Proving Mathematical Statements

When it comes to proves, there are some key terms to be defined.

First, a proposition. A proposition is a claim that needs to either be proven or disproven.

There are three main kinds of proposition:

1. Theorem: These are strong propositions.
2. Lemma: These are subsidiary propositions of a theorem, used to proof the theorem.
3. Corollary: These are propositions that follow almost immediately from a theorem.

Direct Proofs#

Choose a representatives $x, y \ldots$ from sets $S, T, \ldots$

Use the representatives to show that the open sentence $P(x)$ is true. Starting from left hand side and working our way to the right hand side

Since $x, y\ldots$ can be any member of $S, T, \ldots$ respectively, it must be true for all of $S$, therefore, we conclude the proof.

Never begin by assuming that the open sentence is true!

Direct Proofs: An Example#

Proof that for every real number $x, x^2 + 1 \geq 2x$

We start by identifying that if $x\in \mathbb{R}$, then $x - 1\in \mathbb{R}$

$\begin{aligned}(x - 1)^2 \geq 0 \implies &x^2 - 2x + 1 \geq 0 \&x^2 + 1 \geq 2x \text{ (shown)}\end{aligned}$

Notice how we started by assuming the variable was in the specified domain and went from there.

An alternate strategy is to break the given domain up into sets.

Often, this is done in conjunction with trying to prove universally quantified statements

Proving Universally Quantified Statements $\forall$#

Case Analysis, a Direct Proofs Method#

For example, to prove that $\forall k \in \mathbb{Z} k(k - 1) \geq 0$

Consider 3 separate domains:

• $k \leq -1$
  • $k \leq -1 \implies k - 1 \leq -2 \implies k(k-1) \geq 2 \geq 0$
• $k = 0 \cup k = 1$
  • $k = 1, k (k-1) = 0$
  • $k = 0, k (k-1) = 0$
• $k \geq 2$
  • $k\geq 2\implies k-1 \geq 1 \implies k(k-1) \geq 2 \geq 0$

Of course, rather than just writing pure math, fun as it is, it is important to explain each step in english to make sure that there are no ambiguity.

Like for example, why $k-1 \geq 1 \implies k(k-1) \geq 2$ etc.

Proving identities#

Proving identities is easily done by starting with the left hand side and working your way to the right.

It might be helpful to start from the right and work your way back.

Remember though, proofs should always start with what is given (LHS).

Disproving Universally Quantified Statements#

To disprove a Universally quantified statement, simply find a counter example.

That is, an example within the domain such that the open sentence is false.

Proving Existentially Quantified Statements $\exists$#

To Prove such a statement, often we do what we do when disproving universally quantified statements.

We simply find an example such that the given statement is true.

This can be done is one of two ways:

1. Trial and error (Okay for small domains)
2. Mathematically

To prove the statement mathematically, we often transform the open sentence into some other form. The quadratic form is one example.

Often, these transformations might result in additional solutions to the original equation, otherwise known as extraneous solutions.

There are two ways to deal with extraneous solutions:

• check against the original open sentence
• make sure that transformations are invertible.
  • Something is invertible if the solutions to the transformed equation is exactly the same as the original equation.

The goal is to narrow the values for which we can find an $x$ such that $P(x)$ is true.

We are not assuming that $P(x)$ is true for all $x$!

Disproving Existentially Quantified Statements#

To disprove existentially quantified statements, simply proved the negated statement.

That is, to disprove $\exist x \in \mathbb{R}, P(x)$, prove that $\forall x \in \mathbb{R}, \neg P(x)$

Proving Nested Quantifiers#

If the quantifiers are all of the same kind, either direct proof, counter-example, or example will be sufficient.

If they are of the different kind, then direct proof will be considered with counter-example or example.

Proving Implications#

To prove an implication, $P(x) \implies Q(x)$, assume that the hypothesis $P(x)$ is true.

Use $P(x)$ to show that the conclusion $Q(x)$ is true.

Proving Implications: trivial Example#

Let’s attempt to prove $\forall x \in \mathbb{R}, x > 4 \implies x^2 > 9$ (choosing representative in domain)

Let’s assume $x$ to be an arbitrary number $\in \mathbb{R}$.

If $x > 4$, then $x$ is positive. Multiplying both sides by $x$, we get $x^2 >4x$. Multiplying the same equation $x > 4$ by 4 on both sides, we get $4x > 16$. (using the representative $x$ to show $P(x)$ is true)

Therefore, $x^2 > 4x > 16 > 9$. Hence we conclude that $x^2 > 9$ (concluding the proof)

Notice how we used the hypothesis. (In this case at the start)

Proving harder examples#

Often, implications have a domain, definition, or open sentence as the hypothesis.

The conclusion will then often be an open statement.

In these cases, we start with the left hand side of the conclusion’s open statement and weave the assumption of the hypothesis inside.

$\forall s, t, x, y \in \mathbb{R}, x^2+y^2=1 \implies (sx+ty)^2 \leq s^2+t^2.$

The above quantified statement is one such example. We start with $(sx+ty)^2$ and note how $(sx+ty)^2 + (sx-ty)^2 = s^2 + t^2$

The assumption will also be used in the process.

We then conclude that since $(sx-ty)^2 \geq 0$, $(sx+ty)^2 + 0 \leq (sx+ty)^2 + (sx-ty)^2k = s^2 + t^2$

In cases like these, algebraic manipulation is the name of the game.

Proving Implications with $\lor$ in hypothesis#

When there is a $\lor$ in the hypothesis, we need to prove the implication is true for each case.

That is to say, to prove $a \lor b \implies c$, we need to prove $a \implies c$ and $b \implies c$.

If the cases of the proof are similar, we can write something to the effect of:

• The case to proof is similar, and is omitted.

Proof By Contrapositive#

If the hypothesis is too complicated for the given implication, but the conclusion looks easy, a proof by Contrapositive might be in line. Alternatively, when the conclusion is disjunct, as in $A \implies B \lor C$. Then flipping might help too.

Proving by contrapositive lets us flip the conclusion and our hypothesis. Its a free win!

Proof By Contradiction#

Proof by contradiction is based on the law of excluded middle.

That means that for a given statement $A$, either $A$ or $\neg A$ must be false. The compound statement $A \land \neg A$ then is always false.

Thus having the statement $A \land \neg A = True$ is called a contradiction.

There is a strong link between a proof by contradiction and a proof by contrapositive. In the former, when proving $A\implies B$, we try to show $A \land \neg B$. For the latter, we show $\neg B \lor \neg A$

In either case, we use the $\neg B$ to go about our proof.

Proving for Uniqueness#

Uniqueness is like a subset of $\exists$. Instead of one or more, however, we have exactly one element.

There are two main ways to prove for Uniqueness:

• Assume that $P(x)$ and $P(y)$ are true for $x, y \in \mathbb{S}$, then show how this assumption leads to $x=y$
• Assume that $P(x)$ and $P(y)$ are true for distinct $x, y \in \mathbb{S}$, then show how this assumption leads to a contradiction

Definitions#

Using facts about the equation, hypothesis, domain, can get us only so far. Sometimes we have to use a little extra. This is where definitions come into play.

Even and Odd#

An integer is said to be even if it can be written in the form $2k$ where $k$ is some integer.

An integer is said to be odd if it can be written in the form $2k + 1$ where $k$ is an integer.

The usage of if here actually means if and only if. Often, this is implicitly implied in definitions.

Using definitions to solve proofs#

We can apply the definition by saying:

By definition, $\exists k \in \mathbb{Z}$ s.t. $variable = 2k$ (even definition applied)