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show that \(\frac{1}{1+\frac{1}{\sqrt{}2}}\) can be written as \(2-\sqrt{2}\)

Guest Nov 27, 2017

Best Answer 

 #1
avatar+5587 
+3

     \(\frac{1}{1+\frac{1}{\sqrt2}}\)

                     Rewrite the  1  in the denominator as  \(\frac{\sqrt2}{\sqrt2}\) .

\(\frac{1}{\frac{\sqrt2}{\sqrt2}+\frac{1}{\sqrt2}}\)

                     Add the fractions in the denominator together.

\(\frac{1}{\frac{\sqrt2+1}{\sqrt2}}\)

                                That is the same as...

\(1\div\frac{\sqrt2+1}{\sqrt2}\)

                                Invert the second fraction and multiply.

\(1\,\cdot\,\frac{\sqrt2}{\sqrt2+1}\)

 

\(\frac{\sqrt2}{\sqrt2+1}\)

                                Multiply the numerator and denominator by  \(\sqrt2-1\) .

\(\frac{\sqrt2}{\sqrt2+1}\,\cdot\,\frac{\sqrt2-1}{\sqrt2-1}\)

 

\(\frac{2-\sqrt2}{2-1}\)

 

\(\frac{2-\sqrt2}{1}\)

 

\(2-\sqrt2\)

hectictar  Nov 27, 2017
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2+0 Answers

 #1
avatar+5587 
+3
Best Answer

     \(\frac{1}{1+\frac{1}{\sqrt2}}\)

                     Rewrite the  1  in the denominator as  \(\frac{\sqrt2}{\sqrt2}\) .

\(\frac{1}{\frac{\sqrt2}{\sqrt2}+\frac{1}{\sqrt2}}\)

                     Add the fractions in the denominator together.

\(\frac{1}{\frac{\sqrt2+1}{\sqrt2}}\)

                                That is the same as...

\(1\div\frac{\sqrt2+1}{\sqrt2}\)

                                Invert the second fraction and multiply.

\(1\,\cdot\,\frac{\sqrt2}{\sqrt2+1}\)

 

\(\frac{\sqrt2}{\sqrt2+1}\)

                                Multiply the numerator and denominator by  \(\sqrt2-1\) .

\(\frac{\sqrt2}{\sqrt2+1}\,\cdot\,\frac{\sqrt2-1}{\sqrt2-1}\)

 

\(\frac{2-\sqrt2}{2-1}\)

 

\(\frac{2-\sqrt2}{1}\)

 

\(2-\sqrt2\)

hectictar  Nov 27, 2017
 #2
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+2

Great job, hectictar !. Thank you.

Guest Nov 27, 2017
edited by Guest  Nov 27, 2017

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