Intro to Proofs: Greatest Common Divisors

  • Thread starter Thread starter rdr3
  • Start date Start date
  • Tags Tags
    Intro Proofs
Click For Summary
SUMMARY

The discussion focuses on proving properties of the greatest common divisor (gcd) in the context of integer solutions to linear equations. It establishes that if gcd(a, b) = d, then the equation xa + yb = d has integer solutions given by x = m + k(b/d) and y = n - k(a/d) for some integer k. Additionally, it confirms that the equation xa + yb = c has integer solutions if and only if d divides c (d | c). These results are foundational in number theory and are essential for understanding linear combinations of integers.

PREREQUISITES
  • Understanding of greatest common divisors (gcd) and their properties.
  • Familiarity with linear Diophantine equations.
  • Basic knowledge of integer arithmetic and modularity.
  • Experience with mathematical proofs, particularly "if and only if" statements.
NEXT STEPS
  • Study the Extended Euclidean Algorithm for finding integer solutions to gcd-related equations.
  • Explore linear Diophantine equations in greater depth, focusing on their applications.
  • Learn about modular arithmetic and its relationship with gcd and integer solutions.
  • Review proof techniques for "if and only if" statements in mathematics.
USEFUL FOR

Students of mathematics, particularly those studying number theory, as well as educators and anyone interested in the properties of integers and their linear combinations.

rdr3
Messages
3
Reaction score
0

Homework Statement



(a) Let a and b be integers with gcd(a,b)=d, and assume that ma+nb=d for integers m and n. Show that the solutions in Z to

xa+yb=d

are exactly

x=m+k(b/d), y=n-k(a/d)

where k∈Z.

(b) Let a and b be integers with gcd(a,b)=d. Show that the equation

xa+yb=c

has a solution (x,y)∈ ZxZ if and only if d|c.

Z refers to integers.

Homework Equations





The Attempt at a Solution



Ok so I'm pretty sure I figured out part (a).

Part (b) on the other hand is causing me problems. I know since it's an "if and only if" question, I have to prove it both ways.

This is what I have so far.

So we have some x and y, such as

x0a+y0b=d

Then we multiply both sides by some arbitrary element, such as n. Giving us,

(nx0)a+(ny0)b=nd=c

So that's the first part. Now I'm supposed to go the other way with it, showing that
xa+yb=c -> c=nd, n∈ Z.
 
Physics news on Phys.org
How about the contrapositive for the other direction? That might be easier.
 
So maybe this?

Suppose d∤c...

Then c≠nd

In which case, c≠nd≠(nx0)a+(ny0)b.

So like that? Or am I way off? Sorry I'm not very good at this... The teacher explains things way over everyone's head.
 

Similar threads

Replies
2
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
6K
Replies
3
Views
2K
Replies
1
Views
1K
Replies
4
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
8
Views
2K