5 divided by 0 is 0.

Thoughts?

Related: http://rationalexpressions.blogspot.com/2012/11/how-not-to-teach-it-division-by-zero.html

  • Fact families! Practice: 2 times 3 is 6. 3 times 2 is 6. 6 divided by 2 is 3. 6 divided by 3 is 2.

    Practice backwards: 8 divided by 4 is 2. 8 divided by 2 is 4. 4 times 2 is 8. 2 times 4 is 8.

    Now 5 divided by 0 is 0. 0 times 0 is 5. Wait a minute… no, that should be 0 times 5 is 0. That means 0 divided by 5 is 0, good. Oh, wait, that also means that 0 divided by 0 is 5. Interesting …

  • So 5 divided by 5 is? 1. Because if you split 5 things into 5 groups each group contains? 1. So if you five things into 0 groups each group contains? How do you split 5 things into 0 groups? Wouldn’t you have to make then disappear? Well, that’s interesting…

    • What is 2 to the third power? It’s 2 times 2 times 2. What’s 2 to the fourth power? 2 times 2 times 2 times 2. What’s 2 to the -1st power? Wait, you’d have to negative copies of 2? Huh, that’s interesting…

      It’s a bit cheeky, but what’s the difference between our two arguments?

      • I think this is where a table comes in handy: 2^3=8, 2^2=4, 2^1=2, 2^0=1, 2^-1=1/2 and so forth.

        • mpershan

          Now we’re getting closer to a real argument in favor of the idea that 2^-1 should be 1/2. So let’s focus on arguments against the notion of 2/0 instead of the failure of a particular conceptual model.

          • If we think of division as repeated subtraction, the following can be explored: 8/2–> -2-2-2-2=0 –> 4
            2/0–> -0-0-0-0… Never get to 0 so it’s undefined. (BTW, this notation is probably going to render ugly)

          • Hold on. I don’t buy into the idea that using a conceptual model in that way isn’t a “real” argument. An argument that holds up when values are positive shouldn’t just be thrown out because of the existence of negatives. To be a bit cheeky myself, we have these exponential graphs for b^x. But oh, they don’t work with a negative base because the answer would keep switching signs creating a totally different kind of graph. Back to the drawing board then! Come up with a better way of graphing that works for ALL values! Invent a new graphing system maybe!

            I think the difference between the two “arguments” is more a difference in perspective. The implication is that splitting things into 0 groups is impossible, and therefore we are done. mpershan is protesting saying we are NOT done, so present a better case. But why can’t the implication of ‘splitting things into 0 groups’ act as the starting point for some other line of reasoning? Why can’t we say “we are done… FOR NOW”? Those who want to take it further will take it further. We don’t necessary want to overload a younger student with high grade complex numbers just to prove the non-existence of 0^0. Motivate them to want to look into it themselves, maybe, but for every student who would be motivated I suspect there would be another saying “You’re just making things too complicated – this is why I hate math!” And we (presumably) don’t want to cause that reaction.

  • I just had this come up with a 5th grade teacher. I showed him two different models he could use when discussing this with his students, including the limit argument (what happens as we reduce the size of the denominator?), and the intuitive argument above (how can you take 16 things and break them into equal groups so that everyone involved gets nothing).

    It’s nice when our models are consistently vertically (as in the same model in one grade works to extend the understanding in a later grade) but I don’t think it’s always possible. The proof model you shared in your related post only really works if the students get that equal operations on either side of an equal sign maintain the equality – which is a fundamental property of algebra.

    • That’s a fair criticism, David. But I think that the “fact families” approach that Josh presents could be offered to 5th graders. Not that I’ve ever taught 5th graders.

  • Jasper

    Two answers:

    * OK, let’s do a check-by-substitution. Does 0 times 0 equal 5? Uh-oh.

    * Sure, I can graph y = 5 / x. And when I get to zero, how do I connect the two branches of the hyperbola? Yes, zero is about half-way between minus infinity and plus infinity — but compared to infinity, every real number is about half-way between minus infinity and plus infinity. And I still have not connected the branches of the hyperbola. Or in other words, if I connected the branches of the hyperbola by drawing along the y-axis, then 5 divided by 0 is every number. Uh-oh.

    Conclusion: So long ago, the mathematicians decided that they would just never divide by zero — because it just doesn’t work right.

    For the really clever students: The mathematicians did figure out how to divide by numbers that are really really close to zero — but they had to make sure that the numerator was also really really close to zero. You’ll find out more about that when you take calculus.