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?
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Copyright © 1996-2012 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.
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Copyright © 1996-2012 Alexander Bogomolny
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