WEBVTT 00:00:00.540 --> 00:00:02.990 The following content is provided under a Creative 00:00:02.990 --> 00:00:04.500 Commons license. 00:00:04.500 --> 00:00:06.840 Your support will help MIT OpenCourseWare 00:00:06.840 --> 00:00:11.210 continue to offer high quality educational resources for free. 00:00:11.210 --> 00:00:13.820 To make a donation, or view additional materials 00:00:13.820 --> 00:00:17.716 from hundreds of MIT courses, visit MIT OpenCourseWare 00:00:17.716 --> 00:00:18.341 at ocw.mit.edu. 00:00:23.310 --> 00:00:26.490 PROFESSOR: So welcome to this week. 00:00:26.490 --> 00:00:28.980 We are going to talk about number theory. 00:00:28.980 --> 00:00:30.560 Actually, before I forget, there are 00:00:30.560 --> 00:00:32.450 some handouts at the very back. 00:00:32.450 --> 00:00:35.670 Please raise you hand if you don't have any, then one of us 00:00:35.670 --> 00:00:37.802 can actually come over and hand you 00:00:37.802 --> 00:00:41.070 out this sheet, which contain some facts 00:00:41.070 --> 00:00:43.900 about the visibility. 00:00:43.900 --> 00:00:45.824 Thanks a lot. 00:00:45.824 --> 00:00:50.780 And we will be using these throughout the lecture. 00:00:50.780 --> 00:00:53.330 So today we're going to talk about number theory. 00:00:53.330 --> 00:00:56.420 And this is a really different way of thinking, actually. 00:00:56.420 --> 00:00:58.170 But we will use the same concepts 00:00:58.170 --> 00:01:01.280 as you have learned before, like induction, and invariance, 00:01:01.280 --> 00:01:04.179 stuff like that, to prove whole theorems. 00:01:04.179 --> 00:01:05.220 So what is number theory? 00:01:09.070 --> 00:01:12.050 Well, first of all, it's a very old science. 00:01:12.050 --> 00:01:14.940 One of the oldest mathematical disciplines. 00:01:14.940 --> 00:01:19.340 And only recently it actually got 00:01:19.340 --> 00:01:21.480 to have some more practical applications. 00:01:21.480 --> 00:01:23.950 So what this number theory-- it's actually 00:01:23.950 --> 00:01:28.200 the study of the integers. 00:01:31.580 --> 00:01:33.160 And what are the integers? 00:01:33.160 --> 00:01:38.950 Well, these are the numbers 0, 1, 2 3, and so on. 00:01:38.950 --> 00:01:44.430 So number theory got-- Oh, there's some more over here. 00:01:44.430 --> 00:01:45.555 Another handout over there. 00:01:48.530 --> 00:01:53.870 So number theory got used actually in cryptography 00:01:53.870 --> 00:01:56.890 only about 40 years ago. 00:01:56.890 --> 00:01:59.880 And at the end of the second lecture, 00:01:59.880 --> 00:02:03.150 we will be talking about this application into cryptography. 00:02:03.150 --> 00:02:05.970 There are many application in cryptography. 00:02:05.970 --> 00:02:07.700 But we'll be talking about one of them 00:02:07.700 --> 00:02:10.500 to show you how useful this actually is. 00:02:10.500 --> 00:02:16.110 Now cryptography is the study and practice of hiding numbers. 00:02:16.110 --> 00:02:19.230 So you can imagine how important that is. 00:02:19.230 --> 00:02:23.090 We have like medical data that we need 00:02:23.090 --> 00:02:25.360 to store outside in the cloud. 00:02:25.360 --> 00:02:25.860 Right? 00:02:25.860 --> 00:02:27.500 So, gee. 00:02:27.500 --> 00:02:29.490 Do we really want that? 00:02:29.490 --> 00:02:33.200 We actually want to hide our information. 00:02:33.200 --> 00:02:35.680 We do not want others who are not 00:02:35.680 --> 00:02:39.030 allowed to see my private information to see it. 00:02:39.030 --> 00:02:40.850 So this art of hiding information 00:02:40.850 --> 00:02:44.080 is extremely important, especially nowadays. 00:02:44.080 --> 00:02:47.020 And number theory actually will help us with this. 00:02:47.020 --> 00:02:52.530 So number theory is something, you'll be very surprised, 00:02:52.530 --> 00:02:57.520 that can be used to save-- oops. 00:02:57.520 --> 00:03:00.110 I have to put this on. 00:03:00.110 --> 00:03:04.550 To save New York City in the Die Hard number 3, I believe. 00:03:04.550 --> 00:03:05.960 So let me start up again. 00:03:16.339 --> 00:03:17.505 So let's see where it plays. 00:03:20.780 --> 00:03:22.265 Maybe not. 00:03:22.265 --> 00:03:25.240 [VIDEO PLAYBACK] 00:03:25.240 --> 00:03:27.193 -Yeah, go ahead and grab it. 00:03:27.193 --> 00:03:28.140 -You're the cop. 00:03:28.140 --> 00:03:30.306 -Simon said you're supposed to be helping with this. 00:03:30.306 --> 00:03:30.930 -I'm helping. 00:03:30.930 --> 00:03:32.350 -Well, when you going to start helping? 00:03:32.350 --> 00:03:33.349 -After you get the bomb. 00:03:51.740 --> 00:03:53.220 Careful. 00:03:53.220 --> 00:03:54.000 -You be careful. 00:03:54.000 --> 00:03:55.040 -Don't open it. 00:03:55.040 --> 00:03:56.100 -What? 00:03:56.100 --> 00:03:57.539 I got to open it. 00:03:57.539 --> 00:03:58.830 And it's going to be all right. 00:04:08.790 --> 00:04:10.284 [BEEPING] 00:04:10.284 --> 00:04:11.778 [ELECTRONIC CHIRPING] 00:04:11.778 --> 00:04:13.190 Shit. 00:04:13.190 --> 00:04:13.690 -Shit! 00:04:13.690 --> 00:04:15.421 I told you not to open it. 00:04:15.421 --> 00:04:16.834 [PHONE RINGING] 00:04:16.834 --> 00:04:18.718 [PHONE RINGING] 00:04:19.660 --> 00:04:22.215 -I thought you'd see the message. 00:04:22.215 --> 00:04:25.380 It has a proximity circuit, so please don't run. 00:04:25.380 --> 00:04:26.142 -Yeah, I got it. 00:04:26.142 --> 00:04:27.100 We're not going to run. 00:04:27.100 --> 00:04:28.908 How do we turn this thing off? 00:04:28.908 --> 00:04:30.830 -On the front there should be two jugs. 00:04:30.830 --> 00:04:32.270 Do you see them? 00:04:32.270 --> 00:04:34.850 A give gallon, and a three gallon. 00:04:34.850 --> 00:04:38.510 Fill on of the jugs with exactly four gallons of water 00:04:38.510 --> 00:04:41.880 and place it on the scale, and the timer will stop. 00:04:41.880 --> 00:04:43.970 You must be precise. 00:04:43.970 --> 00:04:46.920 One ounce or lower less will result in demolition. 00:04:46.920 --> 00:04:49.405 If you're still alive in five minutes, we'll speak again. 00:04:49.405 --> 00:04:49.905 -Wait! 00:04:49.905 --> 00:04:50.730 Wait a sec. 00:04:54.410 --> 00:04:55.390 I don't get it. 00:04:55.390 --> 00:04:55.930 You get it? 00:04:55.930 --> 00:04:56.429 -No. 00:04:59.067 --> 00:04:59.650 -Get the jugs. 00:05:02.407 --> 00:05:04.490 Obviously, we can't fill the three gallon jug will 00:05:04.490 --> 00:05:06.096 four gallons of water, right? 00:05:06.096 --> 00:05:06.690 -Obviously. 00:05:06.690 --> 00:05:07.190 -I know. 00:05:07.190 --> 00:05:08.610 There we go. 00:05:08.610 --> 00:05:11.210 We fill the three gallon jug exactly to the top, right? 00:05:11.210 --> 00:05:11.710 -Uh-huh. 00:05:11.710 --> 00:05:12.210 -OK. 00:05:12.210 --> 00:05:14.760 Now we pour that three gallons into the five gallon jugs, 00:05:14.760 --> 00:05:17.180 giving us exactly 3 gallons in the five gallon jug, right? 00:05:17.180 --> 00:05:17.680 -Right. 00:05:17.680 --> 00:05:18.640 Then what? 00:05:18.640 --> 00:05:20.799 -Now, we take the three gallon jug, 00:05:20.799 --> 00:05:22.090 fill it a third of the way up-- 00:05:22.090 --> 00:05:22.190 -No, no. 00:05:22.190 --> 00:05:23.190 He said be precise. 00:05:23.190 --> 00:05:23.830 Exactly four gallons. 00:05:23.830 --> 00:05:25.746 -Every cop in 50 miles is running his ass off, 00:05:25.746 --> 00:05:27.970 and I'm out here playing kids games in a park. 00:05:27.970 --> 00:05:29.950 -Hey. 00:05:29.950 --> 00:05:31.820 You want to focus on the problem at hand? 00:05:31.820 --> 00:05:32.740 [END PLAYBACK] 00:05:32.740 --> 00:05:35.040 [LAUGHING] 00:05:35.040 --> 00:05:36.370 PROFESSOR: All right. 00:05:36.370 --> 00:05:39.690 You can imagine what we are going to do right here, right? 00:05:39.690 --> 00:05:42.510 So. 00:05:42.510 --> 00:05:45.594 You can imagine what's below this table is a bomb. 00:05:45.594 --> 00:05:47.910 [LAUGHING] 00:05:47.910 --> 00:05:49.610 You guys have to save 6042. 00:05:49.610 --> 00:05:50.940 [LAUGHING] 00:05:50.940 --> 00:05:52.890 So we have the fountain here. 00:05:52.890 --> 00:05:55.030 Each tennis ball is one gallon of water. 00:05:55.030 --> 00:05:58.050 We have a big jug, five gallons and three gallons. 00:05:58.050 --> 00:06:00.040 So you all got to help me out here. 00:06:00.040 --> 00:06:02.380 So who has an idea of what we can do? 00:06:02.380 --> 00:06:02.880 So. 00:06:05.720 --> 00:06:07.196 AUDIENCE: [INAUDIBLE] 00:06:07.196 --> 00:06:08.680 PROFESSOR: All right. 00:06:08.680 --> 00:06:09.690 Let's first do that. 00:06:09.690 --> 00:06:11.180 Fill up the three gallons. 00:06:11.180 --> 00:06:12.680 AUDIENCE: And pout it into the five. 00:06:12.680 --> 00:06:16.170 PROFESSOR: Let's pour it into five. 00:06:16.170 --> 00:06:19.010 Maybe someone else can-- can continue. 00:06:19.010 --> 00:06:20.700 Over there. 00:06:20.700 --> 00:06:22.450 AUDIENCE: If we do the same again, 00:06:22.450 --> 00:06:26.000 we'll end up with just one gallon in the three gallon. 00:06:26.000 --> 00:06:26.750 PROFESSOR: Uh-huh. 00:06:26.750 --> 00:06:28.405 So, let's do that. 00:06:28.405 --> 00:06:29.530 Because that's true, right. 00:06:29.530 --> 00:06:32.010 You can only fill it up to five gallons. 00:06:32.010 --> 00:06:34.030 So only, at more, two gallons can add 00:06:34.030 --> 00:06:36.190 to this, exactly two gallons. 00:06:36.190 --> 00:06:37.780 And one gallon is left. 00:06:37.780 --> 00:06:38.877 All right, next one. 00:06:38.877 --> 00:06:39.377 You? 00:06:39.377 --> 00:06:40.168 Would you like to-- 00:06:40.168 --> 00:06:41.335 AUDIENCE: Take out the five. 00:06:41.335 --> 00:06:42.542 PROFESSOR: Take out the five. 00:06:42.542 --> 00:06:43.120 All right. 00:06:47.300 --> 00:06:48.424 And then what? 00:06:48.424 --> 00:06:49.840 AUDIENCE: Pour the one over there. 00:06:49.840 --> 00:06:51.256 PROFESSOR: Pour the one over here? 00:06:55.142 --> 00:06:56.769 AUDIENCE: [INAUDIBLE] 00:06:56.769 --> 00:06:58.310 AUDIENCE: Then fill the three gallon, 00:06:58.310 --> 00:07:00.042 and put it into the five. 00:07:00.042 --> 00:07:01.399 PROFESSOR: All right. 00:07:01.399 --> 00:07:01.940 That's great. 00:07:01.940 --> 00:07:03.380 And I fill it up right here. 00:07:03.380 --> 00:07:04.720 Fantastic. 00:07:04.720 --> 00:07:07.690 So we actually have four gallons here. 00:07:07.690 --> 00:07:08.870 And luckily, they are safe. 00:07:08.870 --> 00:07:09.370 Right? 00:07:09.370 --> 00:07:10.730 So you say, thank god. 00:07:10.730 --> 00:07:13.354 6042 00:07:13.354 --> 00:07:14.145 So we can continue. 00:07:17.400 --> 00:07:20.230 So this is actually pretty amazing, though. 00:07:20.230 --> 00:07:22.770 How can we get four gallon out of three gallon jug, 00:07:22.770 --> 00:07:23.750 and a five gallon jug? 00:07:23.750 --> 00:07:26.370 And that's what we are going to talk about in more 00:07:26.370 --> 00:07:28.190 generality, actually. 