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Proving $\sqrt{100,001}-\sqrt{100,000} < \frac{1}{2\sqrt{100,000}}$

I squared both sides of the equation to get

$100,001 + 100,000+-200\sqrt{10}\sqrt{100,001} < \frac{1}{400,000}$.

I am just not sure how to justify this. I've tried multiplying both sides by -1, but it still would not hold.

Daniel W. Farlow
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user3699546
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2 Answers2

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$$\sqrt{a} - \sqrt{b} = \frac{a-b}{\sqrt{a} + \sqrt{b}}$$

rubik
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Olórin
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    I don't follow. So would we change the left hand side of the equation to 1/((sqrt(100,001)-sqrt(100,000))? I don't see how this makes sense because the denominator would be extremely small which would make the left hand side of the equation greater than the right. – user3699546 Jan 30 '15 at 14:45
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    @user3699546 I answered 17 minutes ago. Give yourself maybe more time for thinking about it, and it will maybe make sense. – Olórin Jan 30 '15 at 14:47
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    Oh, okay I think I got it. I have to justify that the sqrt(100,000)+sqrt(100,001) is greater than 2*sqrt(100000), which would mean that the right hand side is indeed larger. – user3699546 Jan 30 '15 at 15:14
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    correct ! you're right – Olórin Jan 30 '15 at 15:14
  • Thanks again! Also, is there a name for that formula you used? Where is it derived from? If you hadn't told me I never would have seen it. – user3699546 Jan 30 '15 at 15:16
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    It is derived from $(X-Y)(X+Y) = X^2 - Y^2$, with $X = \sqrt{a}$ and $Y = \sqrt{b}$. In french these kinds of equalities are called "remarkable identities/equalities." You have others : $(X+Y)^2 = X^2 + 2XY + Y^2$, etc etc. If you're happy with my answer, do not hesitate to validate it; so that the "case" is closed. ;-) – Olórin Jan 30 '15 at 15:18
  • +1 I've never seen that inequality before! Why not?!? – Mathemagician1234 Feb 23 '15 at 22:54
  • Remember this trick for the future! A trick used 3 times or more is a technique. – GEdgar Feb 23 '15 at 22:55
  • The "worse" one I know it this one : $X^4 + 4Y^4 = (X^2 - 2XY + 2Y^2) (X^2 + 2XY + 2Y^2)$. Very useful if you want to show that $n^4 + 4^n$ is not prime as soon as $n\in \mathbf{N}$ is $n\geq 2$. ;-) – Olórin Feb 23 '15 at 23:13
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Another way:

$$\sqrt{a+1}-\sqrt a<\frac1{2\sqrt a}.$$ Multiplying by $2\sqrt a$ and moving the terms,

$$2\sqrt{(a+1)a}<1+2a.$$

Squaring,

$$4a^2+4a<1+4a+4a^2.$$

This works for any $a>0$ !