Methods of proof is a subject relevant to all fields in mathematics, since it establishes a framework by which typical rigorously proven tools are utilized to systematically decide upon the validity of a statement as being either true or false.

Different methods of proof are useful depending on the scope of the proposition as well as on the elements the proposition applies to.

For example:

Propositions about $\N$ will typically ask for the use of induction.
Propositions about finite or infinite groups of elements with clear cut subgrouping will invite the use of proof by cases. In the following sections we will show the mechanism of each method, and provide an example applied to a proposition.

Many times, a theorem or a set of theorems previously shown to be true, can be utilized directly in order to prove a newly formulated proposition.
The main work to be done in a direct proof is the correct utilization of the tools for manipulation in the math domain and the presentation and ordering of theorems utilized to establish the truth of the new proposition to be established as true. Example:

Proposition

$a(b-c) = ab-ac$

Direct proof

We proved in the section about the field axioms of the real numbers the theorem $(-1)a = -a$, and applied in our case, $(-1)ac= -ac$ and $(-1)c=-c$.
Thus,
$ab-ac=ab+(-1)ac=a(b+(-1)c)=a(b-c) \quad \Box$
Here, In this example we utilized directly, a previously proven theorem and one axiom (namely, distributivity and the property of algebra mentioned).

Sometimes the proposition stated targets a finite or infinite set of elements, that can be subgrouped easily and then a proof by cases (where we probe each case and check the truth of the proposition) is viable and recommended.

Example:

Proposition

if $|x-4| + |3-x| > 13$,
then $x<-3$ or $x>10$.

Proof by cases

Because of the subgrouping into cases nature of the absolute value definition, we should break the inequality into three cases and analyze each.
If we reach the same intervals $x$ must be included in for satisfying the inequality as the proposition, then the proposition must be true (a more rigorous explanation comes from symbolic logic. Since the structure of the proposition in symbolic logic is $ p \to q$, its negation is $\neg (p \to q) \equiv p \wedge \neg q$.
Now, if we assume the truth of p , namely, the given inequality, and we get to different intervals for $x$ than the ones stated, then $\neg q$ is satisfied and the negation of the entire proposition is true, so the the proposition itself is false).

1) $4 \leq x$:

$|x-4| + |3-x| > 13 \rightarrow x-4 + [-(3-x)] > 13$
$x-4-3+x>13 \rightarrow 2x > 20 \rightarrow x>10$

2) $3\leq x<4$:

$|x-4| + |3-x| > 13 \rightarrow -x+4 + [-(3-x)] > 13$
$-x+4-3+x>13 \rightarrow 1 > 13 \rightarrow false $

3) $x<3$:

$|x-4| + |3-x| > 13 \rightarrow -x+4 + 3-x > 13$
$ \rightarrow 2x < -6 \rightarrow x < -3 $

The breaking into 3 subgroups for analysis covers all of $\R$, and thus allows a complete proof for the truth of the proposition.

Sometimes the proposition is not easily proven by using the previous two methods, but can be demonstrated to be true using a binary approach: Show the contradiction of the proposition is false, thus, the proposition must be true. This means that, for such propositions, the ability to demonstrate the falseness of their contradiction is easier, maybe utilizing the other methods.

Example:

Proposition

if $x^2 = 2$, then $x \notin \mathbb{Q}$

Proof by contradiction

This is a classic example of using the binary nature of the statement in the proof tactic. In this case , we can assume the validity of the equation, and that either $x \notin \mathbb{Q}$ or $x \in \mathbb{Q}$.
Since the definition for being part of the rational set is normative, we can use it directly. Namely, assume $x \in \mathbb{Q}$, then $x=\frac{n}{m}$, where $n, m \in \Z$ and that $n,m$ have no common factors (since if there are, we can always cancel).

