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Unusualus
Unusualus
Unusualus

Station  10B
A Troubling Number our Usual Mathematics Rejects 

The number 0.99990.9999\ldots (if you choose to believe it is a one) has infinitely many 99s to the right of the decimal point. What if we consider the “number” with infinitely many 99s to the left of the decimal point instead?

9999\ldots9999

This is a number that ends with nine. Actually it ends with ninety-nine. Actually it ends with nine-hundred-and-ninety-nine. And so on.

Let’s apply our algebraic argument to see what value it must have.

STEP 1: Give the quantity a name.

We’ll call it AA for Allistaire.

A=9999A = \ldots9999.

STEP 2: Multiply by ten

We obtain

10A=9999010A = \ldots99990.

STEP 3: Subtract

We see that AA and 10A10A differ by nine (it is only their final digits that differ). Looking at A10AA-10A we get

9A=9-9A=9

giving

A=1A=-1.

That is, our mathematics establish that

9999=1\ldots9999 = -1.

Apparently, if we pulled out a calculator and computed the sum 9+90+900+9000+9+90+900+9000+\cdots the calculator will show at the end of time the answer 1-1!

Do you believe that?

Putting it another way: On a number line, do you believe that the numbers 99, 9999, 999999, …. are marching closer and closer to the number 1-1?

I10S10B - Image01

Challenge: Let’s make matters worse! Consider the “number” with infinitely many 99s both to the left and to the right of the decimal point: 9999.9999\ldots9999.9999\ldots. Use the same algebraic argument to show that this equals zero. (And this makes sense, because 9999.9999=9999+.9999=1+1=0\ldots9999.9999\ldots = \ldots9999+ .9999\ldots = -1 + 1 = 0.)

It is hard to believe that 9999\ldots9999 is a meaningful number and, moreover, it has the value 1-1, at least in our usual way of think about arithmetic. But remember, all we proved here is that IF we choose to say that 9999\ldots9999 is a meaningful number, then it has value 1-1. It is up to us to decide whether or not it is meaningful quantity in the first place. Most people say it is not and stop there and that is fine.

But this begs the question:

Is there an UNUSUAL system of arithmetic for which 9999\ldots9999 is meaningful (for which it has value 1-1)?

Challenge: One might be able to argue that 9999\ldots9999 does behave like 1-1 in ordinary arithmetic to some degree. For example, consider performing the (very) long addition shown. Do you see the answer zero results?

I10S10B - Image02

If you prefer, imagine what happens if you add one dot to this loaded 1101 \leftarrow 10 machine.

I10S10B - Image03

Challenge: Try this (very) long multiplication problem. Do you see that 66667\ldots66667 is behaving like the fraction 13\dfrac{1}{3}?
I10S10B - Image04

If you prefer, imagine what happens if you triple the count of dots in each of these boxes.

I10S10B - Image05

Extra: What “number” behaves like 23\dfrac{2}{3}?

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