Educerie · IB Diploma · Mathematics: analysis and approaches
Topic 1 Number and algebra · 1.6 Simple deductive proof
What you must be able to do
| You must be able to | Level | What it looks like in the exam |
|---|---|---|
| Tell an equation from an identity, and use = and ≡ correctly | SL, HL | "Show that … ≡ …", where you must treat the statement as true for every value |
| Prove a numerical result by a chain of exact steps | SL, HL | "Show that 1/4 + 1/20 = 3/10" or a surd or index result (2 marks), no calculator |
| Generalise a numerical result into algebra and prove it | SL, HL | "Show that the algebraic generalisation of this is …" (3 to 4 marks) |
| Lay out a proof from the left-hand side to the right-hand side, or the reverse | SL, HL | Any "show that" or "prove": one chain, ending "= RHS" |
| Write whole numbers in general form (2k, 2k + 1, n, n + 1) and prove a statement about them | SL, HL | "Prove that the sum of any three consecutive integers is divisible by 3" (3 marks) |
| Check a result, including your own, by substitution, by graphing or by working backwards | SL, HL | A check line in your working; a Paper 2 part that asks you to "verify" |
Before you start
You need to expand brackets, collect like terms, factorise, and add fractions by finding a common denominator, all without a calculator. You need the laws of indices from 1.5 and simplifying surds such as √50 = 5√2. Nothing else: this subtopic is about how you write, not about new algebra.
1The idea in one paragraph
A proof is an argument that shows a statement must be true, not just that it has been true in the cases you tried. A deductive proof starts from something you know and reaches the statement through a chain of steps, each one following from the one before by a rule everybody accepts: expanding a bracket, adding fractions, a law of indices. At standard level the chain almost always runs from one side of an identity to the other, left-hand side (LHS) to right-hand side (RHS). The marks are for the chain. The final answer is already printed on the question paper, so a "show that" is marked entirely on whether every link is visible and every link is legal.
2Equals, identical, and the difference
Two symbols look alike and mean different things.
= (equals) in an equation says two expressions have the same value for the values of x that make it true, and part of your job may be to find those values. x² = 2x + 3 is an equation. It is true when x = −1 and when x = 3, and false for every other x.
≡ (is identical to) in an identity says two expressions have the same value for every value of the variable. (x + 1)² ≡ x² + 2x + 1 is an identity. It is true for x = 7, for x = −0.3, for x = π, for anything. An identity is really one expression written in two forms.
Figure 1 shows the difference as graphs. In panel (a) the two sides of the equation draw two different curves that cross in two places, and those two crossings are the only solutions. In panel (b) the two sides of the identity draw exactly the same curve, lying on top of each other, because they are the same function.
This matters for proof for one reason. When a question says "show that (x − 5)(x + 2) ≡ x² − 3x − 10", you are not solving anything. There is no x to find. You are showing that the left-hand expression can be turned into the right-hand expression by legal algebra, which is what makes it true for every x at once.
Two more symbols finish the set. ≠ means "is not equal to", and you will use it to say which values are excluded, such as m ≠ 0 when m sits in a denominator. ≈ means "is approximately equal to" and has no place inside a proof: a proof is exact from start to finish.
An equation is true for some values and asks you to find them. An identity (≡) is true for every value and asks you to show it.
3Why checking examples is not a proof
Try this on paper.
Figure 2 draws the first two as bars. Each time, the difference of two neighbouring unit fractions is one over their product: 1/6 is 1/(2 × 3), 1/12 is 1/(3 × 4), 1/20 is 1/(4 × 5). You could check fifty more and every one would work.
Fifty checks are still not a proof. They tell you the pattern is worth trying to prove; they cannot tell you it never fails, because there are infinitely many cases and you have checked fifty. Mathematics has famous patterns that held for millions of cases and then broke. The only way to cover every case is to write the case as a letter, m, and do the algebra once for all m together. That is the algebraic generalisation, and proving it is section 4.
The guide asks for both kinds of proof: numerical, where you show one specific fact exactly (1/2 − 1/3 = 1/6, without a calculator), and algebraic, where you show the general statement. Exam questions often ask for the numerical one first and then the generalisation, so that the first part gives you the shape of the second.
4The layout: left-hand side to right-hand side
Every standard level proof uses the same layout.
- Write LHS = and copy the left-hand side exactly.
- Transform it one step per line, each line starting with =, each step a known rule.
- Stop when the expression is the right-hand side, and write = RHS.