00:07:28.190 --> 00:07:32.250 And if you would just change it a little bit, right? 00:07:32.250 --> 00:07:36.060 Then things would get more difficult. For example, 00:07:36.060 --> 00:07:40.850 if you would change the five gallon jug into a six gallon 00:07:40.850 --> 00:07:44.320 jug, can we still get four gallons? 00:07:44.320 --> 00:07:44.820 No. 00:07:44.820 --> 00:07:45.319 Why not? 00:07:47.695 --> 00:07:49.690 AUDIENCE: [INAUDIBLE] 00:07:49.690 --> 00:07:51.826 PROFESSOR: Everything has to be multiples of three. 00:07:51.826 --> 00:07:54.410 That's exactly right. 00:07:54.410 --> 00:07:55.510 This is a multiple of 3. 00:07:55.510 --> 00:07:56.190 1 times 3. 00:07:56.190 --> 00:07:57.280 This is 2 times 3. 00:07:57.280 --> 00:07:59.770 So if I do combinations with those, 00:07:59.770 --> 00:08:03.440 like pouring one into the other completely, or emptying, 00:08:03.440 --> 00:08:05.350 of filling up, we always will have 00:08:05.350 --> 00:08:10.160 a multiple of three gallons in either one of those, or both. 00:08:10.160 --> 00:08:12.630 So we can never have four gallons. 00:08:12.630 --> 00:08:15.540 So this is something that we would like 00:08:15.540 --> 00:08:16.960 to analyze a little bit more. 00:08:16.960 --> 00:08:20.420 And to do that, we're going to first a all 00:08:20.420 --> 00:08:22.080 start with a definition. 00:08:22.080 --> 00:08:25.080 Actually you can put up the screen over here. 00:08:28.480 --> 00:08:29.930 So let me take that out. 00:08:33.840 --> 00:08:37.610 Can someone up there pull up the screen? 00:08:37.610 --> 00:08:38.520 Maybe not? 00:08:38.520 --> 00:08:40.190 Maybe later. 00:08:40.190 --> 00:08:40.840 All right. 00:08:40.840 --> 00:08:42.524 So let's go with a definition. 00:08:48.150 --> 00:08:59.340 We say n denote by m and a bar, and a, we mean m defines a. 00:08:59.340 --> 00:09:00.920 And how do you define this? 00:09:00.920 --> 00:09:04.910 Well, we say that n defines a, if and only 00:09:04.910 --> 00:09:11.260 if there exists an integer k, such that a can 00:09:11.260 --> 00:09:18.100 be written as some multiple m, mainly k times m. 00:09:18.100 --> 00:09:21.980 So if you look at this definition 00:09:21.980 --> 00:09:25.100 we, for example, have that 3 divides 6, 00:09:25.100 --> 00:09:27.645 like what we just discussed. 00:09:27.645 --> 00:09:29.270 There's something interesting going on. 00:09:29.270 --> 00:09:33.160 Suppose a is equal to 0. 00:09:33.160 --> 00:09:38.030 Well, any integer will define a, will define 0. 00:09:38.030 --> 00:09:39.030 Why is that? 00:09:39.030 --> 00:09:42.090 Because I can't take k to be equal to 0, 00:09:42.090 --> 00:09:45.190 so this is equal to 0 times any integer m. 00:09:45.190 --> 00:09:48.910 So m defines 0 for all integers. 00:09:48.910 --> 00:09:51.440 So this is kind of the exception, right? 00:09:51.440 --> 00:09:56.840 And we are going to use this to set up a theorem, 00:09:56.840 --> 00:10:02.090 and analyze this whole situation over here. 00:10:02.090 --> 00:10:08.310 Now in order to do that, we will need to sort of define 00:10:08.310 --> 00:10:09.790 what we can do with all this. 00:10:09.790 --> 00:10:12.710 So there are states. 00:10:12.710 --> 00:10:14.940 We will define a state machine. 00:10:14.940 --> 00:10:18.060 We will see what kind of possible transitions 00:10:18.060 --> 00:10:19.230 we can have. 00:10:19.230 --> 00:10:22.120 And once we have modeled all this very precisely, 00:10:22.120 --> 00:10:25.520 we can start proofing stuff. 00:10:25.520 --> 00:10:30.970 Now let me first of all write out what our assumptions are. 00:10:30.970 --> 00:10:36.980 So suppose we have an a gallon jug. 00:10:40.110 --> 00:10:44.500 So in our case, a equals 3. 00:10:44.500 --> 00:10:46.480 And we have also b gallon jug. 00:10:50.140 --> 00:10:53.740 And in our case, b equals 5, right? 00:10:53.740 --> 00:10:58.180 And we issue that a is at most b. 00:10:58.180 --> 00:11:01.169 That is sort of the situation that we are working with. 00:11:01.169 --> 00:11:02.710 And he would like to prove a theorem. 00:11:05.220 --> 00:11:08.330 Exactly what we notice over here, that three defines both. 00:11:08.330 --> 00:11:10.580 The three gallon jug, and the six gallon jug, 00:11:10.580 --> 00:11:13.740 we would like to prove something like this. 00:11:13.740 --> 00:11:20.270 If m defines a, and also m defines be, 00:11:20.270 --> 00:11:26.720 well, then m should define any results 00:11:26.720 --> 00:11:32.670 that I can get with the pouring, and emptying and filling 00:11:32.670 --> 00:11:34.042 those jugs. 00:11:34.042 --> 00:11:36.000 So this is the theorem, if you'd like to prove. 00:11:36.000 --> 00:11:38.080 And we can only do that if you start to have 00:11:38.080 --> 00:11:40.380 a proper model for this. 00:11:40.380 --> 00:11:43.020 So let's go for that. 00:11:43.020 --> 00:11:44.060 And-- 00:11:47.490 --> 00:11:52.570 And, well, the state machine that we're going to use here 00:11:52.570 --> 00:11:53.380 looks like this. 00:12:00.570 --> 00:12:05.060 First of all, the states that we have 00:12:05.060 --> 00:12:08.750 are the number of gallons that are in these two jugs. 00:12:08.750 --> 00:12:10.575 So we will denote those by pairs. 00:12:13.810 --> 00:12:16.700 Pairs x, comma y. 00:12:16.700 --> 00:12:24.020 And x denotes the number of gallons in the a gallon jug. 00:12:26.640 --> 00:12:33.560 The number of gallons m m that we abbreviate as by the a jug, 00:12:33.560 --> 00:12:41.970 and y is the number of gallons in the b jug. 00:12:41.970 --> 00:12:43.960 So these are the states. 00:12:43.960 --> 00:12:49.760 And the start state it exactly as it is right there. 00:12:49.760 --> 00:12:51.717 We have nothing in either of the jugs. 00:12:55.300 --> 00:12:57.980 So that's the pair 0, comma 0. 00:12:57.980 --> 00:13:00.590 So now we start to build up some mathematics here, right? 00:13:00.590 --> 00:13:04.380 So we express the state of this whole situation 00:13:04.380 --> 00:13:05.980 by a pair of number. 00:13:05.980 --> 00:13:08.877 Now we need to find out what they can do with it. 00:13:08.877 --> 00:13:10.043 So what are the transitions? 00:13:18.480 --> 00:13:20.650 The transitions are, as we have seen, right? 00:13:20.650 --> 00:13:22.830 We can just fill one of the jugs. 00:13:22.830 --> 00:13:23.881 We can empty those. 00:13:23.881 --> 00:13:25.380 And the other possibility is that we 00:13:25.380 --> 00:13:29.020 can pour one jug over into the other one as much as we can. 00:13:29.020 --> 00:13:30.590 So let's write all of those out. 00:13:34.320 --> 00:13:36.010 We can do emptying. 00:13:36.010 --> 00:13:38.080 Well, how does that change the state? 00:13:40.820 --> 00:13:43.970 If we have x gallons in this jug, 00:13:43.970 --> 00:13:46.620 and y-- and y gallons in that one, 00:13:46.620 --> 00:13:49.460 we can transition this into, for example, 00:13:49.460 --> 00:13:51.420 emptying the a gallon jug. 00:13:51.420 --> 00:13:53.540 So be y of 0. 00:13:53.540 --> 00:14:00.262 Or we can empty the b jug. 00:14:00.262 --> 00:14:01.720 Well, filling is something similar. 00:14:05.710 --> 00:14:10.860 But now we are actually pouring more water from the fountain, 00:14:10.860 --> 00:14:11.430 essentially. 00:14:11.430 --> 00:14:11.930 Right? 00:14:11.930 --> 00:14:13.930 All those tennis balls here. 00:14:13.930 --> 00:14:20.430 And we can fill up say the a gallon up to a gallons, 00:14:20.430 --> 00:14:23.410 and leave the b jug as it is. 00:14:23.410 --> 00:14:30.410 Or w we can fill up the b gallon jug, and leave the a gallon jug 00:14:30.410 --> 00:14:32.520 as it is. 00:14:32.520 --> 00:14:34.670 So these are these two transitions. 00:14:34.670 --> 00:14:40.570 And the pouring of one-- of one jug into the other 00:14:40.570 --> 00:14:42.570 is actually a little bit more complex. 00:14:42.570 --> 00:14:43.850 So let's have a look. 00:14:47.370 --> 00:14:49.770 So how does pouring work? 00:14:49.770 --> 00:14:55.000 Well, suppose we start with x and y. 00:14:55.000 --> 00:14:57.562 So let's have a look here. 00:14:57.562 --> 00:14:58.270 Um, I don't know. 00:14:58.270 --> 00:15:02.450 Suppose we have 2 balls in here, and 2 balls in here. 00:15:02.450 --> 00:15:06.400 Well, in that case, I can say pour all of these over in here. 00:15:06.400 --> 00:15:07.670 Right? 00:15:07.670 --> 00:15:08.930 So that's easy. 00:15:08.930 --> 00:15:11.920 But there's also another possibility, better 00:15:11.920 --> 00:15:13.840 when I pour all of these over in here. 00:15:13.840 --> 00:15:16.340 But hey, I can only put in 1 ball, 00:15:16.340 --> 00:15:19.000 because it's only a three gallon jug. 00:15:19.000 --> 00:15:22.170 So I'm left with only 1. 00:15:22.170 --> 00:15:25.410 A gallon in this jug. 00:15:25.410 --> 00:15:28.600 So these are two-- these are two situations that we 00:15:28.600 --> 00:15:30.080 need to explain. 00:15:30.080 --> 00:15:34.400 So let's first do the first example that I just did. 00:15:34.400 --> 00:15:37.210 I pour everything over into the other jug. 00:15:37.210 --> 00:15:41.440 So we have 0 gallons left in here, 00:15:41.440 --> 00:15:45.800 and x plus y gallons left in the other jug. 00:15:45.800 --> 00:15:50.160 And this can happen if there's sufficient space, right? 00:15:50.160 --> 00:15:55.320 So this can only happen if x plus y is at most b. 00:15:55.320 --> 00:15:59.530 Which is the capacity of this b gallon jug. 00:15:59.530 --> 00:16:01.170 Now if that's not the case, then I 00:16:01.170 --> 00:16:04.680 can pour in just a little bit, like just say 1 ball. 00:16:04.680 --> 00:16:08.480 Like just one of these can go in here. 00:16:08.480 --> 00:16:10.090 So that's the other case. 00:16:10.090 --> 00:16:16.620 So x, y we'll actually go to-- well, 00:16:16.620 --> 00:16:19.580 let's just see how this works. 00:16:19.580 --> 00:16:25.260 How many gallons are left in this b gallon jug to fill up? 00:16:25.260 --> 00:16:28.600 Well, we have b minus y gallons left, right? 00:16:28.600 --> 00:16:30.080 Space left. 00:16:30.080 --> 00:16:33.790 So we can take b minus y gallons out of this one 00:16:33.790 --> 00:16:35.410 to fill up this one. 00:16:35.410 --> 00:16:37.170 So let's do it. 00:16:37.170 --> 00:16:42.580 We take b minus y gets out of the a jug, 00:16:42.580 --> 00:16:46.300 and put it all in here, and it makes it completely filled up. 00:16:46.300 --> 00:16:48.800 So we have b gallons over here. 00:16:48.800 --> 00:16:56.350 So this is really equal to x plus y minus b, comma b. 00:16:56.350 --> 00:17:03.010 And this only is possible if-- if you are essentially 00:17:03.010 --> 00:17:05.119 in the complimentary case. 00:17:05.119 --> 00:17:12.980 So we have that x plus y is at least b, 00:17:12.980 --> 00:17:19.720 such that there is enough gallons in the a jug 00:17:19.720 --> 00:17:22.460 to be poured over to fill up the b jug. 00:17:22.460 --> 00:17:24.430 So these are the two kinds of cases. 