
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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May 6th, 2014, 04:07 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 04:28 AM. 
May 6th, 2014, 05:53 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, 06:05 AM  #3 
Newbie Joined: May 2014 From: India Posts: 6 Thanks: 0  ???? 
May 6th, 2014, 06:28 AM  #4 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 Math Focus: Elementary mathematics and beyond 
Choice c seems to suggest $\displaystyle \sqrt{2}=\sqrt2$ which isn't true.

May 6th, 2014, 06:57 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”. Last edited by Olinguito; May 6th, 2014 at 07:17 AM. 
May 6th, 2014, 07:18 AM  #6 
Global Moderator Joined: Oct 2008 From: London, Ontario, Canada  The Forest City Posts: 7,814 Thanks: 1046 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 07:37 AM. 
May 6th, 2014, 07:56 AM  #7 
Senior Member Joined: Apr 2014 From: Glasgow Posts: 2,115 Thanks: 708 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!


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