Proofs Discrete Maths

Question

For the following ‘proof’, find the mistake and explain how it can be rectified.

Theorem: For every natural number x, x is even if and only if x^2x2 is even.

Proof: Let x be an even number. Then, by the definition of 'even', there exists a natural number kk such that x = 2kx=2k. Therefore, x^2 = (2k)^2 = 4k^2 = 2(2k^2).x2=(2k)2=4k2=2(2k2). Hence there exists a natural number, namely l = 2k^2l=2k2, such that x^2 = 2lx2=2l. Therefore, by the definition of 'even', x^2x2 is even.

Answer 1

There isn't much wrong with this except for the fact that the proof is incomplete. This is an if and only if statement. We proved that if x is even then x² is even, but we didn't prove that if x² is even, then x is even.

You need to prove both ways.