One problem that occurs to me now is that, when discussing it extemporaneously, I myself tend to forget that there's two very separate and distinct cases involved: dividing any run-of-the-mill number by zero, versus dividing zero by zero. I think the following is a pretty complete proof:

All variables below are in some set U with multiplication.

Define: Zero (0) as, for all x: 0x = 0.

Define: Division (a/b) as the solution of a = bx (that is, a/b=x iff a=bx).

To prove that division by zero is undefined, let's assume the opposite: there exists some number n such that n/0 = x, that is, n = 0x (def. of division), where x is a unique element of U.

Case 1: Say n≠0. Then n = 0x => n=0 (def. of zero). This is a contradiction that shows this case cannot really occur.

Case 2: Say n=0. So 0 = 0x (def. of division), which is true for all x (def. of zero). Thus x is not a unique number, contradicting our initial assumption.

To summarize: If you try to divide any

*nonzero*number by zero, the answer is "Not A Number". However, if you try to divide

*zero*by zero, the answer is "All Numbers Simultaneously".

(Side note: I guess there's an interesting gap in what I just wrote there, that maybe in Case #2 if the set U just has one single element [zero], then actually you can identify x=0, and do in fact have definable division by zero for that one, degenerate case. Hmm, never noticed that.)

Anyway, that's a fairly intricate amount of basic logic for my intermediate "didn't even know about division by zero" algebra students (2 kind-of-exotic definitions, two cases from an implied "or" statement, and doubled proof by contradiction). So I think what I wind up doing, to give them at least some sense of it, is to consider a single numerical example and gloss over the secondary n=0 case:

Say x = 6/0.

So: x*0 = 6 <-- Multiply both side by zero. Violates rule (for all x: x*0 = 0), so x is not any number (NAN). (Of course, normally you can't multiply both sides of an equation by zero. But you

*could*if division by zero was defined [see degenerate case above, for example] and that actually is the assumption here. It's a sneaky proof-by-contradiction, for one numerical case, without calling it out as such.)

Now, I pick the number "6" above for a very specific reason; if that

*still*leaves someone unconvinced, it gives me the option to do the following. Take six bits of chalk (or torn-up pieces of paper, or whatever), put them in my hand, and write 6/2 = ? on the board. Ask one student to take two pieces out, then another student, and another, until they're all gone. How many students can I do to with before the chalk is gone? That's the answer to the division problem: 3 (obviously enough).

Now take back the chalk and write 6/0 = ? on the board. Ask one student to take

*zero*pieces out, and then another, and another. How many students can I do

*this*to before the chalk is gone? That's the answer to this division problem: Infinitely many, which is (again)

*not a number*defined in our real number system. (If we have a short discussion about infinity at this point, and even if I leave students with the message "I want to say this should be infinity", I think that's okay.)

This is all part of an early lecture on basic operations with 1 and 0, serving to (a) remind students who routinely trip up over them, (b) serve as a foundation for simplifying exercises (why you're expected to simplify x*1 but not x+1), and (c) serve as an example for how much more convenient/fast it is to express rules in algebraic notation, versus regular English. It's also around that time that we have to categorize different sets of numbers, sometimes resulting in the question "what's

*not*in the set of Real numbers?", for which I recommend the examples of infinity (∞) and a negative square root (like √(-4)). So, a lot of those ideas segue together.

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