show that \(\frac{1}{1+\frac{1}{\sqrt{}2}}\) can be written as \(2-\sqrt{2}\)
+9481 \(\frac{1}{1+\frac{1}{\sqrt2}}\)
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 \(\sqrt2-1\) .
= \(\frac{\sqrt2}{\sqrt2+1}\,\cdot\,\frac{\sqrt2-1}{\sqrt2-1}\)
= \(\frac{2-\sqrt2}{2-1}\)
= \(\frac{2-\sqrt2}{1}\)
= \(2-\sqrt2\)
+9481 \(\frac{1}{1+\frac{1}{\sqrt2}}\)
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 \(\sqrt2-1\) .
= \(\frac{\sqrt2}{\sqrt2+1}\,\cdot\,\frac{\sqrt2-1}{\sqrt2-1}\)
= \(\frac{2-\sqrt2}{2-1}\)
= \(\frac{2-\sqrt2}{1}\)
= \(2-\sqrt2\)
