Equality, identity, and what a proof is doing

An equation is a mathematical statement saying that two expressions have the same value for particular values of the variable. For example,

, is true only when $x=3$.

An identity is a mathematical statement saying that two expressions have the same value for every allowed value of the variable. We use $\equiv$ for an identity. For example,

, is true for all real values of $x$.

The equals sign $=$ states equality on a particular line of working. The identity sign $\equiv$ says that the two sides are the same expression written in different forms, across the stated domain. The domain is not a small detail. For example, $\frac{1}{m}$ is not even defined when $m=0$, where $m$ is a dimensionless variable, so any proof using it must silently or explicitly exclude values that make denominators zero.

LHS to RHS proof

A deductive proof is a chain of valid mathematical steps that starts from known facts or from one side of a statement and reaches the required conclusion. In a left-hand side (LHS) to right-hand side (RHS) proof, start with the LHS and transform it, line by line, until it is clearly the RHS. You can also go RHS to LHS if that side is easier. What you must not do is write the whole statement at the start as though it has already been proved.

Here is the kind of layout I want you to copy. To show

, begin with one side:

Each line follows from the one before it. That’s the point. You are not just showing algebra; you are showing the reasoning.

Numerical deductive proof

Deductive proof is not only algebraic. You can justify a numerical statement by exact manipulation as well. For example,

This counts as a proof because it uses exact equivalent fractions. A decimal check such as $0.25+0.0833\ldots\approx 0.333\ldots$ can build confidence, but the exact fraction working is the proof.

Algebraic identity proof

For an identity such as

, the quickest route is usually to expand the more compact side:

Checking $x=0$ or $x=10$ would not prove the identity. It would only show that the two sides match for those particular values. A proof of an identity has to handle all allowed values at the same time.

Checking your own result

A check is not usually the proof itself, but it is part of good mathematical practice. After proving an identity, substitute a convenient value to catch arithmetic slips. In the previous identity, $x=3$ gives LHS $=5$ and RHS $=9-18+14=5$, so the check supports the proof.

You can also check by reversing the algebra in your head: if the proof used expanding, try factorising or completing the square afterwards. And check the restrictions every time. If your proof involved cancelling $m+1$, then $m=-1$ cannot be in the domain. This habit is one place where mathematical reasoning differs from scientific reasoning: in mathematics, once the logic is valid and the assumptions are clear, the conclusion is forced; in science, evidence supports a conclusion but remains open to revision.

The idea of induction

Proof by mathematical induction is a proof method for statements indexed by consecutive integers. You prove a first case, then prove that any one true case makes the next case true. That locks in every later case. It is built for statements such as

The structure matters. In fact, the structure is the logic. You need three pieces:

• a base case, which proves the first statement, usually $P(1)$;
• an induction hypothesis, which assumes $P(k)$ is true for an arbitrary positive integer $k$;
• an induction step, which proves that the assumption $P(k)$ being true forces $P(k+1)$ to be true.

Once those pieces are in place, you can write the conclusion: since $P(1)$ is true and $P(k)\Rightarrow P(k+1)$, by mathematical induction $P(n)$ is true for all

Don’t confuse this with the inductive method in science. Scientific induction uses many observations to suggest a general rule. Mathematical induction proves a logical chain: the first link is true, every link forces the next link, so the whole infinite chain is secure.

A standard sum proof

Let

For the base case, when $n=1$,

$\text{LHS}=1,\qquad \text{RHS}=\frac{1(1+1)}{2}=1$

so $P(1)$ is true.

Now assume $P(k)$ is true for some $k\in\mathbb{Z}^+$. So

$\sum_{r=1}^{k} r = \frac{k(k+1)}{2}$

Use that assumption to prove $P(k+1)$:

That is exactly the RHS of $P(k+1)$. Therefore $P(1)$ is true and $P(k)\Rightarrow P(k+1)$, so the formula is true for all $n\in\mathbb{Z}^+$.

Divisibility proofs by induction

Induction also works cleanly for divisibility. To prove $3$ divides $4^n-1$ for all $n\in\mathbb{Z}^+$, let $P(n)$ be the statement $4^n-1=3q$ for some integer $q$.

Base case: $4^1-1=3$, so $P(1)$ is true.

Assume $P(k)$ is true, so

Then $4^k=3q+1$, and

Since $4q+1$ is an integer, $3$ divides $4^{k+1}-1$. The induction step is complete.

