01 · An identity
Prove (a + b)³ + (a − b)³ = 2a(a² + 3b²) for real a, b.
Hint
Expand both cubes; odd powers of b cancel.
Worked solution
LHS = a³ + 3a²b + 3ab² + b³
+ a³ − 3a²b + 3ab² − b³
= 2a³ + 6ab² = 2a(a² + 3b²)
Understand · explore · practise
Proof by deduction
Learn to write algebraic and geometric proofs using identities, integer definitions, inequalities and coordinate geometry. Includes Pythagoras, parameter conditions and geometric bounds for pi.
Before you startAlgebraic manipulation, quadratics, straight lines and circles
01 / A proof’s structure
Start with what is given and justify each step.
A proof explains why a claim follows from stated assumptions. Begin with definitions or facts you know, make valid deductions, and finish with the exact claim.
State the domain. “For every integer n” is different from “for every real n”. One example can illustrate a proof, but cannot establish a claim about infinitely many values.
Assumptions → justified deductions → conclusion
To prove an identity, a reliable method is to start from one side and transform it into the other. Do not simply assume the identity you are trying to establish.
Can I work backwards while planning?
Yes. Working backwards can help you discover a route. In the final proof, either write the justified forward argument or clearly show that every backward step is reversible. Squaring and dividing by a possibly zero quantity need particular care.
02 / Algebraic identities
Transform one expression without changing its value.
To prove (u + v)² − (u − v)² = 4uv for real u and v, expand the left side:
u² + 2uv + v² − (u² − 2uv + v²)
= 4uv
The calculation works for all real u and v. Substituting a few numbers would establish only those particular cases.
Rational identity: state exclusions
For x ≠ ±2:
1/(x − 2) − 1/(x + 2)
= [(x + 2) − (x − 2)]/(x² − 4)
= 4/(x² − 4)
The common denominator is non-zero on the stated domain.
Surd identity
(√7 + √3)² − (√7 − √3)²
= (10 + 2√21) − (10 − 2√21)
= 4√21
This is also the general identity above with u = √7 and v = √3.
Fractional powers: a safe real domain
For x > 0, put t = x¹ᐟ³. Then t² = x²ᐟ³ and t³ = x:
(x¹ᐟ³ + 1)(x²ᐟ³ − x¹ᐟ³ + 1)
= (t + 1)(t² − t + 1)
= t³ + 1 = x + 1
The stated positive domain avoids any ambiguity about real fractional-power conventions.
03 / Integer proofs
Write parity and divisibility as algebra.
An even integer is 2k and an odd integer is 2k + 1 for some integer k. To prove divisibility by d, write the result as d times an integer.
(2k + 1)² − 1
= 4k(k + 1)
One of the consecutive integers k and k + 1 is even, so k(k + 1) = 2r for an integer r. Therefore every odd square is one more than a multiple of 8.
Every odd integer is a difference of squares
(k + 1)² − k² = 2k + 1
This works for every integer k, including negative values. It therefore also applies to every odd prime; primality is not needed.
Consecutive even and odd numbers
(2k + 2)² − (2k)² = 4(2k + 1)
(2k + 3)² − (2k + 1)² = 8(k + 1)
Both differences are divisible by 4. The difference for consecutive odd numbers is divisible by 8 as well.
04 / Inequalities
Use a non-negative square and track equality.
A square is never negative for real inputs. Completing the square can turn that fact into a useful bound:
3x² − 12x + 17 = 3(x − 2)² + 5 ≥ 5
Equality holds exactly when x = 2. A complete proof gives the equality case when it matters.
For positive x and y, start with (x − y)² ≥ 0. Expand, then divide by xy, which is positive:
x² + y² ≥ 2xy
x/y + y/x ≥ 2
Equality holds exactly when x = y. The positivity assumption justifies preserving the inequality direction during division.
Squaring an inequality
If A and B are both non-negative, A ≥ B is equivalent to A² ≥ B². Without the sign condition, squaring can lose information: −3 < 2 but 9 > 4. Also, √(x²) = |x|, not always x.
