Prove that the additive inverse of an odd integer is an odd integer

This is a homework problem, but I don't want the answer, just a little guidance:

Prove that the additive inverse of an odd integer is an odd integer.

When approaching a problem like this, how much is it safe to assume? Is it safe to assume that "the additive inverse of an integer is an integer?" Or does that need to be proven first, before we can start talking about odds and evens?

I have two ideas about how to approach this, and that is to either:

1) Use absolute value to negate the fact that something is negative so that the absolute values of something like $4$ and $-4$ are both $4$. But is it safe to assume something like "the absolute values of any integer positive or negative are equal?"

2) Do something like subtract $2$ times a number to get the negative or positive: e.g. the additive inverse of $4$ is $(4 - 2(4))$. The additive inverse of $-4$ is $(-4 -(2(-4))$.

Exactly where I would follow those ideas to, I'm not sure yet, but I'd like to at least know I'm on the right track and not completely going off in the wrong direction.

Solution 1:

The sum of an odd and an even integer is odd, since $(2m + 1) + 2n = 2(m + n) + 1$.

Let $x$ be any odd integer. Then $-x$ is odd, since otherwise, $x + (-x) = 0$ is odd.

Solution 2:

Hint: Start your proof like this:

Let $k$ be any arbitrary odd integer. Then by the definition of an odd integer, we have $k=2a+1$ for some integer $a$. Thus...

Then consider $-k=-(2a+1)$ and perform some algebraic manipulations. Your final step should look something like:

...hence, since $-k=2b+1$ where $b$ is an integer, it follows by definition that $-k$ is also an odd integer, as desired.

Solution 3:

An integer $n$ is odd if and only if there exists an integer $k$ such that $n = 2k+1$. (Note, $2$ does not divide $n = 2k + 1$: $2$ divides $2k$, but not $1$, hence $n = 2k + 1$ is not even, therefore is odd).

So let $n$ be an arbitrary odd integer; i.e. $n = 2k+1$ where $k$ is some integer.

Then

$\begin{align} \;-n & = -(2k+1) \\ & = -2k -1 \\ & = -2k + (- 2 + 2) - 1 \\ & =(-2k - 2) + (2 - 1) \\ &=2(-k-1) + 1\end{align}$.

Now, $\;j = (-k - 1)\,$ is some integer (because $k$ is some integer). So have that $-n = 2(-k -1) = 2j + 1$, which by definition, is an odd integer.

Solution 4:

One could prove this inductively:

Assume that the $n^\text{th}$ odd positive integer, $2n-1$, has an odd negation. Then $-2n+1$ is odd, so $$\underbrace{-2n+1}_{\text{odd}}-\underbrace{2}_{\text{even}}=\underbrace{-2n-1}_{\text{odd}}.$$ Thus the statement holds for the $(n+1)^\text{th}$ positive integer, $2n+1$.