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Station  10D
A Serious Flaw of Our Ten-adic Numbers 

With the ten-adic numbers we can represent all integers and fractions. But did you notice that all the numbers we presented so far have repeating cycles?

We can also consider numbers in this system, with infinitely many digits to the left, that don’t fall into repeating cycles. These must correspond to irrational numbers. (And possibly other new types of numbers?)

We can add and multiply ten-adic numbers. For example, we have seen 66667\ldots66667 is 13\dfrac{1}{3} and 9999\ldots9999 is 1-1. We can compute their sum to see an answer indeed one less than =6667\ldots=6667. (And 6666\ldots6666 is 69=23=131-\dfrac{6}{9} = -\dfrac{2}{3} = \dfrac{1}{3}-1.)

I10S10D - Image01

And we can compute their product to see the answer 3333\ldots3333, which is indeed 13-\dfrac{1}{3}.

I10S10D - Image02

And since we know how to make negative ten-adic numbers and fractions as ten-adic numbers, we can also subtract (add the negative) and divide (multiply by a fraction) ten-adic numbers.

Well, almost. We can’t divide by some ten-adic numbers. There’s a flaw in the ten-adic system.

Here’s an example of the problem. We have that

2×5=102 \times 5=10 is a number close to zero.

22×52=1002^{2} \times 5^{2}=100 is a number closer to zero.

23×53=10002^{3} \times 5^{3}=1000 is a number even closer to zero.

And so on.

It is possible to construct a non-zero ten-adic number NN that behaves like an infinitely large power of two and a non-zero ten-adic number MM that behaves like an infinitely large power of five, so their product then, N×MN \times M, is so close to zero that it actually is zero!

N×M=0N \times M = 0

Non-zero numbers that multiply to zero don’t exist ordinary arithmetic, but they do in the ten-adic system. This means we can’t divide by some non-zero numbers, like NN and MM in this arithmetic. (If dividing by NN is possible, then divide the equation N×M=0N \times M = 0 through by NN and get the contradictory statement that M=0M=0.)

How might one construct these numbers NN and MM? It’s a bit tricky, but here’s the gist of it.

Here are the first few powers of five:

55, 2525, 125125, 625625, 31253125, 1562515625, 7812578125, …

All of these powers end in 55.

Infinitely of them actually end in 2525.

Actually, one can verify that infinitely of then actually end in 125125. (Multiply any one that does by 2525 to get another one.)

Actually, infinitely many of them end in 31253125. (Multiply any on that does by 625625 to find another one that does.)

And so on.

In principle, we can construct a ten-adic number M=3125M=\ldots3125 for which there are infinitely many powers of five that end with the same set of digits as MM does, for any size set of digits you want. (There are infinitely many powers of five that end with the final set of one-hundred digits as MM does, and there are infinitely many powers of five that end with the same million final digits, and so on.)

We can do the same construction for the powers of two and construct a ten-adic number N=832N=\ldots832 for which there exist infinitely many powers of two that end with any final set of digits of NN.

Now look what happens when you multiply NN and MM. You do indeed get zero.

I10S10D - Image03

The problem is that 1010 is a composite number. One can prove that this annoyance will never arise if one works in base that is a prime number instead!

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