Deductive Proof

In science, you become more confident in an idea each time an experiment agrees with it. Mathematics asks for more: a proof shows that a statement is true in every case, with no exceptions, using only facts you already know and logical steps. This page shows you how to write simple proofs clearly, which is a skill you’ll use all through IB Mathematics.

Key ideas

What a deductive proof is

A deductive proof starts from things that are already known to be true (definitions, algebra rules, earlier results) and moves forward one justified step at a time until it reaches the statement you want. Each step must follow from the ones before it.

A proof can be numerical (showing a particular calculation is true, like $\dfrac{1}{5} - \dfrac{1}{6} = \dfrac{1}{30}$) or algebraic (showing a general result with letters, true for every value).

Equality and identity: = and ≡

The two symbols mean different things:

• An equation uses $=$ and is true only for some values of the variable. For example, $3x - 2 = 10$ is true only when $x = 4$.
• An identity uses $\equiv$ (“is identically equal to”) and is true for every value of the variable. For example, $3(x - 2) \equiv 3x - 6$ is true whatever $x$ is.

Statement	Type	True for
$x^2 = 4x$	equation	$x = 0$ or $x = 4$ only
$(x + 1)^2 \equiv x^2 + 2x + 1$	identity	every real $x$

When you’re asked to “show that” an identity holds, you are proving that it is true for all allowed values. (If there are values where an expression is undefined, like $x = 1$ in $\dfrac{1}{x - 1}$, the identity holds for all the other values.)

Laying out an LHS-to-RHS proof

To prove that $A \equiv B$:

1. Start with one side only, usually the more complicated one. Call it the left-hand side (LHS).
2. Transform it using correct algebra steps: expanding, factoring, finding a common denominator, simplifying.
3. Stop when you reach exactly the other side, the right-hand side (RHS).
4. Finish with a sentence: “LHS $=$ RHS, so the identity is proved.” (or $\blacksquare$, or “as required”).

You may also start from the RHS and work to the LHS. What you must not do is write the whole statement down and work on both sides at once, as if it were an equation to solve. That assumes the result before you’ve proved it.

Writing integers algebraically

Many proofs are about whole numbers. To cover every case, use letters. Let $n$ and $m$ be integers.

Type of number	Algebraic form
even number	$2n$
odd number	$2n + 1$
multiple of $3$	$3n$
consecutive integers	$n,\ n + 1,\ n + 2, \dots$
consecutive even (or odd) numbers	$2n,\ 2n + 2$ (or $2n + 1,\ 2n + 3$)
two different odd numbers	$2m + 1$ and $2n + 1$

To show a number is even, write it as $2 \times (\text{an integer})$. To show it’s a multiple of $3$, write it as $3 \times (\text{an integer})$.

Checking is not proving

Trying a few examples is a great way to test a claim and to check your own algebra. For instance, substitute $x = 2$ into both sides of an identity you’ve just proved: if they don’t match, you’ve made a mistake. But no number of examples is a proof, because there could always be a case you didn’t try. Practice question 9 shows a pattern that works nine times in a row and then fails.

Worked examples

Example 1: A numerical result and its generalization

(a) Show that $\dfrac{1}{5} - \dfrac{1}{6} = \dfrac{1}{30}$.

(b) Show that the general result $\dfrac{1}{n} - \dfrac{1}{n + 1} \equiv \dfrac{1}{n(n + 1)}$ is true for all $n \ne 0, -1$.

Solution.

(a) Start with the LHS and use a common denominator of $30$:

$$\text{LHS} = \frac{1}{5} - \frac{1}{6} = \frac{6}{30} - \frac{5}{30} = \frac{1}{30} = \text{RHS}$$

(b) Start with the LHS. The common denominator is $n(n + 1)$:

$$\begin{aligned} \text{LHS} &= \frac{1}{n} - \frac{1}{n + 1} \\ &= \frac{n + 1}{n(n + 1)} - \frac{n}{n(n + 1)} \\ &= \frac{n + 1 - n}{n(n + 1)} \\ &= \frac{1}{n(n + 1)} = \text{RHS} \end{aligned}$$

So the identity is true for all $n \ne 0, -1$. Part (a) is the case $n = 5$.

Example 2: Expanding to prove an identity

Show that $(2x - 1)^2 + 4x \equiv 4x^2 + 1$.

Solution.

$$\begin{aligned} \text{LHS} &= (2x - 1)^2 + 4x \\ &= 4x^2 - 4x + 1 + 4x \\ &= 4x^2 + 1 = \text{RHS} \end{aligned}$$

LHS $=$ RHS, as required.

Check with $x = 3$: LHS $= 5^2 + 12 = 37$ and RHS $= 36 + 1 = 37$ ✓. The check doesn’t prove anything, but it would catch an algebra slip.

Example 3: Three consecutive integers

Prove that the sum of any three consecutive integers is a multiple of $3$.

Solution. Let the three consecutive integers be $n$, $n + 1$ and $n + 2$, where $n$ is an integer. Their sum is

$$n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1)$$

Since $n + 1$ is an integer, $3(n + 1)$ is a multiple of $3$. So the sum of any three consecutive integers is a multiple of $3$. $\blacksquare$

Notice the last sentence. A proof ends by saying clearly what has been shown.

Example 4: The product of two odd numbers

Prove that the product of any two odd numbers is odd.

Solution. The two odd numbers might be different, so use different letters. Let them be $2m + 1$ and $2n + 1$, where $m$ and $n$ are integers.

$$\begin{aligned} (2m + 1)(2n + 1) &= 4mn + 2m + 2n + 1 \\ &= 2(2mn + m + n) + 1 \end{aligned}$$

Since $2mn + m + n$ is an integer, the product has the form $2(\text{integer}) + 1$, so it is odd. $\blacksquare$

If you had used $2n + 1$ for both numbers, you’d only have proved that the square of an odd number is odd.

