# How to report a child predator

I will give a different answer than the answer I gave in the other thread which tries to appeal to intuition. I am sure your daughter has no problem accepting that $2\times 0 = 0$. Intuitively this is because if you add $2$ to itself zero times, you get zero. Or, to be concrete, if someone gives you two apples zero times, you have zero apples.

For repeatedly adding $2$, talking about collections of apples is a good model. But for repeatedly multiplying by $2$, it isn't necessarily, since you can't multiply apples and apples (at least, not in a way that makes sense to a child). But you can multiply apples by numbers; that is, you can start with $1$ apple, then double the number of apples you have to get $2$ apples, then double the number of apples you have to get $4$ apples, and so forth. In general if you double your apples $n$ times, you have $2^n$ apples.

What happens if you double your apples zero times? Well, that means you haven't started doubling them yet, so you still have $1$ apple. If you want your notation to be consistent, then you should say $2^0 = 1$.

This is a subtly different argument from the argument I gave before. It's intuitive what it means to add different amounts of apples, and it's intuitive what it means to have zero apples. But the twos I am now

working with aren't numbers of apples, they're just abstract numbers; in other words, they're unitless, so it's harder to get a grip on them. What $2^n$ really represents above is an endomorphism of the free commutative monoid on an apple, which is much less concrete than an apple.

There is a way to gain intuition here which sort of involves units, but I don't know if you can convince your daughter that it makes sense. One way to interpret $2^n$ is that it is the "size" of an $n$-cube of side length $2$ in dimension $n$. For example, the length of a segment of side length $2$ is $2$, the area of a square of side length $2$ is $4$, and so forth. One way to say this is that $2^n$ is the number of $n$-cubes of side length $1$ that fit into an $n$-cube of side length $2$.

To get a meaningful interpretation of the above when $n = 0$ we need to decide what $0$-dimensional objects are. Well, if $2$-dimensional space is a plane and $1$-dimensional space is a line, then $0$-dimensional space must be. a point. In particular, a $0$-cube, of any side length, is a point, and so exactly one $0$-cube of side length $1$ fits into a $0$-cube of side length $2$. Hence $2^0 = 1$.

(I'm really curious what her response to this argument will be, actually. Could you report back on this?)

Source: math.stackexchange.comCategory: Bank

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