GCSE Ordering and comparing numbers Practice Questions

Spot the negative numbers

Negative numbers are always smaller than $0$ and any positive number. On a number line they sit to the left of zero.

Order the negatives

For negatives, the bigger the digit the smaller the number. Think of $-8^{\circ}$C being colder than $-3^{\circ}$C.

Write the full ascending list

Place the negatives first, then $0$, then the positives in normal order. This is smallest to largest.

Both are negative

Both numbers are below zero. For negatives, the one with the bigger digit is actually the smaller number.

Compare positions

$-15$ is further down the number line, so it is the smaller value.

Write with the symbol

So $-15 < -13$.

Choose one unit

To compare fairly, put every length into the same unit — here, metres.

Convert the centimetres

There are $100$ cm in a metre, so divide by $100$.

Convert the fraction

Three-quarters of a metre is $0.75$ m.

List all in metres

Now every length is a decimal number of metres.

Order the decimals

Compare the decimals from smallest to largest.

Rewrite in original form

Swap the decimals back for the original lengths, shortest to longest.

Write down what is given

We have three 'less than or equal to' statements linking the numbers in a loop.

Chain the first two

If $a$ is at most $b$, and $b$ is at most $c$, we can link them.

Use transitivity

Following the chain, $a$ must be at most $c$. This is the transitive property.

Bring in the third fact

We were also told directly that $c$ is at most $a$.

Combine the two facts about a and c

So $a$ is at most $c$ AND $c$ is at most $a$ at the same time.

What that forces

The only way two numbers can each be less-than-or-equal to the other is if they are equal.

Substitute back

Since $c=a$, the chain $a\le b\le c$ becomes $a\le b\le a$.

Squeeze b

So $b$ is squeezed between $a$ and $a$.

Conclude about b

By the same reasoning, $b$ must equal $a$.

State the conclusion

All three numbers must be equal — the only way to satisfy the loop of inequalities.

Write the starting fact

All four numbers are positive and the first fraction is smaller than the second.

Turn it into a product statement

Cross-multiplying (safe because $b,d>0$) turns $\frac{a}{b}<\frac{c}{d}$ into $ad<bc$.

Introduce the middle fraction

This is the 'mediant' — add the tops and add the bottoms. We must show it lies between the two.

Aim for the left inequality

First prove the mediant is bigger than $\frac{a}{b}$.

Cross-multiply the left

Compare the two by cross-multiplying (denominators positive).

Expand both sides

Multiply out the brackets.

Cancel the common term

Take $ab$ off both sides, leaving $ad$ against $bc$.

Use the known fact

We already know $ad<bc$, so the left side is smaller.

Conclude the left inequality

Therefore the mediant is greater than $\frac{a}{b}$.

Aim for the right inequality

Now prove the mediant is smaller than $\frac{c}{d}$.

Cross-multiply the right

Again compare by cross-multiplying.

Expand and cancel

Multiplying out and cancelling $cd$ again leaves $ad$ against $bc$.

Conclude the right inequality

Since $ad<bc$, the mediant is less than $\frac{c}{d}$.

Combine the two halves

Both parts proved, so the mediant lies strictly between the two fractions.

Verify with numbers

With $\frac{1}{3}$ and $\frac{1}{2}$ the mediant is $\frac{2}{5}=0.4$, and $0.333<0.4<0.5$, confirming the result.