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    Suppose \frac{m}{n}(m and n both integers) is a good approximation for \sqrt{2}. Then show that \frac{m+2n}{m+n} is a better approximation and that the sign of the error is reversed.

    I'm not really sure how to do this. I started by deciding that
    1<\frac{m}{n}<2, so n<m<2n (loosely interpreting "good approximation." )

    Beginning with \sqrt{2}-\frac{m}{n}=\delta, I want to end up with \frac{m+2n}{m+n} - \sqrt{2}< \delta. I tried using the inequality in the previous paragraph to manipulate these, but I'm struggling to get anywhere. I don't see how I'm going to get from a plus root 2 to a minus root 2. Thanks for any help.

    The inequality isn't really relevant. You have an expression for m/n in terms of sqrt(2) and delta. Note that (m+2n)/(m+n) can be rewritten in terms of (m/n) by dividing top and bottom by n. Then it's just a bit of algebra to get an expression for the 2nd term in terms of sqrt{2} and delta. Once you've done that, the rest should be straightforward.
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Updated: January 27, 2010
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