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# Dare enough to Prove !

Using Principle of Mathematical Induction :

Prove that for all $$n \in \mathbb{N}$$ : $$(\sqrt{2}-1)^n$$ can be written in the form of $$\sqrt{m}-\sqrt{m-1}$$, where $$m$$ is a positive integer.

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Note by Sandeep Bhardwaj
2 years, 9 months ago

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Base case: When $$n = 1$$, $$(\sqrt{2} - 1)^{1} = \sqrt{2} - \sqrt{2 - 1}$$. $$m = 2$$ satisfies the equation. $$P(1)$$ is true.

Inductive step: Assume $$P(k)$$ is true, where $$k \in \mathbb{N}$$. Then we must prove that $$P(k + 1)$$ is true as well. Let $$l$$ be a positive integer satisfying the following equation.

$(\sqrt{2} - 1)^{k} = \sqrt{l} - \sqrt{l - 1}$

\begin{align} (\sqrt{2} - 1)^{k + 1} & = (\sqrt{l} - \sqrt{l - 1})(\sqrt{2} - 1) \\ & = (\sqrt{2l} + \sqrt{l - 1}) - (\sqrt{2(l - 1)} + \sqrt{l}) \end{align}

Since both the terms are positive, we can square them and then sqaure root them.

\begin{align} (\sqrt{2} - 1)^{k + 1} & = \sqrt{2l + l - 1 + 2\sqrt{ (2l)(l - 1)}} - \sqrt{2(l - 1) + l + 2 \sqrt{(2 (l -1) ) (l) }}\\ & = \sqrt{3l + 2 \sqrt{2l^2 - 2l} - 1} - \sqrt{(3l + 2 \sqrt{2 l^2 - 2l} - 1) -1}\end{align}

[It only remains to prove that $$3l + 2 \sqrt{2l^2 - 2l} - 1$$ is a positive integer, which I shall prove later.]

Since the inductive hypothesis is true for $$k + 1$$, it is true for all integers $$n \geq 1$$. Thus we have proved that for every $$n$$, $$(\sqrt{2}-1)^n$$ can be written in the form of $$\sqrt{m}-\sqrt{m-1}$$, where $$m$$ is a positive integer. $$_\square$$

- 2 years, 9 months ago

So basically it remains to prove that sqrt (2m(m-1)) is an integer, sorry for the l changed to m, but in my post, small/lowercase L could be thought as capital/uppercase i

- 2 years, 7 months ago

Sorry for Pranshu Gaba, but I'll take the last part. (changed L to m) sqrt (m) - sqrt (m-1) = [sqrt (2) - 1]^k. By Newton's binomial theorem, then one of the sqrt (m) and sqrt (m-1) is in form xsqrt(2) for a positive integer x while the other one is an integer, so obviously, this comes sqrt(m(m-1)) is in form ysqrt(2) for a positive integer y, thus, sqrt(2m(m-1)) is an integer. Proof done. What else? Thanks so much for Pranshu Gaba to largely simplify the problem into a lesser form.

- 2 years, 7 months ago

Can we generalize it into sqrt(m+1) - sqrt(m) for every m, element of R+, instead of only for m = 1?

- 2 years, 7 months ago