Proving Euclid's Lemma: The Role of Primality in Divisibility

  • Context: High School 
  • Thread starter Thread starter Suyogya
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around Euclid's Lemma, specifically the assertion that if a prime \( p \) divides the product \( ab \) of two integers \( a \) and \( b \), then \( p \) must divide at least one of those integers. Participants explore the implications of this lemma, attempting to prove it and examining cases where \( p \) is both prime and non-prime.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant attempts to prove Euclid's Lemma by assuming \( p \) is any integer and concludes that either \( a/p \) or \( b/p \) must be an integer, suggesting \( p \) divides \( a \) or \( b \) or both.
  • Another participant suggests testing with specific numbers, proposing \( a = 4 \), \( b = 15 \), and \( p = 6 \) to illustrate a counterexample with a non-prime.
  • A participant reiterates the proof attempt and questions the necessity of \( p \) being prime, referencing a counterexample with \( a = 2 \), \( b = 3 \), and \( p = 6 \).
  • Some participants emphasize that the proof should focus on showing \( p \) as a divisor of \( a \) or \( b \) rather than proving \( p \) itself is prime.
  • There is a recognition that the initial proof attempt incorrectly assumes the lemma applies to non-prime integers.

Areas of Agreement / Disagreement

Participants generally agree that the lemma is specific to prime numbers and that the initial proof attempt fails when \( p \) is not prime. However, there is no consensus on the validity of the proof presented or the exact nature of the error in reasoning.

Contextual Notes

The discussion highlights the limitations of applying the lemma to non-prime integers and the need for careful consideration of definitions and assumptions in the proof process.

Who May Find This Useful

This discussion may be useful for those interested in number theory, particularly in understanding the properties of prime numbers and divisibility.

Suyogya
Messages
14
Reaction score
0
"If a prime p divides the product ab of two integers a and b,then p must divide at least one of those integers a and b."(its euclid lemma true for primes only) when i tried to prove it as:
let for any integer p divides ab (ab)=(pn) ;for some integer n (a*b)/p=n
since RHS is integer, therefore for LHS to be integer either (a/p) or (b/p) or both must be integer, which means either p divides a or b or both.(hence proved)
But there is no restriction on p to be prime yet? is there any mistake in it?
 
Mathematics news on Phys.org
Try some sample numbers with p being prime and nonprime to see how it plays out. Here's one to try with a nonprime.
If a is 4 and b is 15, and p is 6, see how that works and go from there.
 
Suyogya said:
"If a prime p divides the product ab of two integers a and b,then p must divide at least one of those integers a and b."(its euclid lemma true for primes only) when i tried to prove it as:
let for any integer p divides ab (ab)=(pn) ;for some integer n (a*b)/p=n
since RHS is integer, therefore for LHS to be integer either (a/p) or (b/p) or both must be integer, which means either p divides a or b or both.(hence proved)
But there is no restriction on p to be prime yet? is there any mistake in it?

If ##a=2, b=3, p=6##, then you have a counterexample when ##p## is not prime.

In other words, in your "proof" you have said that as ##6## divides ##2 \times 3##, then either ##6## divides ##2## or ##6## divides ##3##.

It is necessary, therefore, that ##p## is prime.
 
Last edited:
Rereading it looks like they telling you that p is prime. Your job is not to prove p is prime, but to prove it is a divisor of a or b. Try representing a and b as product of prime factors.
 
scottdave said:
Rereading it looks like they telling you that p is prime. Your job is not to prove p is prime, but to prove it is a divisor of a or b. Try representing a and b as product of prime factors.
I proved it by assuming p an integer, hence should be valid for composites as well as primes, but its only for primes actually. so where its wrong
 
Suyogya said:
I proved it by assuming p an integer, hence should be valid for composites as well as primes, but its only for primes actually. so where its wrong
See post #3.
 
You have 2 examples of nonprimes, showing it doesn't necessarily work for nonprimes. I gave one in Post #2, and @PeroK gave one in Post #3.
 

Similar threads

Replies
7
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
10
Views
3K
  • · Replies 16 ·
Replies
16
Views
5K
  • · Replies 25 ·
Replies
25
Views
8K