Trouble setting up to prove unique factorization

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Discussion Overview

The discussion revolves around a lemma related to unique factorization in number theory, specifically addressing the existence of integers q and r that satisfy a particular equation involving integers a and b. Participants explore the implications of the lemma and its proof, while also questioning specific examples.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant expresses confusion regarding the lemma stating that for integers a and b (with b > 0), there exist integers q and r such that a = qb + r and 0 ≤ r < b.
  • The same participant questions the validity of the lemma by providing an example with a = 3 and b = 8, suggesting that no suitable q and r can be found.
  • Another participant counters this by demonstrating that for a = 3 and b = 8, one can take q = 0 and r = 3, which satisfies the conditions of the lemma.
  • A later reply indicates frustration, suggesting a lack of understanding or agreement with the discussion.

Areas of Agreement / Disagreement

Participants do not reach a consensus, as one participant questions the lemma's applicability to their example while another provides a counterexample that adheres to the lemma's conditions.

Contextual Notes

The discussion highlights potential misunderstandings regarding the definitions and conditions set forth in the lemma, particularly in relation to the values of q and r.

lants
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This lemma the book states, I can't make sense of it.

Lemma: If a,b[itex]\in[/itex] [itex]Z[/itex] and b > 0, there exist q,r [itex]\in[/itex] [itex]Z[/itex] such that a = qb + r with 0 [itex]\leq[/itex] r < b.

Proof: Consider the set of all integers of the form a-xb with x [itex]\in[/itex] [itex]Z[/itex]. This set includes positive elements. Let r = a - qb be the least nonnegative element in this set. We claim that 0[itex]\leq[/itex] r < b. If not, r = a - qb [itex]\geq[/itex] b and so 0[itex]\leq[/itex] a-(q+1)b<r, which contradicts the minimality of r.

Can someone help explain this to me? Why can't a = 3, and b = 8. Then no q,r could exist so that r is less than b, right?
 
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Is this considered a textbook style question? Sorry, how can I move it? I just saw number theory and posted
 
What is wrong with [itex]a=3[/itex] and [itex]b=8[/itex]? Here you simply take [itex]q=0[/itex] and [itex]r=3[/itex] (there is no condition that says either q or r have to be nonzero) to get [itex]3=(0)(8) +3[/itex] and [itex]0\leq 3< 8[/itex] as required.
 
Wow I'm quitting now

thanks
 

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