The symmetry principle

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Some problems become easy the moment you notice that they look the same after a swap: of sin x and cos x, of two variables, or of two objects in a random arrangement. The symmetry principle turns that observation into an argument: if a problem is unchanged when its parts are interchanged, then quantities that the interchange swaps must be equal. This lesson uses symmetry to evaluate integrals that resist direct methods, to cut a proof down to one case “without loss of generality”, and to find probabilities without listing outcomes. In each use, the proof must name the symmetry and show that it really leaves the problem unchanged.

Symmetry can evaluate what direct methods cannot

A substitution that maps the interval of integration to itself can turn an integral into a second form of itself. Adding the two forms may give something simple.

Find I = ∫₀^(π/2)(sin³ x)/(sin³ x + cos³ x) dx.

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Reflect the interval
Method
Substitute u = π/2-x.
Reason
It swaps sin and cos and maps [0,π/2] to itself.
Working
Answer
I = ∫₀^(π/2)(cos³ u)/(cos³ u + sin³ u) du.
Add the two forms
Method
Add the original and reflected integrals.
Reason
The integrands add to exactly 1.
Working
Answer
2I = ∫₀^(π/2)1 dx = π/2, so I = π/4.

Quick check

A numerical integration gives 0.785398… = π/4. The same argument works with any positive power in place of 3, because only the sine–cosine symmetry is used.

The same idea gives a quick rule for odd and even integrands on an interval symmetric about 0. The substitution u = -x shows

∫₋ₐ^af(x) dx = { 0, f odd: f(-x) = -f(x),; 2∫₀^af(x) dx, f even: f(-x) = f(x).

For example, ∫_(-π)^π(x⁵ + 3)dx = 0 + 2∫₀^π3 dx = 6π, and ∫₋₁¹x³ cos x dx = 0 although it has no simple antiderivative.

Symmetry can reduce the cases

When a statement is unchanged by permuting its variables, you may assume an order without loss of generality (WLOG), such as a ≥ b. The proof must say why: every other ordering is obtained by relabelling, and relabelling does not change the statement.

Prove that for all real numbers a and b,

max(a,b) = (a + b + |a-b|)/2.

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Justify the WLOG
Method
Both sides are unchanged when a and b are swapped, since |a-b| = |b-a|.
Reason
So proving the case a ≥ b proves the case b ≥ a as well, by relabelling.
Working
Answer
Without loss of generality, assume a ≥ b.
Prove the one remaining case
Working
Answer
If a ≥ b then |a-b| = a-b, so
(a + b + (a-b))/2 = a = max(a,b). blacksquare

Quick check

Replacing + |a-b| by -|a-b| gives min(a,b) by the same argument. A WLOG assumption is only valid when you have checked the symmetry; it would fail for a statement such as a-b ≥ 0, which changes when a and b are swapped.

Symmetry also guides algebra. A difference of two symmetric expressions is itself symmetric, so its factors are often symmetric too, such as a + b or (a-b)².

Prove that for all real numbers a and b,

a³ + b³ ≥ a²b + ab² whenever a + b ≥ 0.

[4]

Write your answer, then compare it with the solution

Your proof

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Factorise the difference
Method
a³ + b³-a²b-ab² = a²(a-b)-b²(a-b) = (a-b)²(a + b).
Reason
Both sides are unchanged when a and b are swapped, so look for symmetric factors such as a + b.
Working
Answer
(a-b)² ≥ 0 and a + b ≥ 0, so the product is at least 0. blacksquare
Use the symmetry to explain equality
Method
Equality needs a = b or a + b = 0.
Reason
Swapping a and b leaves both conditions unchanged, as it must.
Working
Answer
So equality holds exactly when a = b or b = -a.

Self-mark with the mark scheme

Tick each mark point that your answer includes.

Tick a point only if your written proof contains it. This self-check does not record a mark.

Mark points

1 mark: Considers the difference of the two sides.
1 mark: Factorises the difference as (a − b)²(a + b).
1 mark: Uses (a − b)² ≥ 0 and a + b ≥ 0 to conclude the difference is non-negative.
1 mark: States the equality cases a = b or a + b = 0.

Common mistake

Symmetry lets you assume a ≥ b without loss of generality, because swapping a and b leaves the claim unchanged, but it does not let you assume signs such as a ≥ 0. The hypothesis a + b ≥ 0 is essential: a = 0, b = -1 gives a left side of -1 and a right side of 0.

Symmetry gives probabilities without counting

If a random arrangement is equally likely to be any ordering, then swapping two objects pairs each outcome with another of the same probability. Events that the swap exchanges must therefore have equal probabilities.

The numbers 1,2,…,10 are arranged in a random order, all orders equally likely.

What is the probability that 1 comes before 2?
What is the probability that 1, 2 and 3 appear in increasing order (not necessarily next to each other)?

Both probabilities and the symmetry you used

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Swapping the positions of 1 and 2 is a one-to-one pairing of orders with 1 first and orders with 2 first, so the probability is 1/2.
The 3! = 6 relative orders of 1, 2, 3 are equally likely by the same symmetry, and exactly one is increasing, so the probability is 1/6.

Common mistakes

Writing “by symmetry” without saying which interchange leaves the problem unchanged.
Assuming an order WLOG when a condition in the question is not symmetric, such as a ≥ 0 or a constraint on only one variable.
Using a reflection substitution that does not map the interval of integration to itself.
Treating outcomes as equally likely when the swap changes their probabilities, for example when the objects are not chosen at random.

Before you move on

Find ∫₀^(π/2)(square root of (sin x))/(square root of (sin x) + square root of (cos x)) dx, giving your answer to 3 decimal places.
The numbers 1,2,…,6 are arranged in a random order. Find the probability that 1 comes before 2 and 3 comes before 4.

1. Value of the integral, 3 decimal places

2. Probability

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With u = π/2-x the integral becomes the same expression with sin and cos exchanged. Adding the two forms gives 2I = π/2, so I = π/4 ≈ 0.785.
Swapping 1 and 2 pairs the orders with 1 first and with 2 first, and swapping 3 and 4 does the same for that pair. The four combinations of the two relative orders are equally likely, so the probability is 1/4.

Syllabus and review details

GCE A-Level H3 Mathematics 2027 · 2027

Last reviewed: 2026-11-07

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Guided steps

Reading settings

Symmetry can evaluate what direct methods cannot

An integral found by reflection

Reflect the interval

Add the two forms

Symmetry can reduce the cases

A formula for the larger of two numbers

Justify the WLOG

Prove the one remaining case

An ordering inequality

Factorise the difference

Use the symmetry to explain equality

Symmetry gives probabilities without counting

Random orders

Common mistakes

Before you move on

Two answers from symmetry

Lesson done

Symmetry can evaluate what direct methods cannot

An integral found by reflection

Symmetry can reduce the cases

A formula for the larger of two numbers

An ordering inequality

Symmetry gives probabilities without counting

Random orders

Common mistakes

Before you move on

Two answers from symmetry

Questions