At first thought, it might seem that all prime numbers are odd. This is because it seems that all even numbers are not prime as 2 is a factor. However, by definition, 2 is a prime number but it is not odd and so we have found an example of when the statement is not true. This disproves the statement by counterexample.

# What is Disproof by Counterexample? – Examples & Videos

## Disproof by Counter-Example Notes

**Disproof by counterexample** is the technique in mathematics where a statement is shown to be wrong by finding a single example for when it is not satisfied. Not surprisingly, disproof is the opposite of proof so instead of showing that something is true, we must show that it is false. Any statement that makes inferences about a set of numbers can be disproved by finding a single example for which it does not work.

## Examples of Disproof by Counterexample

Prove or disprove the statement that all prime numbers are odd.

Disprove by counterexample that for any $a,b\in{\mathbb Z}$, if $a^2=b^2$, then $a=b$.

Note that ${\mathbb Z}$ is the set of all positive or negative integers. Finding an $a$ and $b$ such that $a\ne b$ but $a^2=b^2$, then the statement is disproved. Choosing any integer for $a$ and then choosing $b=-a$ will accomplish this. For example, let $a=4$ and $b=-4$. In this case $a^2=16$ and $b^2=16$ and so we have found an example where $a^2=b^2$ but $a\ne b$ and thus disproving the statement.

Find any values for $p$ and $q$ for which the following statement isn’t true:

$\sqrt{pq}\leq \frac{1}{2}(p+q)$

The tendency in this situation is to try various positive values of $p$ and $q$ until the statement isn’t true. However, the statement IS true for all positive values and so negative values of $p$ and $q$ must be tried. Bear in mind that you can only square root a positive number (without complex numbers) and so both $p$ and $q$ must both be negative. Try, for example, $p=-2$ and $q=-2$. The left hand side is the square root of $pq=4$ which is $\pm 2$ whereas the right hand side is $-2$. Since 2 is not less than or equal to $-2$, we have found a counterexample.

### Exam-Style Proof Examples

### More on Proof

- Go back to PURE MATHS
- SeeĀ QUESTIONS BY TOPIC
- Go to PAST and PRACTICE PAPERS