### Equal Equilateral Triangles

Using the interactivity, can you make a regular hexagon from yellow triangles the same size as a regular hexagon made from green triangles ?

### The Square Hole

If the yellow equilateral triangle is taken as the unit for area, what size is the hole ?

### An Introduction to Irrational Numbers

Tim Rowland introduces irrational numbers

# Rationals Between

##### Stage: 4 Challenge Level:

We had lots of interesting responses to this problem. Ellie and Georgia from Melbourn Village College had a systematic way of finding fractions:

$\sqrt{58} = 7.6158$ to four decimal places.
$\sqrt{56} = 7.4833$ to four decimal places.
We then converted the decimals between$\sqrt{56}$ and $\sqrt{58}$ into fractions.
For example, $7.55 = \frac{755}{100}$. We simplified it by dividing the denominator and the numerator by $5$, because $5$ goes into both numbers. So one correct anwser is $\frac{151}{20}$.

Ellie and Georgia went on to find other fractions by converting decimals between 7.4833 and 7.6158.

Ashley from Durrington High School and Ha Young Jung from Wesley College used a similar method to Ellie and Georgia to find lots of fractions between $\sqrt {56}$ and $\sqrt {58}$.

John from Takapuna Grammar School showed that all denominators apart from 1 and 3 will work (although he didn't explain how he knew 1 and 3 wouldn't work):

First we calculate $\sqrt{56}$ and $\sqrt{58}$, which turn out to be about $7.483$ and $7.615$ respectively. We can also eliminate quite a lot of denominators considering that any number $x$ in which $\frac{1}{x} < \sqrt{58} - \sqrt{56}$ (which is about $0.132$) has to have a fraction between the two numbers.

Therefore we only need to test $1, 2, 3, 4, 5, 6$ and $7$.

$2, 4$ and $6$ can be easily explained off since $\frac{15}{2}$ (or $7.5$) is between $7.483$ and $7.615$, and $\frac{15}{2}$ is the same as $\frac{30}{4}$ and $\frac{45}{6}$. $5$ is also explained off because $\frac{38}{5}$ (or $7.6$) is also between $7.483$ and $7.615$. $\frac{53}{7}$ lies in the range.

Eduardo from the British School in Manila explained clearly why a denominator of 1 wouldn't work:
$\sqrt{56} < \frac{p}{q} < \sqrt{58}$ so $56q^2 < p^2 < 58q^2$
We know the denominator 1 will not work because there is no square number between 56 and 58, so when $q=1$, there is no value of $p$ to satisfy the inequality.

Harry from the Beacon School used a similar algebraic approach to give a very clear and full solution.
Let the fraction be $\frac{x}{y}$.
Now $$\sqrt56< \frac{x}{y}< \sqrt58$$
$$56< \frac{x^2}{y^2}< 58$$
$$56y^2< x^2< 58y^2$$

Therefore, as $x$ and $y$ are whole numbers, there must be a square between $56y^2$ and $58y^2$ for a fraction between $\sqrt56$ and $\sqrt58$ to exist for denominator $y$.

If there is no square in this interval, then the difference between the largest square below $56y^2$ and the smallest square above $58y^2$ must be greater than $2y^2$.

The square root of the former must be the largest integer below $\sqrt{56y^2}$

The difference between $n^2$ and the square immediately above it is:
$$(n+1)^2-n^2=2n+1$$

So there is a solution in the interval when $2n+1< 2y^2$, where $n$ is the largest integer below $\sqrt{56y^2}$.

$2\sqrt{56y^2}+1< 2y^2$ for $y=8$ but not $y=7$, so for $y\geq8$ there will always be a square between $56y^2$ and $58y^2$.

For $y = 1$, $56y^2 = 56$ and $58y^2 = 58$. There are no squares in this range, so $1$ cannot be the denominator.

For $y = 2$, $56y^2 = 224$ and $58y^2 = 232$. There is a square in this range $(225 = 15^2)$, so $2$ can be the denominator. The numerator $x$ is $\sqrt{225}$ so the fraction is $\frac{15}{2}$

For $y = 3$, $56y^2 = 504$ and $58y^2 = 522$. There are no squares in this range, so $3$ cannot be the denominator.

For $y = 4$, $56y^2 = 896$ and $58y^2 = 928$. There is a square in this range $(900 = 30^2)$, so $4$ can be the denominator. The fraction is $\frac{30}{4}$, which is equal to $\frac{15}{2}$.

For $y = 5$, $56y^2 = 1400$ and $58y^2 = 1450$. There is a square in this range $(1444 = 38^2)$, so $5$ can be the denominator. The fraction is $\frac{38}{5}$.

For $y = 6$, $56y^2 = 2016$ and $58y^2 = 2088$. There is a square in this range $(2025 = 45^2)$, so $6$ can be the denominator. The fraction is $\frac{45}{6}$, which is equal to $\frac{15}{2}$.

For $y = 7$, $56y^2 = 2744$ and $58y^2 = 2842$. There is a square in this range $(2809 = 53^2)$, so $7$ can be the denominator. The fraction is $\frac{53}{7}$.

There is a fraction between $\sqrt{56}$ and $\sqrt{58}$ for all whole denominators except 1 and 3.

Well done Harry!