# Explain Why a Root is Close to...

#### huruff

Show that

Use this to explain why is close to, but slightly less than, 10.05.

~ I know why the first part equals the other, but what relevance has that to the explanation?

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#### Plato

MHF Helper
Show thatView attachment 31607

Use this to explain whyView attachment 31609 is close to, but slightly less than, 10.05.
$\sqrt{N+1}-\sqrt{N}=\dfrac{1}{\sqrt{N+1}+\sqrt{N}}$
$\sqrt{N+1}=\dfrac{1}{\sqrt{N+1}+\sqrt{N}}+\sqrt{N}$
$\sqrt{101}=\dfrac{1}{\sqrt{101}+\sqrt{100}}+\sqrt{100}$

#### JeffM

Show thatView attachment 31607

Use this to explain whyView attachment 31609 is close to, but slightly less than, 10.05.

~ I know why the first part equals the other, but what relevance has that to the explanation?
Let's deal with the general proposition first.

$x = \sqrt{N + 1} - \sqrt{N} \implies x\left(\sqrt{N + 1} + \sqrt{N}\right) = \left(\sqrt{N + 1} - \sqrt{N}\right) \left(\sqrt{N + 1} + \sqrt{N}\right) = \left(\sqrt{N + 1}\right)^2 -\left(\sqrt{N}\right)^2 \implies$

$x\left(\sqrt{N + 1} + \sqrt{N}\right) = (N + 1) - N = 1 \implies x = \dfrac{1}{\sqrt{N + 1} + \sqrt{N}} \implies \sqrt{N + 1} - \sqrt{N} = \dfrac{1}{\sqrt{N + 1} + \sqrt{N}} \implies$

$\sqrt{N + 1} = N + \dfrac{1}{\sqrt{N + 1} + \sqrt{N}} < N + \dfrac{1}{2\sqrt{N}}.$

Good now?

#### huruff

lol, I'm even more lost now. I know how to make one equal the other by multiplying top and bottom by one so the appearance is altered,
but all of that, I'm lost in and don't know how to explain the root 101 still. sorry.

#### romsek

MHF Helper
lol, I'm even more lost now. I know how to make one equal the other by multiplying top and bottom by one so the appearance is altered,
but all of that, I'm lost in and don't know how to explain the root 101 still. sorry.
let $N=100$

$\sqrt{101}-\sqrt{100}=\dfrac{1}{\sqrt{101}+\sqrt{100}}$

$\sqrt{101}=\sqrt{100}+\dfrac{1}{\sqrt{101}+\sqrt{100}}$

$\sqrt{101}\approx \sqrt{100}+\dfrac{1}{2\sqrt{100}}$

it's a bit less than this since we decreased the denominator slightly in the approximation

$\sqrt{101}\approx 10 + \dfrac{1}{20}=10.05$

1 person

#### Melody2

Okay, I expect you are already overloaded but I might try anyway. LOL

You already know that the first part equals the second part so I will start from there.

Consider
$\sqrt{101}+\sqrt{100}>\sqrt{100}+\sqrt{100}=10+10=20$

so
$\sqrt{101}+\sqrt{100}>20$ Just by a really little bit

so
$\frac{1}{\sqrt{101}+\sqrt{100}}<\frac{1}{20}$

$\frac{1}{\sqrt{101}+\sqrt{100}}<0.05$ Just by a really little bit

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$\begin{array}{rlll}\sqrt{101}-\sqrt{100}&=&\frac{1}{\sqrt{101}+\sqrt{100}}<0.05&\\\\ \sqrt{101}-\sqrt{100}&<&0.05 \qquad &\mbox{Here I just left the middle bit out}\\\\\sqrt{101}-10&<&0.05 &\mbox{Now I can just solve it as an inequality}\\\\\sqrt{101}&<&0.05+10 &\\\\\sqrt{101}&<&10.05\qquad &\mbox{But just by a tiny bit}&\\\\\end{array}$

1 person