Algebra: show that x > 1 is prime

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To show that x > 1 is prime if and only if there is no divisor t such that 1 < t ≤ √x, one must demonstrate two implications. First, if x is prime, then it cannot be divided by any t in the specified range. Conversely, if no such t divides x, then x must be prime. The discussion highlights the importance of understanding the relationship between prime numbers and their factors, particularly in relation to their square roots. Participants suggest starting with the assumption that x is not prime to explore the implications of its factors.
Vince00

Homework Statement


2.4 Show that x > 1 is prime, iff x doesn't have any divisor t; where 1 < t \leq \sqrt{x}. It is given that x,t \in N.

Homework Equations


?

The Attempt at a Solution


The "iff" thing makes me think; what can I do to show this?
I have to show that x (x can be 2, 3, 4, 5, 6, ...) is prime if there is no "t", 1<t\leq \sqrt{x} (t can be 1, 2, 3, 4, ...) that divides x. AND that that there is no t that divides x if x is prime.
So I have to show it 2 ways.
First: x is prime if t doesn't divide x
Second: t doesn't divide x if x is prime

And well, that's all I got! Please help.
Vince, fresmen physics.
 
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Welcome to PF!

Hi Vince! Welcome to PF! :smile:

(have a square-root: √ and a ≤ :wink:)

Try starting with the opposite …

suppose x is not prime, and all its factors (two or more) are > √x. :wink:
 


Thanks for the welcome tim!
Okay, so you said: suppose x is not prime, and all its factors are > √x

I really have no idea what I can do with that...I tried, for a few days, but I just don't get it!
Maybe you can explain it a bit more?
Tnx!
 
Hi Vince00! :smile:

Take 103 and 105 …

how many factors can they have > 10 ? :wink:​
 

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