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Prove that √ab =√a × √b

  1. Nov 23, 2016 #1
    1. The problem statement, all variables and given/known data

    Prove that √ab = √a × √b is possible for all the value of a and b except when both are less than 0 at the same time .


    2. Relevant equations

    √(-a)b =(√a)i × b = (√ab)i
    3. The attempt at a solution
    I don't know how to start, to prove this type of question, but i think that there must be an answer for this.
     
  2. jcsd
  3. Nov 23, 2016 #2

    BvU

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    So what goes wrong when both a and b are < 0 ?
     
  4. Nov 23, 2016 #3

    andrewkirk

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    Divide it into four cases according to whether each of a and b is <0 or ##\geq 0##. For the case where both are ##\geq 0## it is easy. For the two where only one of them is ##\geq 0##, use the formula to prove the equation true. For when both are negative, prove it false.
     
  5. Nov 23, 2016 #4
    This happens if a and b are less than 0, √36 = √(-4)(-9) = √(-4) × √(-9) = 2i × 3i = -6(which is wrong)
    Why √ab = √a × √b not works when both are less than 0. It works when only one of them is less than 0, ex: √-36 = √(-9)(4) = (3i) × 2 = 6i, then why it not works when both of them are less than 0
     
  6. Nov 23, 2016 #5

    BvU

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    We've had this before, haven't we ?
    You did check fresh's suggested link, I suppose ? Nothing unclear any more in that insight ?
    This is understandable if you are a bit familiar with complex numbers. You understand it has nothing to do with the modulus of the numbers (the absolute value), only with the arguments. Our definition of square root of -1 = ##i## and not ##-i## has a lot to do with it.

    You repeat the question: Why ?

    So apparently the answer isn't fully understandable for you. However, repeating those answers won't help. Perhaps you can get us out of this loop by specifying what exactly it is that's hard to understand ?
     
  7. Nov 23, 2016 #6
    Yes i checked his link and i got everything which was written in his link.


    I repeated this question because, i fully understood fresh's link and it was written there:
    Rule 2
    If x,y≥0 and n∈R, then and only then (xy)^n = x^n × y^n and therefore we can't write √(-a)(-b) =√(-a) . √(-b), note here a and b are positive numbers.
    But this is not my question, my question is why the above rule is only applicabe if x,y>0 why it is not apllicable when x,y<0
     
    Last edited by a moderator: Nov 28, 2016
  8. Nov 28, 2016 #7

    Mark44

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    The reason is that, for a < 0 and b < 0, ##\sqrt{(-a) \cdot (-b)} = \sqrt{a \cdot b}##, which is a real, positive number. However, ##\sqrt{-a} \cdot \sqrt{-b} = (-1)\sqrt{ab}##, which is a real, negative number. Obviously a positive number can't also be negative. This is the reason for the restriction that both a and b must be nonnegative.
     
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