- \(F\) 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 \(\frac{1}{2}\).)
- \(T\) 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.
