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A good problem

  1. Nov 19, 2005 #1
    If m^2<a<b, & a*b=a perfect square, TPT: b>(m+1)^2
  2. jcsd
  3. Nov 19, 2005 #2


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    a*b=a means a is zero or b=1?? This is surely not what you mean?

    What does TPT mean? "To Prove True"?
  4. Nov 19, 2005 #3


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    Huh, on furthur reflection I guess "& a*b=a perfect square" means "and a*b is a perfect square".

    Is this a homework problem? In case it is, just a simple hint for now- treat the cases where "a" is a perfect square and "a" is not a perfect square seperately.
  5. Nov 19, 2005 #4
    I've understood the trouble... and I've got my picture but I think that I can't explain in a simple way... in sense that I'm not able to make it easy...
  6. Nov 21, 2005 #5
    Sorry for the confusion!

    Sorry, a*b= k^2, k belongs to the naturals. Not that a*b=a. Really sorry!
    Thanks for the hint, Shmoe! I may be clser to the solution by an another method. I'll let you know if I get it. BTW, TPT is "to prove that". :wink:
  7. Nov 21, 2005 #6
    Hey does anybody know how to create mathematical symbols while posting?
  8. Nov 21, 2005 #7

    matt grime

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    THnigs like [tex]ab=k^2, k \in \mathbb{N}[/tex]?

    Try looking up the LaTeX sticky post in the physics section. Or trying seraching the site using the search function for such "frequently asked questions"
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