Friday, November 09, 2007
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:
- one third (1/3) can be written as "0.333..." (3s to infinity)
- three times one third, 3 * 1/3 = 1.
- since 1/3 is "0.333...", we can also write this as 3 * "0.333...", which is "0.999..."
- so, 1 = "0.999..."; both are equal.
- let a = "0.999..."
- 10 * a = "9.999..."
- 10 * a - a = "9.999..." - "0.999..." = 9
- 9 * a = 9
- a = 1
- 1 = "0.999..."; both are equal
- represent "9.9999...." as an infinite series; 9* Sum (xk for k = 0..Inf) - 9; with x = 0.1
- since x < 1, this series converges to ( 9 / (1 - x)) - 9 = ( 9 / 0.9) - 9 = 10 - 9 = 1
- so again, 1 = "0.999..."; both are equal.
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.
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11 comments:
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?
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.
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.
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.
1/3 + 1/3 + 1/3 = 1 NOT 0.999(9)
enough 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...
that does clear things up. thanks.
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
k, if I had a buck for every time I spoke hastily...well, you get the picture. :-)
granskyline, forgive me if I am not convinced by your thesis of the math working out "somehow".
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...
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