00:17:24.430 --> 00:17:26.180 And, of course, by symmetry we can 00:17:26.180 --> 00:17:29.830 do also the pouring from the other jug into the first. 00:17:29.830 --> 00:17:32.650 So let's write all those out, as well. 00:17:32.650 --> 00:17:38.530 So x, y can actually go to x plus y, comma 0. 00:17:38.530 --> 00:17:41.750 I pour everything from here to there. 00:17:41.750 --> 00:17:47.435 And this only holds if x plus y is at most a. 00:17:54.970 --> 00:17:59.720 The other possibility is where, exactly as in this case, 00:17:59.720 --> 00:18:06.170 we can only pour a minus x gallons over from y 00:18:06.170 --> 00:18:08.220 into this particular jug. 00:18:08.220 --> 00:18:10.300 And then this one is completely filled up. 00:18:10.300 --> 00:18:12.740 And then I have a few gallons left over here. 00:18:12.740 --> 00:18:14.910 So how does that look? 00:18:14.910 --> 00:18:21.460 Well, we completely fill this up to its capacity. 00:18:21.460 --> 00:18:24.697 And what is left this is y minus-- how much did 00:18:24.697 --> 00:18:26.990 we have to pour in here? 00:18:26.990 --> 00:18:31.180 Well, that's a minus x. 00:18:31.180 --> 00:18:34.920 And we again have a similar formula. 00:18:34.920 --> 00:18:36.870 But it now looks a little bit different. 00:18:36.870 --> 00:18:38.720 X plus y minus a. 00:18:38.720 --> 00:18:45.060 And this is only for the case where x plus y is at least a. 00:18:47.891 --> 00:18:48.390 OK. 00:18:48.390 --> 00:18:50.050 So these are all the cases. 00:18:50.050 --> 00:18:52.990 So maybe there are some questions about this. 00:18:52.990 --> 00:18:56.940 Is this clear, that we have these different possibilities? 00:18:56.940 --> 00:18:59.090 Like when we look at these jugs we can either 00:18:59.090 --> 00:19:00.840 empty them, filling them up. 00:19:00.840 --> 00:19:04.260 Or we can pour say only 1 ball over up 00:19:04.260 --> 00:19:05.940 to the full capacity of this jug. 00:19:05.940 --> 00:19:09.660 Or we can just pour everything over into, say, this jug. 00:19:09.660 --> 00:19:12.660 So those are the different cases that are now fully described 00:19:12.660 --> 00:19:14.380 by this state machine. 00:19:14.380 --> 00:19:18.100 So now we can start to prove this theorem over here. 00:19:18.100 --> 00:19:20.410 So how do we go ahead? 00:19:20.410 --> 00:19:24.100 How are we going to use what you've learned like induction, 00:19:24.100 --> 00:19:25.660 and invariance? 00:19:25.660 --> 00:19:27.880 So let's do it. 00:19:27.880 --> 00:19:30.470 But before, actually, we do this, 00:19:30.470 --> 00:19:34.940 let's take this example that we had 00:19:34.940 --> 00:19:39.088 and see how we can describe all the transitions that we just 00:19:39.088 --> 00:19:41.920 did, as far as I remember them. 00:19:41.920 --> 00:19:45.850 So we have that a equals 3, b equals 5. 00:19:45.850 --> 00:19:47.320 Right? 00:19:47.320 --> 00:19:51.760 We start with empty jugs. 00:19:51.760 --> 00:19:54.980 We need to filled up the five gallon jug, right? 00:19:59.060 --> 00:20:03.330 Then we started pouring the five gallon jug as much 00:20:03.330 --> 00:20:05.240 as we could into the three gallon jug. 00:20:05.240 --> 00:20:07.100 So it's one of those rules. 00:20:07.100 --> 00:20:09.860 We've got 3 into 2. 00:20:09.860 --> 00:20:13.080 Then we emptied the three gallon jug. 00:20:13.080 --> 00:20:14.330 We got 0 and 2. 00:20:16.940 --> 00:20:20.070 Then we did-- What did we did next? 00:20:20.070 --> 00:20:24.750 Oh yeah, we poured everything into this one. 00:20:24.750 --> 00:20:28.890 So we have 2, 0 as the next state. 00:20:28.890 --> 00:20:35.550 We filled up-- actually I forgot exactly what we did next. 00:20:35.550 --> 00:20:40.790 But I think we filled up the five gallon jug. 00:20:40.790 --> 00:20:43.170 And then we simply poured over as much 00:20:43.170 --> 00:20:45.400 as he could from the five gallon jug. 00:20:45.400 --> 00:20:49.040 And we got 3 and 4, and here we are. 00:20:49.040 --> 00:20:51.440 We got 4 gallons. 00:20:51.440 --> 00:20:55.265 So what we just did is fully describe this state machine. 00:20:55.265 --> 00:20:56.890 So let's not try to prove this theorem. 00:21:02.769 --> 00:21:04.560 So as I said, we're going to use induction. 00:21:09.660 --> 00:21:13.840 So you always would like to write this out 00:21:13.840 --> 00:21:15.085 if you solve your problems. 00:21:17.880 --> 00:21:19.810 What are we going to assume? 00:21:19.810 --> 00:21:25.170 Well, we assume actually that m defines a, 00:21:25.170 --> 00:21:27.919 and m defines also b. 00:21:27.919 --> 00:21:29.460 That's the assumption of the theorem, 00:21:29.460 --> 00:21:32.880 and now we need to prove that defies any result that you can 00:21:32.880 --> 00:21:36.340 achieve in this state machine. 00:21:36.340 --> 00:21:39.560 So what's the invariance that we are thinking about? 00:21:43.090 --> 00:21:46.670 Invariance is going to be-- 00:21:47.401 --> 00:21:47.900 Oops. 00:21:50.890 --> 00:21:53.530 It's a predicate. 00:21:53.530 --> 00:22:00.000 And it says something like, if the state xy-- 00:22:00.000 --> 00:22:06.034 if this is the state after n transitions-- 00:22:14.300 --> 00:22:20.900 Then we would like to conclude that m the fights both x, 00:22:20.900 --> 00:22:23.220 and m defines y. 00:22:23.220 --> 00:22:27.150 So this is our-- our invariance. 00:22:27.150 --> 00:22:29.690 And we like to use this to prove our theorem. 00:22:29.690 --> 00:22:31.280 So how do we start usually, right? 00:22:31.280 --> 00:22:34.500 So we always start with-- with a base state. 00:22:34.500 --> 00:22:35.690 Great. 00:22:35.690 --> 00:22:36.880 So let's do it. 00:22:40.690 --> 00:22:46.880 The base case is-- well, we start with the all 0s, 00:22:46.880 --> 00:22:49.400 like the empty jugs. 00:22:49.400 --> 00:22:53.650 It's-- well, and we also-- have paid a little bit of extra 00:22:53.650 --> 00:22:56.120 attention to what we mean by division over here. 00:22:56.120 --> 00:23:00.400 We said that all integers actually divide 0. 00:23:00.400 --> 00:23:02.790 So in particular, m. m divides 0. 00:23:02.790 --> 00:23:04.450 m, 0. 00:23:04.450 --> 00:23:07.790 So the very initial state, 0 comma 0, 00:23:07.790 --> 00:23:13.760 is indeed complying to is particular invariant. 00:23:13.760 --> 00:23:16.070 So let's write it out. 00:23:16.070 --> 00:23:19.000 So we have the initial state 0, 0. 00:23:19.000 --> 00:23:21.360 We know that m divides 0. 00:23:21.360 --> 00:23:27.060 And therefore, we know that p 0 is true. 00:23:27.060 --> 00:23:28.920 So that's great. 00:23:28.920 --> 00:23:31.380 So the inductive step. 00:23:31.380 --> 00:23:34.720 How do we start the inductive step-- step all the time? 00:23:38.720 --> 00:23:42.420 And we will assume, actually, p of n, right? 00:23:42.420 --> 00:23:45.230 So lets assume that. 00:23:45.230 --> 00:23:51.310 And now we would like to prove p, and then n plus 1. 00:23:51.310 --> 00:23:53.620 So what do we really want to do? 00:23:53.620 --> 00:23:57.910 We want to say, well, we know that we reached a certain state 00:23:57.910 --> 00:24:03.810 x comma y, for which m divides x, and m divides y. 00:24:03.810 --> 00:24:06.370 Now we would like to show that if we transition 00:24:06.370 --> 00:24:10.050 to a next state, we again have that same property, that m 00:24:10.050 --> 00:24:14.820 divides the number of gallons in both jugs once more. 00:24:14.820 --> 00:24:17.700 And then we can con-- can conclude p, n plus 1. 00:24:17.700 --> 00:24:21.020 So that's how we always proceed. 00:24:21.020 --> 00:24:22.900 So let's see where we can write it out 00:24:22.900 --> 00:24:24.330 in a bit more formal way. 00:24:29.121 --> 00:24:29.620 OK. 00:24:29.620 --> 00:24:33.270 So how do we go ahead? 00:24:33.270 --> 00:24:45.175 Suppose that x, y is the state after n transitions. 00:24:50.350 --> 00:24:53.150 Well, what can we conclude? 00:24:53.150 --> 00:24:57.160 Well, we have the predicate pn, the invariant. 00:24:57.160 --> 00:25:06.390 So we know that n divides x, and n divides y. 00:25:06.390 --> 00:25:08.780 And we concluded that because pn is true. 00:25:13.320 --> 00:25:15.030 So after another transition-- what 00:25:15.030 --> 00:25:17.580 happens after another transition? 00:25:17.580 --> 00:25:25.280 So we can conclude that the jugs are 00:25:25.280 --> 00:25:30.625 filled by the different types of numbers 00:25:30.625 --> 00:25:32.840 that we see here this is state machine. 00:25:32.840 --> 00:25:34.660 So let's write them out. 00:25:34.660 --> 00:25:44.060 So after another transition, um, each of the jugs 00:25:44.060 --> 00:25:53.150 is actually filled-- Um, are filled with-- well, 00:25:53.150 --> 00:26:01.990 either if I've emptied it, say a 0, 0 gallons, a, b, x and y. 00:26:01.990 --> 00:26:04.060 I see appearing over here. 00:26:04.060 --> 00:26:06.810 And I also notice that I see x plus y. 00:26:06.810 --> 00:26:08.330 And x plus y, minus b. 00:26:08.330 --> 00:26:09.980 And x plus y, minus a. 00:26:09.980 --> 00:26:11.740 Those are all the different number 00:26:11.740 --> 00:26:13.040 of gallons that can be in jug. 00:26:13.040 --> 00:26:14.161 Yes, please? 00:26:14.161 --> 00:26:15.604 AUDIENCE: [INAUDIBLE] 00:26:18.216 --> 00:26:20.590 PROFESSOR: In our example-- Yeah, that's a good question. 00:26:20.590 --> 00:26:23.450 So in our example problems of 3 and 5, 00:26:23.450 --> 00:26:26.480 it turns out that the only number that 00:26:26.480 --> 00:26:29.765 divides 5 both the three gallon jug, and the five gallon jugs 00:26:29.765 --> 00:26:31.390 is actually one. 00:26:31.390 --> 00:26:35.790 So in our example, we would have that m equals 1. 00:26:35.790 --> 00:26:44.330 So over here we have that only 1 divides a, 00:26:44.330 --> 00:26:47.030 as well as 1 divides b. 00:26:47.030 --> 00:26:49.800 So m equals 1 in our case. 00:26:49.800 --> 00:26:52.990 But for example, in the three gallon jug, and the six 00:26:52.990 --> 00:26:54.650 gallon jug-- Right? 00:26:54.650 --> 00:27:01.320 We have that m equals 3, like 3 divides 3, And 3 divides 6. 00:27:01.320 --> 00:27:04.630 So those are the two cases that you sort of look at right now. 00:27:04.630 --> 00:27:06.870 But you put into a much more general setting, right, 00:27:06.870 --> 00:27:09.270 we are distracted away from the actual numbers. 00:27:09.270 --> 00:27:13.355 And use a and b as representations. 00:27:15.950 --> 00:27:19.130 Are any other questions? 00:27:19.130 --> 00:27:22.790 So after another transition, each of the jugs 00:27:22.790 --> 00:27:25.890 are filled with, well, either 0 gallons, if we 00:27:25.890 --> 00:27:28.120 have a completely emptied them. 00:27:28.120 --> 00:27:33.170 Or we have filled the first a gallon jug, or it can be b. 00:27:33.170 --> 00:27:37.650 We also noticed that it can be-- it can be of x, of course. 