My Math Forum Why is option b alone true?

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 May 6th, 2014, 05:07 AM #1 Newbie   Joined: May 2014 From: India Posts: 6 Thanks: 0 Why not these surds are equal? Okay, there is a question in my text book. Which one is true? A)$\displaystyle \sqrt{2}=\sqrt{(-1)(-2)}=\sqrt{(-1)}.\sqrt{(-2)}$ B)$\displaystyle \sqrt{4}=\sqrt{(-2)(-2)}=\sqrt{2}.\sqrt{2}=2$ C)$\displaystyle \sqrt{4}=\sqrt{(-2)(-2)}=\left (- \sqrt{2} \right )\left (- \sqrt{2} \right )=2$ D)$\displaystyle \sqrt{4}=\sqrt{(-2)(-2)}=\sqrt{2*2}=2$ Okay, Option a is false because square root of negative number results in imaginary number and goes on. Option b. Both I and text book accept it true. Option c. I accept even it is also true along with b. $\displaystyle \\ \sqrt{4}=\sqrt{(-2)(-2)}=\left (- \sqrt{2} \right )\left (- \sqrt{2} \right )=2\\ =\sqrt{4}=\sqrt{(+4)}=+\left (\sqrt{2} \right )^2=2\\ =\sqrt{4}=\sqrt{4}=2=2$ I think even option d is also right. $\displaystyle \\ \sqrt{4}=\sqrt{(-2)(-2)}=\sqrt{2*2}=2\\ =\sqrt{4}=\sqrt{+(4)}=\sqrt{4}=2\\ =\sqrt{4}=\sqrt{4}=\sqrt{4}=2\\$ But my text book says only option b is right. There's no explanation. Is only option b right?Or is option b, c, d right? Or is all of them right? Last edited by GovindBalaji; May 6th, 2014 at 05:28 AM.
 May 6th, 2014, 06:53 AM #2 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra In my opinion, only option A is false. Options B, C, D are all true.
May 6th, 2014, 07:05 AM   #3
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????

Quote:
 Originally Posted by Olinguito In my opinion, only option A is false. Options B, C, D are all true.
That's what I also think.
But why only option b is there in book.

 May 6th, 2014, 07:28 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1093 Math Focus: Elementary mathematics and beyond Choice c seems to suggest $\displaystyle \sqrt{-2}=-\sqrt2$ which isn't true.
 May 6th, 2014, 07:57 AM #5 Senior Member     Joined: Apr 2014 From: Greater London, England, UK Posts: 320 Thanks: 155 Math Focus: Abstract algebra Nevertheless $\sqrt{(-a)(-a)}=(-\sqrt a)(-\sqrt a)$ ($a\in\mathbb R^+\cup\{0\}$) is true. Whether a statement is true or not has nothing to do with what it “seems to suggest”. Thanks from CRGreathouse Last edited by Olinguito; May 6th, 2014 at 08:17 AM.
 May 6th, 2014, 08:18 AM #6 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,900 Thanks: 1093 Math Focus: Elementary mathematics and beyond Sure, but it's not clear what is intended (at least to me). Even though what you say is true, $\displaystyle \sqrt{-2}=-\sqrt{2}$ is not true. Most equalities are constructed to show a mathematical progression from one idea to another, are they not? Last edited by greg1313; May 6th, 2014 at 08:37 AM.
 May 6th, 2014, 08:56 AM #7 Senior Member   Joined: Apr 2014 From: Glasgow Posts: 2,133 Thanks: 719 Math Focus: Physics, mathematical modelling, numerical and computational solutions It's a rubbish question in my opinion. Poke your teacher and get him/her to set you better questions or get a different textbook! Thanks from Olinguito

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