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# [Calvin] For an integer $$k$$, does $$0 \mid k$$ or does $$k \mid 0$$?

I've been getting a lot of clarification questions from users of all levels asking

"Does $$0 \mid k$$? Does $$k \mid 0$$?"

So, how do you determine if either of these statements are true? And how do you help someone else remember which one is true?

Note by Calvin Lin
5 years ago

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Well, the term "a | b" just means that a divides b evenly (or b is a multiple of a). So, this means that b/a is an integer.

Thus, our first term is k/0, which (hopefully obviously) leads to wonky things. Dividing by zero is a iffy business, and usually leads to mass confusion unless you know limits, so I'm just going to say it doesn't work.

Our second term is 0/k. This always equals zero (unless k is zero, then it's even more scary). Thus, 0 | k = 0, but k | 0 is undefined (again, if you know limits, this doesn't exactly apply).

The way I always remembered it was thinking of the | as a reverse fraction of sorts. Usually, when I see a fraction written like 3/4, I read it left to right. The | just means that I read it right to left. But that's just me.

- 5 years ago

That's a good way to remember it Steven. I remember it as a rotation of $$90^\circ$$ anti clockwise to go from $$a \mid b$$ to $$\frac {b}{a}$$.

Of course, division by 0 leads to "wonky things", and this idea still works for $$k\neq 0$$. $$\frac {0}{k}$$ makes sense, so $$k \mid 0$$, but $$\frac {k}{0}$$ doesn't make sense and $$0 \nmid k$$. However, it breaks down at $$\frac {0}{0}$$ not making sense while $$0 \mid 0$$.

If you refer back to the definition, the reason why this happens is because fractions is defined as in terms of division (can't divide by 0) and the divide notation is defined in terms of multiplication (can multiply by 0).

Staff - 5 years ago

The way I learned divisibility from my abstract algebra book states in a ring a|b implies there is some k in the ring such that ak = b. By that definition k|0 for any integer k.

- 5 years ago

This is what I think

- 5 years ago

I think only $$k | 0$$ where $$k \neq 0$$

- 5 years ago