I want to prove $ \sqrt{2} $ is irrational using nested intervals.Assume, for contradiction, that $\sqrt{2}$ is rational. This means$$\sqrt{2} = \frac{a}{b}$$for some integers $a$ and $b$$(b /= 0)$, with $\frac{a}{b}$ in its lowest terms.Let's construct a sequence of nested intervals [an, bn] as follows:$$a_1 = 1, b_1 = 2$$For $n >= 1$:If (an + bn)/2 < √2, then an+1 = (an + bn)/2 and bn+1 = bnIf (an + bn)/2 > √2, then an+1 = an and bn+1 = (an + bn)/2Note that:
Each interval contains $\sqrt{2}$All an and bn are rationalThe length of each interval is halved at each step: bn - an = 1/2^(n-1)
Since √2 is assumed rational, it equals a/b for some integers a and b. This fraction a/b must be in one of our intervals [an, bn] for some large n.But for large enough n, the length of the interval bn - an = 1/2^(n-1) will be less than 1/b\
the idea here is that if √2 were rational, it would have to fit into one of our very small intervals. But these intervals eventually become too small to contain any rational number >1/b
but i don't know how to prove it could you help me please
