I am working on a proof in advanced mathematics, and I believe my professor may have overlooked something. The proof is that if $d|a$, $d|b$, and $d\not|c$, then $ax+by=c$ has no integer solutions for $x$ and $y$. However, if $x$, $y$, and $c$ are $0$, that would be an integer solution.
If an integer $a$ is not divisible by an integer $b$, could this imply that $b = 0$?
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0Without knowing the specifics, it's very hard to say. However, it seems unlikely to be the right argument to make. – 2017-02-10
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0b could be zero. But it is not implied. And therefore can't be implied by that alone. So the answer to your question is technically No. – 2017-02-10
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0I can imagine a case where one concludes $b\not \mid a $ but it's never stated b is nonzero. And I suppose one might have a case where $b\not \mid a $ and b nonzero are incompatible . I think $\gcd (a,b) = a $ and $b \not \mid a $ and $|a| \ge |b|$ would together imply that $b=0$. But it's late and I make dumb mistakes when it is late. – 2017-02-10
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0Everything divides 0. – 2017-02-10
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0If $c=0$ then $d|c $. – 2017-02-10
1 Answers
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$d|0$. Everything divides $0$.