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.

That is true. Even the more strict [TEX]F \le sqrt{C}[/TEX], where F is the smallest prime factor of C, is known to be true. It's easy to prove, since if [TEX]F > sqrt{C}[/TEX], 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. :smile:

First of all, a technical nitpick: the symbol [I]i[/I] has a special, specific meaning in mathematics: namely the imaginary unit [TEX]i = sqrt{-1}[/TEX]. So instead of [I]i[/I] for integer, let's use [I]n[/I]. We can still use [I]C[/I] to represent your composite number, also, of course, an integer.

So you're saying that [I]C [/I]is composite, and that [TEX]n < C < n^2[/TEX] for some integer [I]n[/I]. Well, let's take the square root on all sides of this inequality, which would yield [TEX]sqrt{n} < sqrt{C} < n[/TEX] (*).

Now we know that the maximum possible size of any factor of [I]C[/I] is going to be [TEX]sqrt{C}[/TEX] (think about this for a while until it is clear, if it is not already!).

Therefore, [B]any[/B] factor [I]F[/I] of [I]C[/I] will be such that [TEX]F \leq sqrt{C}[/TEX]. But then, by the inequality (*) above, we know that [TEX]sqrt{C} < n[/TEX], which implies the strict inequality [TEX]F < n[/TEX](**) for [B]all [/B]factors [I]F[/I] of [I]n[/I].

Note that if [I]n[/I] is indeed composite, as we have assumed, we are guaranteed the existence of a nontrivial [I]F[/I], and hence, along with inequality (**), we are guaranteed the existence of a nontrivial factor [TEX]F < n[/TEX].

[B][I]Q. E. D.[/I][/B]

[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.]

[QUOTE=NBtarheel_33;334944]Now we know that the maximum possible size of any factor of [I]C[/I] is going to be [TEX]sqrt{C}[/TEX] (think about this for a while until it is clear, if it is not already!).
[/QUOTE]

Huh?! Isn't 50 a factor of 100? And isn't 50 > 10?

[QUOTE=NBtarheel_33;334944]
Therefore, [B]any[/B] factor [I]F[/I] of [I]C[/I] will be such that [TEX]F \leq sqrt{C}[/TEX]. But then, by the inequality (*) above, we know that [TEX]sqrt{C} < n[/TEX], which implies the strict inequality [TEX]F < n[/TEX](**) for [B]all [/B]factors [I]F[/I] of [I]n[/I].

Note that if [I]n[/I] is indeed composite, as we have assumed, we are guaranteed the existence of a nontrivial [I]F[/I], and hence, along with inequality (**), we are guaranteed the existence of a nontrivial factor [TEX]F < n[/TEX].

[B][I]Q. E. D.[/I][/B]

[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.][/QUOTE]

If you change your uses of "any factor" and "all factors" to "the minimum nontrivial factor" then I agree with your argument.

[QUOTE=jyb;334945]Huh?! Isn't 50 a factor of 100? And isn't 50 > 10?
[/QUOTE]


I believe strictly speaking only primes are factors, the rest ( like 50) are divisors.

[QUOTE=science_man_88;334947]I believe strictly speaking only primes are factors, the rest ( like 50) are divisors.[/QUOTE]

There's a term for factors that are prime: prime factors. Factors and divisors are basically synonymous in this usage.

[QUOTE=Mini-Geek;334955]There's a term for factors that are prime: prime factors. Factors and divisors are basically synonymous in this usage.[/QUOTE]

[i]Strictly[/i] Speaking, Mini-Geek is correct. Of course we often just write "factor" when we mean prime factor.

However interpreting NBtarheel_33's remark as refering to prime factors only isn't enough to save it. For example 5|15 and 5 > [tex]\sqrt{16}[/tex] > [tex]\sqrt{15}[/tex].

[QUOTE=Mr. P-1;334958]However interpreting NBtarheel_33's remark as refering to prime factors only isn't enough to save it. For example 5|15 and 5 > [tex]\sqrt{16}[/tex] > [tex]\sqrt{15}[/tex].[/QUOTE]

the keyword missing is least.

[QUOTE=NBtarheel_33;334944]
Now we know that the maximum possible size of any factor of [I]C[/I] is going to be [TEX]sqrt{C}[/TEX] (think about this for a while until it is clear, if it is not already!).[/QUOTE]

This is a false statement, as has been pointed out already. Also as has been pointed out, there can even be prime factors greater than [tex]\sqrt {C}[/tex].

Nitpick the second: i is a perfectly good variable to use; its use is in no way at all limited to that of [tex]\sqrt{-1}[/tex] (which is certainly its "default" meaning wihtout context). In this case he clearly and unambiguously defined what i is.

[TEX]C < i^2[/TEX] so [TEX]sqrt{C} < i[/TEX].

[B]At least one[/B] factor of C is <= [TEX]sqrt{C}[/TEX] and therefore < i which is what OP asked.

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?