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How to write $\frac{1}{1+\sqrt{3}+\sqrt{5}+\sqrt{15}}$ with a rational denominator?

There is an included hint: factorize the denimator

Edit: There has been some confusion on this question, the first "1" means "1 over 1+√3+√5+√15" Sorry, I can see how it could be perceived as 1/1 (1)

KCd
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blah
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  • Which bit are you stuck on - how to factorize the denominator, or what to do next? If the first, remember that $\sqrt{a} × \sqrt{b} = \sqrt{ab}$. For the second, use the difference of two squares to figure out what to multiply the top and bottom by – IanF1 Mar 15 '15 at 10:48

3 Answers3

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$$\dfrac{1}{1+\sqrt3+\sqrt5+\sqrt{15}}=\dfrac{1}{(1+\sqrt3)(1+\sqrt{5})}=\dfrac{(1-\sqrt3)(1-\sqrt{5})}{(1+\sqrt3)(1+\sqrt{5})(1-\sqrt3)(1-\sqrt{5})}=\dfrac{(1-\sqrt3)(1-\sqrt{5})}{(1-3)(1-5)}=\dfrac{(1-\sqrt3)(1-\sqrt{5})}{8}$$

ASB
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  • Absolutely perfect, thanks a ton! Just a question, how did you manage the fraction lines? – blah Mar 15 '15 at 10:58
  • You're welcome. it's not difficult, i think :) – ASB Mar 15 '15 at 11:04
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    @blah: There are two ways to see the LaTeX source of an answer. 1. If the "edit" button is visible to you, click on it (I don't know if you are able to do this with your reputation). 2. Right-click on the text, and select "Show Math As" -> "TeX Commands". – TonyK Mar 15 '15 at 11:17
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As $\sqrt{ab}=\sqrt a\cdot\sqrt b$ for $a,b\ge0,$

$$1+\sqrt3+\sqrt5+\sqrt{15}=1+\sqrt3+\sqrt5(1+\sqrt3)=(1+\sqrt3)(1+\sqrt5)$$

Now $\sqrt5+1=\dfrac{5-1}{\sqrt5-1}$

lab bhattacharjee
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Consider: $(1+\sqrt{3}+\sqrt{5}+\sqrt{15})(1+\sqrt{3}-(\sqrt{5}+\sqrt{15}))=-16+2(\sqrt{3}-\sqrt{65})$. So you end up with only 2 roots. Do this proces again to end up with only 1. Again and only rational numbers appear.

abcdef
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