Proving that even numbers equal the sum of two odd numbers.

Define E to be the set of even integers; E = {$x$ $\in$ $\mathbb{Z}$ : $x$ = 2$k$, where $k$ $\in$ $\mathbb{Z}$}.

Define F to be the set of integers that can be expressed as the sum of two odd numbers.

Prove E = F.

My attempt: The only way I can figure out the solution is by providing numbers and examples. It's easy to see that two odd numbers will always equal an even integer. I just don't know how to write the proof for it.

• You also have to show that every even number can be written as the sum of two odd numbers, but this is obvious (For even $n$, just choose $n-1$ and $1$). If you actually want to prove that the sum of two odd numbers is even, use the modulo-arithmetic : $a,b \equiv 1\mod 2\implies a+b\equiv 1+1=2\equiv 0\mod 2$ – Peter Sep 12 '17 at 21:38
• try writing them like (2k+1)+(2j+1) what does this equal ?? – user451844 Sep 12 '17 at 21:41

Hint: $2k = (2k - 1) + 1$.

$1$ is an odd number. Is $2k-1$ odd? Even? (Both? neither? impossible to tell?)

This is a quick and simple way to look at it.

Let $F$ be the set of integers that can be represented as $(2a+1)+(2b+1)$ for $a,b \in \mathbb{Z}$. See that $(2a+1)+(2b+1) = 2(a+b+1)$ and that $(a+b+1)$ is an integer.

Thus the RHS is an even integer.

You have two show both inclusions, i.e. $E \subset F$ and vice versa.

Let $a = 2n + 1$ and $b = 2m + 1$ be odd numbers for $n, m \in \mathbb Z$. Then $a + b = 2n + 1 + 2m + 1 = 2(n + m + 1)$ is even.

If $x = 2k$ is even, we consider two cases. If $k$ is odd, i.e. $k = 2n + 1$ for some $n$, then $x = 2k = k + k$ is a trivial sum of odd numbers. If $k$ is even, i.e. $k = 2n$, we have $x = 2k = k + k = (k - 1) + (k + 1)$, where $k-1$ and $k+1$ are odd numbers, since $k$ is even. QED

The motivation is: If you have for example $10$, you can write it as $10 = 5 + 5 = 4 + 6$, as you probably have seen yet. Of course you could also write $10 = 9 + 1$ or so.

You can show that the sum of 2 odd numbers is always an even number as follows.

We have two odd integers $a$ and $b$, which can be written as $a = 2 n + 1$ and $b = 2 m + 1$, for some integers $n$ and $m$, so $$a + b = 2n + 2m + 2 = 2(n+m+1).$$