Mp2 - Conjectures and counterexamples | OCR A-Level Mathematics B (MEI)

Counterexamples test the reach of a mathematical claim. You will choose admissible inputs, show an exact failure and explain what has been disproved. You will also read an argument critically: an unconvincing proof, a false conjecture and a proved replacement are three different outcomes.

Conjectures depend on their domain

Can squaring a positive number make it smaller? The values $2^2=4$ and $3^2=9$ suggest the opposite, but $(1/2)^2=1/4<1/2$ . Before trusting a pattern, ask which inputs the claim includes.

You already know

The integers $\mathbb Z$ include negative whole numbers, zero and positive whole numbers. Squaring a negative input gives a non-negative result. A prime is an integer greater than $1$ with exactly two positive divisors; $1$ is not prime.

Even many consecutive successes can disguise a failure. The expression $n^2+n+11$ gives $11,13,17,23$ at $n=0,1,2,3$ . It gives primes right through $n=9$ , but at $n=10$ it gives $121=11^2$ , which is composite.

Conjecture

A conjecture is a mathematical claim believed to be true but not yet proved. Its domain specifies the cases it claims to cover. For example, “ $x^2\ge x$ for every integer $x$ ” and “ $x^2\ge x$ for every real $x$ ” are different conjectures, even though their displayed inequalities match.

The first of those is true. For an integer $x\le0$ , $x^2\ge0\ge x$ ; for an integer $x\ge1$ , multiplying $x\ge1$ by positive $x$ gives $x^2\ge x$ . The cases cover all integers. The real-number version is false at $x=1/2$ . A non-integer cannot disprove the integer version.

A rational number has the form $a/b$ , where $a,b\in\mathbb Z$ and $b\ne0$ ; $-7$ and $1/8$ qualify. An irrational number is real and has no such representation: $\sqrt{2}$ and $\pi$ are examples. The real numbers $\mathbb R$ contain both kinds. Terminating and recurring decimals are rational, but a finite display can approximate an irrational exact input. The display $1.414213562$ does not show that $\sqrt{2}$ equals that fraction.

Key idea

Successful checks provide evidence. A proof must cover every case in the domain; a disproof needs just one allowed failure.

Technology can search for a failure efficiently. It cannot turn a finite set of successful checks into an argument about every positive integer. Exact arithmetic also keeps a rounded near-equality from hiding a counterexample.

A true premise and a false conclusion

A useful counterexample is a small piece of reasoning, not just a number. It checks the hypothesis and then shows the conclusion fail.

For a claim about positive inputs, choosing a negative value misses the point: that input was never promised to work.

Counterexample

A counterexample to “if $P(x)$ , then $Q(x)$ ” is an input in the stated domain for which $P(x)$ is true and $Q(x)$ is false. A strict inequality also fails when the two sides are equal.

Reciprocals give quick, exact tests of a plausible-looking claim.

Worked example 1: testing a reciprocal claim

Question. Disprove the conjecture that $1/x<x$ for every positive real $x$ .

Plan. Choose a positive fraction less than $1$ ; its reciprocal will be greater than $1$ .

\begin{aligned} x&=\frac14>0 && \text{check the premise} \\ \frac1x&=4 && \text{evaluate the reciprocal} \\ 4&>\frac14 && \text{so the claimed }<\text{ fails} \end{aligned}

Why this step? Checking the positive-input condition makes the failed comparison a valid disproof of this particular implication.

Answer. $x=1/4$ is a counterexample.

Check. $(1/4)\times4=1$ , so the reciprocal is correct.

Example 2 changes only the denominator's exponent. You still need a true premise and a false conclusion, rather than a general theory of the function.

Worked example 2: squaring the denominator

Question. Disprove the conjecture that $1/x^2<x$ for every positive real $x$ .

Plan. Try $x=1/2$ and keep the square inside the denominator.

\begin{aligned} x&=\frac12>0 && \text{allowed input} \\ x^2&=\frac14 && \text{square the input} \\ \frac1{x^2}&=4>\frac12 && \text{compare the two sides} \end{aligned}

Why this step? $1/x^2$ means the reciprocal of the square, not the square of a rounded display.

