The mediant of two fractions

The mediant of two fractions

IP 1
7 marks

For two positive fractions \(\dfrac{a}{b}\) and \(\dfrac{c}{d}\), where \(a\), \(b\), \(c\) and \(d\) are positive integers, the fraction \(\dfrac{a+c}{b+d}\) is called their mediant.

  1. Find the mediant of \(\dfrac37\) and \(\dfrac49\), and show that it lies between \(\dfrac37\) and \(\dfrac49\).[3]
  2. Given that \(\dfrac ab<\dfrac cd\), show that \(\dfrac ab<\dfrac{a+c}{b+d}\).[3]
  3. Hence write down a fraction that lies strictly between \(\dfrac{5}{12}\) and \(\dfrac37\).[1]

Solution:

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Answer:(a) \(\dfrac{7}{16}\); \(48<49\) and \(63<64\) give \(\dfrac37<\dfrac7{16}<\dfrac49\) (b) difference \(=\dfrac{bc-ad}{b(b+d)}>0\) (c) \(\dfrac{8}{19}\) (any fraction strictly between is acceptable)

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