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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.

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Do you know the definition of an odd integer? –  Vectk Jun 23 '13 at 1:02
    
We were told that we could just use "2a + 1" for odd integers. –  CptSupermrkt Jun 23 '13 at 1:03
    
Can you figure out a way to write $-(2a + 1)$ in the form $2x + 1$ where $x$ is an integer? –  Vectk Jun 23 '13 at 1:05
    
Sadly, no. That's how much I suck at math :( I just tried -(2a+1) = (2x+1), which ultimately gives that a = -x, or -a = x. But I'm not sure how to interpret that into anything meaningful. –  CptSupermrkt Jun 23 '13 at 1:10
    
@CptSupermrkt You should get: $$ \begin{align*} 2x+1&=-(2a+1) \\ 2x+1&=-2a-1 \\ 2x&=-2a-2\\ 2x&=2(-a-1)\\ x &= -a-1 \end{align*} $$ This is good! Since the set of all integers is closed under addition and multiplication, $x$ must also be an integer. –  Adriano Jun 23 '13 at 1:12
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7 Answers

up vote 11 down vote accepted

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.

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Thanks, @Amzoti! –  amWhy Jun 23 '13 at 2:57
    
All were great answers. I learned a lot, and I wish I could choose more than one. But this one was the easiest for me to wrap my non-math inclined mind around. –  CptSupermrkt Jun 25 '13 at 1:17
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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.

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This is cute. ${}{}$ –  Pedro Tamaroff Jun 23 '13 at 1:56
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but non-constructive –  Henry Jun 23 '13 at 13:02
    
This blew my mind, but is unfortunately not what I needed. Thanks anyway :) –  CptSupermrkt Jun 25 '13 at 0:23
    
@Henry How so? ${}{}{}$ –  Pedro Tamaroff Jun 25 '13 at 0:49
    
@Peter Presumably because it does not construct the additive inverse –  tba010 Jun 25 '13 at 1:42
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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.

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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, $2n+1$.

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I like the simplicity of this approach. Pedantic: you still also need to prove the base case. –  Lie Ryan Jun 23 '13 at 17:12
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Here's my proof of the equivalent statement for even integers. I'll let you figure out what to do for odds.

Let $n$ be an even integer. By definition of "even", $\exists m \in \mathbb{Z}: n=2m$. Then $-n = -2m = 2(-m)$. Because $\mathbb{Z}$ is closed under additive inverse, $-m$ is an integer. By the definition of "even", $2(-m) = -n$ is also even, Q.E.D.

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Let our odd number be $o$. Since $-1\equiv 1 \pmod{2}$, $-o\equiv o \pmod{2}$. By the definition of odd numbers, $o\equiv 1 \pmod{2}$. Combining this with the earlier statement, one gets

$-o\equiv o \equiv 1 \pmod{2}$ Thus, $-o\equiv 1 \pmod{2}$, which means that $-o$ is odd, using the definition of odd numbers.

NOTE:This is my first answer on math.SE, apologies for any newbie mistakes.

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When approaching a problem like this, how much is it safe to assume?

When proving basic/obvious properties of integers, it's usually safest to assume just the properties of addition and multiplication on integers and axioms of equality. If you're doing this as part of a course, your lecturer would usually also say that you can use (without proofs) most results that have been proven in the lecture/class.

Is it safe to assume that "the negation of an integer is an integer?"

Yes, because the axiom "Existence of inverse elements" states that for every integer $n$, there exist an integer $(-n)$ such that $n + (-n) = 0$.

Prove that the negation of an odd integer is an odd integer.

Definition

The set of Odd numbers is defined by $\{2x + 1 \mid x \in \mathbb{Z}\}$

Goal

Prove:

Theorem 1: $-(2x + 1) = 2y + 1$, where $x,y \in \mathbb{Z}$


Before we start proving Theorem 1, we first will need to prove the following lemmas:

Lemma 1: $-(x + y) = -x + -y$ for $x,y \in \mathbb Z$

Lemma 2: $-1 \times x = -x$ for $x \in \mathbb Z$

Proof of Lemma 1

$$ \begin{eqnarray} -x + -y & = & -x + -y + 0 & (1)\\ & = & -x + -y + ((x + y) + -(x + y)) & (2) \\ & = & (x + -x) + (y + -y) + -(x + y) & (3) \\ & = & 0 + 0 + -(x + y) & (4) \\ & = & -(x + y) & (5) \\ \end{eqnarray} $$

Therefore $-(x + y) = -x + -y$

Explanations or rules used on each step:

  1. $a + 0 = 0$
  2. $a + (-a) = 0$
  3. commutative and associative law
  4. $a + (-a) = 0$
  5. $a + 0 = 0$

Proof of Lemma 2

Fill this part yourself.

Proof of Theorem 1

This part is essentially the same as @amWhy's answer with a little bit more care to use only the basic axioms and the lemmas

$$ \begin{eqnarray} -(2x + 1) & = & -2x + -1 & (1) \\ & = & -2x + 0 + -1 & (2) \\ & = & -2x + (2 + -2) + -1 & (3) \\ & = & (-2x + -2) + (2 + -1) & (4) \\ & = & -2 \times (x + 1) + 1 & (5) \\ & = & (-1 \times 2) \times (x + 1) + 1 & (6) \\ & = & 2 \times (-1 \times (x + 1)) + 1 & (7) \\ & = & 2y + 1 & (8) \\ \end{eqnarray} $$

Because $-(2x + 1)$ can be written in the form $2y + 1$, therefore $-(2x + 1)$ is an odd integer for any $x \in \mathbb Z$.

Explanations or rules used on each step:

  1. Lemma 1
  2. a + 0 = a
  3. a + -a = 0
  4. commutative and associative law
  5. distributive law
  6. Lemma 2
  7. commutative and associative law
  8. Let $y = -1 \times (x + 1)$, note that $y \in \mathbb Z$ because $x \in \mathbb Z$ and closure of addition and multiplication
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protected by Alexander Gruber Jun 23 '13 at 14:56

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