1.2 Proof Methods

A Bogus Proof by Cases

Which step(s) contain the logical error?

Bogus Claim: For any integer \(a\), \(\;\;\;2a^2 > a\).

Exercise 1

Case 1: \(a\) is positive.

Case 2: \(a\) is negative.



These two cases are NOT exhaustive, they leave out the case where \(a=0\), in which case \(2a^2=2\cdot 0^2=0=a\).

Line 2 is not logically erranous because saying "There are two cases" implies the existence of at least two cases.

Line 3 claims that those are the ONLY two cases, and is therefore incorrect.

You should note that this proof can easily be modified to prove the correct claim that for any integer \(a\), \(2a^2 \geq a\)