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If 0<x<y and 0<a<b then ax<yb proof.

  1. Sep 4, 2014 #1
    I need to prove that If 0<x<y and 0<a<b then ax<yb. I've been stuck going in circles for a while now and think I'm missing something really obvious.
     
  2. jcsd
  3. Sep 4, 2014 #2
    Are you able to write down any "interesting" inequalities given 0<x<y and 0<a<b?
     
  4. Sep 4, 2014 #3
    Are you looking for a formal proof, or do you simply need to explain why that is the case?
     
  5. Sep 4, 2014 #4
    I think I'm supposed to be able to do it off the definition we have of x<y mean y-x is positive. So I was able to easily do 0<x<y and 0<a<b mean x+a<y+b. I've tried writing everything I can think of that follows from multiplying various ways but I can't seem to get just ax or yb without something on the otherside messing it up.
     
  6. Sep 4, 2014 #5
    I need the formal proof. It's easy enough to know intuitively but I can't seem to find what's going to make it work for the proof.
     
  7. Sep 4, 2014 #6

    Fredrik

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    The first thing you need to know is that there are only three axioms that involve the inequality in the definition of the real numbers. The one about least upper bounds is irrelevant. The other two look like this:

    (1) If x<y then x+z<y+z.
    (2) If x>0 and y>0, then xy>0.

    Edit: I fixed the typo in (1).
     
    Last edited: Sep 4, 2014
  8. Sep 4, 2014 #7
    Do you mean for (1) If x<y then x+z<y+z?

    If so I've been using both of those and we were told we can use anything else we've already proved as well. I still don't see a way to get the ax<by using just those two axioms or the other properties we have.
     
  9. Sep 4, 2014 #8
    I would assume that z > 0 and try axiom 2.
     
  10. Sep 4, 2014 #9

    Fredrik

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    Yes, you only need axiom 2, if you're allowed to use elementary theorems like 1>0. Here's a hint: Consider y/x.

    Sorry about the typo in axiom 1. I have edited it now.
     
  11. Sep 4, 2014 #10
    We have not defined division so I don't know for sure I could use y/x. I know that ax>0 and by>0 by closure under addition for positive numbers. But I still can't find a way to link ax and yb so that ax<yb.
     
  12. Sep 4, 2014 #11
    By knowing ay-ax is positive and by-bx is positive is there a way to add/multiply that is going to show yb-xa is positive?
     
  13. Sep 4, 2014 #12
    Nope, I'm still going in circles :(
     
  14. Sep 4, 2014 #13

    Fredrik

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    Is there no list of axioms for real numbers in your book? The axioms that define the real numbers include division (by saying that every real number other than 0 has a multiplicative inverse). The axioms for addition and multiplication (including the one about multiplicative inverses) would typically be covered before the axioms that involve the order relation.

    Even if you think that you aren't allowed to use that x>0 implies the existence of 1/x, you should still try to solve the problem this way. Then you will at least have found a proof. (And I still think that this is how you're supposed to do it).
     
  15. Sep 4, 2014 #14
    We have but since we are working in only the positive natural numbers for these proofs there are no multiplicative inverses. So far I know the relations ax<ay and bx<by hold. As does ax+bx<ay+by. I've been trying to find a way to add/multiply them so that I can find one that will show ax<by.
     
  16. Sep 4, 2014 #15
    Would this be reasonable: Since a<b mean b-a is positive then x(b-a)<y(b-a) by axiom. then I get xb-xa<yb-ya then because of this definition of < and the face all the terms are positive get xa<xb<ya<yb and from there that xa<yb.
     
  17. Sep 4, 2014 #16

    Fredrik

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    You say that xa<yb follows from xa<xb<ya<yb, but why is xb<ya?

    A minor issue in the first line of your proof: The intermediate result x(b-a)<y(b-a) doesn't follow immediately from the axioms. There's no axiom that says that you can multiply both sides of an inequality by a positive number. But it's easy to prove this theorem: For all a,b,c, if a<b and c>0, then ac<bc.
     
  18. Sep 5, 2014 #17

    Fredrik

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    I found a simple proof of the theorem you're trying to prove that uses only 0<x<y, 0<a<b and the theorem I mentioned in the quote. I had to use the theorem more than once.
     
    Last edited: Sep 5, 2014
  19. Sep 5, 2014 #18
    That is probably a poor way to do it, but here is what I would have done with that problem. First rewrite it in this form:
    [itex]x<y[/itex], [itex]a<b[/itex], where x,y,a,b are positive; Prove [itex]by-ax>0[/itex]
    Let's introduce a function [itex]f(y,b)=by-ax[/itex]. We can say that:
    [tex]
    \lim_{y\rightarrow x, b\rightarrow a} {f(y,b)}=0
    [/tex]
    Then let's increase y by a positive constant C:
    [tex]
    \lim_{y\rightarrow x, b\rightarrow a} {f(y+C,b)}=bC>0
    [/tex]
    Do the same for b:
    [tex]
    \lim_{y\rightarrow x, b\rightarrow a} {f(y,b+C)}=Cy>0
    [/tex]
    So for any positive value of [itex]y>x[/itex] and [itex]b>a[/itex] we have proven that [itex]by-ax>0[/itex].
     
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