How Do You Prove That sa + tb Divides gcd(a, b) in Bezout's Lemma?

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Discussion Overview

The discussion revolves around proving Bezout's Lemma, specifically the assertion that \( sa + tb \) divides \( \gcd(a, b) \). Participants explore the implications of the definition of the greatest common divisor and the properties of divisibility in relation to linear combinations of integers.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that since \( \gcd(a, b) \) divides both \( a \) and \( b \), it follows that \( \gcd(a, b) \) divides \( sa + tb \).
  • Another participant questions the reasoning behind why \( sa + tb \) divides \( \gcd(a, b) \), indicating a lack of clarity on this point.
  • A third participant emphasizes the definition of \( \gcd(a, b) \) as the largest common divisor, arguing that since \( sa + tb \) is a common divisor of \( a \) and \( b \), it cannot exceed \( \gcd(a, b) \).
  • This participant also notes that every number divides itself, reinforcing the argument that \( sa + tb \) must divide \( \gcd(a, b) \) based on its definition.

Areas of Agreement / Disagreement

Participants express differing views on the proof's validity, with some asserting that \( sa + tb \) must divide \( \gcd(a, b) \) based on its properties, while others remain uncertain about the reasoning behind this conclusion. The discussion does not reach a consensus on the proof.

Contextual Notes

Participants reference the well-ordering principle and Euclid's Algorithm, but there are unresolved questions regarding the implications of these methods in proving the divisibility of \( \gcd(a, b) \) by \( sa + tb \). The definitions and properties of divisibility are also central to the discussion.

nicodemus1
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Good Day,

I have to prove Bezout's Lemma.

I have proven that since gcd (a, b) divides a and gcd (a, b) divides b, gcd (a, b) divides sa + tb.

I've made use of the well-ordering principle and Euclid's Algorithm to show that sa + tb divides a and sa + tb divides b.

What I can't prove is that sa + tb divides gcd (a, b).

Please help me here.

Thank you.
 
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nicodemus said:
Good Day,

I have to prove Bezout's Lemma.

I have proven that since gcd (a, b) divides a and gcd (a, b) divides b, gcd (a, b) divides sa + tb.

I've made use of the well-ordering principle and Euclid's Algorithm to show that sa + tb divides a and sa + tb divides b.

What I can't prove is that sa + tb divides gcd (a, b).

Please help me here.

Thank you.

since d = gcd(a,b) is the largest positive integer that divides both a and b,

and d divides sa + tb, d ≤ sa + tb, whence sa + tb = d, since sa + tb divides both a and b.
 
Good Day,

Thank you for your reply.

However, I still don't get why sa + tb divides gcd(a, b).

Thanks & Regards,
Nicodemus
 
the "g" in gcd stands for "greatest". the "cd" stands for "common divisor".

since sa + tb is a common divisor of a and b, it cannot be "greater than" the greatest common divisor of a and b, can it?

EVERY number divides it self.

specifically, the definition of gcd(a,b) is:

d: d|a and d|b and (for all c: if c|a and c|b then c|d).

we have shown that c = sa + tb divides both a and b. therefore c|d.

do you have a different definition of gcd(a,b)?
 

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