K5MA1 · Proof algebra equations and inequalities · Edexcel A-level Mathematics

Revise the key ideas

Logical statements deduction exhaustion and counterexamples

Statements and assumptions — A mathematical statement has a truth value once its variables and domain are specified. The domain is the set of values allowed for the variables; for example, integers exclude fractions such as 1/2. The claim that x² is positive differs from nonnegative: x = 0 disproves positivity over all real x. State whether variables are real, integer, positive or restricted before manipulating an expression or assessing a proof.
Universal and existential — For every and there exists express different claims. Showing one integer with a property proves existence but does not establish that every integer has it. Conversely, one valid counterexample disproves a universal statement even if many other examples support it. A numerical pattern is evidence for a conjecture, a proposed statement that has not yet been proved, not a proof for an infinite domain.
Implication and converse — An implication P ⇒ Q means that whenever P holds, Q follows. Its converse Q ⇒ P needs separate justification: a number divisible by 6 is even, but an even number need not be divisible by 6. Equivalence P ⇔ Q requires both directions, which matters when a proposed algebraic rearrangement can introduce extra solutions.
Necessary and sufficient — If P implies Q, then Q is necessary for P and P is sufficient for Q. Necessary means P cannot hold without Q; sufficient means P alone guarantees Q. Being even is necessary but insufficient for an integer to be divisible by 6. Use a concrete test case to keep the direction clear rather than swap the words because both properties appear in one sentence.
Deductive proof — A deduction starts from stated assumptions or established results and proceeds through justified steps. To show n² − 4n + 7 is positive for every real n, complete the square: (n − 2)² + 3 ≥ 3 > 0. The reasoning covers the whole domain rather than a list of trial values.
Parity representation — An even integer is 2k and an odd integer is 2k + 1 for some integer k. If a = 2r + 1 and b = 2s + 1, then a + b = 2(r + s + 1), proving two odd integers have an even sum. Use independent integer letters for potentially different numbers rather than silently assume a = b.
Consecutive integers — Among n and n + 1, one is even, so their product is even for every integer n. For a divisibility proof, explain why the complete required factors occur; showing a product is even alone does not prove it is divisible by 6. A statement about consecutive positive integers may need adjustment if the stated domain is broader.
Proof by exhaustion — Exhaustion checks every possibility in a genuinely finite complete set. For integers n with 1 ≤ n ≤ 4, the values of n² + n are 2, 6, 12 and 20, all even; that exhausts this stated set only. The argument does not establish the same property for all integers unless a separate general deduction is supplied.
Cases and completeness — A proof can separate all integers into even and odd cases, or all real x into x < 0, x = 0 and x > 0. The cases must cover the whole domain. Omitting a boundary value can invalidate an otherwise sensible argument, especially where division, sign changes or equality conditions differ.
Counterexample — The claim that n² + n + 11 is prime for every nonnegative integer fails at n = 10: 100 + 10 + 11 = 121 = 11². It is unnecessary to classify all other inputs once one allowed failure is found. A counterexample outside the stated domain proves nothing about the claim.
Proof versus verification — Substituting a proposed root into the original equation verifies that particular candidate, whereas a derivation must also justify why no valid solutions were missed. Graphs and technology can guide a conjecture and reveal likely errors. A required proof still needs a complete mathematical argument with appropriate exact reasoning.

