20071204

1 ≈ .9999999999999...

For anyone who is familiar with math headaches, this is one of the major migraines. You can easily find all manner of argument over the internet as to the matter. I am not going to claim that in one short blog entry I can once and for all close the matter, but I can at least refute one argument.

Specifically, the following argument.

The precise mathematical proof consists of assuming first a number N which is defined by:

1) N = 0.999999999...

If we now multiply both sides of the equation by 10, we obtain:

2) 10N = 9.999999999...

Now substracting N from each side of the equation, we obtain:

3) 10N - N = 9.999999999... - N
4) = 9.999999999... - 0.999999999...
5) 9N = 9

or

6) N = 1 = 0.999999999...


It seems pretty conclusive, but there is a subtle mistake that is made at 5, specifically in the subtraction on the right hand of the equation. It should read as follows.

5) 9N ≈ 9

or

5) 9N = 8.9999999...9991

This is because trailing the end of 9.999... is an additional 0 (due to multiplying by 10), while trailing the end of 0.999... is a 9. That single digit difference is what unravels the issue. Either it must be admitted that step 5 is only an inaccurate approximation made because we are too lazy to actually travel down the path of infinity to find the "last" digit, or we have to account for the difference and display the equation appropriately.

And in case you were wondering, .8999...9991 / 9 = 0.999...

Another common argument is the table of nines.

1 / 9 = 0.111...
2 / 9 = 0.222...
3 / 9 = 0.333...
...
8 / 9 = 0.888...
9 / 9 = 1 = 0.999... = 0.111 * 9 = 1 / 9 * 9


Again, this is slightly disingenuous but in a slightly different fashion. Rather than ignoring the relationship of two separate infinites, this one ignores a fundamental concept learned in grade school. Specifically, the concept of remainders.

Trailing the edge of 0.111... isn't a pure 1, it's a 1 R(1/9). Evaluating the remainder is what gives us our next decimal place in infinity, but it will also, always have it's own remainder of 1/9. This remainder is unrepresentable in machine terms/thinking without evaluating it, hence infinity and the erroneous concept that 1 / 9 "ends" in a simple 1.

It is this remainder that, when multiplying 0.111... by 9, evaluates to 0.000...0001 and ticks the value over from 0.999... to 1.

Q.E.D. I am a nerd, thanks Dad.

2 comments:

Phyvo said...

"This is because trailing the end of 9.999... is an additional 0 (due to multiplying by 10), while trailing the end of 0.999... is a 9."

Except that the number 0.999 repeating doesn't *have* an end for the zero to go on. Pick any 9 you want, but there's always a 9 after it, not a zero. It's like the proof of limits in calculus, where you prove it by showing that no one can specify a place where the function will suddenly deviate from its course towards your limit. 0.999 repeating is the same thing, except that it's by definition of what repeating is.

Your argument against the rule of nines is similarly ridiculous because, again, THERE IS NO EDGE FOR SOMETHING TO TRAIL ON. Infinity doesn't "end". That's why it's infinity, that's why infinities have to be dealt with completely differently than normal numbers. Just because .111 repeating is too long of a number for us to divide by brute force doesn't mean we have to resort to remainders to say that it's equal to 1/9. I'm pretty sure it's provable that 1.0 divided by 9.0 will be .111 repeating. You want to say that somewhere along the line you'll suddenly get a different correct answer when you do the division, 10/9 = 1 with 1 remaining. Obviously, it won't, as division is a function and will give the same answer when given the same input, and in this case the input for each small division in the long chain is the same as the output. 1/9 will divide into .111 repeating And that's what .111 repeating MEANS: 1/9. There IS NO TRAILING EDGE. IT GOES ON FOREVER.

There is no trailing edge. There is no trailing edge. There is *no* trailing edge. Are you saying there's some number which is at the top of infinity, which you can then put another number on top of? No, you can't do that, because infinity has no top. In the same way, repeating numbers HAVE NO TRAILING EDGE. -- To have an edge, an end, is to not be infinite --

Phyvo said...

Well, minus special cases like all the real numbers between 0 and 1, but in this example .111 and .999 repeating resemble an infinity like the natural numbers rather than all the reals between two integers.