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:
- $a + 0 = 0$
- $a + (-a) = 0$
- commutative and associative law
- $a + (-a) = 0$
- $a + 0 = 0$
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:
- Lemma 1
- a + 0 = a
- a + -a = 0
- commutative and associative law
- distributive law
- Lemma 2
- commutative and associative law
- Let $y = -1 \times (x + 1)$, note that $y \in \mathbb Z$ because $x \in \mathbb Z$ and closure of addition and multiplication