1.01c - Conjectures and counterexamples | OCR A-Level Mathematics A

A single well-chosen example can defeat a claim that survives hundreds of successful trials. Learn to read the domain, keep a conditional claim's premise true, and demonstrate exactly where its conclusion fails. You will also distinguish disproving a claim from proving a repaired version of it.

Conjectures depend on their domain

Could a formula keep producing primes and then suddenly fail? The values of $n^2+n+11$ at $n=0,1,2,3$ are $11,13,17,23$ , all prime. In fact it produces primes for every integer from $0$ to $9$ , then gives $121=11^2$ at $n=10$ . A very promising pattern has just lost its claim to be universal.

You already know

A prime is an integer greater than $1$ with exactly two positive divisors. In particular, $1$ is not prime. A composite integer has a divisor other than $1$ and itself.

A claim tells you which inputs you are allowed to test. That is its domain, and a counterexample must stay inside it.

Conjecture and counterexample

A conjecture is a mathematical statement believed to be true but not yet proved. A counterexample is an allowed instance in which the claimed conclusion is false. One counterexample disproves a universal claim; a collection of successful instances does not prove it unless the collection covers the entire domain.

The word “integer” can change everything. The integers $\mathbb Z$ include negatives and zero. A rational number is a ratio $a/b$ of integers with $b\ne0$ , such as $-7$ or $1/8$ . The real numbers $\mathbb R$ contain both rational and irrational numbers, which cannot be expressed as such a ratio; $\sqrt{2}$ and $\pi$ are irrational.

A terminating or recurring decimal is rational, but a finite calculator display is often an approximation to another exact number. Displaying $\sqrt{2}$ as $1.414213562$ does not turn the exact square root into that rational decimal.

For example, $x^2\ge x$ holds for every integer $x$ . If $x\le0$ , then $x^2\ge0\ge x$ . If $x\ge1$ , multiply $x\ge1$ by positive $x$ to get $x^2\ge x$ . Those cases cover every integer. But over the reals, $x=1/2$ gives $x^2=1/4<x$ . The fractional value defeats the real-number claim and says nothing against the integer claim.

Key idea

Read the domain before testing the claim. An example outside the domain cannot disprove a statement about inputs inside it.

A spreadsheet is a useful search tool, but its last checked row is not the end of an infinite domain. Keep exact fractions when a rounded decimal could hide equality or a sign.

A true premise and a false conclusion

To disprove “if $P$ , then $Q$ ”, find an input with $P$ true and $Q$ false. Showing only that $Q$ is false is not enough if your input never satisfied $P$ .

For instance, the claim $x>0\Rightarrow x^2>x$ concerns positive real inputs. Negative $x$ cannot refute it, even if they make an expression behave unexpectedly.

Counterexample to an implication

A counterexample to $P(x)\Rightarrow Q(x)$ is an admissible value for which $P(x)$ is true and $Q(x)$ is false. State the value, verify the premise, then evaluate the conclusion exactly.

Start near a boundary: a strict inequality fails at equality as well as when the opposite strict inequality holds.

Worked example 1: equality can disprove a strict claim

Question. Disprove the conjecture that $x^2>x$ for every positive real $x$ .

Plan. Try the positive boundary value $x=1$ , then compare the two sides.

\begin{aligned} x&=1>0 && \text{the premise holds} \\ x^2&=1^2=1 && \text{evaluate the square} \\ x^2&=x && \text{so }x^2>x\text{ is false} \end{aligned}

Why this step? The claim uses $>$ , so equality is already a failure; the square need not be smaller.

Answer. $x=1$ is a counterexample, so the conjecture is false.

Check. The chosen value is a positive real number, as required.

Example 2 changes the exponent from $2$ to $3$ . A number between zero and one gives a more informative failure than equality.

Worked example 2: a fractional input

Question. Disprove the conjecture that $x^3>x$ for every positive real $x$ .

Plan. Choose $x=1/2$ , whose positive powers become smaller.

\begin{aligned} x&=\frac12>0 && \text{admissible input} \\ x^3&=\left(\frac12\right)^3=\frac18 && \text{exact arithmetic} \\ \frac18&<\frac12 && \text{the claimed conclusion fails} \end{aligned}

Why this step? Exact fractions show the comparison without any rounding uncertainty.

