Proof that an even number plus an odd number is equals to an odd number

Question

How can I prove using an induction technique that an even number plus an odd number is always equals to an odd number?

Thanks in advance!

Answer 1

Induction is not needed.

An integer $m$ is even iff $m = 2n$ for some integer $n$.

Similarly, an integer $m$ is odd iff $m = 2n+1$ for some integer $n$.

The sum of an even and odd integer is of the form $2n+2m+1 =2(n+m)+1 $ so it is odd.

Exercise: Prove that an integer can not both be even and odd.

Answer 2

No need for induction. Consider $2n$ and $2s+1$, numbers that are even and odd for $n,s\in\Bbb Z$ respectively. Then the sum is given by $2n+2s+1=2(n+s)+1$ which is odd.

Answer 3

The use of induction here is somewhat contrived but here goes . . .

Let $2m$ be an arbitrary even number.

Add $1$. Then $2m+1$ is odd.

Assume $2m+(2n+1)$ is odd for all odd numbers less than or equal to $2n+1$.

Consider the next odd number. We have $$2m+(2(n+1)+1)=2(m+n+1)+1$$ is odd by definition (as it is one away from a multiple of two).

But $2m$ was arbitrarily even. $\square$