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This question already has an answer here:

Let a, b, c ∈ N such that (a, b) = 1 and a | bc. Prove that a | c.

I'm a little confused about if I'm doing this proof right.

I know that $\exists p, q \in \mathbb{Z}$. Such that $pa = bc$ and $qa=c$. Re-arranging the first equation.

$$a = \frac{bc}{p} $$

Substituting this into the second equation.

$$ qa = c$$ $$ q(\frac{bc}{p}) = c$$ $$ \frac{qb}{p} c =c $$ $$ \frac{qb}{p} c =c $$

Thus this equation divides c. Therefore a | c. Does this proof make sense.

Thank you for any guidance.

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marked as duplicate by Arturo Magidin, Community Sep 30 '18 at 23:51

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If $c=qa$ then $a|c$ and you are done. This is not the hypothesis. It is what you need to show. $\endgroup$ – mfl Sep 30 '18 at 23:45
  • $\begingroup$ I'm not sure what you mean. Could you explain further? $\endgroup$ – Safder Sep 30 '18 at 23:47
  • $\begingroup$ "I know that $\exists p, q \in \mathbb{Z}$ such that $pa = bc$ and $qa=c$." No. You don't know the existence of $q.$ $\endgroup$ – mfl Sep 30 '18 at 23:48
  • $\begingroup$ It is. Thank you for pointing it out. $\endgroup$ – Safder Sep 30 '18 at 23:51
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You can't assume $c=qa$ because that if what you need to prove.

Hint for a solution: if $gcd(a,b)=1$ then there are numbers $k,l\in\mathbb{Z}$ such that $ak+bl=1$. Multiply this equation by $c$ and see what you get from there.

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You do not know there is such a $q$. That is in fact exactly what you are trying to prove: it says $a$ divides $c$.

The standard argument for this theorem begins with the existence of integers $r$ and $s$ such that $$ ar + bs = 1 . $$

Multiply that equality by $c$ and try to conclude what you hope to prove.

Note: you should strive to write number theory proofs without fractions. Even when their use is correct it's confusing.

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