00:27:37.650 --> 00:27:41.610 It can be y, because that's the state that we are in. 00:27:41.610 --> 00:27:47.950 And we can have x plus y, minus a, which appears over here. 00:27:47.950 --> 00:27:49.570 And x plus y, minus b. 00:27:53.770 --> 00:27:58.801 So these are all the different number-- possible number 00:27:58.801 --> 00:27:59.300 of gallons. 00:28:02.262 --> 00:28:04.280 The x plus y. 00:28:04.280 --> 00:28:06.520 That's also present. 00:28:06.520 --> 00:28:08.473 Is that true? 00:28:08.473 --> 00:28:08.973 Yeah. 00:28:08.973 --> 00:28:10.370 That's right. x plus y. 00:28:10.370 --> 00:28:13.280 So we also have x plus y. 00:28:13.280 --> 00:28:14.980 Actually, it's good to check that again. 00:28:14.980 --> 00:28:18.340 So we have 0, x, y, a, b. 00:28:18.340 --> 00:28:19.280 Got those. 00:28:19.280 --> 00:28:24.090 X plus y, and x plus y, minus p, and x plus y minus b. 00:28:24.090 --> 00:28:26.090 Yeah. 00:28:26.090 --> 00:28:32.120 So now we can start using our-- our assumptions. 00:28:32.120 --> 00:28:33.600 So what our they? 00:28:33.600 --> 00:28:36.980 We have that in order to prove this-- right? 00:28:36.980 --> 00:28:40.980 At the top over here, we assume that m divides, and m divides 00:28:40.980 --> 00:28:42.420 b. 00:28:42.420 --> 00:28:46.900 So we know that first of all, m divides 0, of course. 00:28:46.900 --> 00:28:49.460 But we know that m divides a. 00:28:49.460 --> 00:28:52.000 We know that m divides b. 00:28:52.000 --> 00:28:54.970 We have concluded that m divides x. 00:28:54.970 --> 00:28:56.370 And also m divides y. 00:28:58.980 --> 00:29:02.790 So if you now use some facts about divisibility 00:29:02.790 --> 00:29:05.770 on your handout, which we will not prove now. 00:29:05.770 --> 00:29:10.520 But I think most of them will be on your problem set, actually. 00:29:10.520 --> 00:29:16.270 We can conclude that also linear combination of a, b x and y 00:29:16.270 --> 00:29:18.660 will be divisible by m. 00:29:18.660 --> 00:29:21.850 In particular, m will divide x plus y. 00:29:21.850 --> 00:29:26.400 m will divide x plus y minus a, and also x plus y, minus b. 00:29:26.400 --> 00:29:34.040 So we will conclude that m actually 00:29:34.040 --> 00:29:37.960 divides any possible results. 00:29:37.960 --> 00:29:41.680 So divides any of the above. 00:29:41.680 --> 00:29:44.380 And now we're done. 00:29:44.380 --> 00:29:45.120 Why is that? 00:29:45.120 --> 00:29:48.510 Because we have shown now that after the next transition-- 00:29:48.510 --> 00:29:52.870 after we have reached x, y after n steps, 00:29:52.870 --> 00:29:58.980 then in our n plus 1-th step, all that you can achieve 00:29:58.980 --> 00:30:01.180 is divisible by m. 00:30:01.180 --> 00:30:03.810 So that's exactly the invariance. 00:30:03.810 --> 00:30:08.620 So we conclude that p, n plus 1 is true. 00:30:08.620 --> 00:30:09.600 And so now we're done. 00:30:12.230 --> 00:30:14.900 Are any questions about this proof? 00:30:14.900 --> 00:30:16.540 So this is like the standard technique 00:30:16.540 --> 00:30:20.200 that we tried to use all the time here in this class. 00:30:20.200 --> 00:30:22.850 We will use it in all the other areas, as well. 00:30:22.850 --> 00:30:26.440 In graph theory, in particular. 00:30:26.440 --> 00:30:30.430 And especially in number theory, will also use it, 00:30:30.430 --> 00:30:33.220 especially in this class. 00:30:33.220 --> 00:30:33.720 OK. 00:30:33.720 --> 00:30:36.660 So let's apply this to theorem. 00:30:39.870 --> 00:30:43.790 Let's I think about this movie that we saw, 00:30:43.790 --> 00:30:46.150 this Die Hard number 3. 00:30:46.150 --> 00:30:48.120 Die Hard number 4 came out. 00:30:48.120 --> 00:30:51.890 And then the cast got stuck in Die Hard number 5. 00:30:51.890 --> 00:30:55.730 There's was a problem, because the rumors 00:30:55.730 --> 00:31:01.810 were that in Die Hard number 5, they had like a 33 gallon jug. 00:31:01.810 --> 00:31:02.760 That's a lot. 00:31:02.760 --> 00:31:05.680 And a 55 gallon jug. 00:31:05.680 --> 00:31:13.252 So Bruce has in training his muscles, 00:31:13.252 --> 00:31:15.210 because you can imagine those are pretty heavy. 00:31:15.210 --> 00:31:17.560 So if you want the pour one into the other, my goodness. 00:31:17.560 --> 00:31:22.810 So-- but the question is, is he training the right muscles? 00:31:22.810 --> 00:31:27.450 So can we apply this theorem now, and showed that-- 00:31:27.450 --> 00:31:31.100 Oh, I should to tell you what is the problem. 00:31:31.100 --> 00:31:35.980 Well, again, he has to get say 4 gallons out 00:31:35.980 --> 00:31:39.750 of this-- out of these two jugs. 00:31:39.750 --> 00:31:42.260 So is that possible? 00:31:42.260 --> 00:31:42.760 It's not. 00:31:42.760 --> 00:31:44.280 I see someone shaking his head. 00:31:44.280 --> 00:31:48.030 Do you want to explain why? 00:31:48.030 --> 00:31:50.874 AUDIENCE: A and b are both divisible by 11. 00:31:50.874 --> 00:31:51.540 PROFESSOR: Yeah. 00:31:51.540 --> 00:31:54.640 AUDIENCE: So any other configuration will also 00:31:54.640 --> 00:31:56.360 have to be divisible by 11. 00:31:56.360 --> 00:31:58.590 And 4 is not divisible by 11. 00:31:58.590 --> 00:31:59.810 PROFESSOR: Exactly. 00:31:59.810 --> 00:32:02.540 4 is not divisible by 11, so the whole cast got blown up 00:32:02.540 --> 00:32:04.310 in Die Hard number 5. 00:32:04.310 --> 00:32:08.510 And so we have no Die Hard number 6, as well. 00:32:08.510 --> 00:32:13.520 OK, so-- so now all of this stuff 00:32:13.520 --> 00:32:17.860 actually helps us to define a new concept, as well. 00:32:17.860 --> 00:32:20.580 So let's do that. 00:32:20.580 --> 00:32:21.485 I'll put it up here. 00:32:29.630 --> 00:32:42.770 We will use the terminology GCD of a and b 00:32:42.770 --> 00:32:59.530 as being the greatest common divisor of a and b. 00:32:59.530 --> 00:33:04.380 So, for example, if we are looking at a equals 3, 00:33:04.380 --> 00:33:11.030 and b equals 5, well then the GCD of 3 and 5 00:33:11.030 --> 00:33:13.360 is actually equal to 1. 00:33:13.360 --> 00:33:20.450 There's no other larger integer that divides both 3 and 5. 00:33:20.450 --> 00:33:25.250 In other examples are, for example, if we have the GCD of 00:33:25.250 --> 00:33:29.410 say 52 and 44. 00:33:29.410 --> 00:33:31.550 Well, what's this equal to? 00:33:31.550 --> 00:33:35.630 Well, this actually is 4 times 13. 00:33:35.630 --> 00:33:38.170 This is 4 times 11. 00:33:38.170 --> 00:33:41.350 So 4 divides both this, and both this one. 00:33:41.350 --> 00:33:44.220 But nothing larger can divide both of those. 00:33:44.220 --> 00:33:45.940 So we have that this is equal to 4. 00:33:49.330 --> 00:33:52.110 We will have a separate definition 00:33:52.110 --> 00:33:54.650 that talks about this very special case where 00:33:54.650 --> 00:33:59.630 two numbers-- if you look at their greatest common divisor-- 00:33:59.630 --> 00:34:02.445 when that greatest common divisor is equal to 1, 00:34:02.445 --> 00:34:05.130 we actually define those two numbers to be relatively 00:34:05.130 --> 00:34:08.310 prime to one another. 00:34:08.310 --> 00:34:10.979 So let's put that out over here. 00:34:20.550 --> 00:34:23.630 So that's another definition. 00:34:23.630 --> 00:34:28.929 We say that a and b are relatively 00:34:28.929 --> 00:34:39.610 prime if the greatest common divisor is actually equal to 1. 00:34:42.840 --> 00:34:45.810 Now today we will not really use his definition so much, 00:34:45.810 --> 00:34:47.510 but it's actually very important. 00:34:47.510 --> 00:34:49.714 And we'll come back to this next lecture. 00:34:53.520 --> 00:34:56.600 So if we now look at this particular thing them 00:34:56.600 --> 00:35:02.490 over here, can we see a nice corollary of this? 00:35:02.490 --> 00:35:04.460 Like a result, if you think about this greatest 00:35:04.460 --> 00:35:06.160 common divisor. 00:35:06.160 --> 00:35:09.200 Well, the greatest common divisor off a an b 00:35:09.200 --> 00:35:11.670 divides both a and b. 00:35:11.670 --> 00:35:14.080 So the greatest common divisor of a and b 00:35:14.080 --> 00:35:18.930 will divide any result that we can generate by playing 00:35:18.930 --> 00:35:20.650 this game with the jugs. 00:35:20.650 --> 00:35:28.060 So the corollary here is that the GCD of a and b 00:35:28.060 --> 00:35:31.751 divides any result. 00:35:34.440 --> 00:35:37.200 OK, so that's really cool. 00:35:37.200 --> 00:35:40.950 So this already tells us quite a bit about this game 00:35:40.950 --> 00:35:42.590 that we have here. 00:35:42.590 --> 00:35:46.830 So now what we would like to do is to find out 00:35:46.830 --> 00:35:49.050 what exactly we can be reached? 00:35:49.050 --> 00:35:51.760 We have a property that we have shown here. 00:35:51.760 --> 00:35:55.300 But what else can we do here? 00:35:55.300 --> 00:35:58.640 Now it turns out that you can say much more, 00:35:58.640 --> 00:36:01.600 and we would like to prove the following theorem 00:36:01.600 --> 00:36:05.920 to make-- to analyze this whole thing much better. 00:36:05.920 --> 00:36:09.400 I don't think I need the state machine anymore. 00:36:09.400 --> 00:36:10.709 So let's take that off. 00:36:19.810 --> 00:36:21.580 The theorem that we would like to prove 00:36:21.580 --> 00:36:29.022 is that any linear combination of the-- let's 00:36:29.022 --> 00:36:32.210 change this into the 3 and 5 again. 00:36:32.210 --> 00:36:35.180 Any linear combination of 3 and 5, 00:36:35.180 --> 00:36:41.280 I can make with these 3 and the 5 a gallon jug. 00:36:41.280 --> 00:36:42.660 So let's write it out. 00:36:42.660 --> 00:36:55.680 So any linear combination l, which 00:36:55.680 --> 00:37:01.330 we writes as some integer s times a, plus some integer 00:37:01.330 --> 00:37:02.360 t times b. 00:37:05.170 --> 00:37:15.760 So any linear combination of a and b, with-- well, of course, 00:37:15.760 --> 00:37:19.850 the number of gallons should fit the largest the jug. 00:37:19.850 --> 00:37:25.430 So with 0 is, at most l. 00:37:25.430 --> 00:37:27.620 Is it mostly can be reached. 00:37:34.520 --> 00:37:37.920 So this theorem we would like to prove now. 00:37:37.920 --> 00:37:39.610 And in order to do that, we would 00:37:39.610 --> 00:37:44.260 like to already think about some kind of a property 00:37:44.260 --> 00:37:45.250 that we have. 00:37:45.250 --> 00:37:49.090 So when we talk about linear combinations, the s and the t 00:37:49.090 --> 00:37:50.820 can be negative, or positive. 00:37:50.820 --> 00:37:51.760 We really don't care. 00:37:51.760 --> 00:37:54.110 So for example, we could have like, 00:37:54.110 --> 00:37:58.720 I don't know, minus 2 times-- so for example, 00:37:58.720 --> 00:38:06.140 4 is equal to minus 2, times 3, plus-- actually, is that true? 00:38:06.140 --> 00:38:06.820 Yeah. 00:38:06.820 --> 00:38:10.590 Plus 2, times 5. 