Is Every Prime Factor of a Composite Number Less Than or Equal to sqrt(n)?

[annoymage]

Homework Statement

If n is a composite number, the n has prime factor nor exceeding sqrt(n)

Homework Equations

n/a

The Attempt at a Solution

what did it mean by "nor exceeding sqrt(n)", is it, it must be less than sqrt(n)?

 

[JonF]

i'd gues it meant "not"

 

[annoymage]

is it suppose to mean

If n is a composite number, then n has prime factor not exceeding sqrt(n) ??

 

[Dick]

is it suppose to mean

If n is a composite number, then n has prime factor not exceeding sqrt(n) ??

Yes. If p is the prime factor show p<=sqrt(n).

 

[annoymage]

can you check my proof please,

since n is composite number, so n>3, so by fundamental theorem of arithmetic n can be written by product of primes, say pp_1p_2...p_k=n, so let p_1p_2...p_k=a\ ,\ where\ a&lt;n and assume p \leq a without loss generality, so we get pa=n\ ,\ 3&lt;p \leq a&lt;n, so suppose p&gt;\sqrt{n}, then a&gt;\sqrt{n} then ap&gt;\sqrt{n}\sqrt{n}=n contradiction. is it okay?

but it seems something wrong when i assume p \leq a, because if a \leq p we get 3&lt;a \leq p&lt;n and i can't conclude a&gt;\sqrt{n} right? help T_T

 

[Dick]

can you check my proof please,

since n is composite number, so n>3, so by fundamental theorem of arithmetic n can be written by product of primes, say pp_1p_2...p_k=n, so let p_1p_2...p_k=a\ ,\ where\ a&lt;n and assume p \leq a without loss generality, so we get pa=n\ ,\ 3&lt;p \leq a&lt;n, so suppose p&gt;\sqrt{n}, then a&gt;\sqrt{n} then ap&gt;\sqrt{n}\sqrt{n}=n contradiction. is it okay?

but it seems something wrong when i assume p \leq a, because if a \leq p we get 3&lt;a \leq p&lt;n and i can't conclude a&gt;\sqrt{n} right? help T_T

I don't why or how you want to assume p<=a. Or why the condition 3<p. Look, you got that if pa=n the either p<=sqrt(n) or a<=sqrt(n). Split into cases.

 

[annoymage]

if pa=n the either p<=sqrt(n) or a<=sqrt(n). Split into cases.

i don't even tried/think to prove that, but thanks, that's a hint for me ;P, so I start to prove that first

if p \leq \sqrt{n} then it is proven, if not, suppose a&gt;\sqrt{n} then we have ap&gt;\sqrt{n}\sqrt{n}=n, contradiction. then

if pa=n the either p<=sqrt(n) or a<=sqrt(n)

.

so If p \leq \sqrt{n} then the question is proven,

and If a \leq \sqrt{n}, so p_1p_2...p_k \leq \sqrt{n} then
p_1 \leq \sqrt{n}, then finished, is it ok now?

 

[Dick]

i don't even tried/think to prove that, but thanks, that's a hint for me ;P, so I start to prove that first

if p \leq \sqrt{n} then it is proven, if not, suppose a&gt;\sqrt{n} then we have ap&gt;\sqrt{n}\sqrt{n}=n, contradiction. then .

so If p \leq \sqrt{n} then the question is proven,

and If a \leq \sqrt{n}, so p_1p_2...p_k \leq \sqrt{n} then
p_1 \leq \sqrt{n}, then finished, is it ok now?

Yes, now it looks ok.

 

[annoymage]

k thank you very much ^^