- Close with a short sentence, such as "so LHS ≡ RHS, as required".
Figure 3 sets the right layout beside the one that loses marks.
The layout on the right of Figure 3 starts by writing down the statement you were asked to prove, as though it were already true, then does the same things to both sides until it reaches 0 = 0 or RHS = RHS. That argument runs in a circle: it assumes the result in its first line. It can also "prove" things that are false, because some steps (squaring both sides, multiplying both sides by zero) turn false statements into true ones. An argument that assumes its conclusion is not a proof, so expect to lose at least the final mark for it, and more in a question that says "prove". Work on one side only.
Worked example 1. Show that (x + 4)² − 7 ≡ x² + 8x + 9.
So (x + 4)² − 7 ≡ x² + 8x + 9, as required. That is two marks of working: one for the correct expansion, one for reaching the RHS with nothing skipped.
Start from whichever side has work to do. The guide allows "LHS to RHS or vice versa". If the left-hand side is already simple and the right-hand side is the one with brackets, start from the right.
Worked example 2. Show that x² + 6x + 11 ≡ (x + 3)² + 2.
The left side cannot be simplified. The right side can be expanded, so start there.
Worked example 3. Show that 1/m − 1/(m + 1) ≡ 1/(m(m + 1)), for m ≠ 0, m ≠ −1.
This is the generalisation from section 3. The method is exactly the numerical one, with letters.
The two excluded values are there because m and m + 1 sit in denominators, and division by zero has no meaning. When a question states them, it is telling you the identity holds everywhere else.
Worked example 4. Show that (n + 1)² − (n − 1)² ≡ 4n.
The bracket in the second line is where this proof is usually lost. Without it, students write −n² − 2n + 1 and end with 2 instead of 4n. Keep the bracket until you remove it deliberately.
5Numerical proofs, and generalising them
A numerical proof shows an exact fact about numbers without a calculator. The layout is the same; the rules are the laws of fractions, indices and surds.
Worked example 5. Show that √50 − √18 = 2√2.
A calculator gives 2.828… for both sides, and that is a check, not a proof: two decimals that agree to ten places are not shown to agree for ever. The exact steps are the proof.
Worked example 6. Show that 3⁵ + 3⁵ + 3⁵ = 3⁶, and then that 3ⁿ + 3ⁿ + 3ⁿ ≡ 3ⁿ⁺¹.
Now replace the 5 with n. Every step still holds, because none of them used the fact that the index was 5.
That last observation is the whole skill of generalising. Look at the numerical proof and ask which numbers were special to the example and which were doing the work. The 5 was only an example, so it becomes n. The 3 was doing the work (three copies of a power of three), so it stays.
6Proving statements about whole numbers
Some statements are in words: "the sum of three consecutive integers is always a multiple of 3". To prove one, first turn the words into algebra, using a letter for "any integer".
| In words | In algebra (n and k are integers) |
|---|---|
| any integer | n |
| three consecutive integers | n, n + 1, n + 2 |
| any even number | 2k |
| any odd number | 2k + 1 |
| a multiple of 3 | 3 × (an integer) |
Worked example 7. Prove that the sum of any three consecutive integers is a multiple of 3.
The last line is the reasoning line, and it is the one that earns the R mark. "3(n + 1)" is not yet a proof; the sentence saying that n + 1 is an integer, so the whole thing is a multiple of 3, is.
Worked example 8. Prove that the square of any odd number is odd.
Notice what the proof did with the answer. It did not stop at 4k² + 4k + 1. It rearranged into the form the statement talks about, "2 × integer + 1", because that form is the definition of odd.
Worked example 9. Show that the difference between the squares of two consecutive integers equals the sum of the two integers.
Figure 4 draws the case n = 4: a 4 by 4 square of dots, and the L-shaped border of 9 dots that turns it into a 5 by 5 square. The picture makes the fact obvious for n = 4 and suggests why it always works. It is not a proof, because it shows one case. The three lines of algebra above are the proof, because n is any integer.
7Checking a result, including your own
The guide expects you to check results, and to check your own. A check does not replace a proof, but it tells you before you move on whether the algebra you just wrote is worth trusting. Three checks cover almost everything.
Substitute a number into both sides. Pick a value that is not special. x = 0 and x = 1 are poor choices on their own, because they hide mistakes: 0 kills every term with x in it, and 1 makes x² look the same as x. Figure 5 shows the danger. A student who writes (x + 4)² − 7 = x² + 9, squaring each term separately, gets 9 on both sides at x = 0 and walks away happy. At x = 2 the left side is 29 and the claimed right side is 13, and the error is caught.