05 / Parameter proofs
Check whether a quadratic remains quadratic.
Consider kx² + 2x − k = 0 for real k. If k ≠ 0, its discriminant is 4 + 4k² > 0. It therefore has two distinct real roots.
If k = 0, the equation becomes 2x = 0 and has exactly one real root. It would be wrong to claim two roots for every k using the discriminant without checking the leading coefficient.
Prove a condition for a circle and line
The equation x² + y² + 2x − 4y = k becomes (x + 1)² + (y − 2)² = k + 5. It represents a positive-radius circle only if k > −5.
On y = 1: (x + 1)² = k + 4
The line misses the genuine circle exactly when −5 < k < −4. At k = −4 it is tangent; for k > −4 it cuts twice. The circle-existence condition must accompany the intersection condition.
06 / Coordinate proofs
Choose the calculation that proves the required property.
A sketch helps organise a proof but does not establish equal lengths, right angles or parallel lines. Use gradients, squared distances and midpoints.
Prove a square, including a right angle
Let A(0,0), B(2,1), C(1,3), D(−1,2). The consecutive side displacements are (2,1), (−1,2), (−2,−1), (1,−2). Their squared lengths are all 5. Opposite sides are parallel, and adjacent gradients 1/2 and −2 have product −1.
Thus the quadrilateral is a parallelogram with four equal sides and a right angle: a square. The different adjacent gradients also establish that A, B and C are not collinear.
A parallelogram and a rhombus
For A(0,0), B(4,2), C(3,5), D(−1,3), AB and DC both have displacement (4,2); BC and AD both have displacement (−1,3). The opposite sides are parallel and equal, proving a parallelogram. Alternatively, both diagonals have midpoint (3/2,5/2).
For P(−5,0), Q(0,3), R(5,0), S(0,−3), all four side lengths are √34 and the diagonals share midpoint (0,0), so this is a rhombus. It is not a square: the diagonals have unequal lengths 10 and 6.
Isosceles and cyclic arguments
The triangle with vertices (−3,0), (3,0), (0,4) has squared side lengths 36, 25, 25, proving it is isosceles.
The square A, B, C, D above is cyclic because all four points satisfy (x − 1/2)² + (y − 3/2)² = 5/2. Checking the same positive-radius circle equation at every vertex proves concyclicity.
07 / A tangent proof
Show that the line has exactly one point on the circle.
Prove that 3x + 4y = 25 is tangent to x² + y² = 25 at P(3,4). P satisfies both equations. Every point on the line can be written as:
x = 3 + 4t, y = 4 − 3t, t ∈ ℝ
This changes position along direction (4,−3). Substitute into the circle:
(3 + 4t)² + (4 − 3t)²
= 25 + 25t²
This equals 25 only when t = 0. Therefore P is the unique intersection and the line is tangent. This agrees with the gradient test: the radius gradient is 4/3 and the line gradient is −3/4.
08 / Pythagoras
Account for the same square’s area in two ways.
Take four identical right triangles with positive leg lengths a, b and hypotenuse c. Arrange them inside a square of side a + b as shown.
The central quadrilateral has four sides of length c. Each of its angles is 180° minus the two complementary acute triangle angles, so every angle is 90°. It is therefore a square of area c².
(a + b)² = 4 × ½ab + c²
a² + 2ab + b² = 2ab + c²
a² + b² = c²
The algebra uses arbitrary positive a and b, so it proves the result generally. Changing the model’s numbers illustrates the arrangement; it is not the proof itself.
One area, two calculationsExplore at your pace
For a = 3 and b = 2, the outer square has area 25. Four right triangles have total area 12. The central square has area c² = 13.
Watch the four triangles reveal Pythagoras
Pause, replay or seek freely. The notes explain the same idea and stay in view.
09 / Geometry extension
Bound a curved perimeter with regular polygons.
