Let c be 0.9999~. The ~ means "extending infinitely"; just as 1/3 equals 0.3333~.

Then 10c = 9.9999~.

10c - c = 9.9999~ - 0.9999~.

Solving the right side of this equation is easier if you line the numbers up like so:

9.9999~

0.9999~

---------

9.0000~

So 10c - c = 9.

So 9c = 9.

So c = 1.

Therefore, 0.9999~ = 1.

It doesn't "approximate" 1, it's not just "really close to" 1, it's exactly equal. They are two different expressions in numeral form of the same number, the same way 1/2 = 2/4.

It's remarkably simple and can be understood by anyone who's taken 7th grade algebra, and yet it's so counterintuitive that at least half of people, having seen the proof, refuse to believe it.

Here's another approach.

1/3 = 0.3333~ (nearly everyone believes that one)

Multiply both sides by 3, and you get...

1 = 0.9999~

I love explaining this to people and watching them get upset at how they can't disprove it, but they want to so badly. It's a mind-bending and hopefully mind-broadening experience for them.

## Monday, February 12, 2007

### A delightfully counterintuitive bit of math

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## 2 comments:

So is that actually a proof? Can you legitimately subtract 0.9999~ from 9.9999~ and get 9?

Hmph. Math is tricksy.

Sure you can. If the infinite expansion worries you, consider this.

0.9999~ is another way of writing this: 0 + 9/10 + 9/100 + 9/1000 + 9/10000 + ...

Alternately, write it like this:

0 + the sum as n goes from 1 to infinity of 9/(10^n)

9.9999~ is another way of writing this: 9 + 9/10 + 9/100 + 9/1000 + 9/10000 + ...

Alternately, write it like this:

9 + the sum as n goes from 1 to infinity of 9/(10^n)

Does that reassure you that when you subtract, the infinite series of sums, being identical, cancels out?

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