00:38:10.590 --> 00:38:18.710 So here we have s to be equal to minus 2, and t is equal to 2. 00:38:18.710 --> 00:38:20.430 And of course, a is equal to 3, right? 00:38:20.430 --> 00:38:23.210 And be is equal to 5. 00:38:23.210 --> 00:38:27.160 So 4 is a linear combination of these two. 00:38:27.160 --> 00:38:31.070 And according to the theorem, we can create that number 00:38:31.070 --> 00:38:33.280 of gallons in this jug. 00:38:33.280 --> 00:38:36.480 And we already saw that, because we did it. 00:38:36.480 --> 00:38:40.430 But for our theorem, in order to prove this, 00:38:40.430 --> 00:38:43.020 we really would like s to be positive. 00:38:43.020 --> 00:38:45.390 So how can we do that? 00:38:45.390 --> 00:38:48.150 If anybody has an indea what we could do? 00:38:48.150 --> 00:38:51.224 AUDIENCE: Let's assume that b is greater than m. 00:38:51.224 --> 00:38:51.890 PROFESSOR: Yeah. 00:38:51.890 --> 00:38:56.160 We have still that a is supposed to be-- We will assume that 00:38:56.160 --> 00:38:57.821 throughout the whole lecture. 00:38:57.821 --> 00:38:58.320 Thanks. 00:39:00.731 --> 00:39:02.230 So in order to prove this, we really 00:39:02.230 --> 00:39:04.110 would like to have s to be positive. 00:39:04.110 --> 00:39:05.867 So let's just play around a little bit 00:39:05.867 --> 00:39:07.700 with linear combinations to get a little bit 00:39:07.700 --> 00:39:09.280 of feeling for that. 00:39:09.280 --> 00:39:11.510 How could we write 4 differently, 00:39:11.510 --> 00:39:14.150 as a linear combination of 3 and 5, 00:39:14.150 --> 00:39:19.150 such that we have actually a positive number over here? 00:39:19.150 --> 00:39:23.895 Does anybody see another way to see that? 00:39:23.895 --> 00:39:25.937 AUDIENCE: [INAUDIBLE] 00:39:25.937 --> 00:39:27.145 PROFESSOR: Yeah, that's true. 00:39:27.145 --> 00:39:31.270 3 times 3, minus-- minus 5. 00:39:31.270 --> 00:39:33.460 So-- and how did we do that? 00:39:33.460 --> 00:39:38.580 Well, we can just say 5 times 3 to this one, 00:39:38.580 --> 00:39:42.090 and then subtract the same again, minus 3 times 5, 00:39:42.090 --> 00:39:43.320 over here. 00:39:43.320 --> 00:39:46.170 And if he adds those things together, 00:39:46.170 --> 00:39:50.870 he will see 5 minus 2, is 3 times 3, as you said. 00:39:50.870 --> 00:39:55.580 And we have minus 3 plus 2 is actually minus 1 times 5. 00:39:55.580 --> 00:39:58.900 And this will be a different linear combination of 4. 00:39:58.900 --> 00:40:02.930 So what we can do here, we can sort of play around and make 00:40:02.930 --> 00:40:07.260 this s over here, which we now say call s prime, is positive. 00:40:10.200 --> 00:40:11.640 Actually, it's larger than 0. 00:40:17.340 --> 00:40:22.960 So let's start the proof for this theorem. 00:40:22.960 --> 00:40:27.595 It's pretty amazing to me, actually, 00:40:27.595 --> 00:40:31.470 that you can do so much a game like this, 00:40:31.470 --> 00:40:35.020 and see so much happening. 00:40:35.020 --> 00:40:40.651 So let's figure out how this works. 00:40:40.651 --> 00:40:41.150 OK. 00:40:45.300 --> 00:40:51.850 So let's first formalize this particular trick over here. 00:40:51.850 --> 00:40:54.030 And how do we go ahead with it? 00:40:54.030 --> 00:40:59.990 Ah, well, notice that we can rewrite 00:40:59.990 --> 00:41:05.780 L, which is equal to s times a, plus t times b. 00:41:05.780 --> 00:41:13.130 s, you know, we can just add a multiple of b over here. 00:41:13.130 --> 00:41:16.880 n times b, say m times a. 00:41:16.880 --> 00:41:22.140 And we can subtract the same amount over here, 00:41:22.140 --> 00:41:24.370 minus n times a, times b. 00:41:24.370 --> 00:41:26.530 So do you see what I did over here? 00:41:26.530 --> 00:41:30.570 I have added n times b, times a, and subtracted n, times a times 00:41:30.570 --> 00:41:32.760 b. 00:41:32.760 --> 00:41:35.140 And we did something similar over here, not exactly 00:41:35.140 --> 00:41:35.750 the same. 00:41:35.750 --> 00:41:37.990 But that's what we did. 00:41:37.990 --> 00:41:42.390 And you can imagine that we can choose m, 00:41:42.390 --> 00:41:45.970 such that s plus n times b will be larger than 0. 00:41:45.970 --> 00:41:47.030 We can do that. 00:41:47.030 --> 00:41:53.450 So essentially this proved to us that there exists an x prime, 00:41:53.450 --> 00:41:57.050 and also the t prime, such that L 00:41:57.050 --> 00:41:59.610 can be rewritten as a linear combination, 00:41:59.610 --> 00:42:04.600 s prime, times a, plus t prime, times b. 00:42:04.600 --> 00:42:08.710 But now with you extra property, that s prime 00:42:08.710 --> 00:42:13.010 is actually positive. 00:42:13.010 --> 00:42:15.690 Now this is really important, because we're 00:42:15.690 --> 00:42:18.840 going to create an algorithm of playing 00:42:18.840 --> 00:42:22.410 with those jugs that can achieve this particular linear 00:42:22.410 --> 00:42:23.280 combination. 00:42:23.280 --> 00:42:26.700 And that's how we're going to prove this theorem. 00:42:26.700 --> 00:42:36.215 So let's assume that 0 is less than L, is less than b. 00:42:36.215 --> 00:42:39.690 I know that we, in the theorem, we also 00:42:39.690 --> 00:42:42.472 consider the case is L equals 0, and L equals b. 00:42:42.472 --> 00:42:43.680 But those are obvious, right? 00:42:43.680 --> 00:42:46.020 You could either empty the jugs, or just 00:42:46.020 --> 00:42:49.130 fill up with the bigger one. 00:42:49.130 --> 00:42:51.260 So we will consider just this case. 00:42:54.580 --> 00:42:55.080 All right. 00:42:55.080 --> 00:42:57.243 So what's the algorithm going to do for us? 00:43:03.159 --> 00:43:12.740 The algorithm is going to repeatedly fill and pour 00:43:12.740 --> 00:43:16.160 our jugs in a very special way. 00:43:16.160 --> 00:43:20.470 And miraculously we will be able to get 00:43:20.470 --> 00:43:23.520 the desired linear combination every single time. 00:43:26.229 --> 00:43:28.270 And of course, we're going to use induction again 00:43:28.270 --> 00:43:30.611 to prove this property. 00:43:30.611 --> 00:43:31.110 OK. 00:43:31.110 --> 00:43:33.330 So how does the algorithm work? 00:43:33.330 --> 00:43:45.980 Well, to obtain L gallons we're going to repeat 00:43:45.980 --> 00:43:50.047 s prime times, which is the number that we have over here. 00:43:53.320 --> 00:44:00.260 The following algorithm-- we first of all, 00:44:00.260 --> 00:44:02.160 we will fill the a jug. 00:44:06.800 --> 00:44:08.609 This one. 00:44:08.609 --> 00:44:10.150 After we have done this, we are going 00:44:10.150 --> 00:44:12.380 to pour this into the b jug. 00:44:12.380 --> 00:44:15.090 So how do we go ahead? 00:44:15.090 --> 00:44:18.470 We pour- oops. 00:44:18.470 --> 00:44:19.615 This into the b jug. 00:44:23.450 --> 00:44:28.960 And when this b jug becomes full, we are going pour it out. 00:44:28.960 --> 00:44:31.040 So let's write it out. 00:44:31.040 --> 00:44:45.080 So when it becomes full, it will actually empty it out. 00:44:49.080 --> 00:44:53.610 And we will continue pouring the a jug into the b jug. 00:44:53.610 --> 00:44:56.060 So we'll continue this process. 00:45:00.120 --> 00:45:04.060 So let's take an example to see how that works. 00:45:04.060 --> 00:45:10.680 So we keep on doing this until the a jug is actually empty. 00:45:13.920 --> 00:45:15.070 So let's take an example. 00:45:18.436 --> 00:45:20.890 So let's see. 00:45:20.890 --> 00:45:22.555 Let's do that over here. 00:45:26.560 --> 00:45:29.960 Actually we can do the tennis balls, too. 00:45:29.960 --> 00:45:32.100 Let's do that first. 00:45:32.100 --> 00:45:34.810 See how that works. 00:45:34.810 --> 00:45:41.510 So essentially, in order to get 4 gallons, 00:45:41.510 --> 00:45:45.100 we just fill up the three gallon jug. 00:45:45.100 --> 00:45:47.230 We empty it all in here. 00:45:49.780 --> 00:45:51.670 We fill it up again. 00:45:51.670 --> 00:45:54.110 You pour in as much as we can. 00:45:54.110 --> 00:45:56.480 That's-- that's it. 00:45:56.480 --> 00:45:58.501 We have to empty this one. 00:45:58.501 --> 00:45:59.000 Oops. 00:46:01.760 --> 00:46:04.040 We have to keep on pouring. 00:46:04.040 --> 00:46:05.740 Put this in here. 00:46:05.740 --> 00:46:12.900 Fill this one up, and then pour over into the five gallon jug. 00:46:12.900 --> 00:46:15.640 And now we've got 4 gallons over here. 00:46:15.640 --> 00:46:17.160 So what did we do? 00:46:17.160 --> 00:46:19.490 So let's write it out. 00:46:19.490 --> 00:46:23.180 So for our special linear combination over here, 00:46:23.180 --> 00:46:29.040 we have that 4 equals 3, times 3, minus 1, times 5. 00:46:29.040 --> 00:46:33.620 So we need to repeat this process three times. 00:46:33.620 --> 00:46:35.740 So let's do that. 00:46:35.740 --> 00:46:42.150 In our first loop we will do the following. 00:46:42.150 --> 00:46:47.490 We start with the start state, the pair 0, 0. 00:46:47.490 --> 00:46:50.720 We're going to fill up the very first jug all the way up 00:46:50.720 --> 00:46:52.950 to its capacity, 3. 00:46:52.950 --> 00:46:57.860 And we put it all over into the b jug. 00:46:57.860 --> 00:46:59.414 What happens in the second loop? 00:47:04.850 --> 00:47:09.200 The second loop, we again fill up the a jug. 00:47:09.200 --> 00:47:13.320 So we have-- we start at 0, 3. 00:47:13.320 --> 00:47:15.730 We fill it up. 00:47:15.730 --> 00:47:18.430 We get 3, 3, the pair 3, 3. 00:47:18.430 --> 00:47:22.330 We pour everything in here, as much as we can. 00:47:22.330 --> 00:47:24.320 That give us 1, 5. 00:47:24.320 --> 00:47:29.180 Only 2 gallons are poured into the bigger gallon. 00:47:29.180 --> 00:47:32.230 We empty the bigger gallon, the bigger jug. 00:47:32.230 --> 00:47:34.940 We get 1, 0. 00:47:34.940 --> 00:47:39.110 And we keep on pouring, and you get 0, 1. 00:47:39.110 --> 00:47:41.660 So now in the third loop-- and that's 00:47:41.660 --> 00:47:43.276 where we should get the 4 gallons. 00:47:47.110 --> 00:47:51.440 We start off with 0,1. 00:47:51.440 --> 00:47:57.150 Um, we fill up the a jug. 00:47:57.150 --> 00:48:03.910 We pour everything over into the bigger jug, and we get 0, 4. 00:48:03.910 --> 00:48:06.210 And that's the end result. 00:48:06.210 --> 00:48:09.970 So this algorithm seems to work for this particular example. 00:48:09.970 --> 00:48:13.489 Of course we would like to prove it for the general situation. 00:48:13.489 --> 00:48:14.280 So how do we do it? 00:48:16.800 --> 00:48:22.780 Well, we're going to just to analyze the algorithm 00:48:22.780 --> 00:48:23.660 in the following way. 00:48:27.000 --> 00:48:32.530 We can notice that in this algorithm, 00:48:32.530 --> 00:48:36.020 we fill up s prime times the a jug, 00:48:36.020 --> 00:48:42.420 and we essentially pour everything out into the b jugs, 00:48:42.420 --> 00:48:45.660 and we sometimes empty the b jug. 00:48:45.660 --> 00:48:48.170 So let's try to think about this a little bit, 00:48:48.170 --> 00:48:51.030 and see how we could try to formalize this. 00:48:51.030 --> 00:48:52.160 So let's write it out. 00:48:54.700 --> 00:49:02.720 We have filled the a gallon jug s prime times. 