middle line:90% 01:09:25.740 --> 01:09:26.729 align:middle line:90% Another piece of chalk. 01:09:26.729 --> 01:09:41.350 align:middle line:90% 01:09:41.350 --> 01:09:42.340 align:middle line:90% Now the case that-- 01:09:42.340 --> 01:09:45.950 align:middle line:90% 01:09:45.950 --> 01:09:48.130 align:middle line:84% so you might be thinking about what 01:09:48.130 --> 01:09:51.819 align:middle line:84% about x negative and y non-negative, so y bigger 01:09:51.819 --> 01:09:54.130 align:middle line:84% than or equal to 0, x less than or equal to 0. 01:09:54.130 --> 01:09:56.200 align:middle line:84% It's the same proof, just exchange x and y. 01:09:56.200 --> 01:09:57.790 align:middle line:84% So I'm not going to do that case. 01:09:57.790 --> 01:10:03.160 align:middle line:84% And if x is less than or equal to 0, 01:10:03.160 --> 01:10:05.950 align:middle line:90% y is less than or equal to 0. 01:10:05.950 --> 01:10:11.770 align:middle line:84% Then this implies that minus x is bigger than or equal to 0. 01:10:11.770 --> 01:10:14.800 align:middle line:84% And minus y is bigger than or equal to 0. 01:10:14.800 --> 01:10:17.740 align:middle line:84% Which by this first case, which we've proven, 01:10:17.740 --> 01:10:20.440 align:middle line:84% now applied to minus x and minus y, 01:10:20.440 --> 01:10:26.780 align:middle line:84% I get that the absolute value of minus x times minus y, 01:10:26.780 --> 01:10:31.480 align:middle line:84% which is equal to x times y, so minus x times minus y is equal 01:10:31.480 --> 01:10:32.200 align:middle line:90% to x times y. 01:10:32.200 --> 01:10:36.340 align:middle line:84% This is equal to the absolute value of minus x times 01:10:36.340 --> 01:10:39.280 align:middle line:90% the absolute value of minus y. 01:10:39.280 --> 01:10:42.393 align:middle line:84% And by number two, which we've already proven, 01:10:42.393 --> 01:10:44.560 align:middle line:84% which is that the absolute value of minus the number 01:10:44.560 --> 01:10:49.040 align:middle line:84% is equal to the absolute value of the number again, 01:10:49.040 --> 01:10:49.630 align:middle line:90% we get that. 01:10:49.630 --> 01:10:52.360 align:middle line:90% 01:10:52.360 --> 01:10:53.260 align:middle line:90% So that's three. 01:10:53.260 --> 01:10:56.280 align:middle line:90% 01:10:56.280 --> 01:11:01.140 align:middle line:84% And for four, this is just a special case of three. 01:11:01.140 --> 01:11:04.340 align:middle line:90% 01:11:04.340 --> 01:11:08.030 align:middle line:90% Take y equals x in three. 01:11:08.030 --> 01:11:11.700 align:middle line:90% 01:11:11.700 --> 01:11:17.090 align:middle line:84% Now, for number five, it's an if and only if. 01:11:17.090 --> 01:11:20.800 align:middle line:84% So we need to prove two directions. 01:11:20.800 --> 01:11:24.740 align:middle line:84% Namely, we'll assume this, and then prove this, 01:11:24.740 --> 01:11:28.300 align:middle line:84% and then assume this, and then prove that. 01:11:28.300 --> 01:11:33.100 align:middle line:84% So suppose-- so this is this direction, meaning suppose-- 01:11:33.100 --> 01:11:36.940 align:middle line:84% our assumption is going to be, suppose the absolute value of x 01:11:36.940 --> 01:11:38.695 align:middle line:90% is less than or equal to y. 01:11:38.695 --> 01:11:41.690 align:middle line:90% 01:11:41.690 --> 01:11:48.140 align:middle line:84% So then, x is bigger than or equal to 0. 01:11:48.140 --> 01:11:50.160 align:middle line:90% Then this means that-- 01:11:50.160 --> 01:11:53.270 align:middle line:84% so if the absolute value of x is less than or equal to y, 01:11:53.270 --> 01:11:56.550 align:middle line:84% that automatically tells you y is non-negative. 