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September 13th, 2018, 09:14 PM  #1 
Member Joined: Jun 2017 From: Lima, Peru Posts: 61 Thanks: 1 Math Focus: Calculus  How to prove this rationalization?
I am stuck at this seemingly easy problem: It states this: $$\sqrt{2\sqrt{3}}=\frac{\sqrt{3}1}{\sqrt{2}}$$ So far I tried all the methods I know but I've only was able to get to: $$\sqrt{\frac{42\sqrt{3}}{2}}$$ This was done by multiplying numerator and denominator inside of the square root by two. But I don't know what else can be done to get to the desired result. 
September 13th, 2018, 09:24 PM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,107 Thanks: 1907 
4  2√3 = (3  2√3 + 1) = (√3  1)² 
September 14th, 2018, 09:35 AM  #3  
Senior Member Joined: May 2016 From: USA Posts: 1,253 Thanks: 519  Quote:
$x = \sqrt{2  \sqrt{3}} \implies x^2 = 2  \sqrt{3}.$ $y = \dfrac{\sqrt{3}  1}{\sqrt{2}} \implies y^2 = \dfrac{(\sqrt{3})^2 + 2(\ 1)\sqrt{3} + (\ 1)^2}{(\sqrt{2})^2} = \dfrac{3  2\sqrt{3} + 1}{2} \implies$ $y^2 = \dfrac{4  2\sqrt{3}}{2} = 2  \sqrt{3}.$ So you know that the equality is true. And now you can put together a formal proof by going in reverse. $2  \sqrt{3} \equiv 2  \sqrt{3} \implies$ $2  \sqrt{3} \equiv \dfrac{2(2  \sqrt{3})}{2} \implies$ $2  \sqrt{3} \equiv \dfrac{4  2\sqrt{3}}{2} \implies$ $2  \sqrt{3} \equiv \dfrac{3  2\sqrt{3} + 1}{2} \implies$ $2  \sqrt{3} \equiv \dfrac{(\sqrt{3})^2 + 2(\ 1)(\sqrt{3} + (\ 1)^2}{2} \implies$ $2  \sqrt{3} \equiv \dfrac{(\sqrt{3}  1)^2}{2} \implies$ $\sqrt{2  \sqrt{3}}\equiv \sqrt{\dfrac{(\sqrt{3}  1)^2}{2}} \implies$ $\sqrt{2  \sqrt{3}} \equiv \dfrac{\sqrt{(\sqrt{3}  1)^2}}{\sqrt{2}} \implies$ $\sqrt{2  \sqrt{3}} \equiv \dfrac{\sqrt{3}  1}{\sqrt{2}}. \text { Q.E.D.}$ This going in reverse is frequently very useful in finding proofs: start with what is to be proved as assumed and work backward to something already known. Then if all the steps can be reversed, you have your proof. (If I remember correctly, this way to find a proof is called the method of Proclus.) Last edited by JeffM1; September 14th, 2018 at 10:16 AM. Reason: LaTeX error  
September 14th, 2018, 10:42 AM  #4  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 287 Thanks: 88 Math Focus: Number Theory, Algebraic Geometry  Quote:
Last edited by cjem; September 14th, 2018 at 10:46 AM.  
September 14th, 2018, 10:58 AM  #5  
Senior Member Joined: May 2016 From: USA Posts: 1,253 Thanks: 519  Quote:
 

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