A good problem

1. Nov 19, 2005

If m^2<a<b, & a*b=a perfect square, TPT: b>(m+1)^2

2. Nov 19, 2005

shmoe

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

What does TPT mean? "To Prove True"?

3. Nov 19, 2005

shmoe

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.

4. Nov 19, 2005

armandowww

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...

5. Nov 21, 2005

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".

6. Nov 21, 2005

Hey does anybody know how to create mathematical symbols while posting?

7. Nov 21, 2005

matt grime

THnigs like $$ab=k^2, k \in \mathbb{N}$$?

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"