Proof: integers divisibility property

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SUMMARY

The discussion centers on proving the divisibility property of integers, specifically that if an integer \( a \) divides \( b \) but does not divide \( c \), then \( a \) does not divide \( b + c \). The proof approach suggested involves contradiction, assuming \( a \) divides \( b + c \) and deriving a contradiction based on the definitions of divisibility. Additionally, the converse statement is identified as false, with suggestions to find counterexamples to illustrate this point.

PREREQUISITES
  • Understanding of integer divisibility and definitions of divisibility.
  • Familiarity with proof techniques, particularly proof by contradiction.
  • Basic knowledge of mathematical logic and counterexamples.
  • Experience with algebraic manipulation of equations involving integers.
NEXT STEPS
  • Study the concept of integer divisibility in number theory.
  • Learn about proof techniques, focusing on proof by contradiction.
  • Explore examples of counterexamples in mathematical proofs.
  • Investigate the properties of integers and their implications in algebra.
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Students of mathematics, particularly those studying number theory, educators teaching proof techniques, and anyone interested in the properties of integers and divisibility.

Prathep
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Someone please help me with this qiestion:

Prove that for all integers a, b, and c, if a divides b but not c then a does not
divide b + c, but the converse is false.

Thanks.
 
Last edited:
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Looks like a homework problem to me! First write out the definitions: If "a divides b" then b= an for some integer n. I would use "proof by contradiction". Suppose you know that a divides b+c, that is, that b+c= am for some integer m and that b= an for some integer n. Can you use that to contradict "a does not divide c"?

The "converse" of that statement is, of course, "if a does not divide b+ c, then a divides b but not c". It should be fairly simple to find a counter example for that, or the more general "converse", "if a does not divide b+ c, then it must divide one but not the other".
 
I imagine the question does refer to the more general converse. The specific converse can be disproved simply by symmetry arguments, without using a shred of number theory.
 

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