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Is this a legal operation? Gelfand Argh.

  1. Jan 20, 2008 #1
    Is this a legal operation? Gelfand!!! Argh.

    [tex]a-b=(\sqrt a)^2-(\sqrt b)^2=(\sqrt a + \sqrt b)(\sqrt a - \sqrt b)[/tex]

    Yes, no? :-\
  2. jcsd
  3. Jan 20, 2008 #2
    Sure, why not? a.k.a. Yes.
  4. Jan 20, 2008 #3
    Looks pretty reasonable .... I guess you could make the point that if you're taking a negative root of either a or b, then you have to be careful to be consistent to stick with that root in the final expression.
  5. Jan 20, 2008 #4
    Gelfand ...

    "So we have factored a-b, haven't we? No, we haven't because [tex]\sqrt a - \sqrt b[/tex] is not a polynomial; taking the square root is not a legal operation for polynomials - only addition, subtraction and multiplication are allowed."

  6. Jan 20, 2008 #5
    I see your reasoning in which a, b, or c > 0 would need to be stated.
  7. Jan 20, 2008 #6
    No, that wasn't my point. I meant that for any positive real, there are two roots, e.g. the square roots of 4 are 2 and -2. So when you write those square roots, you could be thinking of the negative root, which is fine, you just have to stick with your choice throughout.
  8. Jan 20, 2008 #7
    What??? Since when?
  9. Jan 20, 2008 #8
    LOL, that's why I posted it to find out.
  10. Jan 20, 2008 #9


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    Dearly Missed


    We have:
    which is the same thing.
  11. Jan 20, 2008 #10
    Your question was not a legal one, because you did not say what were "a" and "b" ... :rofl:

    Even your last equality

    [tex](\sqrt a)^2-(\sqrt b)^2=(\sqrt a + \sqrt b)(\sqrt a - \sqrt b)[/tex]

    is wrong in non-commutative algebra
    Last edited: Jan 20, 2008
  12. Jan 20, 2008 #11
    It's not my question, it's word for word by Gelfand :p
  13. Jan 20, 2008 #12
    Truth, only truth, but not all the truth! :smile:

    May be you cited Gelfand (I.M. ?) word for word, but you forget mention what he said earlier, like ... "let a and b be polynomial"...
  14. Jan 20, 2008 #13
    Unless you have the book Algebra by Gelfand, then you can accuse me of not including more information. I quoted word for word, so Idc cuz I'm not going to type up the whole book for you. And no, he did not state your last "quote."
  15. Jan 20, 2008 #14
    You quoted some fragment of text from page 54 (2003 edition) of the book by IM Gelfand and A Shen. This fragment was part of the section "32 Factoring". In the beginning of the section (p. 51) authors explained that they dealt with POLYNOMIALS...
  16. Jan 20, 2008 #15
    Oh, fooey. I just meant that one had to be careful not to write:
    (note the change of sign on the last [tex]\sqrt{a}[/tex]).

    Since the original expression was fairly trivial for real-valued a and b, I was just trying to imagine how someone learning elementary algebra might make a mistake.
  17. Jan 20, 2008 #16

    Gib Z

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    The identity is correct for all complex numbers (taking the right branch cuts). Stating that, we don't have to talk about non-commutative algebras.

    However, if factoring a polynomial, you can only factor it into integer powered terms ie smaller polynomials. Since this new expression is not a polynomial, you didn't actually factor the polynomial, but your expression is still true.
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