## Friday, November 09, 2007

### The Two Faces of One

Do you always write the number one as "1"? Did you know there's another, exactly equal way to do it? What do you think about "0.99999999..."? (9s repeating to infinity) It turns out, they are exactly the same thing, and it's easy to show why.

Simple proof:

1. one third (1/3) can be written as "0.333..." (3s to infinity)
2. three times one third, 3 * 1/3 = 1.
3. since 1/3 is "0.333...", we can also write this as 3 * "0.333...", which is "0.999..."
4. so, 1 = "0.999..."; both are equal.
Slightly less simple proof:
1. let a = "0.999..."
2. 10 * a = "9.999..."
3. 10 * a - a = "9.999..." - "0.999..." = 9
4. 9 * a = 9
5. a = 1
6. 1 = "0.999..."; both are equal
Convergence of infinite series proof:
1. represent "9.9999...." as an infinite series; 9* Sum (xk for k = 0..Inf) - 9; with x = 0.1
2. since x < 1, this series converges to ( 9 / (1 - x)) - 9 = ( 9 / 0.9) - 9 = 10 - 9 = 1
3. so again, 1 = "0.999..."; both are equal.
When learning this for the first time, many people don't believe it, but this is not a trick. It's difficult to get a visceral understanding about infinity, so this seems counter intuitive. On polling a few students, I found they felt that "it never quite reaches 1" because there will always be a small difference left between 1 and "0.999...", regardless of how many digits you use. This is reasonable and common sense, given our experience in the world, but it's a misunderstanding. The problem is that our experience in the world doesn't prepare us for infinity. There is no difference between 1 and "0.999..." -- none at all! The digits have no end. Ever! Any difference is filled with an infinite regress of smaller segments.

I still don't think it'll work on my cheques, though. Infinity is a harsh mistress.

Burton MacKenZie www.burtonmackenzie.com

Update: There is more explanation of the why and what's going on in these proofs in the comments, below.

slyght said...

i wrote a post about your "slightly less simple proof" several months ago here. after posting it, i think i found a flaw in the proof. in step 3, you subtract the variable, a, from the left side, but the value from the right. i don't think that is allowable is it? i think for consistency, you have to subtract either the variable from both sides or the value from both sides. is this assumption correct?

burton mackenzie said...

I was at your blog within the past week! I recognize your profile shot because I looked closely at the building! Small world. :-)

With regard to step 3, don't worry, you're fine. In the first step the variable was defined as equal to the given value, so they are equal to each other by definition and you can substitute one for the other. As long as you add, subtract, multiply, or divide by equivalent values on each side of the equals sign, you can do whatever you want, which is how algebraic equations are solved. For instance, if x^2 - 2 = 0, then I can add two to each side to get x^2 - 2 +2 = 0 + 2, which then reduces to x^2 = 2.

Similarly, if you used a = 2, then we'd have x^2 - a = 0 to start with. Add a to the left side, and 2 to the right. You still wind up with x^2 = 2.

slyght said...

that is pretty "small world", nice. not to beat a dead horse, but are there any flaws in this line of reasoning, then, using the variable instead until the end? i believe each line is mathematically correct.

10*a - a = 9.999 - a
9a = 9.999 - a
10a = 9.999
a = .999

so we get back to where we started, no? that's why i was concerned about using a variable on one side and a value on the other. it seems to me that though they represent the same thing, somewhere in the equation they change representation. granted, i haven't taken a math course (which i love) in 6 years, but i don't think i'm THAT rusty.

burton mackenzie said...

What you are doing is correct, but stops short of doing a necessary transform for the proof. What your steps are saying (your steps follow in brackets, but i added "..." to them):

2 = 2_______(a = .999...) and (10*a = 9.9...)
2-1 = 2-1___(10*a - a = 9.999... - a)
1 = 1_______(9a = 9.999... - a)
1+1 = 1+1___(10a = 9.999...)
2 = 2_______(a = .999...)

This doesn't really prove anything because you're simply adding a number and then subtracting it. At no point do you remove the infinite mantissa from the equation.

In the 2nd proof I presented, algebraic manipulation is used to remove the infinite mantissa representation, but the finitely represented number remaining is one in the same. (pun intended! :)

The 3rd proof is mostly the same as your proof (in your comment), but I switch (or transform) representations, from the infinite to the finite, by using the sum of a convergent series.

The proofs aren't groundbreaking math, but they do show us how both numbers are actually the same, regardless of the choice of numeric representation.

Thanks for giving me the chance to elaborate on this. I think it makes the post clearer.

k said...

1/3 + 1/3 + 1/3 = 1 NOT 0.999(9)
enough said...

k said...

hmm ok I feel stupid now for shouting out loud my first reaction;
upon closer examination of the phenomenon
http://en.wikipedia.org/wiki/0.999...
I guess [1/3 + 1/3 + 1/3] must be 0.999...

but now I'm interested in the proof that:
[1/2 + 1/2] is also equal to 0.9...

slyght said...

that does clear things up. thanks.

GranSkyline said...

1/3 may be .33333 repeating, but, somehow, there is a little bit more than .3333 repeating, so that when you times is by 3, it's 1...

it doesn't prove that 1 = .999999 repeating

burton mackenzie said...

k, if I had a buck for every time I spoke hastily...well, you get the picture. :-)

burton mackenzie said...

granskyline, forgive me if I am not convinced by your thesis of the math working out "somehow".

GranSkyline said...

well... .333 repeating is basically that... an infinite amount of threes following a decimal... of course, in common sense, you would think that it adds up to .999 repeating.. but .3333 repeating just has a 1/3 at the end.. meaning even more 3's... and 2/3, it's .6666, but it has a 2/3 at the end... when you finally make it to 3/3, it is, in fact, one, but you could say that it is .9999, but then it ends in 3/3... which is one... which makes the nine in front of it, a ten, so it keeps going like that, making it 1...