00:49:05.360 --> 00:49:09.500 We also know that the b jug has been emptied 00:49:09.500 --> 00:49:10.880 a certain number of times. 00:49:10.880 --> 00:49:22.450 So let's-- let's just assume-- suppose that the b jug is 00:49:22.450 --> 00:49:24.375 actually emptied, say, u times. 00:49:30.700 --> 00:49:32.800 I do not know how many times. 00:49:32.800 --> 00:49:34.900 But I say, well, let's assume it's u times, 00:49:34.900 --> 00:49:37.340 and try to figure out whether we can 00:49:37.340 --> 00:49:40.470 find some algebraic expression. 00:49:40.470 --> 00:49:44.690 So at the very end of the algorithm, 00:49:44.690 --> 00:49:48.270 let r be what is in the b jug. 00:49:48.270 --> 00:49:58.235 So let r be the remainder, in the b gallon jug. 00:50:03.500 --> 00:50:06.220 So now we can continue. 00:50:06.220 --> 00:50:09.940 We know if r is what left in the b gallon jug, well, 00:50:09.940 --> 00:50:13.780 we know already some property of it. 00:50:13.780 --> 00:50:17.040 Actually, let's put that on the next board. 00:50:17.040 --> 00:50:22.580 We know that 0 is at most r, and at most b, 00:50:22.580 --> 00:50:25.170 because that's what's left in the b gallon jug, right? 00:50:25.170 --> 00:50:27.740 So we know these bounds. 00:50:27.740 --> 00:50:33.700 We have assumed that 0 is less than L, is less than b, 00:50:33.700 --> 00:50:36.620 which we put over there. 00:50:36.620 --> 00:50:40.650 We know that r must be equal to what 00:50:40.650 --> 00:50:46.720 kind of linear combination of s prime, and u? 00:50:46.720 --> 00:50:52.240 So-- Well, we have been filling of s prime times. 00:50:52.240 --> 00:50:57.020 So this is what we added in water to the whole system, 00:50:57.020 --> 00:50:59.660 you can say, s prime times a. 00:50:59.660 --> 00:51:02.060 And we poured out water. 00:51:02.060 --> 00:51:06.890 Well, we did that u times from the b gallon jug. 00:51:06.890 --> 00:51:09.830 So we poured out u times b gallons. 00:51:09.830 --> 00:51:13.520 So this is the remainder that this left in this bigger jug, 00:51:13.520 --> 00:51:14.910 right? 00:51:14.910 --> 00:51:17.940 So are there any questions about this? 00:51:17.940 --> 00:51:22.150 So-- OK. 00:51:22.150 --> 00:51:27.470 So we also know that L is equal to s prime, 00:51:27.470 --> 00:51:31.220 times a, plus t prime, times b. 00:51:31.220 --> 00:51:33.500 And this is the linear combination 00:51:33.500 --> 00:51:36.820 that we would try to prove of, that it 00:51:36.820 --> 00:51:39.240 is left at the very end. 00:51:39.240 --> 00:51:42.440 So what we want to show is that r equals L. 00:51:42.440 --> 00:51:44.290 So how do we do that now? 00:51:44.290 --> 00:51:51.110 How are we going to show that r can be expressed 00:51:51.110 --> 00:51:52.770 in L, in a special way. 00:51:52.770 --> 00:51:53.710 So let's have a look. 00:51:53.710 --> 00:51:58.150 So these are all tricks in the sense 00:51:58.150 --> 00:52:01.000 that I'm giving you this proof, but how do 00:52:01.000 --> 00:52:03.340 you come up with this yourself? 00:52:03.340 --> 00:52:05.990 Sometimes you play a lot with these kinds of things, 00:52:05.990 --> 00:52:12.250 and you get a feeling of what kind of-- sort of pattern 00:52:12.250 --> 00:52:16.080 exists, and what kind of intuition 00:52:16.080 --> 00:52:19.070 you need in order to write down a proof like this. 00:52:19.070 --> 00:52:22.910 So let's rewrite this. 00:52:22.910 --> 00:52:26.030 I'm going add t prime times b. 00:52:26.030 --> 00:52:27.570 And I'm going to subtract it again. 00:52:30.530 --> 00:52:34.650 So I have s prime times a, plus t prime times b. 00:52:34.650 --> 00:52:38.240 I subtract it again, and I still have this amount left open 00:52:38.240 --> 00:52:39.820 here. 00:52:39.820 --> 00:52:41.755 So what is this equal to? 00:52:41.755 --> 00:52:48.840 Well this part is equal to L. So this is equal to minus-- 00:52:48.840 --> 00:52:52.140 and I have a multiple of b, which is t prime, 00:52:52.140 --> 00:52:53.708 plus u times b. 00:52:56.636 --> 00:52:59.072 Hm. 00:52:59.072 --> 00:53:00.280 Now this is very interesting. 00:53:00.280 --> 00:53:02.760 Does anybody see how we could continue here? 00:53:02.760 --> 00:53:08.260 So we have r expressed as L, minus a multiple of b. 00:53:08.260 --> 00:53:10.630 And I also know that L is in this range. 00:53:10.630 --> 00:53:13.489 I also know that r is in this range. 00:53:13.489 --> 00:53:15.030 So that's kind of interesting, right? 00:53:15.030 --> 00:53:18.730 So how can that be? 00:53:18.730 --> 00:53:20.560 What should be the case here? 00:53:20.560 --> 00:53:25.660 Does anybody see what kind of property t prime plus u 00:53:25.660 --> 00:53:30.490 must have in order to make that happen? 00:53:30.490 --> 00:53:31.960 So let's have a look here. 00:53:31.960 --> 00:53:35.300 We have L. It's in this range. 00:53:35.300 --> 00:53:37.720 So let's just draw an axis. 00:53:37.720 --> 00:53:40.340 So at 0, we have b. 00:53:40.340 --> 00:53:48.300 And somehow in this range, we have L. Now 00:53:48.300 --> 00:53:53.590 if I subtract like actually b, or something more than b, 00:53:53.590 --> 00:53:55.470 or I add more than b. 00:53:55.470 --> 00:53:59.010 I will jump out of this range, and I go somewhere over here, 00:53:59.010 --> 00:54:01.510 or I go somewhere over there. 00:54:01.510 --> 00:54:02.160 Right? 00:54:02.160 --> 00:54:06.050 So if I said suppose L is over here, 00:54:06.050 --> 00:54:09.960 then L minus b would be over here, which would be negative. 00:54:09.960 --> 00:54:12.900 Or if I add b, it will be over here, 00:54:12.900 --> 00:54:15.670 which would be more than b. 00:54:15.670 --> 00:54:20.120 Now we know that this is equal to r, but r is in this range. 00:54:20.120 --> 00:54:21.810 So that's not really possible. 00:54:21.810 --> 00:54:23.780 So let's write it out. 00:54:23.780 --> 00:54:29.370 So if t prime plus u is unequal to 0, 00:54:29.370 --> 00:54:31.420 so we're actually really subtract 00:54:31.420 --> 00:54:36.300 or add a multiple of b. 00:54:36.300 --> 00:54:41.770 Then I know that r is either smaller than 0, 00:54:41.770 --> 00:54:45.110 or r is larger than b. 00:54:45.110 --> 00:54:46.910 Now we know that cannot be the case, 00:54:46.910 --> 00:54:51.600 so we can conclude that t prime plus u equals 0. 00:54:51.600 --> 00:54:58.520 Now that implies that t prime equals minus u, 00:54:58.520 --> 00:55:01.040 or maybe other way around, because that's easier 00:55:01.040 --> 00:55:02.200 to see what's happening. 00:55:02.200 --> 00:55:06.300 So u equals minus t prime. 00:55:06.300 --> 00:55:09.720 If you plug that in here, well, we 00:55:09.720 --> 00:55:11.510 get exactly the same expression. 00:55:11.510 --> 00:55:12.590 You see? 00:55:12.590 --> 00:55:16.640 Minus, minus t prime is equal to plus t prime. 00:55:16.640 --> 00:55:18.560 And we get the exact same linear combination. 00:55:18.560 --> 00:55:21.910 So we conclude that r equals L. 00:55:21.910 --> 00:55:23.370 And now we're done. 00:55:23.370 --> 00:55:24.420 Why is that? 00:55:24.420 --> 00:55:29.034 Well, we have shown that the very last number of gallons 00:55:29.034 --> 00:55:30.450 that is left after this procedure, 00:55:30.450 --> 00:55:32.670 after this algorithm, is actually 00:55:32.670 --> 00:55:35.090 exactly equal to the linear combination 00:55:35.090 --> 00:55:36.630 that we wanted to achieve. 00:55:36.630 --> 00:55:40.720 So now we got the proof for this theorem that tells us 00:55:40.720 --> 00:55:50.370 that any linear combination is actually-- of a and b 00:55:50.370 --> 00:55:54.250 can actually be reached by pouring gallons over and back, 00:55:54.250 --> 00:55:56.980 and emptying and filling those jugs. 00:55:56.980 --> 00:55:58.995 All right let's continue. 00:55:58.995 --> 00:56:00.370 So there was a question over here 00:56:00.370 --> 00:56:02.915 that I would like to-- that I would like to address. 00:56:05.990 --> 00:56:10.600 So maybe I did not make so clear what the s prime, 00:56:10.600 --> 00:56:12.900 and the t prime is over here. 00:56:12.900 --> 00:56:16.222 And in this proof, we started off 00:56:16.222 --> 00:56:17.430 with this linear combination. 00:56:17.430 --> 00:56:20.270 I would like to have an algorithm 00:56:20.270 --> 00:56:25.990 of pouring that creates L gallons in say the bigger jug. 00:56:25.990 --> 00:56:30.870 So in order to do that, I want to find, say, 00:56:30.870 --> 00:56:35.040 a linear combination that makes this L such that this 00:56:35.040 --> 00:56:38.160 s prime is an integer-- positive integer. 00:56:38.160 --> 00:56:40.900 Why do I want to have a positive integer? 00:56:40.900 --> 00:56:44.420 Because in this algorithm, I'm going to repeat something 00:56:44.420 --> 00:56:45.190 s prime times. 00:56:45.190 --> 00:56:47.670 If s prime is negative, I cannot do it, right? 00:56:47.670 --> 00:56:50.710 So s prime has to be a positive integer. 00:56:50.710 --> 00:56:55.200 In order to create such a positive integer, 00:56:55.200 --> 00:56:59.450 I can just add like 1,000 times b times, 00:56:59.450 --> 00:57:03.160 and subtract 1,000 times a times b. 00:57:03.160 --> 00:57:06.080 That's OK I could just add a lot. 00:57:06.080 --> 00:57:11.220 And if I add enough, I can make s plus n times b positive. 00:57:11.220 --> 00:57:17.620 Even if s is, say, minus 100, well, if I add 1,000 times 5, 00:57:17.620 --> 00:57:19.770 I will get a positive number. 00:57:19.770 --> 00:57:22.520 So that's sort of the reason this proof 00:57:22.520 --> 00:57:25.310 that we want to rewrite the linear combination 00:57:25.310 --> 00:57:29.040 to a new one, such that s prime is positive. 00:57:29.040 --> 00:57:31.800 And if we have s prime positive, then we 00:57:31.800 --> 00:57:34.070 can actually talk about this algorithm, 00:57:34.070 --> 00:57:37.890 because we can only repeat something s prime times, 00:57:37.890 --> 00:57:41.435 if s prime is say 1, or 2, or 3, or something positive. 00:57:44.970 --> 00:57:46.330 All right. 00:57:46.330 --> 00:57:51.470 So let's-- I'll talk about say the next part. 00:57:51.470 --> 00:57:56.480 So we have gone-- We have proved two theorems. 00:57:56.480 --> 00:58:01.529 But in the end we would like to have a characterization 00:58:01.529 --> 00:58:02.820 of the greatest common divisor. 00:58:02.820 --> 00:58:05.410 That's the goal of this lecture. 00:58:05.410 --> 00:58:07.250 So let's do it. 00:58:07.250 --> 00:58:10.000 Um. 00:58:10.000 --> 00:58:14.750 In order to do this, let's first of all 00:58:14.750 --> 00:58:18.290 look at our five gallon, and three gallon example. 