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Homework Help: Verification of Simple Proof

  1. Apr 15, 2012 #1
    1. The problem statement, all variables and given/known data

    I am using Spivak's Calculus and just finished the third exercise in part 1.
    It was a very easy exercise but it seems that Spivak makes some assumptions.
    The problem is as stated:

    If x[itex]^{2}[/itex]=y[itex]^{2}[/itex], then either x=y or x=-y. Prove it.

    The proof was relatively simple (by factoring out (x-y)(x+y) from x[itex]^{2}[/itex]
    -y[itex]^{2}[/itex] and showing that either (x+y)=0 or (x-y)=0 or both). What I had problems with was that it was assumed that a(0)=0. So, I will now propose a proof that a(0)=0 for all real a using the properties listed in the book. All I need is a confirmation that my proof haw no flaw. Thank you.
    2. Relevant equations

    3. The attempt at a solution

    Introduce (a*0).


    By the distributive law


    By the existence of the additive identity


    Subtracting (a*0) from both sides yields


    And by the existence of additive identity


    Which is true. Is this valid?
    Last edited: Apr 15, 2012
  2. jcsd
  3. Apr 15, 2012 #2


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    "Introduce (a*0)" in what??.

    That's not the distributive law- that's the fact that 0+ 0= 0. And from that it follows that a(0)+ a(0)= a(0)

    That is certainly NOT true! And I don't see any point in talking about the "existence of the additive identity" when you have already been using "0" which is defined as the additive identity.

    No, it is not. Instead, from the distributive law a(0+ 0)= a(0)+ a(0). And since 0 is additive identity, x+ 0= x for any x. In particular 0+ 0= 0 so the left side is just a(0). Once you have a(0)= a(0)+ a(0) use the fact that a(0) has an additive inverse.
    Last edited by a moderator: Apr 16, 2012
  4. Apr 16, 2012 #3
    Please reread the edited proof. I made an error when I first posted.
  5. Apr 16, 2012 #4


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    Same errors I noted before. You seem to have payed no attention to my first post.
  6. Apr 16, 2012 #5
    The easiest way to do it would probably be just to give it some modus tollens imo

    I'm not quite sure what you're trying to do by playing about with 0 here...
  7. Apr 16, 2012 #6
    Okay. I will redo it here and explain myself better. In Spivak's book, he lists 12 properties of the real numbers. I will use those properties to prove that for any real a


    Begin with the assumption that the above statement is false.

    Since (a*0) is in R, then by the closure under addition property, (a*0)+(a*0) must be in R. Therefore, we can add (a*0) to both sides of the above equation and we shall have

    (a*0)+(a*0)=0+(a*0). (1)

    Now, the left side of the equation can be factored out by the distributive law to yield


    Thus, (1) becomes:

    (a(0+0))=0+(a*0). (2)

    But, 0+0=0 and the left side becomes


    Thus, for (2), we have:

    (a(0))=0+(a(0)). (3)

    But, by property 3 in Spivak's book, for every real k, there exists it's additive inverse denoted -k such that


    Therefore, by adding the negative of (a*0) to both sides of equation (3), we get

    (a(0))+(-(a(0)))=0+(a(0))+(-(a(0))). (4)

    Again, by Spivak's third stated property, we have:

    (a(0))+(-(a(0))=0 and equation (4) becomes

    0=0+0. (5)

    But it is clear from Spivak's second property that


    and equation (5) becomes


    But, this is obviously true, and since valid manipulation was made to our initial equation


    then our initial assumption was false and, therefore,

    (a*0)=0 must be true.

    I really hope this clarifies things.
  8. Apr 16, 2012 #7
    Why are you allowed to use the above equation?? You just said you assumed the equation is false.
  9. Apr 16, 2012 #8
    Hello micromass. I did the proof through contradiction. If we begin with the assumption that the equation is false and manipulate the equation so that everything we do is justifiable, we should end up with a result that varifies the assumption. However, what I got (after showing after steps of manipulation) is that 0=0. Now, since this is obviously true, the INITIAL assumption that the equation was false is in its own right false. Therefore, the equation must be true. I believe that I had actually done this problem in Apostol's calculus book. As far as I cam recall, this was an adequate proof.
  10. Apr 16, 2012 #9
    Yes, I get what you were trying to do.

    But if you assume the equation is false, then you cannot use that in your proof. You can only use things in a proof that you know are true or that you assumed are true.
  11. Apr 16, 2012 #10
    It's like this:

    I want to prove that 2+2=5.
    Assume by contradiction that 2+2=5 is not true.

    So take 2+2=5. Multiply both sides with 0, then we get 0*(2+2)=0*5. We get that 0=0.

    This is true, so 2+2=5 must be true.

    This is essentially what you did, but it is of course not valid.
  12. Apr 16, 2012 #11
    I understand what you mean. But I notice that it also does not change the outcome if I change the assumption from false to true. I must have made an error somewhere else because I made a similar proof a while back and had it verified by another forum. I will post another proof once I get back from grocery shopping. Thank you for your help.
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