Where induction turns up

The syllabus expects induction to show up across the course, not just as a one-off trick. It can prove formulae for sums of sequences, divisibility results, statements involving powers of complex numbers, and results linked to differentiation. It can also prove algebraic formulae such as binomial coefficient identities.

Sometimes the first true case is not $n=1$. If the statement begins at $n=d$, where $d$ is an integer, prove $P(d)$ as the base case and then prove $P(k)\Rightarrow P(k+1)$ for $k\ge d$. The conclusion then starts at $n=d$, not at $n=1$.

When proving a sum, don’t just replace every $k$ by $k+1$ in the assumed formula and announce success. The induction step must start from the expression for $P(k+1)$, split off the new final term, use the induction hypothesis, and then simplify to the required form.

The shape of a contradiction proof

A proof by contradiction proves a statement by assuming its negation, then reaching something impossible. That impossible result may go against a known theorem, a definition, or an assumption made earlier in the proof.

The logic is simple: a statement is either true or false. If treating it as false leads to an impossibility, the statement has to be true.

A tidy contradiction proof often follows this pattern:

• State that you are using contradiction.
• Assume the opposite of what you want to prove.
• Deduce consequences carefully.
• Identify the contradiction explicitly.
• Conclude that the original statement is true.

Irrationality of $\sqrt{3}$

A rational number is a real number that can be written as $\frac{a}{b}$, where $a,b\in\mathbb{Z}$ and $b\neq0$. An irrational number is a real number that is not rational.

We’ll prove that $\sqrt{3}$ is irrational. Suppose, for contradiction, that $\sqrt{3}$ is rational. Then

Squaring both sides gives

$3b^2=a^2$

So $3$ divides $a^2$. Since $3$ is prime, $3$ divides $a$, so we can write $a=3c$, where $c\in\mathbb{Z}$. Substitute this into the equation:

$3b^2=(3c)^2=9c^2$

so

$b^2=3c^2$

Now $3$ divides $b^2$, and again $3$ divides $b$. We have shown that $3$ divides both $a$ and $b$, which contradicts the assumption that $\frac{a}{b}$ was in lowest terms. Therefore $\sqrt{3}$ is irrational.

The same style proves that $\sqrt[3]{5}$ is irrational: assume $\sqrt[3]{5}=\frac{a}{b}$ in lowest terms, cube both sides, and use the fact that if prime $5$ divides $a^3$, then it divides $a$.

Rational plus irrational

Here is a neat contradiction proof: if $a$ is rational and $b$ is irrational, then $a+b$ is irrational. Here $a$ and $b$ are real numbers.

Suppose instead that $a+b$ is rational. Since $a$ is rational, the difference

$(a+b)-a=b$

would be rational, because rational numbers are closed under subtraction. But this contradicts the statement that $b$ is irrational. Hence $a+b$ is irrational.

Infinite primes and proof validity

Euclid's proof that there are infinitely many primes is another famous contradiction. Assume there are only finitely many primes. Multiply them together, add $1$, and then ask which prime divides the new number. Any listed prime leaves remainder $1$, so the finite list cannot have been complete.

Proofs like this raise a useful TOK point: a mathematical proof is checked by a community for logical validity, not by repeated measurement. The Four-colour theorem is an interesting modern case because its accepted proof used extensive computer checking; it reminds us that certainty in mathematics depends not only on a conclusion being true, but on the proof being communicated and verified convincingly.

What a counterexample does

A counterexample is a specific case that fits the assumptions of a general statement but makes the conclusion false. One counterexample is enough to disprove a statement that starts with words like 'for all', 'always', or 'no'.

Still, writing down a number by itself won't do. You need to show two things: your example is allowed by the statement, and it makes the statement fail.

Two examples showing the structure of a counterexample.

Showing a number pattern is not always prime

Consider the set

The claim 'all elements of $P$ are prime' is false.

Take $n=41$, which is allowed because $41\in\mathbb{N}$. Then

$n^2+41n+41=41^2+41^2+41=41(41+41+1)=41\cdot83$

This number is not prime because it has factors other than $1$ and itself. So $n=41$ is a counterexample.

Disproving a statement about integer solutions

The statement 'there are no positive integer solutions to $x^2+y^2=10$' is false.

Then

The example is allowed, and it satisfies the equation. So it contradicts the word 'no'. That explanation is the key part.

Why examples cannot prove universal statements

A counterexample can disprove a universal statement, but examples can't prove one. A formula might work for $n=1$, $n=2$, and $n=3$, where $n$ is a natural number, and still fail later. That is why proof methods such as deduction, contradiction, and induction matter: they validate the general claim instead of just giving supporting evidence.