The number \(0.9999\ldots\)  (if you choose to believe it is a one) has infinitely many \(9\)s to the right of the decimal point. What if we consider the “number” with infinitely many \(9\)s to the left of the decimal point instead?

\(\ldots 9999\)

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 \(A\) for Allistaire.

\(A = \ldots 9999\).

STEP 2: Multiply by ten

We obtain

\(10A = \ldots 99990\).

STEP 3: Subtract

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

\(-9A=9\)

giving

\(A=-1\).

That is, our mathematics establishes that

\(\ldots 9999 = -1\).

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

Do you believe that?

Putting it another way: On a number line, do you believe that the numbers \(9\), \(99\), \(999\), …. are marching closer and closer to the number \(-1\)?

It is hard to believe that \(\ldots 9999\) is a meaningful number and, moreover, it has the value \(-1\), at least in our usual way of think about arithmetic. But remember, all we proved here is that IF we choose to say that \(\ldots 9999\) is a meaningful number, then it has value \(-1\).  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 \(\ldots 9999\) is meaningful (for which it has value \(-1\))?

Please join the conversation on Facebook and Twitter and kindly share this page using the buttons below.

Facebook

Twitter

Resources