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I am trying to figure out how to reduce simple square root fractions without a calculator. In my lecturer's notes, for instance, he reduces $1/\sqrt2$ by multiplying with $4/\sqrt2$. Following is his example and another example of him doing this:

Reducing square root fraction

How does he know how to do this? I am allowed to bring notes for my exam, is there some practical table for the most common square root fractions I could bring? Or is there some rule for reducing I can use?

Really hope you can help me out here, thanks!

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user102937
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  • I think he multiplied it by $2/\sqrt4$, not the other way around... – Simply Beautiful Art Dec 29 '16 at 13:32
  • That is generally called "rationalizing the denominator". [tex]\sqrt{a}\times\sqrt{a}= a[/tex] so multiplying both numerator and denominator of a fraction with a square root in the middle moves the square root to the numerator. $ More generally, if you have something of the form $a+ b\sqrt{c}$ in the denominator – user247327 Dec 29 '16 at 13:33
  • $\vec{u}=\left(\begin{matrix}2\ 2 \ 6\end{matrix}\right)$? – Maestro13 Dec 29 '16 at 13:34
  • What do you call "reducing a square root fraction" ? You can always use $1/\sqrt n=\sqrt n/n$. –  Dec 29 '16 at 13:39

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That is generally called "rationalizing the denominator". [tex]\sqrt{a}\times\sqrt{a}= a[/tex] so multiplying both numerator and denominator of a fraction with a square root in the middle moves the square root to the numerator. $\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}= \frac{\sqrt{2}}{2}$

More generally, if you have something of the form $a+ b\sqrt{c}$ in the denominator multiply both numerator and denominator by its "conjugate" $a- b\sqrt{c}$. Since $(a+ b)(a- b)= a^2- b^2$ that will also get rid of the square root in the denominator: $\frac{1}{2+ 3\sqrt{2}}\frac{2- 3\sqrt{2}}{2- 3\sqrt{3}}= \frac{2- 3\sqrt{2}}{4- 9(2)}= -\frac{2- 3\sqrt{2}}{14}$

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