Consider the equation

Solving this equation is no sweat, right? Let’s do it. First, we subtract from both sides:

Now we can factor an out of the left side:

Now, if the product of two things is zero, one of them must be zero. So either , or , that is, .

Easy, right? Sure. But what I’d like to talk about over the course of the next few posts, inspired by this curiosity, is a funny alternate number system called the *10-adic*, or *decadic numbers*. In the decadic numbers, the equation has the usual solutions 0 and 1, but it also has *two other* funny solutions, which don’t correspond to any familiar real numbers.

The first solution, which I will write about in an upcoming post, we’ll call . That is, is some funny sort of number with the property that . (A fancy math word for this is *idempotent*.) Notice that this means is also a solution, since .

Notice also what happens when we multiply these two solutions: . This is quite strange: two numbers, neither of which is zero, which multiply to give zero! Such numbers are called *zero divisors*. This is the step in our solution above which fails for the decadic numbers: once we factored to obtain , we reasoned that if two things multiply to give zero, then one of them must be zero. But I’ve just claimed that this is not true for the decadic numbers.

Often mathematicians stay away from systems with zero divisors, since they ruin some nice properties. For example, not every nonzero number has a *multiplicative inverse*. (Recall that the multiplicative inverse of a number is another number such that ; for example, the multiplicative inverse of is .) For homework, prove that the we discussed above does not have a multiplicative inverse! (Hint: try supposing that it does, and derive a bogus equation…) But we’re not trying to accomplish anything in particular, just to have fun and learn something, so bring on the zero divisors!

OK, so what exactly are these “decadic numbers”? What is ? And what else can be said about the curiosity in my previous post? All this and more, coming soon to a Math Less Traveled near you…

(If you’re impatient, you can read some about p-adic numbers here, and about the decadic numbers here. But as usual, I plan to explain things at a more leisurely pace.)

Let me try the homework problem:

$ uu^{-1} = 1 // u^2u^{-1} = u // uu^{-1} = u // u^{-1) = 1$

Which is absurd because 1 is the multiplicative inverse of 1 and we already assumed u does not equal 1.

Hmm – I enjoyed coming across the word “idempotent” (same power) in your post – it’s been a while since I saw it.

It got me thinking about how so many governments are displaying impotent characteristics (lacking in power, as to act effectively) and many are even “nullipotent” (meaning that the results are the same if executed zero or multiple times, which is synonymous with “no side effects”.)

Responses to the GFC – borrow more money. Nullipotent, indeed.

Very intuitive explanation Brent. I haven’t had any encounter of p-adic numbers in the undergraduate, so it’s a good primer.

I enjoyed this. A basic, but completely new idea for me. So in the domain of decadic numbers, can x(x – 1) = 0 still be solved the way I am used to? I can’t reason that one of the factors must be zero, but can I reason that one of them

maybe zero?That’s right. Zero times anything is still zero, so there will be a solution corresponding to each of the factors being zero. That just isn’t guaranteed to get you *all* the solutions.

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