00:58:18.290 --> 00:58:22.830 We know that the greatest common divisor is equal to 1. 00:58:22.830 --> 00:58:30.740 We know that 1 can be rewritten as a linear combination, as 2 00:58:30.740 --> 00:58:35.960 times 3, minus 1 times 5. 00:58:35.960 --> 00:58:38.530 So that means that according to the theorem that we 00:58:38.530 --> 00:58:44.120 have up here, we can actually make exactly 1 00:58:44.120 --> 00:58:48.140 gallon in one of these jugs. 00:58:48.140 --> 00:58:51.770 So that means that we can also have any multiple of those. 00:58:51.770 --> 00:58:54.970 So we can reach any multiple 1. 00:58:54.970 --> 00:58:56.180 That's very special. 00:58:56.180 --> 00:58:59.730 So this particular case, we know that any multiple 00:58:59.730 --> 00:59:03.100 of 1, any number of gallons can be reached. 00:59:03.100 --> 00:59:06.270 So can we sort of generalize this a little bit 00:59:06.270 --> 00:59:08.050 by using the greatest common divisor? 00:59:08.050 --> 00:59:11.680 So the greatest common divisor 3 and 5 is equal to 1. 00:59:11.680 --> 00:59:16.700 And we have shown that the greatest common divisor defies 00:59:16.700 --> 00:59:19.000 any result. Can we say something more? 00:59:19.000 --> 00:59:21.840 Can we say that the greatest common divisor 00:59:21.840 --> 00:59:25.040 can be maybe written as a linear combination 00:59:25.040 --> 00:59:26.830 of this type over there? 00:59:26.830 --> 00:59:30.610 And that's how we are going to proceed now. 00:59:30.610 --> 00:59:40.300 So let's set talk about the very special algorithm which 00:59:40.300 --> 00:59:43.240 is called Euclid's algorithm. 00:59:43.240 --> 00:59:47.410 And I think in the book it's also called The Pulverizer. 00:59:47.410 --> 00:59:50.940 And you will have a problem on this 00:59:50.940 --> 00:59:55.660 just to see how that works, and to really understand it. 00:59:55.660 --> 00:59:59.100 So let's explain what we want here. 00:59:59.100 --> 01:00:05.390 So first of all, we know that for any b and a, 01:00:05.390 --> 01:00:08.580 there exists a unique quotient and remainder r. 01:00:08.580 --> 01:00:11.740 So let's write it out. 01:00:11.740 --> 01:00:20.110 There exists unique q, which we will call the quotient. 01:00:25.280 --> 01:00:26.720 And r. 01:00:26.720 --> 01:00:27.845 We call this the remainder. 01:00:33.280 --> 01:00:44.670 Such that b equals q times a, plus r. 01:00:44.670 --> 01:00:54.380 With the property that 0 is at least r, and at most a. 01:00:54.380 --> 01:00:58.630 So we're not going to prove this statement. 01:00:58.630 --> 01:01:00.160 It's actually like a theorem, right? 01:01:00.160 --> 01:01:02.387 But let's just assume it for now. 01:01:02.387 --> 01:01:03.970 And in the book you can read about it. 01:01:07.805 --> 01:01:13.110 We're going to use this to prove the following lemma that we 01:01:13.110 --> 01:01:15.030 will need to give a characterization 01:01:15.030 --> 01:01:18.090 of the greatest common divisor, as a linear combination 01:01:18.090 --> 01:01:20.235 of integers. 01:01:23.280 --> 01:01:28.040 Oh, before I forget, you will denote this remainder 01:01:28.040 --> 01:01:34.400 as rem of b, a. 01:01:34.400 --> 01:01:38.930 And this is the notation that we use in this lecture. 01:01:38.930 --> 01:01:40.090 So what's the lemma? 01:01:40.090 --> 01:01:46.730 The lemma is that the greatest common divisor of a and b, 01:01:46.730 --> 01:01:52.790 is equal to the greatest common divisor of the remainder of b 01:01:52.790 --> 01:01:53.441 and a. 01:01:56.090 --> 01:01:56.590 With a. 01:01:56.590 --> 01:01:58.420 So what did we do? 01:01:58.420 --> 01:02:02.810 Let's give an example to see how this works. 01:02:02.810 --> 01:02:06.680 For example, let's take-- actually 01:02:06.680 --> 01:02:08.405 let's do it on this white board. 01:02:14.678 --> 01:02:16.175 So, let's see. 01:02:19.666 --> 01:02:21.040 For example, let's see whether we 01:02:21.040 --> 01:02:27.590 can use this to calculate the greatest common divisor 105, 01:02:27.590 --> 01:02:28.940 and 224. 01:02:31.260 --> 01:02:33.250 So how can we go ahead? 01:02:33.250 --> 01:02:38.000 Well, according to this lemma, we 01:02:38.000 --> 01:02:41.430 can rewrite this as the greatest common divisor 01:02:41.430 --> 01:02:49.720 of first the remainder of 224, after dividing out 01:02:49.720 --> 01:02:54.680 as many multiples of 105 as possible. 01:02:54.680 --> 01:02:55.180 And 105. 01:02:58.000 --> 01:03:01.640 So what are we going to use here? 01:03:01.640 --> 01:03:09.370 We're going to use that 224 is actually equal to 2 times 105, 01:03:09.370 --> 01:03:10.803 plus 14. 01:03:14.070 --> 01:03:18.340 So we had the GCD of 14 and 105. 01:03:18.340 --> 01:03:20.490 Now why can I do this? 01:03:20.490 --> 01:03:28.540 Well, I'm essentially just subtracting like 2 times 135 01:03:28.540 --> 01:03:30.850 from 224. 01:03:30.850 --> 01:03:32.320 Well, the greatest common divisor 01:03:32.320 --> 01:03:40.050 that divides 105 and 224 also divides 105, 01:03:40.050 --> 01:03:43.700 and a linear combination of 105, 224. 01:03:43.700 --> 01:03:46.430 That's essentially what we are using. 01:03:46.430 --> 01:03:49.370 And that's actually stated in this lemma, 01:03:49.370 --> 01:03:52.840 and that's what we would like to prove. 01:03:52.840 --> 01:03:55.370 So let's continue with this process, 01:03:55.370 --> 01:03:58.330 and do the same trick once more. 01:03:58.330 --> 01:04:03.900 So we can say that we can rewrite this as the greatest 01:04:03.900 --> 01:04:14.660 common divisor of, well, the remainder of 105 01:04:14.660 --> 01:04:20.150 after taking out this many multiples of 14 as possible, 01:04:20.150 --> 01:04:22.580 and 14. 01:04:22.580 --> 01:04:25.720 So what are we going to use over here? 01:04:25.720 --> 01:04:34.200 We are going to use that 105 is equal to 7 times 14, plus 7. 01:04:34.200 --> 01:04:40.580 So this is the greatest common divisor of 7, and 14. 01:04:40.580 --> 01:04:45.550 Now if you just continue this process, 01:04:45.550 --> 01:04:49.230 we can see that this is equal to the greatest common divisor, 01:04:49.230 --> 01:04:53.760 again, of the remainder of now 14, 01:04:53.760 --> 01:05:00.090 after dividing out as many multiples of 7 with 7. 01:05:00.090 --> 01:05:05.630 Now this is equal to 0, 7. 01:05:05.630 --> 01:05:06.180 Why is that? 01:05:06.180 --> 01:05:14.630 Because 14 is equal to 2 times 7, plus 0. 01:05:14.630 --> 01:05:18.940 So 0 is the remainder after dividing out 01:05:18.940 --> 01:05:21.510 7 as many possible times as possible. 01:05:21.510 --> 01:05:22.010 OK. 01:05:22.010 --> 01:05:25.180 So we have the greatest common divisor of 0, and 7. 01:05:25.180 --> 01:05:28.950 What's the largest integer that can divide both 0 and 7? 01:05:28.950 --> 01:05:32.590 Well, any integer can divide 0. 01:05:32.590 --> 01:05:36.280 So we know that this is equal to 7. 01:05:36.280 --> 01:05:38.590 So essentially, what we have done here, 01:05:38.590 --> 01:05:42.600 we have repeatedly used this particular lemma 01:05:42.600 --> 01:05:49.360 to compute in the end, the greatest common divisor of 105 01:05:49.360 --> 01:05:51.290 and 224. 01:05:51.290 --> 01:05:57.030 And we have been very methodol-- we have used a specific method. 01:05:57.030 --> 01:06:01.490 We used the lemma, and we worked it out. 01:06:01.490 --> 01:06:04.120 We used the lemma again, and we just 01:06:04.120 --> 01:06:05.880 plugged in the actual numbers. 01:06:05.880 --> 01:06:06.860 Used to lemma again. 01:06:06.860 --> 01:06:09.860 Plugged in the actual numbers, and so on. 01:06:09.860 --> 01:06:14.020 And this is what is called Euclid's algorithm. 01:06:14.020 --> 01:06:16.910 And in the book it's also called The Pulverizer. 01:06:16.910 --> 01:06:20.270 And there's, I think, a few other names. 01:06:20.270 --> 01:06:23.190 But I like this one. 01:06:23.190 --> 01:06:29.010 So this is an example of Euclid's algorithm. 01:06:29.010 --> 01:06:31.370 So now let's have to look whether we 01:06:31.370 --> 01:06:35.660 can have prove this particular lemma, 01:06:35.660 --> 01:06:42.970 and actually I will-- Yep. 01:06:42.970 --> 01:06:44.930 We're going to prove this lemma. 01:06:55.240 --> 01:06:56.030 OK. 01:06:56.030 --> 01:07:01.190 So how do we do the proof? 01:07:01.190 --> 01:07:05.640 Well, first before we know that if- yeah. 01:07:05.640 --> 01:07:06.840 Well, how do we do this? 01:07:06.840 --> 01:07:10.350 You would like to prove that if the great-- well, if n 01:07:10.350 --> 01:07:13.380 divides a and b, in particular, the greatest 01:07:13.380 --> 01:07:17.360 common divisor divides a and b. 01:07:17.360 --> 01:07:21.910 We would like to show that it's dividing also 01:07:21.910 --> 01:07:27.890 the remainder of b, after dividing out a, and a itself. 01:07:27.890 --> 01:07:30.200 If you can show that, then we know 01:07:30.200 --> 01:07:33.330 that the greatest common divisor of this thing 01:07:33.330 --> 01:07:35.740 is at least what we have over here. 01:07:35.740 --> 01:07:37.370 So I said a lot right now. 01:07:37.370 --> 01:07:40.210 So let's try to write it out a little bit. 01:07:40.210 --> 01:07:44.860 So suppose that m is any divisor of a. 01:07:44.860 --> 01:07:48.205 And at the same time, m also divides b. 01:07:51.690 --> 01:07:57.920 Well, then I know that m also divides 01:07:57.920 --> 01:08:04.520 b minus, say, the quotient, q that we had over here, times a. 01:08:04.520 --> 01:08:11.250 And-- and this is actually equal to the remainder of b and a. 01:08:11.250 --> 01:08:14.950 Now we also note that m divides a. 01:08:14.950 --> 01:08:16.210 So what did we show here? 01:08:16.210 --> 01:08:22.439 We showed that if m divides, and m divides b, then m 01:08:22.439 --> 01:08:25.649 also divides the remainder of b and a. 01:08:25.649 --> 01:08:27.700 And n divides a. 01:08:27.700 --> 01:08:29.790 So what does is prove? 01:08:29.790 --> 01:08:32.370 Well, it proves that, in particular, 01:08:32.370 --> 01:08:36.470 the greatest common divisor over here divides this one. 01:08:36.470 --> 01:08:37.859 That's interesting. 01:08:37.859 --> 01:08:42.930 That essentially means that we have shown this inequality. 01:08:42.930 --> 01:08:49.180 Because if this one divides this, 01:08:49.180 --> 01:08:51.970 well, that means that this number over here 01:08:51.970 --> 01:08:55.569 must be at least what we have over here. 01:08:55.569 --> 01:08:56.399 OK. 01:08:56.399 --> 01:08:58.400 So let's continue. 01:09:07.620 --> 01:09:09.080 We consider two cases. 