Answer. The positive input $1/2$ disproves the conjecture.

Check. The denominator is $1/4$ ; multiplying it by the result $4$ gives $1$ .

Number-set conjectures use exactly the same test. Both $\sqrt{2}$ and $-\sqrt{2}$ are irrational, yet their sum is $0$ , which is rational. Also $\sqrt{2}\times\sqrt{2}=2$ . These examples disprove “always irrational” for the sum and the product respectively. They do not prove “always rational”.

Watch out

Wrong: “ $x=-1$ refutes a claim about positive $x$ .”

Right: the premise must hold at the chosen input. A failed conclusion outside the stated domain is irrelevant to that implication.

Counterexample reasoning is not supplied on the formula sheet. Change the exponent once more and make both checks visible.

Diagnosing and repairing a claim

After a disproof, a revised statement has to earn your trust afresh. Changing “for all positive inputs” to a smaller domain can repair a claim, but further sample calculations still do not prove it.

There is also a different kind of repair: replacing a weak argument for a true result. Checking $n=2,4,6$ does not prove $n^2+n$ even for every integer. The conclusion is true because $n(n+1)$ contains an even factor. The incomplete trials neither prove nor disprove it.

Worked example 3: repairing a strict inequality

Question. A student checks $x=2,3,4$ , then claims $x^2>x$ for every positive real $x$ . Explain the gap, give a counterexample and prove the repaired claim for $x>1$ .

Plan. Separate the sample evidence, a valid failure and the general sign argument.

The three trials leave other positive real inputs unchecked. At $x=1/2$ ,

x^2=\frac14<\frac12=x \qquad \text{the claim fails}

For every $x>1$ ,

\begin{aligned} x^2-x&=x(x-1) && \text{factor the difference} \\ &>0 && \text{both factors positive} \end{aligned}

Why this step? A sign argument covers the whole revised domain, including inputs not among the original trials.

Answer. The original claim is false, but $x^2>x$ is true for every $x>1$ .

Check. At the boundary $x=1$ , equality holds, so that boundary cannot enter the repaired strict claim.

Example 4 changes only the required inequality from strict to non-strict. That small change admits the boundary input.

Worked example 4: deciding whether equality is allowed

Question. Disprove $x^2\ge x$ for every positive real $x$ , then prove it for $x\ge1$ .

Plan. Use a fractional failure, then inspect the factors at and above the boundary.

At $x=1/3$ ,

x^2=\frac19<\frac13=x \qquad \text{so the universal claim fails}

For $x\ge1$ ,

\begin{aligned} x&>0\text{ and }x-1\ge0 && \text{domain conditions} \\ x^2-x&=x(x-1)\ge0 && \text{multiply non-negative factors} \end{aligned}

Why this step? The second factor can be zero, so the conclusion must allow equality.

Answer. $x^2\ge x$ for every $x\ge1$ ; equality occurs at $x=1$ .

Check. $1^2=1$ , while $2^2=4\ge2$ , agreeing with the boundary and interior cases.

A reciprocal claim can be repaired with another general difference. For $0<a<b$ , the proposed $1/a<1/b$ fails at $a=1$ , $b=2$ . Its correct replacement follows from

\frac1a-\frac1b=\frac{b-a}{ab}>0

because both numerator and denominator are positive. That establishes $1/a>1/b$ for every allowed pair. A counterexample to the wrong direction and a proof of the right direction answer different questions.

Problem-solving

When critiquing an argument, name the gap rather than guessing that its conclusion is false. When repairing a false claim, state the new conditions and prove the new claim generally.

These comparison results are not supplied on the formula sheet. Try a new exponent, retaining the boundary care from example 4.

Quick Recap

A domain tells you which inputs a claim promises to cover.
A counterexample keeps the premise true and makes the conclusion false.
An incomplete argument can end at a true conclusion; a disproof needs an actual allowed failure.
A revised claim needs its own proof, with strict or non-strict boundaries checked.

Squaring can make a positive number smaller: $(1/2)^2=1/4$ . The opening pattern from larger inputs did not cover the full positive real domain.

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Pure Mathematics

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Mp2 - 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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