1.2 Proof Methods | 1.2 Proof Methods | Unit 1: Proofs | Mathematics for Computer Science | Electrical Engineering and Computer Science

<A Bogus Proof by Cases
1.2.1Proof By Contradiction: Video
1.2.2Proof By Contradiction
1.2.3Proof By Cases: Video
1.2.4When to Prove by Cases
1.2.5Friends and Strangers
1.2.6A Bogus Proof by Cases
1.2.7A Bogus Proof by Contradiction
>Well Ordering Principle

A Bogus Proof by Contradiction

Which step(s) contain the logical error?

Bogus Claim: \(\sqrt{4}\) is irrational.

This proof is by contradiction.

Suppose the claim is false: \(\sqrt{4}\) is rational.

We can write \(\sqrt{4}=\frac{n}{d}\), where \(n\) and \(d\) are integers with no common factors.

Squaring both sides and moving \(d^2\) to the left-hand side, we get \(4d^2 = n^2\).

This implies that \(n^2\) is a multiple of 4, and thus 4 is a factor of \(n\).

Therefore, \(n^2\) is a multiple of 16.

But since \(4d^2 = n^2\), \(d^2\) must also be a multiple of 4, and thus 4 is a factor of \(d\).

The numerator and denominator have a common factor of 4, which is a contradiction. Hence, \(\sqrt{4}\) must be irrational. \(\blacksquare\)

Explanation

This proof makes the same logical error twice. It claims that since \(n^2\) is a multiple of 4, then 4 must also be a factor of \(n\). However, only 2 must be a factor of \(n\).