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I need to show that the following equality holds $$\frac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =\frac{17+15\sqrt{5}} {7+15\sqrt{5}} $$ The only way I could prove it was via cross multiplication using a calculator.

Is there any simpler way to do this? Following prime factorization is given $$1646=2\times 823,736=2^5\times 23,2641=19\times 139$$ and $1181$ is prime. I strictly want to avoid multiplying large numbers. I don't know if that's possible here.

The above simplification is used to obtain a famous approximation to $\pi$ given by Ramanujan.

Paramanand Singh
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  • I would see what happens when you rationalize the denominators by multiplying by the conjugates. You'd still have to multiply four digit numbers. – Ethan Bolker Jan 31 '21 at 01:24
  • @EthanBolker: this is related to a key result of Ramanujan. I really wonder how he figured out that it was possible to simplify the expression. – Paramanand Singh Jan 31 '21 at 01:27
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    how large is large? – dezdichado Jan 31 '21 at 01:33
  • @dezdichado: well that's subjective. I wish not to generate any numbers of 6 or more digits. – Paramanand Singh Jan 31 '21 at 01:50
  • for one thing $\dfrac{2641}{1181}\approx \sqrt{5}\approx\dfrac{1646}{736}$ as far as eyeballing concerned. Besides, even if you cross multiply, you would not $6$ digits it looks like. Otherwise, I would just stick to the good old undetermined coefficient method. – dezdichado Jan 31 '21 at 02:45

2 Answers2

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Note that \begin{eqnarray*} \frac{1646-736\sqrt{5}} {2641-1181\sqrt{5}} =1 -5\frac{199-89\sqrt{5}}{2641-1181\sqrt{5}} \end{eqnarray*} and \begin{eqnarray*} \frac{89\sqrt{5}-199}{2641-1181\sqrt{5}}=\frac{2}{7+15 \sqrt{5}}. \end{eqnarray*}

Donald Splutterwit
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There is another approach which is more in line with the answer linked in the question.

Let $\phi=(1+\sqrt {5})/2$ denote the golden ratio so that $\sqrt{5}=2\phi-1$ and $$\phi^2=\phi+1\tag{1}$$ The fraction in question can now be written as $$\frac{1191-736\phi}{1911-1181\phi}\tag{2} $$ Let us observe that $$ \frac{a-b\phi} {c-d\phi} =\frac{b\phi^2-a\phi}{d\phi^2-c\phi}=\frac{b-(a-b)\phi}{d-(c-d)\phi}\tag{3}$$ and apply this to $(2)$ repeatedly. We get the following sequence of fractions $$\frac{736-455\phi}{1181-730\phi}, \\ \frac{455-281\phi}{730-451\phi},\\ \frac{281-174\phi}{451-279\phi},\\ \frac{174-107\phi}{279-172\phi},\\ \frac{107-67\phi}{172-107\phi},\\ \frac{67-40\phi}{107-65\phi},\\ \frac{40-27\phi}{65-42\phi},\\ \frac{27-13\phi}{42-23\phi},\\ \frac{13-14\phi}{23-19\phi},\\ \frac{14-(-1)\phi}{19-4\phi},\\ \frac{-1-15\phi}{4-15\phi} $$ The last fraction equals $$\frac{15\phi+1}{15\phi-4}=\frac{30\phi+2}{30\phi-8}=\frac{17+15\sqrt{5}}{7+15\sqrt{5}}$$ The application of $(3)$ was stopped at a point when the numbers in the fraction were least in absolute value. Any further application would have led to larger numbers. This approach has advantage that we don't need to know beforehand the final simple answer.

If we have a fraction like $$\frac {a+b\phi} {c+d\phi} $$ then we can use the transformation $$\frac{a+b\phi} {c+d\phi} =\frac{a/\phi+b} {c/\phi+d} =\frac{a(\phi-1)+b}{c(\phi-1)+d}=\frac{b-a+a\phi}{d-c+c\phi}$$ and hope to get an expression with small numbers.

Paramanand Singh
  • 92,128