Disproof by counter exampleEdexcel International A Level Maths: Revision notes

Many statements in mathematics say that something is true for all values in a set, such as 'for all real $x$' or 'for all integers $n\ge2$'. To disprove such a statement you need only one value for which it is false. This value is called a counterexample.

This is why disproof is simpler than proof: a proof must deal with every case, but a disproof needs a single case. A counterexample must satisfy the conditions of the statement, and you must show clearly that the statement fails for it.

Follow the same steps each time:

1. Choose a value that satisfies the conditions of the statement.
2. Substitute it and calculate both sides, or each part of the statement.
3. Show that the statement is false for this value.
4. Write a conclusion: 'so the statement is false'.

Example. Statement: '$n^2-n+1$ is prime for all integers $n\ge2$'. $n=2$: $3$. $n=3$: $7$. $n=4$: $13$ (all prime). $n=5$: $25-5+1=21=3\times7$, which is not prime. So $n=5$ is a counterexample and the statement is untrue.

When the statement is about prime numbers, show the factors of the non-prime value, for example $21=3\times7$.

Try small values first and work upwards, as in the example above. Other useful ideas:

• zero, $1$ and negative numbers, if the statement allows them
• fractions between $0$ and $1$, where squares are smaller, not larger
• the prime number $2$, which is the only even prime
• numbers of opposite sign, such as $a=3$ and $b=-3$, when the statement involves squares.

Example. Statement T: '$x^2>x$ for all real $x$'. Try $x=\frac12$: $x^2=\frac14$, which is not greater than $\frac12$. So T is false. The value $x=1$ also works because $1$ is not greater than $1$.

'If $p$ is prime then $p+2$ is prime.' Try $p=3$: $5$ and $p=5$: $7$ both agree. Then $p=7$: $7+2=9=3\times3$, which is not prime. So $p=7$ is a counterexample.

'If $a^2=b^2$ then $a=b$.' Take $a=3$, $b=-3$. Then $a^2=9=b^2$ but $a\ne b$.

'$2^n+1$ is prime for all positive integers $n$.' $n=1$: $3$, $n=2$: $5$, $n=3$: $9=3\times3$. So $n=3$ is a counterexample.

'$\sqrt{x^2+9}=x+3$ for all real $x$.' Take $x=4$: $\sqrt{25}=5$ but $4+3=7$.

A counterexample can disprove a statement but examples that agree can never prove it. If a statement is true for $n=2,3,4$, there may still be a value, such as $n=5$ in the example above, where it fails.

You need only one counterexample. You do not need to find all of them or to explain why the statement fails. You may also be asked to explain why a student's testing of a few values is not a proof: a finite number of agreeing cases says nothing about the remaining cases.

Sometimes it is useful to find exactly where a statement holds. For $(x+1)^2>x^2+1$, expanding gives $(x+1)^2-(x^2+1)=2x$, so the statement is true only when $x>0$, and $x=-1$ or $x=0$ are counterexamples.