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 September 13th, 2018, 08:14 PM #1 Member     Joined: Jun 2017 From: Lima, Peru Posts: 97 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: $$\sqrt{\frac{4-2\sqrt{3}}{2}}$$ 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, 08:24 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,464 Thanks: 2038 4 - 2√3 = (3 - 2√3 + 1) = (√3 - 1)² Thanks from topsquark
September 14th, 2018, 08:35 AM   #3
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Quote:
 Originally Posted by Chemist116 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: $\sqrt{\frac{4-2\sqrt{3}}{2}}$ 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.
This is not a formal proof.

$x = \sqrt{2 - \sqrt{3}} \implies x^2 = 2 - \sqrt{3}.$

$y = \dfrac{\sqrt{3} - 1}{\sqrt{2}} \implies y^2 = \dfrac{(\sqrt{3})^2 + 2(-\ 1)\sqrt{3} + (-\ 1)^2}{(\sqrt{2})^2} = \dfrac{3 - 2\sqrt{3} + 1}{2} \implies$

$y^2 = \dfrac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}.$

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 09:16 AM. Reason: LaTeX error

September 14th, 2018, 09:42 AM   #4
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Quote:
 Originally Posted by JeffM1 This is not a formal proof. $x = \sqrt{2 - \sqrt{3}} \implies x^2 = 2 - \sqrt{3}.$ $y = \dfrac{\sqrt{3} - 1}{\sqrt{2}} \implies y^2 = \dfrac{(\sqrt{3})^2 + 2(-\ 1)\sqrt{3} + (-\ 1)^2}{(\sqrt{2})^2} = \dfrac{3 - 2\sqrt{3} + 1}{2} \implies$ $y^2 = \dfrac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}.$
Alternatively, you could make use of this directly to get a proof. You've shown that both $\sqrt{2 - \sqrt{3}}$ and $\dfrac{\sqrt{3} - 1}{\sqrt{2}}$ are roots of the polynomial $x^2 - (2-\sqrt{3})$, which means they are equal up to sign. From here you need only note that both numbers are positive to conclude that they are actually equal.

Last edited by cjem; September 14th, 2018 at 09:46 AM.

September 14th, 2018, 09:58 AM   #5
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Quote:
 Originally Posted by cjem Alternatively, you could make use of this directly to get a proof. You've shown that both $\sqrt{2 - \sqrt{3}}$ and $\dfrac{\sqrt{3} - 1}{\sqrt{2}}$ are roots of the polynomial $x^2 - (2-\sqrt{3})$, which means they are equal up to sign. From here you need only note that both numbers are positive to conclude that they are actually equal.
Yes, you are of course quite right. I was interested, however, in showing the method of Proclus, which comes in handy many times. Maybe it was not the best example to pick.

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