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this problem from An Inquiry-Based Introduction to Proofs v1. by Jim Hefferon

I has problem in the second part that only 0 can divide 0 I think it is interminate form and when put 0/0 in wolfram it shows it's undefined

Is book wrong or I miss something ?
and If the book is not wrong how I begin the proof ?

Lingnoi401
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1 Answers1

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Hint:

We say for two integers $a$ and $b$ that "$a$ divides $b$" iff there exists some integer $k$ such that $b=a\cdot k$.

For example $3$ divides $12$ since $12=3\cdot 4$.

Furthermore, $3$ divides $0$ since $0=3\cdot 0$.

See this page.

Upon searching for sources, apparently in some materials the definition above has the added stipulation that $k\neq 0$. This stipulation was not in the definition I was taught and its inclusion would make the statement false, but that stipulation is not included in the book you got the problem from. Using the definition then from your book which agrees with the above, the proof follows rather directly.

What happens if $a=0$ and $b\neq 0$? Does there exist such a $k$?

What happens if $a=0$ and $b=0$? Does there exist such a $k$?

What happens if $b=0$ and $a\neq 0$? Does there exist such a $k$?

What does all of this imply in relation to your question?

JMoravitz
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  • Ok I got that Thank you ^ ^ I just think about only about actual division – Lingnoi401 Sep 07 '17 at 05:50
  • @JMoravitz: AFAICS: for $0|0$ this seems to hinge on what some means in 'some integer $k$ such that $b=a⋅k$'. For every $a \neq 0$, where $a|b$ there is exactly one $k$ such that $b=a⋅k$ (including where $b=0$). For $a=b=0$ there are an infinity of integers $k$ such that $0=0⋅k$'. If the definition said 'an integer $k$', or more strongly 'a single integer $k$', that would rule out $0|0$. The use of some in this context may may be a "clever" way to allow for $a=b=0$ -- granting that some stretches across $[1,\infty)$. But if so, it's not a miracle of clarity, is it ? – Chris Hall Aug 25 '21 at 15:30
  • @ChrisHall "some" in mathematics is always meant as "at least one." If we meant exactly one, we would have said so. If we meant only finitely many, we would have said so. In symbols... $a\mid b\iff (\exists k\in\Bbb Z)(ak=b)$. See Existential quantification. – JMoravitz Aug 25 '21 at 15:32
  • @JMoravitz: so, yes, the definition allows for $0|b$, where $0|b \iff b=0$. [Given the strict meaning of some as you say.] My point was that the definition would be clearer if it treated $0|b$ as the special case that it is. I note that the wikipedia page you referenced starts by defining $m|n$ only for $m\neq 0$ -- which is consistent with the definition of the division algorithm. The extension to $m=0$ is mentioned in passing, but not explicitly defined there. – Chris Hall Aug 25 '21 at 17:49
  • I disagree that it is that special of a case. In a deleted exchange below, BillDubuque said it well; Consider theorems from group theory and ring theory about the quotient ring $\Bbb Z/n\Bbb Z$ and the statement that $x\equiv y\pmod{n}$, that is that $[x]=[y]$ iff $n\mid x-y$. This works just as well for the quotient set $\Bbb Z/n\Bbb Z$ with the (obvious) statement that $x\equiv x\pmod{n}$ including when $n=0$ noting that $\Bbb Z/{0}$ is a perfectly valid quotient of rings (even if trivial). – JMoravitz Aug 25 '21 at 18:14
  • @JMoravitz: OK, I admit knowing next to nothing about group theory (and an approximately equal amount of ring theory) so for all I know you make a convincing case that $0|b$ (in particular $0|0$) is not that special. I give up. FWIW: stipulating $k \neq 0$, as you say, denies $a|0$ -- at least for $a \neq 0$. For $0|0$ there are infinitely many $k \neq 0$ which satisfy $0 = 0 \cdot k$, so perhaps $0|0$ is still allowed -- unless the stipulation excludes all possible $k$ if $k=0$ is one of them. – Chris Hall Aug 25 '21 at 18:40