SOLUTION: hi am stuck on a question on theory of numbers show that, if p/q is a good approximation of √2 then (p²+2q²)/(2pq) is a better one. starting with p=q=1, show that √2

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Question 461622: hi am stuck on a question on theory of numbers
show that, if p/q is a good approximation of √2 then (p²+2q²)/(2pq) is a better one. starting with p=q=1, show that √2 ~ 577/408 and estimate the accuracy of the approximation.

Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
I will only answer the first part of the problem. I leave the iteration part to you.
Let p, q be positive numbers. Without loss of generality, we can let p%2Fq+%3Esqrt%282%29. I will show that sqrt%282%29+%3C=+%28p%5E2+%2B+2q%5E2%29%2F%282pq%29+%3C+p%2Fq. (If p%2Fq+%3C+sqrt%282%29 the arguments are similar.)
(I) p%2Fq+%3Esqrt%282%29 ==> p+%3E+sqrt%282%29%2Aq
==> p%5E2+%3E+2q%5E2 because y+=+x%5E2+is 1-to-1 over the positive numbers.
==> 2p%5E2+%3E+p%5E2+%2B+2q%5E2
==> %282p%5E2%29%2F%282pq%29+%3E+%28p%5E2+%2B+2q%5E2%29%2F%282pq%29
<==> p%2Fq+%3E+%28p%5E2+%2B+2q%5E2%29%2F%282pq%29
(II) By the AM-GM inequality,
%28p%5E2+%2B+2q%5E2%29%2F2+%3E=+sqrt%282p%5E2q%5E2%29
<==> %28p%5E2+%2B+2q%5E2%29%2F2+%3E=+sqrt%282%29pq%29
==> %28p%5E2+%2B+2q%5E2%29%2F%282pq%29+%3E=+sqrt%282%29
Then combining the results of (I) and (II), we get p%2Fq+%3E+%28p%5E2+%2B+2q%5E2%29%2F%282pq%29%3E=+sqrt%282%29, and the proof is complete.