# Who is bigger !!

• Sep 27th 2010, 01:00 PM
yehoram
Who is bigger !!
Hello to all !

I have a little problem !!

Who is bigger tyen who ????????

$(\sqrt 7-\sqrt 3) \ or \ (\sqrt 6-\sqrt 2)$

Show me with the stage of the solution!

Thank you !
• Sep 27th 2010, 02:10 PM
pickslides
I bet your calculator knows. Are you allowed to use it?

Spoiler:
$(\sqrt 6-\sqrt 2)> (\sqrt 7-\sqrt 3)$
• Sep 27th 2010, 03:18 PM
melese
Let $a=\sqrt7+\sqrt3$ and $b=\sqrt6+\sqrt2$. The squares of these two numbers are $a^2=7+\sqrt{21}+3=10+\sqrt{21}$ and $b^2=6+\sqrt{12}+2=8+\sqrt{12}$.
So it's clear that $a^2>b^2$ and therefore $a>b$.

Now we have $a(\sqrt7-\sqrt3)=7-3=4$ and $b(\sqrt6-\sqrt2)=6-2=4$.
Write $a(\sqrt7-\sqrt3)=b(\sqrt6-\sqrt2)$. Since $a>b$ we must have $\sqrt7-\sqrt3<\sqrt6-\sqrt2$.
• Sep 28th 2010, 06:29 AM
Quote:

Originally Posted by yehoram
Hello to all !

I have a little problem !!

Who is bigger than who ????????

$(\sqrt 7-\sqrt 3) \ or \ (\sqrt 6-\sqrt 2)$

Show me with the stage of the solution!

Thank you !

There is no need for any squaring......

$\sqrt{7}>\sqrt{6}$ and $\sqrt{3}>\sqrt{2}\ \Rightarrow\ \left[\sqrt{7}+\sqrt{3}\;\right]>\left[\sqrt{6}+\sqrt{2}\;\right]$

Then use the property of surd conjugates

$\left[\sqrt{a}+\sqrt{b}\;\right]\left[\sqrt{a}-\sqrt{b}\;\right]=a-b$

Since 7-3=6-2, then (surd)(surd conjugate) is equal for both.

Hence, since $\left[\sqrt{7}+\sqrt{3}\;\right]>\left[\sqrt{6}+\sqrt{2}\;\right]$

$\Rightarrow\ \left[\sqrt{7}-\sqrt{3}\;\right]<\left[\sqrt{6}-\sqrt{2}\;\right]$

A more general way to look at this.....

For any positive difference "k" and any positive increment "w"

$\left[\sqrt{x+k}+\sqrt{x}\;\right]<\left[\sqrt{x+k+w}+\sqrt{x+w}\;\right]$

$\left(\sqrt{x+k}+\sqrt{x}\;\right)\left(\sqrt{x+k}-\sqrt{x}\;\right)=k$

$\left(\sqrt{x+k+w}+\sqrt{x+w}\;\right)\left(\sqrt{ x+k+w}-\sqrt{x+w}\;\right)=k$

$\Rightarrow\ \left[\sqrt{x+k+w}-\sqrt{x+w}\;\right]<\left[\sqrt{x+k}-\sqrt{x}\;\right]$

Yet another general way to consider it is as follows.....

$\sqrt{6}-\sqrt{2}$

$\sqrt{7}-\sqrt{3}$

Generally as the numbers increase with a common difference, the difference between their square roots keeps decreasing,

since $\sqrt{x}-\sqrt{x-4}$, defined for x>4, becomes insignificantly small as x increases without bound,

while x-(x-4) is always 4.

This means $\left[\sqrt{6}-\sqrt{2}\;\right]>\left[\sqrt{7}-\sqrt{3}\;\right]>\left[\sqrt{8}-\sqrt{4}\;\right]>.....>\left[\sqrt{n}-\sqrt{n-4}\;\right]$ as "n" keeps increasing.

You can use calculus to show that $f(x)=\sqrt{x}-\sqrt{x-4}$
is a decreasing function as x increases.

$\displaystyle\frac{d}{dx}\left[\sqrt{x}-\sqrt{x-4}\;\right]=\frac{d}{dx}\left(x^{\frac{1}{2}}-(x-4)^{\frac{1}{2}\right)$

$=\displaystyle\frac{1}{2}x^{-\frac{1}{2}}-\frac{1}{2}(x-4)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x-4}}$

$\displaystyle\sqrt{x-4}<\sqrt{x}\;\; \Rightarrow\frac{1}{\sqrt{x-4}}>\frac{1}{\sqrt{x}}$

Therefore the derivative is negative, so the function is a decreasing function.

Therefore $\left[\sqrt{7}-\sqrt{3}\;\right]<\left[\sqrt{6}-\sqrt{2}\;\right]$

You could also examine the function $f(x)=\sqrt{x+4}-\sqrt{x}$

which is defined for x positive.