Contradiction irrationality and divisibility

Contradiction structure — To prove a proposition by contradiction, assume its negation, the statement that it is false, derive an impossibility through valid steps, and conclude that the original proposition holds. State the negation precisely. Merely obtaining an inconvenient expression, or assuming the intended conclusion partway through, does not complete a contradiction.
Irrational square root — Assume √2 = p/q for integers p and q in lowest terms with q ≠ 0. Squaring gives p² = 2q², so p² is even and therefore p is even; write p = 2r. Then 4r² = 2q² gives q² = 2r², so q is even too, contradicting that p/q was in lowest terms. Hence √2 is irrational.
The schematic follows the contradiction proof after assuming an integer ratio in lowest terms. Both numerator and denominator become even, contradicting that assumption. The accompanying text supplies the parity lemma; the diagram is a sequence of implications, not a substitute for justifying them.
Parity lemma — The step that an even square has an even integer root needs justification: if p were odd, p = 2k + 1 would give p² = 4k² + 4k + 1, which is odd. Thus p² even rules out p odd. The argument concerns integers, not a claim that any real number with an even-looking square is an even integer.
Infinitely many primes — Suppose there are only finitely many primes p₁, p₂, …, pₖ. Form N = p₁p₂…pₖ + 1, which is greater than 1 and has a prime divisor. None of the listed primes divides N because each leaves remainder 1. A prime divisor therefore lies outside the supposedly complete list, giving the contradiction.
Product plus one — In the prime proof, N need not itself be prime; it is enough that N has a prime divisor outside the list. For example 2 × 3 × 5 × 7 × 11 × 13 + 1 = 30031 = 59 × 509. Avoid replacing a valid divisibility argument with the false general claim that every prime-product-plus-one is prime.
Negating quantified claims — The negation of every allowed x satisfies P is that at least one allowed x fails P. The negation of there exists an x satisfying P is that none does. Distinguish not all from none when setting up a contradiction or interpreting a claimed disproof. To adapt contradiction to an unfamiliar irrationality claim, suppose √3 = p/q in lowest terms. Integer squares have remainders 0 or 1 on division by 3, so p² divisible by 3 forces p divisible by 3. Substituting p = 3r then forces q divisible by 3 too, contradicting lowest terms. Supply the remainder-case argument rather than simply replacing 2 by 3 in the parity proof.
Cancelling with conditions — From ab = ac one may infer b = c only if a ≠ 0, or handle a = 0 separately. Invalid proofs such as 1 = 2 often conceal division by zero. State any restriction introduced by cancelling, squaring, taking roots or reciprocals, and check the final candidates in the original statement.

Indices surds and exact manipulation

Index laws — For suitable real bases and defined expressions, aᵐaⁿ = aᵐ⁺ⁿ, aᵐ/aⁿ = aᵐ⁻ⁿ with a ≠ 0, and (aᵐ)ⁿ = aᵐⁿ under the relevant domain conditions. Positive bases avoid many complications for arbitrary real exponents. These laws concern multiplication and powers, not a claim that aᵐ + aⁿ = aᵐ⁺ⁿ.
Zero and negative powers — For a ≠ 0, a⁰ = 1 and a⁻ⁿ = 1/aⁿ. Thus 5⁻² = 1/25 rather than −25 or −1/25. Zero denominators remain excluded; the rule does not define 0⁻¹. A negative exponent changes the reciprocal, while a leading minus sign changes the value's sign.
Rational exponents — For a positive base, a^m/n is the nth root of aᵐ, with n positive and the fraction simplified as appropriate. For example 81^3/4 = (⁴√81)³ = 3³ = 27. Over real numbers, even roots of negative quantities are not defined, so domain conditions cannot be discarded when variables replace positive numbers.
Brackets and signs — The expressions (−3)² and −3² differ: the first is 9, while the second means −(3²) = −9. Brackets identify the base. Similarly, √(x²) = the nonnegative magnitude of x, so it equals x only when x ≥ 0; for x < 0 it equals −x.
Collecting surds — A surd is an irrational root left in exact form, such as √2. Factor out square factors before collecting like radicals: √72 − √8 = 6√2 − 2√2 = 4√2. Different square-free radicals cannot generally be combined as though their coefficients multiply. In particular √a + √b does not generally equal √(a + b).
Multiplying surds — For nonnegative real a and b, √a√b = √(ab). Expanding (√5 + 2)(√5 − 2) gives 5 − 4 = 1 because the cross terms cancel. State the applicable real-domain conditions rather than freely multiplying square roots of negative numbers using the same real rule.
Rationalising a monomial — To remove a radical from the denominator of 3/√5, multiply numerator and denominator by √5, obtaining 3√5/5. The value is unchanged because the multiplier is √5/√5 = 1. Rationalising does not mean rounding the irrational number to a decimal.
Conjugates — The conjugate of a + b√c is a − b√c. For 1/(3 + √2), multiply by (3 − √2)/(3 − √2), obtaining (3 − √2)/(9 − 2) = (3 − √2)/7. The original denominator must be nonzero, and conjugate multiplication changes both numerator and denominator.
Exact versus approximate — Retain exact fractions, radicals and symbolic constants until a numerical approximation is required. For √18/3, the exact simplified value is √2; 1.414 is an approximation and should use ≈ rather than = if it is rounded. Premature rounding can alter later roots, intersections and contextual decisions.

