Simplify this division of square roots?

#((sqrt2)/2)/(1+(sqrt2)/2)#

3 Answers
Ratnaker Mehta
Sep 1, 2016

#sqrt2-1#.

Explanation:

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#.

sente
Sep 1, 2016

#(sqrt(2)/2)/(1+sqrt(2)/2)=sqrt(2)-1#

Explanation:

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#

EZ as pi
Sep 1, 2016

#sqrt2-1#

Explanation:

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#

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