A check at two ordinary values is quick. For worked example 1:
| x | LHS (x + 4)² − 7 | RHS x² + 8x + 9 | Agree? |
|---|---|---|---|
| 2 | 36 − 7 = 29 | 4 + 16 + 9 = 29 | yes |
| −1 | 9 − 7 = 2 | 1 − 8 + 9 = 2 | yes |
Graph both sides. On Paper 2 you can enter each side as a function on the GDC. An identity draws one curve, as in Figure 1(b); an error draws two curves that part company, as in Figure 5. The table view of the GDC does the same job as the substitution table above.
Work backwards. If you factorised, expand your answer. If you solved an equation, substitute the solution into the original equation, not into a line you rewrote along the way.
One thing a check can never do: prove. Two values that agree are evidence. A single value that fails, though, is enough to show a claimed identity is false, and at HL that single failing value has a name, a counterexample (1.15).
8Where marks are lost
Working on both sides at once. Writing the statement down first and simplifying both sides until they match assumes what you were asked to show. Start from one side and end at the other.
Skipping the step that carries the mark. "LHS = … = RHS" with the key expansion done in your head scores the last mark at best. On a printed answer, every step must be written, because the marker already knows the answer; what they are marking is how you got there.
Losing a sign when a minus stands in front of a bracket. −(n − 1)² is −(n² − 2n + 1), which is −n² + 2n − 1. Keep the bracket until you remove it on purpose.
Squaring term by term. (x + 4)² is not x² + 16. It is x² + 8x + 16, because the square of a sum has a middle term.
Using the answer to prove itself. In a "show that", you may not use the printed result as a step. If part (b) says "hence", you use part (a); you never use part (b)'s own answer.
Proving by example. Checking n = 1, 2, 3 and writing "so it is always true" scores zero for a "prove" question. Examples suggest; algebra proves.
Stopping one step short. 3(n + 1) is not the end of a proof that the sum is a multiple of 3. The sentence "n + 1 is an integer, so this is a multiple of 3" is where the reasoning mark sits.
Writing = where you mean ≡, or ≈ inside a proof. In an identity the symbol that states the result is ≡, and a proof is exact: no rounded decimal belongs anywhere in it.
9Work it right
- Read the statement and decide: is it an equation (find x) or an identity (show it holds for all x)?
- Choose the side with work to do. Write LHS = or RHS = and copy it exactly.
- One step per line, each line starting with =. Keep brackets until you remove them deliberately.
- Never write the statement you are proving as your first line, and never work on both sides.
- For a statement in words, write the integers in algebra first: n, n + 1, n + 2; 2k; 2k + 1.
- End at the other side, write = RHS, and close with a sentence: "so LHS ≡ RHS, as required", or "… is an integer, so the sum is a multiple of 3".
- Check with two ordinary values (not 0 and 1 alone), or on Paper 2 by graphing both sides.
- State any excluded values (≠) when a letter sits in a denominator.
10Try it
Marks in brackets. Q1 to Q4 are Paper 1 style (no calculator). Q5 is Paper 2 style.
Q1. Paper 1.
(a) Show that 3/4 − 2/3 = 1/12. 2 marks
(b) Show that the algebraic generalisation of this, (m + 1)/(m + 2) − m/(m + 1) ≡ 1/((m + 1)(m + 2)), is true for m ≠ −1, m ≠ −2. 3 marks
Q2. Paper 1. Show that (2x − 1)² − (x − 2)² ≡ 3(x² − 1). 3 marks
Q3. Paper 1. Prove that the sum of any four consecutive integers is 2 more than a multiple of 4. 3 marks
Q4. Paper 1. A student is asked to show that (x + 5)(x − 1) − (x + 2)² ≡ −9. They write:
(x + 5)(x − 1) − (x + 2)² = −9 x² + 4x − 5 − x² − 4x − 4 = −9 −9 = −9 ✓
(a) Explain why this would not earn full marks. 1 mark
(b) Write a correct proof. 3 marks
Q5. Paper 2. A student claims that (x + 3)³ ≡ x³ + 27, and checks it at x = 0, where both sides equal 27.
