All Integers Are Even

We are going to prove by induction that,

For all integer n ≥ 2, n is even.

(This was brought up by Adam Liss at a Linkedin discussion.)

The claim is verified for n = 2; for indeed, 2 is the smallest even number.

Assume the claim holds for all 2 ≤ n < k, that is

All integers below k starting with 2 are even.

From these we now derive that k is also even.

Split it into two parts, m₁ and m₂, respectively. Of course, m₁ + m₂ = k. By the induction hypothesis we know that m₁ is even and m₂ is even. So the total N = m₁ + m₂ must be even.

Induction is complete.

What went wrong?

The error in the proof is subtle. The inductive step had to be formulated as

Assume the claim holds for all 2 ≤ n < k, that is, all integers below k starting with 2 are even, k ≥ 2.

The derivation then clearly fails for k = 3 since it is impossible to split 3 into the sum of two numbers each at least 2.

508 Resource Limit Is Reached

Resource Limit Is Reached

The website is temporarily unable to service your request as it exceeded resource limit. Please try again later.