01:11:56.550 --> 01:11:59.730 align:middle line:84% So therefore, minus y is less than 01:11:59.730 --> 01:12:01.890 align:middle line:84% or equal to 0, which is less than 01:12:01.890 --> 01:12:05.070 align:middle line:84% or equal to x, which equals the absolute value of x, which 01:12:05.070 --> 01:12:06.730 align:middle line:90% is less than or equal to y. 01:12:06.730 --> 01:12:07.230 align:middle line:90% I.e. 01:12:07.230 --> 01:12:11.100 align:middle line:90% 01:12:11.100 --> 01:12:17.680 align:middle line:84% minus y is less than or equal to x is less than or equal to y. 01:12:17.680 --> 01:12:22.410 align:middle line:84% So in the other case, that x is less than or equal to 0, 01:12:22.410 --> 01:12:26.100 align:middle line:84% I can apply basically this part which I've already proven. 01:12:26.100 --> 01:12:29.760 align:middle line:90% 01:12:29.760 --> 01:12:32.760 align:middle line:84% n minus x is bigger than or equal to 0. 01:12:32.760 --> 01:12:36.130 align:middle line:84% And the absolute value of minus x, 01:12:36.130 --> 01:12:38.580 align:middle line:84% which is equal to the absolute value of x, 01:12:38.580 --> 01:12:41.430 align:middle line:84% is less than or equal to y, which 01:12:41.430 --> 01:12:46.680 align:middle line:84% implies by this first case I've proven applied now to minus x 01:12:46.680 --> 01:12:51.890 align:middle line:84% here, that minus y is less than or equal to minus 01:12:51.890 --> 01:12:55.700 align:middle line:90% x is less than or equal to y. 01:12:55.700 --> 01:13:00.890 align:middle line:84% And multiplying through by minus 1 flips all the inequalities 01:13:00.890 --> 01:13:03.140 align:middle line:84% and also flips the sign, which basically 01:13:03.140 --> 01:13:05.210 align:middle line:84% means it leaves this inequality unchanged 01:13:05.210 --> 01:13:08.780 align:middle line:90% if I replace this by x. 01:13:08.780 --> 01:13:16.150 align:middle line:84% So multiply through by minus 1 flips the inequalities 01:13:16.150 --> 01:13:18.820 align:middle line:90% and proves what we want to do. 01:13:18.820 --> 01:13:20.960 align:middle line:90% So in either case-- 01:13:20.960 --> 01:13:23.830 align:middle line:84% so for the first case, we actually wrote down a proof, 01:13:23.830 --> 01:13:28.960 align:middle line:84% and we reduced the second case, x negative, to the first case 01:13:28.960 --> 01:13:30.580 align:middle line:90% that we've already proven. 01:13:30.580 --> 01:13:34.840 align:middle line:84% So we've proven that this inequality, the absolute value 01:13:34.840 --> 01:13:42.100 align:middle line:84% of x less than or equal to y, implies that this. 01:13:42.100 --> 01:13:44.860 align:middle line:84% So now, we need to prove the converse direction. 01:13:44.860 --> 01:13:53.156 align:middle line:90% 01:13:53.156 --> 01:13:54.830 align:middle line:90% So the converse direction. 01:13:54.830 --> 01:14:08.370 align:middle line:90% 01:14:08.370 --> 01:14:15.630 align:middle line:84% So suppose minus y is less than or equal to x is less than 01:14:15.630 --> 01:14:16.770 align:middle line:90% or equal to y. 01:14:16.770 --> 01:14:22.380 align:middle line:84% And I want to prove the absolute value of x is less than 01:14:22.380 --> 01:14:23.820 align:middle line:90% or equal to y. 01:14:23.820 --> 01:14:28.520 align:middle line:90% 01:14:28.520 --> 01:14:34.700 align:middle line:84% So if x is bigger than or equal to 0, 01:14:34.700 --> 01:14:37.590 align:middle line:84% then the absolute value of x is equal to x, 01:14:37.590 --> 01:14:43.270 align:middle line:84% which by this inequality is less than or equal to y. 01:14:43.270 --> 01:14:48.650 align:middle line:84% If x is less than or equal to 0, then minus y 01:14:48.650 --> 01:14:52.880 align:middle line:84% is less than or equal to x, implies that minus x 01:14:52.880 --> 01:14:55.400 align:middle line:90% is less than or equal to y. 01:14:55.400 --> 01:15:00.170 align:middle line:84% And because x is negative, minus x 01:15:00.170 --> 01:15:01.490 align:middle line:90% is equal to its absolute value. 