Prduct of prime (m,m+1) < C(2m,m)

[cantorset1985]

Homework Statement

Prove that the product of primes between m+1 and 2m is less than C(2m,m)

Homework Equations

The Attempt at a Solution

I have that it is less than (2m)!/m! = m!C(2m,m) which is just the product of all of the numbers from m+1 to 2m. Any help is appreciated. Even if you don't know the solution, an idea would be great. (BTW, this isn't a homework problem, I just heard it was true, so being new to the forums, I'm not sure if this is the correct place for this.) Are there any combinatoralists here?

 

[Gokul43201]

Let p be a prime in (m+1,2m). Does p divide the numerator of C(2m,m)? What about the denominator? What can you deduce from that?

 

[cantorset1985]

Let p be a prime in (m+1,2m). Does p divide the numerator of C(2m,m)? What about the denominator? What can you deduce from that?

Ohhhhhhhhh. Wow, that's a lot better than what I was planning to do!

Thanks.

 

[cantorset1985]

Let p be a prime in (m+1,2m). Does p divide the numerator of C(2m,m)? What about the denominator? What can you deduce from that?

Wait, I thought that I had it, but I don't think I do. Yes, the p divides (2m)! but not m!, so it divides the numerator of C(2m,m) but not the denominator. I need some more help :(

 

[Robert1986]

Wait, I thought that I had it, but I don't think I do. Yes, the p divides (2m)! but not m!, so it divides the numerator of C(2m,m) but not the denominator. I need some more help :(

Whoops. I did have it, I just forgot some basic elementary number theory. So, here is my solution:

Let P = the product of primes from (m+1 to 2m). Since (2m!)/(m!) is just the product of all numbers in (m+1,2m) then P | (2m!)/(m!). But, (2m!)/(m!) = m!C(2m,m) so we have that P |m!C(2m,m). Since m < any prime in (m+1,2m) P does not divide m! which means that P must divide C(2m,m), which, in turn imples that P < C(2m,m).

Thanks, that one was driving me crazy.

Also, the starter of this thread is me, through some confusion, I managed to create two accounts, and then use the cantorset account when I meant to use this account. My apologies!

 

[Gokul43201]

Whoops. I did have it, I just forgot some basic elementary number theory. So, here is my solution:

Let P = the product of primes from (m+1 to 2m). Since (2m!)/(m!) is just the product of all numbers in (m+1,2m) then P | (2m!)/(m!). But, (2m!)/(m!) = m!C(2m,m) so we have that P |m!C(2m,m). Since m < any prime in (m+1,2m) P does not divide m! which means that P must divide C(2m,m), which, in turn imples that P < C(2m,m).

That looks good.

Just for completeness, from the previous post:

Wait, I thought that I had it, but I don't think I do. Yes, the p divides (2m)! but not m!, so it divides the numerator of C(2m,m) but not the denominator. I need some more help :(

If a prime p divides N but not D, and N/D is an integer, then it follows from the prime factorization of N and D, that p must divide N/D.

So every prime, p in (m+1,2m) divides C(2m,m). Therefore, the product, P, must too.