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

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.

## 2 Answers

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

• 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. Jan 30 '15 at 14:45
• @user3699546 I answered 17 minutes ago. Give yourself maybe more time for thinking about it, and it will maybe make sense. Jan 30 '15 at 14:47
• 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. Jan 30 '15 at 15:14
• correct ! you're right Jan 30 '15 at 15:14
• 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. ;-) Jan 30 '15 at 15:18

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 $a>0$ !