Simplify this division of square roots?

The Expression`=(sqrt2/2)/(1+sqrt2/2)`

`=(sqrt2/cancel2)/((2+sqrt2)/cancel2)`

`=sqrt2/(2+sqrt2)`

`=sqrt2/(2+sqrt2)`

`=cancel(sqrt2)/(cancelsqrt2(sqrt2+1)`

`=1/(sqrt2+1)xx((sqrt2-1)/(sqrt2-1))`

`=(sqrt2-1)/(2-1)`

`=sqrt2-1`.

We will continue under the assumption that "simplifying" requires rationalizing the denominator.

First, we can remove fractions from the numerator and denominator by multiplying both by `2`:

`(sqrt(2)/2)/(1+sqrt(2)/2) = (sqrt(2)/2)/(1+sqrt(2)/2)*2/2`

`= sqrt(2)/(2+sqrt(2))`

Then, we rationalize the denominator by multiplying by the conjugate of the denominator, and taking advantage of the identity `(a+b)(a-b)=a^2-b^2`

`sqrt(2)/(2+sqrt(2)) = sqrt(2)/(2+sqrt(2))*(2-sqrt(2))/(2-sqrt(2))`

`=(2sqrt(2)-sqrt(2)*sqrt(2))/(2^2-sqrt(2)^2)`

`=(2sqrt(2)-2)/(4-2)`

`=(cancel(2)(sqrt(2)-1))/cancel(2)`

`=sqrt(2)-1`

We will make use of the fact that `(a/b)/(c/d) = (axxd)/(bxxc)`

But before we can do that, we need to add the fractions in the denominator to make one fraction.

`(sqrt2/2)/(1+sqrt2/2)" = " (sqrt2/2)/((2+sqrt2)/2)`

`(color(red)(sqrt2)/color(blue)(2))/(color(blue)((2+sqrt2)/color(red)(2))) " = "(color(red)(cancel2sqrt2))/ (color(blue)(cancel2(2+sqrt2))` Much better!

Now rationalise the denominator:

`sqrt2/((2+sqrt2)) xxcolor(lime)(((2-sqrt2))/((2-sqrt2))) = (2sqrt2-sqrt2^2)/(2^2 - sqrt2^2)`

`(2sqrt2-2)/(4 - 2) = (cancel2(sqrt2 -1))/cancel2`

=`sqrt2 -1`