Review question

# Can we list all $p/q$ between $0$ and $1$ if $2 \le q \le 5$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9354

## Solution

1. $F$ is the set of all the fractions $\dfrac{p}{q}$ which lie between $0$ and $1$, where $p$ and $q$ are positive integers having no common factor and $2 \le q \le 5$. List all the members of $F$ and indicate the smallest and the largest members.

Let $\dfrac{p}{q}$ be any element of $F$. Since $\dfrac{p}{q}$ lies between $0$ and $1$, we know that $0 < p < q$.

Since $p/q\le 1$, we must have $p\le q$, and $p\ne q$ as $p$ and $q$ have no common factor.

We can find the elements of $F$ by considering each value of $q$ in turn: for each $q$, the admissible values of $p$ are the integers with $0 < p < q$ which have no factors in common with $q$.

If $q = 2$, $p = 1$ is the only possibility.

if $q = 3$, then $p$ could be $1$ or $2$.

if $q = 4$, then $p$ can be $1$ or $3$ (but not $2$, since then $p$ and $q$ would share a factor).

If $q = 5$, then $p$ could be $1$, $2$, $3$ or $4$.

Thus, $F = \left\{ \frac{1}{2}, \frac{1}{3}, \frac{2}{3}, \frac{1}{4}, \frac{3}{4}, \frac{1}{5}, \frac{2}{5}, \frac{3}{5}, \frac{4}{5} \right\}.$

Find the mean value of the members of $F$.

The mean of a finite set of numbers is the sum of the numbers, divided by the size of the set.

We add the fractions with equal denominators first to make our calculations simpler:

\begin{align*} \text{sum} &= \frac{1}{2}+\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{3}{4}\right)+\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)\\ &= \frac{1}{2}+1+1+2\\ &= \frac{9}{2}. \end{align*}

Since $F$ has $9$ elements, the mean is therefore $\frac{1}{2}$.

There’s also a quicker way: the elements of $F$ are symmetrical about $\frac12$: every element $x$ in $F$ can be paired with the element $1-x$ which is also in $F$. So the mean must be $\frac{1}{2}$. (More explicitly, these two numbers $x$ and $1-x$ have a mean of $\frac{1}{2}$; also, the mean of the single fraction $\frac12$ is $\frac12$. It follows that the mean of the whole set is also $\frac{1}{2}$.)

1. $T$ is the set of all right-angled triangles with sides of length $x$, $(x+1)$ and $y$, where $x$ and $y$ are integers. If $(x+1)$ is the length of the hypotenuse, find an expression for $y$ in terms of $x$, giving your answer in its simplest form.

By Pythagoras’ theorem we know that, for any right-angled triangle with side lengths $a$, $b$ and $c$, with $c$ being the length of the hypotenuse,

$a^2 + b^2 = c^2.$ In particular, with $a = x$, $b = y$, and $c = (x+1)$, we obtain $x^2 + y^2 = (x+1)^2$. Expanding and rearranging gives $y^2 = 2x + 1$ so that $y = \sqrt{2x+1}.$ (As $y$ is the length of a side, it can’t be negative, so we take the positive square root here.)

Hence find a member of $T$ for which $y > 10$.

Since $y > 10$, we know that $2x + 1 = y^2 > 100$.

We are thus looking for a positive integer $x$ such that $2x + 1$ is a square number that exceeds $100$.

Since $11^2 = 121 = 2 \times 60 + 1$, one suitable member of $T$ has $x = 60$ and $y = 11$.

In fact, $11$ isn’t special here; any odd number larger than $10$ will fit the bill.