Quadratics simultaneous equations and hidden substitutions

Factorising quadratics — Expand to check factorisation: 2x² − x − 6 = (2x + 3)(x − 2). Setting the product to zero gives x = −3/2 or x = 2. The zero-product rule applies to a product equal to zero, not to an arbitrary sum or a product equal to a nonzero number.
Completing the square — For a ≠ 0, ax² + bx + c = a(x + b/(2a))² + c − b²/(4a). For 2x² − 12x + 11, factor the leading 2 before completing: 2(x − 3)² − 7. The coefficient's sign determines whether the turning point is a minimum or maximum.
Discriminant — For ax² + bx + c = 0 with real coefficients and a ≠ 0, Δ = b² − 4ac. Positive Δ gives two distinct real roots, zero gives one repeated real root, where the two quadratic solutions coincide, and negative gives no real roots. The conclusion concerns real roots; it does not claim that complex roots cannot exist.
Quadratic formula — Solving ax² + bx + c = 0 gives x = (−b ± √(b² − 4ac))/(2a). For 3x² − 2x − 2 = 0, x = (2 ± √28)/6 = (1 ± √7)/3. The denominator applies to the entire numerator; retain both signs unless the domain excludes a candidate.
Parameter conditions — For x² + kx + 9 = 0 to have a repeated real root, k² − 36 = 0, so k = ±6. Repeated does not mean the parameter has one possible value. For distinct roots use a strict discriminant inequality and solve it over the stated parameter domain.
Graph interpretation — Completing the square identifies a turning point and the discriminant identifies intersections with the x-axis. For y = −(x − 1)² + 4, the maximum is 4 at x = 1 and roots are −1 and 3. A quadratic's y-intercept is obtained by x = 0, not by setting y = 0.
Linear and quadratic systems — Substitute y = x + 1 into y = x² − 3x + 1: x² − 4x = 0, so x = 0 or 4 and the corresponding pairs are (0,1) and (4,5). Keep each x paired with its own y. Two x-values alone are not the complete solution to a two-variable system.
Elimination and equivalence — Multiplying an equation by a nonzero constant and adding equations can eliminate a variable without changing the common solutions. Dividing by an expression containing an unknown requires care because zero cases can be lost. Check every resulting pair in both original equations when a manipulation may introduce or discard candidates. A system can contain squares of both unknowns: with x − y = 1 and x² + y² = 25, substitute y = x − 1 to get 2x² − 2x − 24 = 0, so x = 4 or −3. The corresponding pairs (4,3) and (−3,−4) satisfy both original equations. This is still a linear-quadratic system even though both variables occur squared in one equation.
Quadratic substitution — An equation can be quadratic in a function of the unknown. For x⁴ − 5x² + 4 = 0, put u = x² to obtain (u − 1)(u − 4) = 0. Then x² = 1 or 4 gives x = −2, −1, 1 or 2; stopping at u = 1,4 does not finish the original problem.
Substitution domains — For 4ˣ − 5·2ˣ + 4 = 0, put u = 2ˣ > 0, giving u² − 5u + 4 = 0. Thus u = 1 or 4 and x = 0 or 2. A negative root for u would be inadmissible for a real exponential even if it solved the auxiliary quadratic.
Squaring equations — For √(x + 5) = x − 1, first require x ≥ 1, then square to get x + 5 = x² − 2x + 1. Solving x² − 3x − 4 = 0 gives x = 4 or −1, but only 4 satisfies the original equation and domain. Squaring preserves necessary consequences but can introduce extraneous candidates, values that solve the squared equation but not the original one.

