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.

Go through more proofs via Proofs - Rigorous Mathematics and enhance your mathematical growth!

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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. \(_\square\)

Log in to reply

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

Log in to reply

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 x

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.Log 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?

Log in to reply