Use a circle of diameter 1, so its circumference is π. We use the geometric fact that, for convex shapes, an enclosing boundary has at least the perimeter of a boundary it encloses. For these straight-sided polygons and the circle, the inequalities are strict.
An inscribed square has diagonal 1 and side 1/√2 by Pythagoras. A circumscribed square has side 1. Their perimeters give:
2√2 < π < 4
An inscribed regular hexagon has side equal to the radius, 1/2. For the outer regular hexagon, split one of its six equilateral centre triangles in half. Its height (apothem) is 1/2. If half its side is t, Pythagoras gives (2t)² = t² + (1/2)², so the side is 1/√3.
3 < π < 2√3
Distance between opposite hexagon edges
A regular hexagon of side s has apothem √(s² − (s/2)²) = √3s/2. The perpendicular distance between opposite edges is twice this: √3s. For side 2√3 the distance is 6, a rational number despite the irrational side length.
Compare convex perimetersExplore at your pace
A circle of diameter 1 lies between two squares. Their perimeters are 2√2 and 4, so 2√2 < π < 4.
10 / Your turn
Write an argument that covers the stated domain.
State what each algebraic or geometric calculation proves. Include exclusions and equality cases.
02 · A rational identity
Prove 1/(x − 1) + 1/(x + 1) = 2x/(x² − 1), stating the domain.
Hint
Use a common denominator.
Worked solution
[(x + 1) + (x − 1)]/[(x − 1)(x + 1)]
= 2x/(x² − 1), x ≠ ±1
03 · Odd integers
Prove the sum of two odd integers is even.
Hint
Use different integer parameters; the numbers need not be consecutive.
Worked solution
(2r + 1) + (2s + 1) = 2(r + s + 1)
Since r + s + 1 is an integer, the sum is even.
04 · A lower bound
Prove 2x² + 8x + 11 ≥ 3 for all real x. When does equality hold?
Hint
Complete the square.
Worked solution
2x² + 8x + 11 = 2(x + 2)² + 3 ≥ 3
Equality holds exactly at x = −2.
05 · Positive variables
For positive u and v, prove (u + v)² ≥ 4uv.
Hint
Subtract 4uv from the left side.
Worked solution
(u + v)² − 4uv = (u − v)² ≥ 0
Equality holds exactly when u = v. This particular squared identity also works for all real u and v.
06 · A degenerate case
How many real roots does kx² + 2x − k = 0 have for each real k?
Hint
Separate k = 0 before using the discriminant.
Worked solution
For k = 0, x = 0 is the only root. For k ≠ 0, Δ = 4 + 4k² > 0, so there are two distinct real roots.
07 · A right triangle
Prove A(1,1), B(5,3), C(4,5) form a right triangle. Identify the right angle.
Hint
Compare the gradients of AB and BC.
Worked solution
mAB = 2/4 = 1/2
mBC = 2/(−1) = −2
mAB × mBC = −1
The sides are perpendicular, so the right angle is at B. Their different gradients also show the points are not collinear.
08 · A cyclic quadrilateral
Prove (5,0), (3,4), (−5,0), (0,−5) lie on one circle.
Hint
Try the origin as centre.
Worked solution
5² + 0² = 3² + 4²
= (−5)² + 0² = 0² + (−5)² = 25
All four satisfy x² + y² = 25, a circle of radius 5, so the quadrilateral is cyclic.
09 · An exact geometric length
A regular hexagon has side 4√3. Prove the distance between opposite parallel sides is 12.
Hint
Bisect an equilateral centre triangle and find the apothem.
Worked solution
Apothem² = (4√3)² − (2√3)²
= 48 − 12 = 36
Apothem = 6; distance = 2 × 6 = 12.
11 / Recap
The reason each step works belongs in the proof.
- State assumptions and the domain.
- Start with a known fact or transform one side of an identity.
- Represent integers using integer parameters.
- Track signs, non-zero denominators and equality cases.
- Check parameter values that change an equation’s degree.
- Use coordinate calculations to establish geometric properties.
Section 1 of 11 · A proof’s structure