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If xy^2 = ab^2 : Does xy = ab?

  1. Jun 20, 2011 #1

    Femme_physics

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    If

    xy2 = ab2

    Does that mean automatically that xy = ab?
     
  2. jcsd
  3. Jun 20, 2011 #2

    Char. Limit

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    Not necessarily. It's possible that xy = -ab.
     
  4. Jun 20, 2011 #3

    Femme_physics

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    Got it :) Thanks
     
  5. Jun 20, 2011 #4

    I like Serena

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    Let's try something.

    [itex]1 \cdot 2^2 = 4 \cdot 1^2[/itex] is true isn't it?

    But is: [itex]1 \cdot 2 = 4 \cdot 1[/itex] ? :confused:
     
  6. Jun 20, 2011 #5

    Char. Limit

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    Oh, I was assuming the OP meant (xy)^2 = (ab)^2. If it's how you're suggesting, then (if x y^2 = a b^2, then xy = ab) only if y=b.
     
  7. Jun 20, 2011 #6

    I like Serena

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    @CL: I suspect that it was what the OP meant too, but I felt the playful need to point out to the OP that round thingies should be used. :smile:
     
  8. Jun 20, 2011 #7

    Char. Limit

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    Oh, of course. Round thingies should always be used. Or square thingies, they work too.

    [xy]^2 = (ab)^2.
     
  9. Jun 20, 2011 #8
    a^2-b^2 =0
    <=> (a-b)(a+b)=0
    <=> a=b or a=-b
     
  10. Jun 20, 2011 #9

    Mark44

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    Femme_physics,
    By chance, you don't mean (xy)2 = (ab)2, do you?
     
  11. Jun 20, 2011 #10
    Assuming you mean (xy)^2 = (ab)^2, xy does not neccesarilly equal ab:

    sqrt[(xy)^2] = sqrt[(ab)^2]
    then
    (+/-)xy = (+/-)ab
    which is true for only two out of four cases:
    +xy=+ab
    -xy=-ab
    But, you wind up with problems when -xy = +ab and +xy=-ab.

    Hope that helps a little.
     
  12. Jun 20, 2011 #11

    Ouabache

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    I did not assume the OP meant {if [itex] (xy)^2 = (ab)^2[/itex], does [itex] xy = ab[/itex]. I approached it as written {if [itex] xy^2 = ab^2 [/itex] does [itex] xy = ab[/itex]} which is false. Without proving it, by testing empirically, it becomes obvious it is false.

    However if we assume it the first way, I don't see 4 cases, only 2, but that is sufficient to show the two equations are not equal (see below cases iii and iv, resolve to i and ii. Below my designation for case and equation numbering are interchangeable).

    (i) xy = ab
    (ii) xy= -ab
    (iii) -xy= ab; multiply eq. by -1, you get xy= -ab (which is eq. ii)
    (iv) -xy = -ab; multiply eq. by -1, you get xy = ab (which is eq. i )
     
  13. Jun 21, 2011 #12
    You're right Ouabache, I didn't bother equating the pairs of equations to reduce it to two cases.

    As written,
    if (xy^2)=(ab^2)
    then (x/a) = (b^2/y^2)

    if xy=ab
    then (x/a)=(b/y)

    Then the original statement is true only iff (b/y)=(b^2/y^2)=((b/y)^2) which, as previously pointed out, is only true iff b=y.

    And apologies for my ignorance of TeX.
     
  14. Jun 21, 2011 #13
    not true if y =0 and x = 0
     
  15. Jun 21, 2011 #14
    That's true. Always good to keep in mind the impact of nul denominators. Anyway, I think we've disproven this more ways than Femme_Physics has time to read about lol
     
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