The following eight unit fractions are given: \[\dfrac12\quad\dfrac13\quad\dfrac14\quad\dfrac15\quad\dfrac16\quad\dfrac17\quad\dfrac18\quad\dfrac19\]

What is the probability that a pair of distinct fractions, chosen at random from these, have a difference that is also a unit fraction?

What can you say about the pairs of fractions that have a difference that is a unit fraction?

Can you generalise your findings?

The fractions in this problem are all unit fractions. Can you find a pair of fractions, \(\frac{a}{b}\) and \(\frac{c}{d}\) where \(a\neq 1\) and \(c\neq 1\), which have a difference that is a unit fraction?

Can you expand on your answer to the previous question – under what conditions will a pair of fractions, \(\frac{a}{b}\) and \(\frac{c}{d}\) where \(a\neq 1\) and \(c\neq 1\), have a difference that is a unit fraction?

How do you subtract two fractions with different denominators?

How many pairs of fractions could you select?

If you have two algebraic fractions, \(\frac{a}{b}\) and \(\frac{c}{d}\), how could you apply your approach to numerical fractions, like those on the cards, to add or subtract the algebraic ones?

If you cannot find a pair of fractions, \(\frac{a}{b}\) and \(\frac{c}{d}\) where \(a\neq 1\) and \(c\neq 1\), which have a difference that is a unit fraction, does this mean that it’s impossible?

If you can find a pair of fractions, can you find another? Can you describe the conditions on the values of \(a\), \(b\), \(c\) and \(d\) that will, or will not, give a pair of fractions which have a difference that is a unit fraction?