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A good problem

by Aditya89
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Aditya89
#1
Nov19-05, 04:35 AM
P: 23
If m^2<a<b, & a*b=a perfect square, TPT: b>(m+1)^2
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shmoe
#2
Nov19-05, 08:18 AM
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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"?
shmoe
#3
Nov19-05, 01:20 PM
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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.

armandowww
#4
Nov19-05, 02:46 PM
P: 78
A good problem

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...
Aditya89
#5
Nov21-05, 04:16 AM
P: 23
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".
Aditya89
#6
Nov21-05, 04:38 AM
P: 23
Hey does anybody know how to create mathematical symbols while posting?
matt grime
#7
Nov21-05, 04:53 AM
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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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