## 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, m_{1} and m_{2}, respectively. Of course, m_{1} + m_{2} = k. By the induction hypothesis we know that m_{1} is even and m_{2} is even. So the total _{1} + m_{2}

Induction is complete.

|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny

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.

|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny

71211225