"Show that if $\;(a, b) = 1\;$, $\;a|c\;$ and $\;b|c$, then $(a · b)|c$." $$$$Show: We know that $$x\mid w \;\;\text{and}\;\; y\mid w \Longleftrightarrow \frac{x\cdot y}{(x,y)}\mid w$$So if$$a\mid c\;\;\text{and}\;\;b\mid c\Longrightarrow\frac{a\cdot b}{(a,b)}\mid c\Longrightarrow \frac{(a\cdot b)}{1}\mid c\Longrightarrow(a\cdot b)\mid c$$$\blacksquare$$$$$*Statement correct?*
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2How do you know what you say "we know"? – Ted Shifrin Aug 04 '13 at 14:12
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@TedShifrin: Theorem is a common wish of proof? – benjamin_ee Aug 04 '13 at 14:16
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@GitGud Fixed, thanks. – benjamin_ee Aug 04 '13 at 14:18
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@marcelolpjunior O que o Ted quis dizer é que não consegues demonstrar esse teorema sem antes demonstrares o que estás a tentar provar. – Git Gud Aug 04 '13 at 14:18
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@GitGud: Fala/escreve português? – benjamin_ee Aug 04 '13 at 14:20
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@marcelolpjunior Não, nem um pouco $\ddot \smile$ – Git Gud Aug 04 '13 at 14:21
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1To me it is important to prove the general case, not just assert it and make the trivial deduction. – Ted Shifrin Aug 04 '13 at 14:21
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@TedShifrin: Because actually, this is a corollary of the theorem "Let $a$, $b$ and $c$ natural numbers. If $a\mid(b · c)$ and $(a, b) = 1$, then $a\mid c$." – benjamin_ee Aug 04 '13 at 14:24
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@marcelolpjunior If you know how to prove the theorem, then you're done. The statement is a particular case of the theorem, you just need to replace the symbol $(a,b)$ with the symbol $1$. – Git Gud Aug 04 '13 at 14:26
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@GitGud Well .. Thanks a lot .. – benjamin_ee Aug 04 '13 at 14:46
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If you'd like to prove the assertion without the "general theorem" which we "all know":
Suppose $\;(a, b) = 1\;$, $\;a|c\;$ and $\;b|c$. We show, then, that $(a · b)|c$.
We take as given that $\gcd(a,b)=1$. So there exist integers $m$ and $n$ such that $am +bn =1$. Multiplying this equation by $c$ gives us $$cam +cbn =c$$
Now, we also take as given that $a \mid c$ and $b \mid c$. So there must exist integers $x$ and $y$ such that $c = xa$ and $c= yb$. Then $$yb a m + xabn = c\tag{$\dagger$}$$ Hence, since $ab$ divides the left hand side of $\dagger$, it must also divide the right hand side: which gives us $ab\mid c$.
amWhy
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True, but I specified a and b natural. Logo should be used $am-bn=1$ For $am+bn=1$ is used in the whole. with $m$ and $n$ both natural. – benjamin_ee Aug 04 '13 at 14:44
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This right, I understood, but there are restrictions for the entire natural. – benjamin_ee Aug 04 '13 at 14:47
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This is a more general proof, which can be easily stated for naturals $a, b, c, m, n, x, y \in \mathbb N$ if we change $am + bn = 1$ to $am - bn = \gcd(a, b) = 1$, etc using $cam - cbn = c,$ and $ybam - xabn = c$ – amWhy Aug 04 '13 at 22:37