Then, we can write:

$x^2 = (\frac{n}{m})^2 = 2 \rightarrow n^2 = 2m^2$

From the properties of odd/even integers (we won't prove them here from first principles) and exponents, we conclude that $n | 2 $ (meaning $n$ is divisible by 2).
Thus, we can write $\exists k \in \Z$, such that $n = 2k$. Substituting, $n^2 = (2k)^2 = 4k^2 = 2m^2 \rightarrow m^2 = 2k^2$
This implies, $m | 2$ as well.

However, this gives a contradiction, since our assumption entailed $n, m$ not having common factors, but our analysis led to $n, m$ having the common factor $2$. A rigorous explanation of why the proof by contradiction works, demands the usage of symbolic logic:

We assumed $\neg (p \to q) \equiv p \wedge \neg q $ is true, and established $\neg (p \to q) \to C \equiv T$ (where $C$ stands for contradiction, and always takes the value of $F$, since in formal logic a statement cannot be false and true and the same time).
This implies that $\neg (p \to q) \equiv F$ so that $(p \to q) \equiv T$ and the proposition is established as true.

Note that the technique utilizes the contradiction tool, but in order to establish the contradiction, the reverse proposition assumed true is analyzed by the direct proof method (namely, using the established theorems from number theory about properties of integers and their powers regarding evenness\oddity)

As mentioned, propositions about the set $\N$ or subsets of $\N$ are welcoming the use of a tool called mathematical induction to establish their truth.

How does mathematical induction work?

1) Establish the proposition truth about a base specific case (induction root):

$$P(k=1)$$
2) Assume the truth of the proposition for a general $n \in \N$:

$$P(k=n)$$
3) Show that given the assumption of $P(k=n) \equiv T$ we can establish that $P(k=n+1) \equiv T$, and a consequence $P(k=n) \to P(k=n+1) \equiv T$.

In essence, steps 1, and 3 are the essential steps (where step 2 is an auxiliary tool to establish step 3), since once we have an anchor (root) case to apply the proposition to in a true sense, then by 3, $P(k = 1) \to P(k = 2)$ must be true, implying $P(k = 2)$ is true as well. Then we can do the same ad infinitum (since $\N$ is unbounded above), and thus covering the proposition for all members of $\N$, we create a complete proof of the proposition.

We are not obliged to anchor ourselves at $k=1$, or even show the implication is true for a single step integer increase in case we want to prove a statement about some subset of $N$ [For example, a statement about the even naturals, starting from 10 included, would invite the root case to be $P(k = 10)$ and the implication to be established to become $P(k = n) \to P(k = n + 2)$, where the generalized $n$ is assumed to satisfy $n|2$.]

It should be noted that there are two variants in use for mathematical induction:

Regular induction and Complete induction

The difference between them is only in step 2, where complete induction assumes the truth of a complete set of cases up to the root case, namely $\{P(k=1), P(k=2), ..., P(k=m)\}$, where $m < n$, to establish the truth of the implication $\{P(k=1), P(k=2), ..., P(k=m)\} \to P(k = n)$.

The complete induction variant exists, because some propositions are more easily and elegantly proven by taking into account an aggregate form implication, instead of step-wise implication form.
However, most propositions calling for proof by induction can be proven in a simple manner by utilizing either variant of the induction method.

Example:

Proposition

Prove: $\forall n \in \N$, $1+2+3+ ... +n = \frac{n(n+1)}{2}$

Proof by induction

1) Base case: $1 = \frac{1 \cdot (1+1)}{2} = \frac{2}{2}= 1$ 2) Assume true for $P(k=n)$. 3) Show that if (2) is true, then $P(k=n) \to P(k=n+1)$:
$1+2+...+n + (n+1) = \frac{n(n+1)}{2} + (n+1) = \frac{n(n+1)+2(n+1)}{2} = \frac{(n+1)(n+2)}{2}$, which is just the form of the proposition right hand side for $k = n + 1$.
Thus, $P(k=n) \to P(k=n+1) \equiv T \quad \Box$.