Dare enough to Prove !

Using Principle of Mathematical Induction :

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

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#Algebra #MathematicalInduction #JEE #IITJEE #SandeepBhardwaj

Note by Sandeep Bhardwaj
6 years ago

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Comments

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

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

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

(2−1)k+1=(l−l−1)(2−1)=(2l+l−1)−(2(l−1)+l)\begin{aligned} (\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{aligned}

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

(2−1)k+1=2l+l−1+2(2l)(l−1)−2(l−1)+l+2(2(l−1))(l)=3l+22l2−2l−1−(3l+22l2−2l−1)−1\begin{aligned} (\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{aligned}

[It only remains to prove that 3l+22l2−2l−13l + 2 \sqrt{2l^2 - 2l} - 1 is a positive integer, which I shall prove later.]

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

Pranshu Gaba - 6 years ago

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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

Gian Sanjaya - 5 years, 10 months ago

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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.

Gian Sanjaya - 5 years, 10 months ago

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

Gian Sanjaya - 5 years, 10 months ago
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