There is an argument to be made that the real numbers, by which I mean the completed reals, does not belong in an algebra course. In the answer above, all of the motivation comes from the real-algebraic-closure of $mathbb{Q}$ (the largest algebraic field extension not containing $sqrt{-1}$).
The reason one might want to introduce the whole continuum comes from numbers like $e$ and $pi$, which are transcendental (the fact that these numbers are transcendental is not immediate and requires a proof that I would consider past middle-school level). If you're willing to state those facts without proof, you can give a moral argument for $e$ by showing that it is the limit of the sequence $((1+1/n)^n)_{nin mathbb{N}}$, which is Cauchy, and its inclusion in the real numbers follows from the completeness of $mathbb{R}$. However, this argument may still be somewhat sophisticated for a middle-school algebra course.
Edit: On Prof. Clark's suggestion, I've copied my comments into the body text (with an additional section as well):
The notion of a sequence converging to a limiting value has a very intuitive geometric interpretation, so it wouldn't be hard to give a geometric argument (say on a graph, for example) that $e$ is a real number, since after relatively few iterations, the graph does level out. Showing that it is not the solution to a polynomial is effectively proving that it is transcendental, and I can't think of an informal argument showing this, but since this answer is community wiki, if someone has an idea, this would give a "moral" argument for the study of the "whole" continuum.
I find this approach useful because it can be introduced using the compound interest formula, which is often taught in an introductory algebra class.
We can see this as follows $$A=P(1+r/n)^{nt}.$$
Let $N:=n/r$. Then we have $$A=P(1+1/N)^{Nrt},$$
which we can rewrite as $$A=P((1+1/N)^N)^{rt}.$$
If we increase $N$ and leave $r$ fixed, this is equivalent to increasing $n$ (this is obvious because $r>0$), which amounts to increasing the number of compounding periods per unit time. Taking the limit (in some informal geometric sense), we can see that as we increase the number of compounding periods, we approach the continuously compounded interest formula $$A=Pe^{rt}.$$
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