Algebra: show that x > 1 is prime

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Homework Help Overview

The problem involves demonstrating the conditions under which a natural number x greater than 1 is considered prime, specifically through the absence of divisors t within a certain range. The discussion centers around the definitions and properties of prime numbers.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand the implications of the "if and only if" condition in the context of prime numbers and their divisors. They express uncertainty about how to approach the proof from both directions: showing that x is prime if no t divides it, and vice versa.

Discussion Status

Participants are engaging with the original poster's confusion by suggesting alternative approaches, such as considering the implications of assuming x is not prime. There is an ongoing exploration of ideas without a clear consensus on the next steps.

Contextual Notes

The discussion includes a focus on the mathematical definitions of prime numbers and their factors, with specific reference to natural numbers and the constraints of the problem statement.

Vince00

Homework Statement


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