20130325, 19:54  #1 
Mar 2013
2^{2} Posts 
Is there any such theorem that states this?
Is there any such theorem that implies/states that if C is a composite number, so that it is greater than an integer i and less than i^2 (i.e. the square of i), then C must have at least one factor (other than 1) which is less than i?
i.e. if i<C<i^2 (where i is any positive integer and C a composite number) then C must have at least one factor (other than 1) say F, so that F<i. 
20130325, 20:16  #2 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts 
That is true. Even the more strict , where F is the smallest prime factor of C, is known to be true. It's easy to prove, since if , the product must be larger than C, which is a contradiction.
I don't know if this fact is called a "theorem", but I know it's proven true. 
20130325, 20:31  #3 
"Nathan"
Jul 2008
Maryland, USA
5·223 Posts 
First of all, a technical nitpick: the symbol i has a special, specific meaning in mathematics: namely the imaginary unit . So instead of i for integer, let's use n. We can still use C to represent your composite number, also, of course, an integer.
So you're saying that C is composite, and that for some integer n. Well, let's take the square root on all sides of this inequality, which would yield (*). Now we know that the maximum possible size of any factor of C is going to be (think about this for a while until it is clear, if it is not already!). Therefore, any factor F of C will be such that . But then, by the inequality (*) above, we know that , which implies the strict inequality (**) for all factors F of n. Note that if n is indeed composite, as we have assumed, we are guaranteed the existence of a nontrivial F, and hence, along with inequality (**), we are guaranteed the existence of a nontrivial factor . Q. E. D. [Not sure if this has a name, e.g. "Blah's Theorem", but it's easy enough to prove to yourself. More likely it is something like an exercise in an elementary number theory course.] Last fiddled with by NBtarheel_33 on 20130325 at 20:33 
20130325, 20:40  #4  
Aug 2005
Seattle, WA
3245_{8} Posts 
Quote:
Quote:


20130325, 21:09  #5 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
I believe strictly speaking only primes are factors, the rest ( like 50) are divisors. Last fiddled with by science_man_88 on 20130325 at 21:10 
20130325, 22:35  #6 
Account Deleted
"Tim Sorbera"
Aug 2006
San Antonio, TX USA
17×251 Posts 

20130325, 23:10  #7  
Jun 2003
1169_{10} Posts 
Quote:
However interpreting NBtarheel_33's remark as refering to prime factors only isn't enough to save it. For example 515 and 5 > > . 

20130325, 23:22  #8 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

20130326, 05:29  #9  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3·29·83 Posts 
Quote:
Nitpick the second: i is a perfectly good variable to use; its use is in no way at all limited to that of (which is certainly its "default" meaning wihtout context). In this case he clearly and unambiguously defined what i is. 

20130326, 15:56  #10 
Einyen
Dec 2003
Denmark
17·181 Posts 
so .
At least one factor of C is <= and therefore < i which is what OP asked. Last fiddled with by ATH on 20130326 at 15:57 
20130327, 19:07  #11 
Mar 2013
2^{2} Posts 
One of my friends suggested a simpler proof which I want to share. Let's presume all factors of C are greater than 'i'. Therefore, all such factors can be written in the form i+k. Let's presume the smallest factor of 'i' is F, which is equal to i+x. So now we can say that C=(i+x) * (i+y) = i^2 +i(x+y)+xy, which is greater than i^2. So our first proposition must be false.
A big "thank you" to all of you. But I would still want to know, if such a theorem already exists? Last fiddled with by soumya on 20130327 at 19:08 
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