Common mistakes

Proving by example. Showing that $3 \times 5 = 15$ is odd and $7 \times 9 = 63$ is odd does not prove that every product of two odd numbers is odd. Examples are for testing and checking; a proof must use letters that cover every case.

Working on both sides at once. Writing $(2x - 1)^2 + 4x = 4x^2 + 1$ and then “doing the same thing to both sides” until you get $0 = 0$ assumes the very thing you’re trying to prove. Start from one side and transform it into the other.

Using the same letter for two different numbers. “Let the two odd numbers be $2n + 1$ and $2n + 1$” makes them equal. Use $2m + 1$ and $2n + 1$, or $2n + 1$ and $2n + 3$ if they’re consecutive.

Stopping one step short. Getting $3n + 3$ isn’t enough to show a multiple of $3$. Factor it to $3(n + 1)$ and say that $n + 1$ is an integer.

Mixing up = and ≡. Use $\equiv$ when a statement is true for every value of the variable, and $=$ for an equation you solve. In a proof, each line of the working can use $=$, because each step is a true equality.

Forgetting the conclusion. End with a sentence such as “so the sum is a multiple of $3$” or “LHS $=$ RHS, as required.” It shows the reader the proof is complete.

Practice

1. (Warm-up) Say whether each statement is an equation or an identity. For each equation, give the value(s) that make it true.

• (a) $3(x - 2) = 3x - 6$
• (b) $3x - 2 = 10$
• (c) $x^2 = 4x$
• (d) $(x + 1)^2 = x^2 + 2x + 1$

Solution

(a) Identity: expanding the left side gives the right side for every $x$, so we can write $3(x - 2) \equiv 3x - 6$.

(b) Equation: true only for $x = 4$.

(c) Equation: $x^2 - 4x = 0$, so $x(x - 4) = 0$, which is true only for $x = 0$ or $x = 4$.

(d) Identity: $(x + 1)^2 \equiv x^2 + 2x + 1$ for every $x$.

2. (Warm-up) Show that $(x + 5)(x - 2) \equiv x^2 + 3x - 10$.

Solution

$$\begin{aligned} \text{LHS} &= (x + 5)(x - 2) \\ &= x^2 - 2x + 5x - 10 \\ &= x^2 + 3x - 10 = \text{RHS} \end{aligned}$$

3. (Warm-up) Show that $8^2 - 6^2 = 4 \times 7$.

Solution

$$\text{LHS} = 8^2 - 6^2 = 64 - 36 = 28 = 4 \times 7 = \text{RHS}$$

4. (Core) Question 3 is one case of a general result. Prove that $(n + 1)^2 - (n - 1)^2 \equiv 4n$.

Solution

$$\begin{aligned} \text{LHS} &= (n + 1)^2 - (n - 1)^2 \\ &= (n^2 + 2n + 1) - (n^2 - 2n + 1) \\ &= 4n = \text{RHS} \end{aligned}$$

Question 3 is the case $n = 7$.

5. (Core) Prove that the sum of any two consecutive odd numbers is a multiple of $4$.

Solution

Let the two consecutive odd numbers be $2n + 1$ and $2n + 3$, where $n$ is an integer.

$$(2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1)$$

Since $n + 1$ is an integer, the sum is a multiple of $4$. $\blacksquare$

6. (Core) Show that $\dfrac{1}{x - 1} - \dfrac{1}{x + 1} \equiv \dfrac{2}{x^2 - 1}$ for $x \ne \pm 1$.

Solution

$$\begin{aligned} \text{LHS} &= \frac{1}{x - 1} - \frac{1}{x + 1} \\ &= \frac{(x + 1) - (x - 1)}{(x - 1)(x + 1)} \\ &= \frac{2}{x^2 - 1} = \text{RHS} \end{aligned}$$

7. (Core)

• (a) Show that $(a + b)^2 - (a - b)^2 \equiv 4ab$.
• (b) Use part (a) to work out $1001^2 - 999^2$ without a calculator.

Solution

(a)

$$\begin{aligned} \text{LHS} &= (a^2 + 2ab + b^2) - (a^2 - 2ab + b^2) \\ &= 4ab = \text{RHS} \end{aligned}$$

(b) Take $a = 1000$ and $b = 1$:

$$1001^2 - 999^2 = 4(1000)(1) = 4000$$

8. (Challenge) Prove that $x^2 - 6x + 11 \gt 0$ for every real number $x$.

Solution

Complete the square:

$$x^2 - 6x + 11 = (x - 3)^2 - 9 + 11 = (x - 3)^2 + 2$$

A square is never negative, so $(x - 3)^2 \ge 0$ for every real $x$. That means

$$x^2 - 6x + 11 = (x - 3)^2 + 2 \ge 2 \gt 0$$

for every real $x$. $\blacksquare$

9. (Challenge) A student notices that $n^2 + n + 11$ is prime for $n = 1, 2, 3, \dots, 9$, and claims that it is prime for every positive integer $n$. Explain why her checks don’t prove the claim, and show that the claim is false.

Solution

Checking nine cases says nothing about the infinitely many other values of $n$; a proof has to cover every case.

In fact the claim is false. Try $n = 10$:

$$10^2 + 10 + 11 = 121 = 11 \times 11$$

$121$ is not prime, so the statement is not true for every positive integer $n$. (A single case like this, which shows a general claim is false, is called a counterexample. You’ll meet it again in proof by contradiction.)