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 January 30th, 2015, 04:49 PM #1 Newbie   Joined: Jan 2015 From: Asia Posts: 3 Thanks: 0 logarithms please help ! how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2? January 30th, 2015, 05:01 PM   #2
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 Originally Posted by hi2you how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?
$\displaystyle \frac{1}{\sqrt{3} + \sqrt{2}}$

Try multiplying the numerator and denominator by the conjugate:
$\displaystyle \frac{1}{\sqrt{3} + \sqrt{2}} \cdot \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}}$

Can you finish it?

-Dan January 31st, 2015, 03:44 PM #3 Newbie   Joined: Jan 2015 From: Asia Posts: 3 Thanks: 0 but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..? January 31st, 2015, 04:03 PM   #4
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Quote:
 Originally Posted by hi2you but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?
much easier the way topsquark has recommended ... and that is working "from the back" January 31st, 2015, 04:04 PM   #5
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Quote:
 Originally Posted by hi2you but I'm supposed to find that square root 3 - square root 2 equates to 1/square root three + square root 2, so I can't work from the back..?

$\displaystyle \color{blue}{\sqrt3-\sqrt2=(\sqrt3-\sqrt2)\cdot \dfrac{\sqrt3+\sqrt2}{\sqrt3+\sqrt2}= \dfrac{(\sqrt3-\sqrt2)(\sqrt3+\sqrt2)}{\sqrt3+\sqrt2}=\dfrac{( \sqrt3)^2-(\sqrt2)^2}{\sqrt3+\sqrt2} = \dfrac{3-2}{\sqrt3+\sqrt2} = \dfrac{1}{\sqrt3+\sqrt2}}$ January 31st, 2015, 04:11 PM   #6
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Quote:
 logarithms please help ! how do I prove that square root 3- square root 2 = 1/square root 3 + square root 2?

Where is the logarithm ...? January 31st, 2015, 06:21 PM #7 Newbie   Joined: Jan 2015 From: Asia Posts: 3 Thanks: 0 sorry! because this question was in my logarithm homework questions so I thought I needed to use logarithms! But thank you for your help !! Tags logarithms, math Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Ian McPherson Algebra 2 January 15th, 2013 03:04 PM vic Algebra 1 January 4th, 2013 02:15 PM empiricus Algebra 8 July 9th, 2010 08:31 AM stekemrt Abstract Algebra 11 November 21st, 2007 10:37 AM Ian McPherson Calculus 2 December 31st, 1969 04:00 PM

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