I understand that if a|b, then amodb=0, thus a = qb (remainder is zero) and b = qc. How can I prove using this that a|c?
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3$a = qb$ and $b = pc$ so $a = qpc$ – Rushabh Mehta Oct 18 '18 at 20:45
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just do it....... – cat Oct 18 '18 at 20:47
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2This is absolutely a duplicate. Anyone wanna help me search for it? If nothing else, it's a no context question. – amWhy Oct 18 '18 at 20:50
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1There is also this question which is itself a duplicate. – user505379 Oct 18 '18 at 20:55
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1$a\mid b$ implies $b\mod a=0$, not $a\mod b=0$. – Oct 18 '18 at 21:00
2 Answers
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Note: From now, we will be assuming $j,k\neq 0.$
Since $a|b,$ we can say $b=ja$ for $j\in \mathbb{Z}.$ Similarly, $b|c$ so $c=bk$ for $k\in \mathbb{Z}.$
Substituting for $b$ yields $c=ajk.$ Clearly, $jk\in \mathbb{Z}$ so, our assertion is true. $\blacksquare$
Prathmesh
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Hmm thats true. But it is just a minor thing. The claim still holds for negative integers. – Prathmesh Oct 18 '18 at 20:59
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Yes, but it might be why you are getting downvoted. You should edit it. – Sean English Oct 18 '18 at 21:01
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@Prathmesh Answering PSQs is obviously against MSE rules, hence your two downvotes – Rushabh Mehta Oct 18 '18 at 21:05
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1Ah sorry I just actively began posting here yesterday :) I was unaware – Prathmesh Oct 18 '18 at 21:06
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By assumption, $na=b$ and $mb=c$ for some integers $m$ and $n$. Therefore $mna=c$ so that $c$ is an integer multiple of $a$ ( and of course $a|c$ by definition). $\blacksquare$
JDivision
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3You knew this was/is a duplicate when you posted this answer. Yet you answered anyway. Do not answer posts that have already been asked and answered four times over, minimally. It becomes just noise. Answer questions needing answers, not those which have already been answered repeatedly. – amWhy Oct 18 '18 at 21:02
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Thanks for the answer. @amWhy Really great attitude by the way, I searched, but I coudn't find anything with the title of this own thread. – J. Doe Oct 18 '18 at 21:45