01:15:01.490 --> 01:15:04.935 align:middle line:90% 01:15:04.935 --> 01:15:07.060 align:middle line:84% And therefore, we've proven the absolute value of x 01:15:07.060 --> 01:15:10.570 align:middle line:90% is less than or equal to y. 01:15:10.570 --> 01:15:14.080 align:middle line:90% So we've proven number five. 01:15:14.080 --> 01:15:23.030 align:middle line:84% And number six, so for number six, what do we do? 01:15:23.030 --> 01:15:33.360 align:middle line:84% We just take y equals the absolute value of x from five, 01:15:33.360 --> 01:15:37.320 align:middle line:84% to conclude that x is less than or equal to the absolute value 01:15:37.320 --> 01:15:43.830 align:middle line:84% of x and bigger than or equal to minus the absolute value of x. 01:15:43.830 --> 01:15:47.315 align:middle line:90% 01:15:47.315 --> 01:15:48.280 align:middle line:90% And that's the proof. 01:15:48.280 --> 01:15:56.820 align:middle line:90% 01:15:56.820 --> 01:16:00.330 align:middle line:84% Now, I'm going to prove one last theorem. 01:16:00.330 --> 01:16:04.980 align:middle line:84% So this one is actually very important, 01:16:04.980 --> 01:16:06.360 align:middle line:90% about the absolute value. 01:16:06.360 --> 01:16:10.590 align:middle line:84% It's probably-- so this inequality 01:16:10.590 --> 01:16:15.600 align:middle line:84% is one of the most important tools in all of analysis. 01:16:15.600 --> 01:16:19.540 align:middle line:84% And you get to see it right here in your first analysis class. 01:16:19.540 --> 01:16:25.460 align:middle line:84% There's two-- basically two other things 01:16:25.460 --> 01:16:30.670 align:middle line:84% we'll prove at some-- at one point, which I don't know, 01:16:30.670 --> 01:16:33.630 align:middle line:84% are the two other most important things in analysis. 01:16:33.630 --> 01:16:36.630 align:middle line:90% This is the triangle inequality. 01:16:36.630 --> 01:16:40.500 align:middle line:84% And the other two are integration by parts and change 01:16:40.500 --> 01:16:41.070 align:middle line:90% of variables. 01:16:41.070 --> 01:16:47.640 align:middle line:84% Those three things fuel the analysis machine. 01:16:47.640 --> 01:16:50.490 align:middle line:84% So this theorem, so this is the triangle inequality. 01:16:50.490 --> 01:17:00.950 align:middle line:90% 01:17:00.950 --> 01:17:09.170 align:middle line:84% And it states for all xy in R, the absolute value of x plus y 01:17:09.170 --> 01:17:12.260 align:middle line:84% is less than or equal to the absolute value of x 01:17:12.260 --> 01:17:13.970 align:middle line:90% plus the absolute value of y. 01:17:13.970 --> 01:17:16.070 align:middle line:84% Why is it called the triangle inequality? 01:17:16.070 --> 01:17:22.850 align:middle line:84% Well, let's try to think, so although x and y are elements 01:17:22.850 --> 01:17:26.240 align:middle line:84% of the real number line, let's instead 01:17:26.240 --> 01:17:30.605 align:middle line:84% try and think of these as two vectors instead. 01:17:30.605 --> 01:17:40.970 align:middle line:90% 01:17:40.970 --> 01:17:42.860 align:middle line:90% So here's the vector x. 01:17:42.860 --> 01:17:45.575 align:middle line:90% And then, let's say-- 01:17:45.575 --> 01:17:50.620 align:middle line:90% 01:17:50.620 --> 01:17:54.640 align:middle line:84% so actually, I'm running out of time. 01:17:54.640 --> 01:18:00.230 align:middle line:90% So I think we'll stop here. 01:18:00.230 --> 01:18:12.000 align:middle line:90%