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The question asked to find the "smallest value of n such that $(1+2^{0.5})^n$ is within 10^-9 of a whole number." I'm unsure of the approach to the question.

The question was in the chapter of 'binomial expansion' in the textbook.

Thanks for your time!

hmakholm left over Monica
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CowNorris
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  • Hint: Compare $(1+\sqrt{2})^2$ with $(\sqrt{2}-1)^2$, and $(1+\sqrt{2})^3$ with $(\sqrt{2}-1)^3$. Expand them out. – Empy2 Aug 28 '15 at 12:06

1 Answers1

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For such problems, the concept of conjugated terms is important. If you look at the binomial expansion of $(1+\sqrt{2})^n$ and $(1-\sqrt{2})^n$ you will see that all the terms including $\sqrt{2}$ in an odd power will have opposite signs. Hence $$(1+\sqrt{2})^n+(1-\sqrt{2})^n$$ is always an integer. But fortunately $|1-\sqrt{2}| \approx 0.4<1$ and so the second term quickly converges to 0 which makes $(1+\sqrt{2})^n$ getting closer to an integer value. Now, to estimate how close this is you have to analyse how quick $(1-\sqrt{2})^n$ converges to 0. Can you continue from here?

Tintarn
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  • thanks! I was able to find the answer (n=24) with your hint. Just wondering, is it possible for the answer to be found without the help of a calculator? – CowNorris Aug 28 '15 at 12:29
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    I'm afraid this is not easily possible. Note that the condition you obtain is $n>-\frac{9\log(10)}{\log (\sqrt{2}-1)} \approx 23.51$. So any "elementary" way would have to make bounds on this number. Of course, this is technically possible but you would have to use really close approximations on the $\log$s to get the exact result. – Tintarn Aug 28 '15 at 12:35