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Prove of Statement, Calculus, Basic Properties of numbers

  1. Jul 6, 2012 #1
    hi there!

    If ab > 0, then (a > 0 and b > 0) or (a < 0 and b < 0). This statment I can prove, just with the basic properties of numbers!

    Then, 1[itex]/[/itex]b is defined as b[itex]^{-1}[/itex] right?

    So, how does one prove that if [itex]\frac{a}{b}[/itex] > 0, then (a > 0 and b > 0) or (a < 0 and b < 0)?

    Can you give me the complete proog of that? Thanks!

    For example, how does one prove that if [itex]\frac{x+1}{x-1}[/itex] > 0, then

    x > 1 or x < -1?

    Regards,
     
  2. jcsd
  3. Jul 6, 2012 #2

    micromass

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    So [itex]\frac{a}{b}>0[/itex] implies that [itex]ab^{-1}>0[/itex]. Now apply the property you could prove.
     
  4. Jul 6, 2012 #3
    So (a > 0 and b[itex]^{-1}[/itex] >0) or (a < 0 and b[itex]^{-1}[/itex] < 0 ). That is easy of course, but how do I know what b[itex]^{-1}[/itex] is?

    In the example I made, how do I relate the variavle x, with that fact (x-1)[itex]^{-1}[/itex] < 0, or (x-1)[itex]^{-1}[/itex] > 0?
     
  5. Jul 6, 2012 #4

    micromass

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    I don't understand your problem. Do you mean that you can't prove that b>0 if [itex]b^{-1}>0[/itex]??
     
  6. Jul 6, 2012 #5
    No, that is not what I am saying. I clearly understand the ab[itex]^{-1}[/itex] > 0.

    Well, I better just try to explain my self using the example.

    So,we have, [itex]\frac{x-1}{x+1}[/itex] > 0 [itex]\Leftrightarrow[/itex] (x-1)(x+1)[itex]^{-1}[/itex] > 0. So [(x-1) > 0 [itex]\vee[/itex] (x+1)[itex]^{-1}[/itex] > 0 ] [itex]\wedge[/itex] [(x-1) < 0 [itex]\vee[/itex] (x+1)[itex]^{-1}[/itex] < 0 ]. Right?

    My question is what good is it to know that for example (x+1)[itex]^{-1}[/itex] > 0 ?

    How do I relate this with x < something or x > something?

    If I was trying to solve this without this without all the ( proof. based on properties), I would just say that if [itex]\frac{x-1}{x+1}[/itex] > 0, then ( x-1) > 0 and ( x+1) > 0( or the other way around).

    Is my doubt clear now?

    regards,

    cteixeira
     
  7. Jul 6, 2012 #6
    Regarding my last post.

    Actually I think that is it! How can I prove that b > 0 given b[itex]^{-1}[/itex] > 0 ?
     
  8. Jul 7, 2012 #7

    micromass

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    If b<0, then from [itex]b^{-1}>0[/itex] would follow [itex]bb^{-1}<0[/itex] or 1<0. This is a contradiction. So b>0 (since b=0 is clearly not allowed).
     
  9. Jul 7, 2012 #8
    That was exactly my doubt! Thanks, my question is answered!

    So, b and b[itex]^{-1}[/itex] have always to have the same sign (given any b), right?

    Regards,
     
  10. Jul 7, 2012 #9

    micromass

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    Right!
     
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