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A key property of the integers is that: if $\gcd(a,b) = 1$ and $a |bc$, then $a|c$. Use this property to prove that: if $a \in\mathbb{Z}_+$ is prime and $b_i \in \mathbb{Z}_+$ for $1 \leq i \leq n$ and $a | \prod_{i = 1}^{n} b_i$, then $a|b_i$ for atleast one $b_i$.

I'm not exactly sure how to do this. I was thinking something like:

If $a$ is prime, then this tells us $\gcd(a,b_i) = 1$. We also know that $a| \prod_{i = 1}^{n} b_i$ and so clearly this is in the form $a | bc$. We can therefore write $b_i = b_1 \cdot b_2 \cdot ... \cdot b_i \cdot b... \cdot b_n$. If we take one of these out, we can see that $a | (b_1 \cdot ... \cdot b_{i-1} \cdot b_{i + 1} \cdot .. \cdot b_n) b_i$ which is know in the form $a | bc$ and we can see from here that using the property we are given, $a|b_i$ fot atleast $1$ $b_i$.

Is this correct? One of my friends mentioned something to do with induction, but I don't get how to prove by induction properly or how to use it in this case.

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Note: "...We can therefore write $b_i = b_1 \cdot b_2 \cdot ... \cdot b_i \cdot b... \cdot b_n$..." Here, you define $b_i$ in terms of $b_i$. – apnorton Feb 9 '13 at 16:27
up vote 1 down vote accepted

It should indeed be an argument by induction. The base case, $n=1$, is completely trivial: if $p\mid b$, then $p\mid b$. For the induction step you want to assume that the result is true for some $n\ge 1$ and prove that it holds for $n+1$. That is, you assume that if $p$ is a prime and $p\mid\prod_{k=1}^nb_i$, then $p\mid b_i$ for some $i\in\{1,\dots,n\}$, and you try to prove from this that if $p\mid\prod_{i=1}^{n+1}b_i$, then $p\mid b_i$ for some $i\in\{1,\dots,n+1\}$.

You actually already have the basic idea, even if you didn’t see how to incorporate it into a proof by induction. Assume that the prime $p\mid\prod_{i=1}^{n+1}b_i$. Let $b=\prod_{k=1}^nb_i$. Of course $$\prod_{k=1}^{n+1}b_i=\left(\prod_{k=1}^nb_i\right)b_{n+1}=bb_{n+1}\;.$$

Now either $p\mid b$, or $p\nmid b$. If $p\nmid b$, then $(p,b)=1$, and $p\mid b_{n+1}$ by the theorem that you were given. And if $p\mid b$, then $p\mid\prod_{i=1}^nb_i$, and your induction hypothesis lets you conclude ... what?

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How do you know its induction? – Kaish Feb 9 '13 at 17:08
Also, the induction hypotheses will let me conclude that $p | b_{n+1}$ as well as $b_{n+1}$ is a multiple of $\prod b_i$? – Kaish Feb 9 '13 at 17:13
@Kaish: First, I see that it’s a statement about each integer $n\ge 1$, and such statements are often proved by induction. Then I see that if I can prove it for $n=2$, I can use that same idea repeatedly to prove it for $n=3,4,5,\dots$ $-$ and that’s a dead giveaway that induction should work. – Brian M. Scott Feb 9 '13 at 19:19
@Kaish: For the other question, the ind. hyp. lets you conclude that if p$\mid\prod_{i=1}^nb_i$, then it divides $b_i$ for some $i\in\{1,\dots,n\}$. Thus, either $p\mid b_{n+1}$, or $p\mid b_i$ for some $i\in\{1,\dots,n\}$, and in either case we have the conclusion that we wanted: $p\mid b_i$ for some $i\in\{1,\dots,n+1\}$. You definitely don’t get that $b_{n+1}$ is a multiple of $b$. – Brian M. Scott Feb 9 '13 at 19:22

Suppose otherwise, that no $b_i$ has $a$ as a factor. Then we have $a|(b_1(b_2\dotsb b_n))$. Since $\gcd(a, b_1) = 1$, $a|(b_2(b_3\dotsb b_n))$. Since $\gcd(a, b_2) = 1$, $a|(b_2(b_3\dotsb b_n))$. We can continue "peeling off" the $b_i$ from the front until we are left with $a|1$, which is absurd.

