GCD Discrete Math: Proving GCD(a,b)=1

Click For Summary

Discussion Overview

The discussion revolves around proving properties of the greatest common divisor (GCD) in the context of discrete mathematics. Participants explore specific cases where GCD(a,b) = 1 and its implications for GCD(a+b, a-b) and GCD(2a+b, a+2b), seeking assistance with proofs and reasoning.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • Post 1 presents two problems involving GCD and asks for assistance in proving them, specifically focusing on the conditions under which GCD(a+b, a-b) equals 1 or 2, and GCD(2a+b, a+2b) equals 1 or 3.
  • Post 2 reiterates the problems from Post 1, emphasizing the need for help and encouraging participants to share their attempts or points of confusion.
  • Post 3 highlights a key fact about divisibility, stating that if a divisor d divides two numbers x and y, it also divides any linear combination of those numbers.
  • Post 4 introduces a substitution method to express a and b in terms of x and y, suggesting that if x and y share a common factor other than 2, it leads to a contradiction regarding the common factors of a and b.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the proofs, and multiple approaches and ideas are presented without resolution. The discussion remains open-ended with various viewpoints and methods being explored.

Contextual Notes

The discussion includes assumptions about the properties of GCD and divisibility, but these assumptions are not universally agreed upon or fully explored in detail. The implications of the proofs are also not settled, leaving room for further exploration.

ssome help
Messages
3
Reaction score
0
Given that GCD(na,nb) = n * GCD(a,b) for a,b,n ∈ Z+

a) Prove that, if GCD(a,b) = 1 then GCD(a+b, a-b) = 1 or GCD(a+b,a-b) = 2
Hint: Let D = GCD(a+b, a-b), show that D | 2a and D | 2b thus D | GCD(2a,2b) then use the given
b) Prove that, if GCD(a,B) = 1, then GCD(2a+b, a+2b) = 1 or GCD(2a+b, a+2b) = 3

Any assistance would be great.
 
Physics news on Phys.org
ssome help said:
Given that GCD(na,nb) = n * GCD(a,b) for a,b,n ∈ Z+

a) Prove that, if GCD(a,b) = 1 then GCD(a+b, a-b) = 1 or GCD(a+b,a-b) = 2
Hint: Let D = GCD(a+b, a-b), show that D | 2a and D | 2b thus D | GCD(2a,2b) then use the given
b) Prove that, if GCD(a,B) = 1, then GCD(2a+b, a+2b) = 1 or GCD(2a+b, a+2b) = 3

Any assistance would be great.

Welcome to MHB, ssome help!

Nice problem. ;)

Did you try anything?
When you show something you tried, or if you explain where you are stuck, we can help you to solve and understand this.
 
The key fact to use here is that if d | x and d | y, then d divides any linear combination of x and y, i.e., d | (mx + ny) for any integer m and n.
 
Setting...

$\displaystyle x = a + b$

$\displaystyle y = a - b$ (1)

... solving (1) we obtain...

$\displaystyle a = \frac{x + y}{2}$

$\displaystyle b = \frac{x - y}{2}$ (2)

Now if x and y have a common factor different than 2, then x + y and x - y have the the same common factor and the same would be for a and b and that is a contradiction...

Kind regards

$\chi$ $\sigma$
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
1
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
2
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
14
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K