Well-ordering proof

What is wrong with this "proof" of the following statement?

For every positive integer n, n, the number n2+n+1 n^2+n+1 is even.

Proof:

Let S S be the subset of positive integers nn for which n2+n+1n^2+n+1 is odd. Assume S S is nonempty.

Let m m be its smallest element.

Then m1S, m-1 \notin S, so (m1)2+(m1)+1 (m-1)^2+(m-1)+1 is even.

But (m1)2+(m1)+1=m2m+1=(m2+m+1)2m,(m-1)^2+(m-1)+1 = m^2-m+1 = (m^2+m+1)-2m, so m2+m+1 m^2+m+1 equals ((m1)2+(m1)+1)+2m, \big((m-1)^2+(m-1)+1\big)+2m, which is a sum of two even numbers, which is even.

So mS; m \notin S; which is a contradiction. Therefore, S S is empty, and the result follows.

×

Problem Loading...

Note Loading...

Set Loading...