arithmetic operations on alephs in WolframAlpha

Question

Can someone please enlighten me if or how both of these facts can be true? I entered $\aleph_0-\aleph_0$ into WolframAlpha, which gave the result as undefined. But it also plotted the result on a number line at zero. Similarly, $\frac{\aleph_0}{\aleph_0}$ is 1 on the number line, but undefined. Is this just a bug?

Answer 1

You ought to consider what the aleph numbers really mean: $\aleph_0$ is the cardinality of the naturals, and is, loosely speaking, a "type of infinity". Normal arithmetic operations are not defined for these numbers because it doesn't make any sense to use them; WA is probably just using a rule (x-x=0, x/x=1) that doesn't apply in this case.

To see what I mean, consider the set of naturals $\mathbb{N}$. It has cardinality $\aleph_0$. Now consider the set of even numbers $\mathbb{2N}$: it also has cardinality $\aleph_0$. If you take $\mathbb{N}$ and exclude (subtract) all even numbers, you get another set, the set of odds, which also has cardinality $\aleph_0$. So $\aleph_0-\aleph_0=\aleph_0$! Obviously this is silly, but you could construct a case where this subtraction gave you any number.

Answer 2

Ignore the number line. That was there just in case "aleph-0" was supposed to represent a real number.

Cardinal subtraction and division are problematic to define, for many reasons, some of which are included here, others here.

Answer 3

Cardinal addition and multiplication (at least for $\aleph$ numbers, which in $\sf ZFC$ means all cardinal arithmetics) has the following property:$$\aleph_\alpha+\aleph_\beta=\aleph_\alpha\cdot\aleph_\beta=\max\{\aleph_\alpha,\aleph_\beta\}=\aleph_{\max\{\alpha,\beta\}}.$$

When an operation can be reversed we usually expect the answer to be uniquely determined. But $\aleph_0+\aleph_0=\aleph_0\cdot\aleph_0=\aleph_0$, and so if we could cancel either of the operations we would get: $$\begin{align} &\aleph_0+0=\aleph_0\\&\aleph_0+\aleph_0=\aleph_0\\&\aleph_0+\diagup\hspace{-.7em}\aleph_0=\diagup\hspace{-.7em}\aleph_0& \implies\aleph_0=0.\end{align}$$

Or similarly, for multiplication:

$$\begin{align}&\aleph_0\cdot1=\aleph_0\\&\aleph_0\cdot\aleph_0=\aleph_0\\&\aleph_0\cdot\diagup\hspace{-.7em}\aleph_0=\diagup\hspace{-.7em}\aleph_0&\implies\aleph_0=1.\end{align}$$

Either result is absurd, of course.