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.
$\endgroup$ 24 Answers
$\begingroup$Hint: $2k = (2k - 1) + 1$.
$1$ is an odd number. Is $2k-1$ odd? Even? (Both? neither? impossible to tell?)
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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 $$a + b = 2n + 2m + 2 = 2(n+m+1).$$
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