Prove that if n is an integer and 3n + 2 is even then n is even.

Solution:

We have to prove that if n is an integer and 3n + 2 is even then n is even.

A proof by contradiction means we will prove that if the given statement is F, then this will lead to a contradiction.

For the given statement 3n + 2 to be F, 3n + 2 has to be even while n is odd.

Assume 3n + 2 is even and n is odd.

Substitute with the assumed n in the given expression,

3n + 2 = 3(2m + 1) + 2

= 6m + 3 + 2

= 6m + 5

Rewriting the expression,

= (6m + 4) + 1

= 2(3m + 2) + 1

Since m is an integer,so 3m + 2 = k.

This means that 3n + 2 is of the form 2l + 1, and hence is odd.

But this contradicts the fact that we are given 3n + 2 is even.

Hence, n cannot be odd.

The expression implies that 3m + 2 is odd which contradicts the assumption 3n + 2 is even

Therefore, it is proved that n is an integer and 3n + 2 is even.

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Prove that if n is an integer and 3n + 2 is even then n is even.

Summary:

It is proved that if n is an integer and 3n + 2 is even then n is even.