Inequalities regions and algebraic division

Inequality direction — Adding the same value preserves an inequality's direction; multiplying or dividing by a negative value reverses it. For −3x + 2 > 11, obtain −3x > 9 and x < −3. If multiplying by an expression of unknown sign, split cases or use a justified sign analysis rather than assume its sign. With brackets on both sides, 2(x − 1) ≤ 3x + 4 gives 2x − 2 ≤ 3x + 4 and hence x ≥ −6. Expand and collect terms before deciding whether a negative division reverses the comparison.
Quadratic sign intervals — For (x − 1)(x + 3) < 0, the roots −3 and 1 split the line into intervals. The upward-opening product is negative between the roots, so −3 < x < 1. Strict inequalities exclude zeros; ≤ would include both endpoints. A sketch or sign table explains the answer rather than merely listing roots.
Outside intervals — For x² − x − 6 ≥ 0, factor (x − 3)(x + 2) and obtain x ≤ −2 or x ≥ 3. The word or joins two disjoint ranges; writing x ≤ −2 and x ≥ 3 would demand the impossible intersection. Include equality precisely where the product is zero and the original expression is defined.
Rational inequalities — For (x − 2)/(x + 1) > 0, split at the zero 2 and undefined point −1. Numerator and denominator have the same sign for x < −1 or x > 2. The value −1 is excluded even in a nonstrict inequality. Cross-multiplying by x + 1 without knowing its sign can give the wrong solution.
The numerator changes sign at 2, while the denominator changes sign and is undefined at −1. The ratio is positive when the signs agree, giving x less than −1 or greater than 2. Neither boundary is included in this strict inequality.
Curve and line comparisons — To find where y = x² − 2x lies below y = x + 4, solve x² − 3x − 4 < 0, giving −1 < x < 4. The intersection x-values are boundary points, while the inequality describes the whole interval. Relate the sign of the difference to which graph lies above the other.
The green curve y=x²−2x and brown line y=x+4 use a common coordinate system with horizontal scale 30 pixels per unit and vertical scale 10 pixels per unit. Their intersection x-values are −1 and 4. The curve lies below the line strictly between them. Unequal horizontal and vertical scales do not change which graph has the greater y-value.
Graphical regions — For y > x + 1, draw a dashed boundary line and shade the side satisfying the inequality; a test point can decide the side. For y ≤ x², use a solid parabola and include the boundary. A combined region must satisfy every required inequality, with and meaning intersection rather than all separate shaded areas indiscriminately.
Set notation — The set x < −1 or x ≥ 2 can be written (−∞,−1) ∪ [2,∞), where ∪ means union, combining the two allowed sets, while −1 < x < 2 is the intersection of x > −1 and x < 2. Square brackets include a finite endpoint and round brackets exclude it; infinity is not a real endpoint to include. Use notation consistently with the stated domain.
Polynomial division — Divide x³ − 2x² − 5x + 6 by x − 1: the quotient is x² − x − 6 and remainder 0, as multiplication verifies. Include zero coefficients for missing powers to align the division. The identity dividend = divisor × quotient + remainder provides an independent check.
Factor theorem — For a polynomial f, f(a) = 0 if and only if x − a is a factor. With f(x) = x³ − 2x² − 5x + 6, f(1) = 0, so divide by x − 1, then factor the quotient as (x − 3)(x + 2). The complete factorisation is (x − 1)(x − 3)(x + 2).
Linear factors and remainders — For divisor ax − b with a ≠ 0, the associated root is b/a, not necessarily b. A remainder on division by x − a equals f(a); for 2x − 3 the constant remainder is f(3/2). Distinguish the root used for substitution from the written coefficients of the factor.
Cancelling factors — The expression (x² − 1)/(x − 1) simplifies to x + 1 only for x ≠ 1. Cancellation removes a common nonzero factor, not the original domain restriction; it does not make the original fraction defined at its hole. Terms in a sum cannot be individually cancelled as if they were factors of the whole numerator.
The straight segment schematically represents y=x+1; the open circle at (1,2) marks the input excluded by the original quotient. Cancelling a nonzero factor preserves equality only for x not equal to 1. Axis scales in this schematic are illustrative rather than calibrated.
Improper rational expressions — Algebraic division gives (x² + 1)/(x − 1) = x + 1 + 2/(x − 1), with x ≠ 1. Multiply the right-hand side by the denominator to check. The polynomial part and remainder retain exact equivalence on the original domain and support later graph, partial-fraction and integration work.
Method and reasonableness — Choose a method matching the structure: factors for visible roots, completing the square for extrema, a discriminant for parameter conditions, substitution for hidden quadratics, and sign analysis for inequalities. State restrictions, retain exact working and check proposed solutions in the original problem. Technology supports checking and modelling but does not replace the reasoning a proof or method question asks to see.

Test yourself

50 questions · Random sets of 10. These quick checks support revision; practise longer explanations and justified judgements too.