IX class ...chapter 1 real numbers
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If 3√7 is rational then it can be expressed as p/q where p and q are integers.

(3√7)²=(p/q)², making p²/q²=63=64-1=64(1-1/64). p/q=8√(1-1/64).

From this p=8q would nearly give us 3√7–but not quite. (1-1/64)^½=1-1/128 approx=127/128.

Therefore p/q=8(127/128)=127/16, a better approximation. (p/q)²=16129/256=63+1/256.

But the expansion of (1-1/64)^½ has an infinite number of terms. The next approximation gives p/q=32515/4096, (p/q)²=1057225225/16777216=63+260617/16777216. And so on. So we never find p and q such that p/q=3√7 exactly.

Another way to prove this is to use a continuous fraction. The solution to the quadratic equation x²-16x+1=0 is x=8±3√7, that is, x-8=3√7 or x-8=-3√7, and we can write x=1/(16-x) as a different form for the same quadratic. This is effectively a continuous fraction. It works iteratively.

We start be substituting x=0 on the right-hand side, and we get x=1/16. Then we substitute x=1/16 into the same expression, so we get 16/255, and if we repeat the process we get 255/4064, then 4064/64769. So we have a series: 1/16, 16/255, 255/4064, 4064/84769, ... Represent the general term as a/b.

Each term in the series can be derived from the one before:


If we subtract each term from 8 we get another series: 127/16, 2024/255, 32257/4064, ... Each of these terms squared gets closer to 63, but, because the series is infinite, it can never reach exactly 63. So 3√7 is irrational. Specifically, ((8b-a)²-1)/63=b² for every term in the series. For example, for a/b=16/255, then ((8×255-16)²-1)/63=((2040-16)²-1)/63=(2024²-1)/63=4096575/63=65025=255².


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