01:09:09.080 --> 01:09:14.439 If the remainder of b and a is unequal to 0, well, 01:09:14.439 --> 01:09:17.600 what can we say now? 01:09:17.600 --> 01:09:21.859 We can say that if I know that m divides 01:09:21.859 --> 01:09:27.569 this remainder of b and a, which can be rewritten as b minus q, 01:09:27.569 --> 01:09:29.819 times a. 01:09:29.819 --> 01:09:37.569 And I also note that-- if I also know that n divides a, 01:09:37.569 --> 01:09:41.710 then this actually implies the reverse of this statement, 01:09:41.710 --> 01:09:45.290 that n divides a, and divides b. 01:09:45.290 --> 01:09:46.910 Now why is that? 01:09:46.910 --> 01:09:49.439 Well, we're actually using the fact 01:09:49.439 --> 01:09:53.960 that if n divides b, minus q, times a, and m divides a, 01:09:53.960 --> 01:09:57.680 then m also defies any linear combination of these two. 01:09:57.680 --> 01:10:04.620 In particular, this plus q, times a, which is b. 01:10:04.620 --> 01:10:05.710 m divides b. 01:10:05.710 --> 01:10:10.780 So maybe I'm going a little bit fast here, I notice. 01:10:10.780 --> 01:10:15.000 This all also has to do with all the lecture handouts. 01:10:15.000 --> 01:10:17.650 You see a few facts on the divisibility. 01:10:17.650 --> 01:10:23.380 And in particular, item number three that talks about the fact 01:10:23.380 --> 01:10:24.380 that I'm using here. 01:10:24.380 --> 01:10:28.610 If a divides b on your handout, and a divides c, then I 01:10:28.610 --> 01:10:33.800 know that a divides any linear combination of b and c. 01:10:33.800 --> 01:10:36.255 So that's essentially what I'm using here repeatedly. 01:10:39.090 --> 01:10:41.250 OK 01:10:43.610 --> 01:10:48.930 So let's look at the other case. 01:10:48.930 --> 01:10:54.620 If the remainder is equal to 0, well, then I actually 01:10:54.620 --> 01:11:00.540 know that b minus q, times a is equal to 0. 01:11:00.540 --> 01:11:07.780 Well, if I know that m divides a, well, 01:11:07.780 --> 01:11:16.380 then since 0 equals b minus q, times a, I know that b equals 01:11:16.380 --> 01:11:17.450 q, times a. 01:11:17.450 --> 01:11:21.260 So if m divides a, I also now that m divides b. 01:11:23.920 --> 01:11:25.570 So this is one argument. 01:11:25.570 --> 01:11:27.240 This is another one. 01:11:27.240 --> 01:11:30.610 And this was-- These are the three arguments that 01:11:30.610 --> 01:11:35.970 now show that anything that divides these two also 01:11:35.970 --> 01:11:37.439 divides a and b. 01:11:37.439 --> 01:11:39.230 So now we have the reverse argument, right? 01:11:39.230 --> 01:11:43.590 So this greatest common divisor divides this one here, 01:11:43.590 --> 01:11:45.340 and this one. 01:11:45.340 --> 01:11:49.002 And we just proved that it divides a and b, 01:11:49.002 --> 01:11:50.460 and so it must divides the greatest 01:11:50.460 --> 01:11:51.940 common divisor of a and b. 01:11:51.940 --> 01:11:55.030 So now we have shown the other inequality, 01:11:55.030 --> 01:11:56.190 and this proves equality. 01:11:56.190 --> 01:11:59.160 So you should definitely look this up in your lecture notes. 01:12:02.110 --> 01:12:05.790 So now we can finally prove this beautiful theorem 01:12:05.790 --> 01:12:12.895 a that will help us to characterize the-- actually, 01:12:12.895 --> 01:12:14.360 let me put this over here. 01:12:21.600 --> 01:12:24.600 So the final theorem that we prove here 01:12:24.600 --> 01:12:30.630 is that the greatest common divisor of a and b 01:12:30.630 --> 01:12:38.280 is actually a linear combination of a and b. 01:12:40.910 --> 01:12:43.280 So we're going to use this algorithm that you have 01:12:43.280 --> 01:12:47.460 over here, Euclid's algorithm. 01:12:47.460 --> 01:12:53.860 And we are going to do a proof, again, by induction. 01:12:53.860 --> 01:12:55.470 And we use an invariance. 01:12:55.470 --> 01:13:03.360 So we use a similar kind of strategy, of course. 01:13:03.360 --> 01:13:08.770 The invariance that we are going to use 01:13:08.770 --> 01:13:26.050 says-- well, if Euclid's algorithm reaches the greatest 01:13:26.050 --> 01:13:30.150 common divisor of x and y-- so for example, 01:13:30.150 --> 01:13:36.350 it's reach, say, 7 or 14, and 105, for example. 01:13:36.350 --> 01:13:47.790 Then, say, after n steps then both x and y are 01:13:47.790 --> 01:13:51.950 linear combinations of a and b. 01:13:51.950 --> 01:14:04.370 So then x and y are linear combinations of a and b. 01:14:04.370 --> 01:14:07.020 And at the same time, we also know 01:14:07.020 --> 01:14:10.270 that the greatest common divisor of a and b 01:14:10.270 --> 01:14:15.020 is equal to the greatest common divisor of x and y. 01:14:15.020 --> 01:14:17.390 So this is my invariance. 01:14:17.390 --> 01:14:24.540 And the way I will go ahead is to simply do what you do always 01:14:24.540 --> 01:14:25.650 in these situations. 01:14:25.650 --> 01:14:28.210 So we start with the base case. 01:14:28.210 --> 01:14:33.220 And we can immediately see that after 0 steps in the Euclidean 01:14:33.220 --> 01:14:35.530 algorithm, I've done absolutely nothing. 01:14:35.530 --> 01:14:42.890 So obviously after 0 steps, x equals a. 01:14:42.890 --> 01:14:44.200 y equals b. 01:14:44.200 --> 01:14:48.060 So of course, they are linear combinations of a and b. 01:14:48.060 --> 01:14:51.440 And this equality holds, as well. 01:14:51.440 --> 01:14:54.840 So for the base case-- 01:14:58.130 --> 01:15:04.790 So after 0 steps, we immediately know that p 0 is true. 01:15:04.790 --> 01:15:08.690 Now for the inductive step, we have to do a little bit more. 01:15:17.780 --> 01:15:19.025 As usual, right? 01:15:19.025 --> 01:15:24.410 We always assume p n. 01:15:24.410 --> 01:15:27.320 And now we would like to prove p n plus 1. 01:15:27.320 --> 01:15:29.572 So how do we do this? 01:15:29.572 --> 01:15:41.510 Well, we notice that there exists a q such 01:15:41.510 --> 01:15:49.730 that the remainder of y and x is equal to y minus q, times x. 01:15:49.730 --> 01:15:51.270 So we assume p n. 01:15:51.270 --> 01:15:54.470 We have reached some state, x, y. 01:15:54.470 --> 01:15:57.910 We know that the remainder of y, x equals y minus q, 01:15:57.910 --> 01:16:02.830 times x, for some quotient q. 01:16:02.830 --> 01:16:06.610 We know that y is a linear combination of a and b, 01:16:06.610 --> 01:16:08.610 and x is, as well. 01:16:08.610 --> 01:16:12.530 So that means that this one is actually also 01:16:12.530 --> 01:16:18.860 a linear combination of a and b. 01:16:18.860 --> 01:16:25.420 So now when we look at this , algorithm we can see that-- 01:16:25.420 --> 01:16:28.320 that if you look at the remainder that appears in here, 01:16:28.320 --> 01:16:30.850 that's still a linear combination of a and b. 01:16:30.850 --> 01:16:35.420 So after a extra step, we notice that what we have reached 01:16:35.420 --> 01:16:38.170 are still in combinations of a and b. 01:16:38.170 --> 01:16:40.890 And of course, the lemme has showed-- 01:16:40.890 --> 01:16:46.392 has shown us that what we reach is still equal-- 01:16:46.392 --> 01:16:48.600 the greatest common divisor is still equal to what we 01:16:48.600 --> 01:16:51.320 originally started out with. 01:16:51.320 --> 01:16:54.300 So this proves p of n plus 1. 01:16:58.800 --> 01:17:03.380 So n-- let's finish this particular proof. 01:17:03.380 --> 01:17:07.140 So for the very last step, if you now 01:17:07.140 --> 01:17:14.470 look at this particular-- so if you look at the very end, 01:17:14.470 --> 01:17:17.950 we notice that in every step the remainder 01:17:17.950 --> 01:17:20.260 is getting smaller, and smaller, and smaller. 01:17:20.260 --> 01:17:22.250 Right? 01:17:22.250 --> 01:17:24.700 And you can use a similar kind of proof technique 01:17:24.700 --> 01:17:28.510 to show that after a finite number of steps, 01:17:28.510 --> 01:17:33.190 we will reach a GDP of 0, y. 01:17:33.190 --> 01:17:35.200 Something like this. 01:17:35.200 --> 01:17:41.030 So in the very last step of Euclid's algorithm 01:17:41.030 --> 01:17:44.890 we achieve something off this form. 01:17:44.890 --> 01:17:49.220 We now use our predicate over here, 01:17:49.220 --> 01:17:53.890 and say that y is a linear combination of a and b, 01:17:53.890 --> 01:17:57.020 but the greatest common divisor of 0, y 01:17:57.020 --> 01:17:59.900 is also equal to the original greatest common divisor 01:17:59.900 --> 01:18:01.860 that we want to characterize. 01:18:01.860 --> 01:18:05.760 So now we have proved the theorem 01:18:05.760 --> 01:18:08.110 that says that the greatest common divisor of a and b 01:18:08.110 --> 01:18:11.830 is actually a linear combination. 01:18:11.830 --> 01:18:13.840 So now we're going to combine all those three 01:18:13.840 --> 01:18:16.969 theorems in one go. 01:18:16.969 --> 01:18:25.580 And that will show us the final result, 01:18:25.580 --> 01:18:35.230 which is that the theorem that the greatest common divisor 01:18:35.230 --> 01:18:42.010 of a and b is actually the smallest 01:18:42.010 --> 01:18:54.000 positive linear combination of a and b. 01:18:54.000 --> 01:18:56.840 So we're going to combine all of these together. 01:18:56.840 --> 01:19:01.970 We know that the greatest common divisor divides any result. 01:19:01.970 --> 01:19:05.160 The theorem up there says that any linear combination 01:19:05.160 --> 01:19:07.160 can be reached. 01:19:07.160 --> 01:19:09.710 And also just showed-- have shown 01:19:09.710 --> 01:19:12.325 that the greatest common divisor is a linear combination of a 01:19:12.325 --> 01:19:13.890 and b. 01:19:13.890 --> 01:19:17.450 So we can combine those three to get this theorem. 01:19:17.450 --> 01:19:19.120 So how do we do it? 01:19:19.120 --> 01:19:25.620 Well, let's just look 0 all the way up to b. 01:19:25.620 --> 01:19:29.770 Suppose these are all the results that we 01:19:29.770 --> 01:19:32.530 can reach in our problem. 01:19:32.530 --> 01:19:38.160 We know that the greatest common divisor divides all of those. 01:19:38.160 --> 01:19:41.390 At the same time, it's also linear combination 01:19:41.390 --> 01:19:42.160 that's over here. 01:19:42.160 --> 01:19:45.860 Since it's a linear combination, it can also be reached, right? 01:19:45.860 --> 01:19:47.840 By the theorem that we have. 01:19:47.840 --> 01:19:51.099 So suppose that this is the greatest common divisor. 01:19:51.099 --> 01:19:53.140 But we also know that the greatest common divisor 01:19:53.140 --> 01:19:57.760 is dividing all of these points here that can be reached. 01:19:57.760 --> 01:20:01.420 So therefore, it must be the smallest one. 01:20:01.420 --> 01:20:04.050 And I will leave you with some homework 01:20:04.050 --> 01:20:06.210 to think about this very carefully. 01:20:06.210 --> 01:20:09.220 And you can show for yourself that you can now 01:20:09.220 --> 01:20:12.660 combine those three arguments together, and see 01:20:12.660 --> 01:20:15.630 that the greatest common divisor must be the smallest positive 01:20:15.630 --> 01:20:18.420 linear combination. 01:20:18.420 --> 01:20:21.770 So, I will see next Thursday.