(a) Use a different substitution to show that the claim is false. 2 marks
(b) Expand (x + 3)³ correctly. 2 marks
Q6. Paper 1. Prove that the square of any odd number is 1 more than a multiple of 8. 4 marks
11In one breath
An equation is true for some values and asks you to find them; an identity, written ≡, is true for every value and asks you to show it, because the two sides are one expression in two forms. Checking examples is evidence, not proof: to cover every case, write the case as a letter and do the algebra once. Lay every proof out the same way: LHS =, one legal step per line, = RHS, and a closing sentence; start from whichever side has work to do, and never write the statement first and work on both sides. For statements about whole numbers, write n, n + 1, n + 2 or 2k and 2k + 1, rearrange into the form the statement talks about, and say in words why that form proves it. Check your own work at two ordinary values, not just 0, or by graphing both sides on Paper 2; a check that passes is not a proof, but one that fails shows the claim is false.
Answers
Q1. (a) LHS = 9/12 − 8/12 = 1/12 = RHS. (b) LHS = (m + 1)/(m + 2) − m/(m + 1) = [(m + 1)² − m(m + 2)] ÷ [(m + 1)(m + 2)] = [m² + 2m + 1 − m² − 2m] ÷ [(m + 1)(m + 2)] = 1/((m + 1)(m + 2)) = RHS. (a) M1 for a common denominator of 12 with both fractions converted, A1 for 1/12 shown. (b) M1 for the common denominator (m + 1)(m + 2) with both numerators correct, M1 for expanding (m + 1)² and m(m + 2), A1 for the numerator reduced to 1 and the RHS reached. A calculator decimal in (a) scores zero; working on both sides in (b) loses the final A1.
Q2. LHS = (4x² − 4x + 1) − (x² − 4x + 4) = 4x² − 4x + 1 − x² + 4x − 4 = 3x² − 3 = 3(x² − 1) = RHS. Alternatively, as a difference of two squares: [(2x − 1) − (x − 2)][(2x − 1) + (x − 2)] = (x + 1)(3x − 3) = 3(x + 1)(x − 1) = 3(x² − 1). M1 for expanding both squares correctly, M1 for removing the bracket with the sign change shown, A1 for 3(x² − 1) reached with = RHS. An answer that loses the sign on the second bracket cannot reach the RHS and scores M1 only.
Q3. Let the integers be n, n + 1, n + 2, n + 3, where n is an integer. Sum = 4n + 6 = 4n + 4 + 2 = 4(n + 1) + 2. Since n + 1 is an integer, 4(n + 1) is a multiple of 4, so the sum is 2 more than a multiple of 4. M1 for the four integers written in general form, A1 for 4(n + 1) + 2, R1 for the sentence that n + 1 is an integer so the sum is a multiple of 4 plus 2. Checking a few examples scores zero.
Q4. (a) The student starts from the statement to be proved and works on both sides, so the argument assumes its own conclusion. (b) LHS = (x + 5)(x − 1) − (x + 2)² = (x² + 4x − 5) − (x² + 4x + 4) = x² + 4x − 5 − x² − 4x − 4 = −9 = RHS. (a) R1 for saying the statement was assumed, or that both sides were worked on. (b) M1 for both expansions, M1 for the bracket removed with signs changed, A1 for −9 = RHS from the LHS alone.
Q5. (a) At x = 1: LHS = 4³ = 64, RHS = 1 + 27 = 28. 64 ≠ 28, so the two expressions are not identical. (Any value other than 0 and −3 works: those are the only two values where the two sides happen to agree.) (b) (x + 3)³ = (x + 3)(x² + 6x + 9) = x³ + 9x² + 27x + 27. (a) M1 for substituting one value into both sides, A1 for both values and the statement that they differ. (b) M1 for a correct method (repeated multiplication, or the binomial coefficients 1, 3, 3, 1 from 1.9), A1 for the full expansion. A graph on the GDC showing two different curves also earns (a) if the student says so.
Q6. Let the odd number be 2k + 1, where k is an integer. (2k + 1)² = 4k² + 4k + 1 = 4k(k + 1) + 1. k and k + 1 are consecutive integers, so one of them is even, so k(k + 1) is even; write k(k + 1) = 2j for an integer j. Then (2k + 1)² = 4(2j) + 1 = 8j + 1, which is 1 more than a multiple of 8. M1 for 2k + 1, A1 for 4k(k + 1) + 1, R1 for the reason k(k + 1) is even (one of two consecutive integers is even), A1 for 8j + 1 and the conclusion. Stopping at 4k² + 4k + 1 scores 2.
Educerie · written from the published IB Diploma Programme Mathematics: analysis and approaches guide, first assessment 2021, section 1.6 Simple deductive proof. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.