Hello to all !
I have a little problem !!
Who is bigger tyen who ????????
$\displaystyle (\sqrt 7-\sqrt 3) \ or \ (\sqrt 6-\sqrt 2)$
Show me with the stage of the solution!
Thank you !
Let $\displaystyle a=\sqrt7+\sqrt3$ and $\displaystyle b=\sqrt6+\sqrt2$. The squares of these two numbers are $\displaystyle a^2=7+\sqrt{21}+3=10+\sqrt{21}$ and $\displaystyle b^2=6+\sqrt{12}+2=8+\sqrt{12}$.
So it's clear that $\displaystyle a^2>b^2$ and therefore $\displaystyle a>b$.
Now we have $\displaystyle a(\sqrt7-\sqrt3)=7-3=4$ and $\displaystyle b(\sqrt6-\sqrt2)=6-2=4$.
Write $\displaystyle a(\sqrt7-\sqrt3)=b(\sqrt6-\sqrt2)$. Since $\displaystyle a>b$ we must have $\displaystyle \sqrt7-\sqrt3<\sqrt6-\sqrt2$.
There is no need for any squaring......
$\displaystyle \sqrt{7}>\sqrt{6}$ and $\displaystyle \sqrt{3}>\sqrt{2}\ \Rightarrow\ \left[\sqrt{7}+\sqrt{3}\;\right]>\left[\sqrt{6}+\sqrt{2}\;\right]$
Then use the property of surd conjugates
$\displaystyle \left[\sqrt{a}+\sqrt{b}\;\right]\left[\sqrt{a}-\sqrt{b}\;\right]=a-b$
Since 7-3=6-2, then (surd)(surd conjugate) is equal for both.
Hence, since $\displaystyle \left[\sqrt{7}+\sqrt{3}\;\right]>\left[\sqrt{6}+\sqrt{2}\;\right]$
$\displaystyle \Rightarrow\ \left[\sqrt{7}-\sqrt{3}\;\right]<\left[\sqrt{6}-\sqrt{2}\;\right]$
A more general way to look at this.....
For any positive difference "k" and any positive increment "w"
$\displaystyle \left[\sqrt{x+k}+\sqrt{x}\;\right]<\left[\sqrt{x+k+w}+\sqrt{x+w}\;\right]$
$\displaystyle \left(\sqrt{x+k}+\sqrt{x}\;\right)\left(\sqrt{x+k}-\sqrt{x}\;\right)=k$
$\displaystyle \left(\sqrt{x+k+w}+\sqrt{x+w}\;\right)\left(\sqrt{ x+k+w}-\sqrt{x+w}\;\right)=k$
$\displaystyle \Rightarrow\ \left[\sqrt{x+k+w}-\sqrt{x+w}\;\right]<\left[\sqrt{x+k}-\sqrt{x}\;\right]$
Yet another general way to consider it is as follows.....
$\displaystyle \sqrt{6}-\sqrt{2}$
$\displaystyle \sqrt{7}-\sqrt{3}$
Generally as the numbers increase with a common difference, the difference between their square roots keeps decreasing,
since $\displaystyle \sqrt{x}-\sqrt{x-4}$, defined for x>4, becomes insignificantly small as x increases without bound,
while x-(x-4) is always 4.
This means $\displaystyle \left[\sqrt{6}-\sqrt{2}\;\right]>\left[\sqrt{7}-\sqrt{3}\;\right]>\left[\sqrt{8}-\sqrt{4}\;\right]>.....>\left[\sqrt{n}-\sqrt{n-4}\;\right]$ as "n" keeps increasing.
You can use calculus to show that $\displaystyle f(x)=\sqrt{x}-\sqrt{x-4}$
is a decreasing function as x increases.
$\displaystyle \displaystyle\frac{d}{dx}\left[\sqrt{x}-\sqrt{x-4}\;\right]=\frac{d}{dx}\left(x^{\frac{1}{2}}-(x-4)^{\frac{1}{2}\right)$
$\displaystyle =\displaystyle\frac{1}{2}x^{-\frac{1}{2}}-\frac{1}{2}(x-4)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}-\frac{1}{2\sqrt{x-4}}$
$\displaystyle \displaystyle\sqrt{x-4}<\sqrt{x}\;\; \Rightarrow\frac{1}{\sqrt{x-4}}>\frac{1}{\sqrt{x}}$
Therefore the derivative is negative, so the function is a decreasing function.
Therefore $\displaystyle \left[\sqrt{7}-\sqrt{3}\;\right]<\left[\sqrt{6}-\sqrt{2}\;\right]$
You could also examine the function $\displaystyle f(x)=\sqrt{x+4}-\sqrt{x}$
which is defined for x positive.