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SpivakProve x^2 - y^2 = (x-y)(x+y)

  1. Nov 18, 2009 #1
    Ok. So know that I am supposed to use one or more of the 12 basic properties of numbers from chapter 1 of Spivak's Calculus to prove this, but I can't see to figure out which one to start with.

    Can someone start me off here?

    Just a 'nudge.'

    edit: it says to 'prove' it, not to 'show' it. How do you 'prove' this? I can't see how to even start the proof.
     
    Last edited: Nov 18, 2009
  2. jcsd
  3. Nov 18, 2009 #2

    Mark44

    Staff: Mentor

    I would start with the right side, (x -y)(x + y) and use the distributive property to get
    (x -y)(x + y) = (x - y)(x) + (x - y)(y).
    Then use the commutative property of multiplication to rewrite (x - y)(y) as y(x - y), and the same for the other product. And so on...

    Eventually you'll get to x^2 - y^2 and will have justified each step along the way by one of the basic properties of real numbers.
     
  4. Nov 18, 2009 #3
    See now, that's what I wanted to do and I know that is a perfectly acceptable way to do it. But I have a question about that:

    Perhaps it is just some false notion that I have conjured up somewhere along the way, but I always thought that when they (the math gods) said "Prove A = B," you were supposed to start with 'A' and successively apply 'basic rules' until you arrive at 'B.'

    And when they say: "Show A = B," you can just manipulate 'A' or 'B' to show that the two are equivalent.


    Is this something that I have made up in my head? Or is there some truth to it?

    Thank you,
    Casey
     
  5. Nov 18, 2009 #4
    'Proving' a statement and 'showing' a statement is true are the same. If you are asked to show that A=B, then you can manipulate both A and B (as long as you're using valid operations, of course). Since A=B and B=C implies that A=C, this strategy is valid.
     
  6. Nov 18, 2009 #5

    Mark44

    Staff: Mentor

    No, you can start with either side. To elaborate on what VeeEight said, you can also start with A = B and apply reversible operations to both sides until you arrive at a new equation that has the same solution set. Because all the operations are reversible, that means that your first equation has the same solution set.

    For a simple example, suppose we start with x + 2 = 8. I can add -2 to both sides (a reversible operation) to get x = 6. The solution set for the second equation is obviously {6}, and because the operation I applied was one-to-one, and therefore invertible, the first equation's solution set is also {6}.

    On the other hand, if the operation is not reversible, then the solution sets of the starting and ending equations do not have to be the same. Starting with x = -2, and squaring both sides, I get x2 = 4. The solution set of this equation is {2, -2}, while that of the original equation is {-2}. The squaring operation is not reversible, for the same reason that the function f(x) = x2 is not 1-to-1.
     
  7. Nov 18, 2009 #6
    Okay. Thanks Mark44 and VeeEight :smile:

    That is very helpful to know! I had a piece of paper in front of me that just had:

    x^2 - y^2 =

    written on it and I was trying to find a way to apply basic properties of numbers to this expression such that I would naturally 'arrive at' (x - y)(x + y). :redface:
     
  8. Nov 18, 2009 #7
    I did that problem starting with just that.

    x2 - y2 = x2 - xy + xy -y2

    x(x - y) + y(x - y) = (x-y)(x+y)
     
  9. Nov 19, 2009 #8

    Mark44

    Staff: Mentor

    If you work it through from the side with (x - y)(x + y), it makes the "trick" of subtracting and adding xy more understandable. Obviously, you can work it from either end.
     
  10. Nov 19, 2009 #9
    How would you start it from the LHS? What basic property can you apply such that:

    x2-y2=x2+xy-xy+y2 ?

    I am missing how one logically deduces the RHS from the LHS?

    How did you arrive at that l'hopital? Without just knowing that the two are equivalent...
     
  11. Nov 19, 2009 #10

    lanedance

    User Avatar
    Homework Helper

    i think you dropped a sign somewhere, but i would go as follows
    x2- y2 = x2- y2 + 0 = x2 - y2 + ((xy)+ -(xy)) = x2 + xy - xy - y2

    but as I think Mark was pointing out, all the steps you took working from RHS to LHS are perfectly valid in the opposite direction
     
  12. Nov 19, 2009 #11
    x2 - y2
    x2 - y2 + 0
    x2 - y2 + (xy -xy)
    ... and so on
     
  13. Nov 19, 2009 #12
    Oh jeesh....how do you guys come up with this stuff?! Who would ever think to do that? I mean...I guess you would :smile: But seriously.....


    Thanks!

    And yes, I missed a '-' sign.


    EDIT: And apparently l'hopital would think to do that. It seems that I missed the memo about randomly adding zero to expressions :rofl:
     
  14. Nov 19, 2009 #13

    lanedance

    User Avatar
    Homework Helper

    zero is often pretty good to use in these types of proofs as you an turn it into a sum or multiplication with almost anything to help
     
  15. Nov 19, 2009 #14
    Thank you. I will need to hold onto these tricks and learn start thinking a little differently if I am to make it through this text by the time classes start up again.
     
  16. Nov 19, 2009 #15

    Mark44

    Staff: Mentor

    Not only is adding zero a good technique to remember, it's one of a very few things you can do to just one side of an equation, with others being multiplying by 1 in some form, simplifying, and expanding.
     
  17. Nov 19, 2009 #16
    Great point! I never thought of it that way.
     
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