Answer. $x=1/2$ disproves the claim.

Check. Multiplying $1/2$ by another factor $1/2$ reduces it; doing so twice gives $1/8$ .

The same logic handles number-set claims. The product of two irrational numbers need not be irrational: $\sqrt{2}\times\sqrt{2}=2$ . Their sum need not be irrational either: $\sqrt{2}+(-\sqrt{2})=0$ . Both inputs in each example are irrational and the result is rational. This does not prove that all such products or sums are rational; it only disproves “always irrational”.

Watch out

Wrong: “The claim fails at $x=-2$ , so a claim about positive $x$ is false.”

Right: first satisfy the positive-input premise. An inadmissible input cannot be a counterexample.

There is no formula-sheet shortcut for this decision. Change the exponent once more and supply both the admissibility check and the failed comparison.

Diagnosing and repairing a claim

A failed proof and a false conclusion are different things. Diagnose the argument first, then decide whether the claim needs disproving or can be repaired.

Checking $n=2,4,6$ cannot prove that $n^2+n$ is even for every integer: it misses odd integers and almost all even integers. The conclusion is still true, because $n^2+n=n(n+1)$ and one consecutive factor is even. An invalid argument does not itself make its conclusion false.

A false claim may become true on a smaller domain. For $x>1$ ,

x^2-x=x(x-1)>0

because both factors are positive. Thus $x>1\Rightarrow x^2>x$ is proved. At $x=1$ , equality means you may assert $x^2\ge x$ , but not $x^2>x$ . A repaired universal claim needs a new general argument.

Reciprocals are a useful place to practise that separation.

Worked example 3: counterexample, then a general repair

Question. A student claims that $1/a<1/b$ whenever $0<a<b$ . Give a counterexample, then prove the correct inequality.

Plan. Test a simple pair, then compare the fractions over their common positive denominator.

With $a=2$ , $b=4$ ,

\frac1a=\frac12>\frac14=\frac1b \qquad \text{a counterexample}

For any $0<a<b$ ,

\begin{aligned} \frac1a-\frac1b&=\frac{b-a}{ab} && \text{common denominator} \\ &>0 && \text{numerator and denominator positive} \end{aligned}

Why this step? Comparing the difference proves the direction for every admissible pair, not just the chosen one.

Answer. The claim is false; the correct general result is $1/a>1/b$ .

Check. At $a=2$ , $b=4$ , the difference is $1/4=(4-2)/(2\times4)$ .

Example 4 changes only the positive numerator. The denominator's sign still controls the reasoning.

Worked example 4: a common positive numerator

Question. For $0<a<b$ , disprove $3/a<3/b$ , then prove its correct replacement.

Plan. Use the same test pair and include the positive numerator in the difference.

At $a=2$ , $b=4$ ,

\frac3a=\frac32>\frac34=\frac3b \qquad \text{the claim fails}

In general,

\begin{aligned} \frac3a-\frac3b&=\frac{3(b-a)}{ab} && \text{combine the fractions} \\ &>0 && \text{all relevant factors positive} \end{aligned}

Why this step? Multiplication by positive $3$ preserves the direction already established for reciprocals.

Answer. $3/a>3/b$ for every $0<a<b$ .

Check. $3/2-3/4=3/4$ , matching $3(4-2)/(2\times4)$ .

Do not transfer this conclusion without its conditions: if a denominator were zero, the reciprocal would be undefined; opposite-sign denominators would require a different sign argument. The reciprocal comparison is not supplied on the formula sheet.

Problem-solving

A counterexample answers “Is the original claim true?” A general sign argument answers “Why is the repaired claim true?” Keep those two jobs separate.

Use a new numerical pair, then supply the general justification rather than another list of examples.

Quick Recap

Check the domain before choosing an input.
A counterexample to an implication makes its premise true and its conclusion false.
One failure disproves a universal claim; finite success may be evidence without being proof.
Prove a repaired claim with its own general argument and conditions.

Yes, a prime-producing pattern can suddenly fail: $n^2+n+11$ gives $121$ at $n=10$ , so the opening conjecture is false.

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Pure Mathematics

Statistics

Mechanics

1.01c - Conjectures and counterexamples

Conjectures depend on their domain

A true premise and a false conclusion

Diagnosing and repairing a claim

Quick Recap

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