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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TopNewestBase 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. \(_\square\) – Pranshu Gaba · 2 years, 3 months ago

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– Gian Sanjaya · 2 years, 1 month 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 iLog in to reply

sqrt(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. – Gian Sanjaya · 2 years, 1 month agoLog in to reply

Can we generalize it into sqrt(m+1) - sqrt(m) for every m, element of R+, instead of only for m = 1? – Gian Sanjaya · 2 years, 1 month ago

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