This is actually induction in action, but phrased differently to prioritize intuition over rigor. Induction is "repeating the same process" over each incremental step, where each "step" involves a decreasing number of $b_i$'s on the right side.

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Use induction. We know $a\big| \prod_{i=1}^2b_i$ implies $a|b_1$ or $a|b_2$. Then look at it like $$\prod_{i=1}^{n+1}b_i=\left(\prod_{i=1}^nb_i\right)b_{n+1}.$$ Now $$a\big|\prod_{i=1}^nb_i\quad\text{or}\quad a\big|b_{n+1}.$$ From the hypothesis, we know if $a\big|\prod_{i=1}^nb_i$, then $a\big|b_k$ for some $1\leq k\leq n$, otherwise $a|b_{n+1}$, as desired.

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The OP said a friend of his already told him to try induction, but the OP didn't know how to apply it... – apnorton Feb 9 '13 at 16:26

Hint $\ $ Specializing the given Euclid's Lemma to the case of prime divisor $\rm\:a = p\:$ yields

$$\begin{eqnarray}\rm a\mid bc\ &\Rightarrow&\rm\ a\mid b\ \ or\ \ (a,c) > 1\quad Euclid's\ Lemma\\ \rm p\mid bc\ &\Rightarrow&\rm\ p\mid b\ \ or\ \ (p,c) > 1\\ \rm p\mid bc\ &\Rightarrow&\rm\ p\mid b\ \ or\ \ \,p\mid c\qquad\quad Prime\ Divisor\ Property\\ \end{eqnarray}$$

The above Prime Divisor Property inductively extends to products of any length:

$$\begin{eqnarray}\rm p\mid b_n b_{n-1}\cdots b_1 &\Rightarrow&\ \rm p\mid b_n\ \ or\ \ \ \ \color{#C00}{p\mid b_{n-1}\cdots b_1}\quad by\ Prime\ Divisor\ Property\\ \\ &\Rightarrow&\rm\ p\mid b_n\ \ or\ \ \color{#C00}{(p\mid b_{n-1}\ \ or\ \cdots\ or\ \ p\mid b_1)}\quad by\ \color{#C00}{induction}\end{eqnarray}$$

Remark $\ $ If you know modular arithmetic then you can interpret it in $\rm\,\Bbb Z/p = $ integers $\rm\,mod\ p.\:$ Then the Prime Divisor Property says: $\rm\, bc\equiv 0\:\Rightarrow\:b\equiv 0\:$ or $\rm\:c\equiv 0\:$ and its $\rm\,n$-ary extension is

$$\rm b_n \cdots b_2 b_1\equiv 0\ \ \Rightarrow\ \ b_n\equiv 0\ \ or\ \cdots\ or\ \ b_2\equiv 0\ \ or\ \ b_1\equiv 0$$

Equivalently, we may express the property as cancellability of nonzero elements

$$\rm\:a\not\equiv 0,\ ab\equiv ac\:\Rightarrow\:a(b\!-\!c)\equiv 0\:\Rightarrow\:b\!-\!c\equiv 0\:\Rightarrow\:b\equiv c$$

The rings like $\rm\,\Bbb Z,\Bbb Q,\Bbb R,\Bbb C\,$ satisfying this property are known as integral domains. They play a fundamental role in number theory and algebra since, as above, by working modulo a prime, one can reduce many problems in rings to problems in integral domains, where the problem may become simpler due to the ability to cancel nonzero elements. Much of interest is a consequence of this property, e.g. a nonzero polynomial over an integral domain has no more roots than its degree.

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