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Spivak: Is his how to approach these proofs?

  1. Sep 1, 2011 #1
    Hello all :smile:

    I have started the problem set for Chapter one (basic properties of numbers) in Spivak's Calculus (self study). I think I am doing these right, but I have some questions.

    As a solid example, problem 1-(iv) says to prove the following:

    [tex]x^3 - y^3 = (x-y)(x^2+xy+y^2)\qquad(1)[/tex]

    My approach was to add and subtract terms to the left hand side of (1) until it could be factored into the desired form:

    [tex]\begin{array}{l}
    x^3 - y^3 &= x^3 - y^3 + (x^2y - x^2y) + (xy^2 - xy^2) \\
    &= x^3 +x^2y+xy^2 - y^3 - x^2y - xy^2 \\
    &=x(x^2+xy+y^2)-y(x^2+xy+y^2) \\
    &= (x-y)(x^2+xy+y^2)
    \end{array}
    [/tex]

    Now this seems correct to me, but I feel a little guilty because I only knew what to add and subtract to the LHS of (1) because I knew what I was trying to achieve- the RHS of (1).

    My second question is similar, but requires some clarifying of Spivak's intent. In problem 1-(vi) he says: Prove the following:

    [tex]x^3 + y^3 = (x+y)(x^2-xy+y^2)\qquad(2)[/tex]

    Then he says, "There is a particularly easy way to do this, using (iv), and it will show you how to find a factorization of [itex]x^n+y^n[/itex] whenever n is odd."

    Well, I solved this one the exact same way I did with (iv) above: I used the right hand side of (2) to infer what terms to add/subtract and obtained the solution. However, the fact that I do not see what he means by the quoted text above leads me to belive that there was some other approach. That is, I have not discovered a way to factorize [itex]x^n+y^n[/itex] whenever n is odd in my procedure.

    Any thoughts are appreciated :smile:
     
  2. jcsd
  3. Sep 1, 2011 #2

    LCKurtz

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    Certainly, if you know (1), you can get (2) by simply substituting -y in for the y.
     
  4. Sep 1, 2011 #3
    Hi LC :smile: Yes. I arrived at the solution, but in particular, I am concerned about my approach as I do not see what he means by

     
  5. Sep 1, 2011 #4

    LCKurtz

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    OK, I see what you mean now. Apparently you need a more clever approach to (1) that easily generalizes to higher odd exponents. I dunno -- I'll think about it if this football game doesn't keep me too distracted. But don't count on me for this one :frown:
     
  6. Sep 1, 2011 #5

    LCKurtz

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    Also, I guess that observing that y = x is a root for odd exponents and factoring (y-x) by long division doesn't count?
     
  7. Sep 1, 2011 #6
    Maybe he made a typo? He says "this can easily be done using (iv)". Maybe he meant (v). In problem (v) we proved that

    [tex]
    x^n-y^n = (x-y)(x^{n-1} + x^{n-2} y+\dots+xy^{n-2} +y^{n-1} )
    [/tex]

    Anyway, no big deal! Get back to your game! :smile:
     
  8. Sep 2, 2011 #7

    micromass

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    My thoughts:

    Spivak is correct, (vi) can be easily done by (iv) by just substituting -y for y.
    This gives you a hint how to obtain the factorization of [itex]x^n+y^n[/itex]. Indeed, take (v) and substitute -y for y again...
     
  9. Sep 2, 2011 #8
    I am pretty sure this is what Spivak intended.

    Also note that you don't have to start from the left side for the first one..

    (x-y)([itex]x^{2}[/itex]+xy+[itex]y^{2}[/itex])=x([itex]x^{2}[/itex]+xy+[itex]y^{2}[/itex])-y([itex]x^{2}[/itex]+xy+[itex]y^{2}[/itex])=([itex]x^{3}[/itex]+[itex]x^{2}[/itex]y+x[itex]y^{2}[/itex])-([itex]x^{2}[/itex]y+x[itex]y^{2}[/itex]+[itex]y^{3}[/itex])=[itex]x^{3}[/itex]-[itex]y^{3}[/itex]

    This approach will come in handy for at least one problem in chapter two, but I will let you figure out which one!
     
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