This may be a dumb question. I'm not a math major, but, since I'm studying logic, I decided to learn a bit of number theory. I've just begun my studies (I'm reading Davenport's The Higher Arithmetic) and one of the first things I came across is the fundamental theorem of arithmetic. Now, in proving the uniqueness claim, I found an assertion that puzzled me. Let $n$ be a natural number, and both $p$ and $p'$ distinct prime factors of $n$. Consider the number $n - pp'$. Since both $p$ and $p'$ are factors of $n$, it follows that they are factors of $n - pp'$, whence they must appear in the number's prime factorization. So far so good. But then, Davenport goes on to say: "This implies that the number $pp'$ is a factor of $n$" (p. 11). I feel like I'm missing something obvious, but how does the preceding reasoning imply this fact?
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1You can find complete presentations of slight variants of that famous proof of Zermelo in answers here and here. – Bill Dubuque Mar 07 '14 at 04:46
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If $p\cdot q\mid(a-p\cdot q)$, then $a-p\cdot q=p\cdot q\cdot l$ for some $l \in \mathbb{Z}$. So we simply have $a=p\cdot q(l+1)$.
$\therefore p\cdot q\mid a$.
I hope this helped!
MJD
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William Chang
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I'm not sure I understood the antecedent of the conditional. Where did you get the assumption that $p.q \mid a = p.q$? – Nagase Mar 07 '14 at 04:40
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Sorry! I meant $p.q|a-p.q$, not $p.q|a=p.q$. I edited my mistake. – William Chang Mar 07 '14 at 04:43
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Yes, yes, that makes total sense now. Bizarrely, I had actually gotten to the point where $a = p.q(l+1)$, but couldn't draw the right conclusion, haha. Thanks a lot. – Nagase Mar 07 '14 at 04:44
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