Prove that Every number divides 0 while the only number that 0 divides is itself. 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 ?

**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 $k\
eq 0$. 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 $b\
eq 0$? 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 $a\
eq 0$? Does there exist such a $k$?

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