Let me just say, wow. I haven't had my brain rattled like this in a while!
A fellow over at another blog made an intriguing post. He argues and claims to "prove" that 0.99999… repeated in fact equals 1. If you are anything like me, your first reaction is that of utter rage. 😛 How can that be? Have my years in math class gone completely to waste? Why was I never informed of this. So I decided to dig a little deeper and see what the guy had to say.
Guess what? I still disagree! He does has some interesting "proofs" though.
Let x = 0.99999…
10 * x = 9.99999...
- x = 0.99999...
9 * x = 9
x = 1
Now as far as this proof is concerned it sure is scary to look at. Here are my thoughts about it.
When you multiply a number by 10 all you're doing is shifting the decimal place, correct? So for example 1.234 * 10 = 12.34. At first there were 3 digits after the decimal, after the multiplication there are only 2…so one less. What is so different here?
0.99999…(infinite number of nines) * 10 = 9.99999…(infinite number of nines – 1 nine)
Now when you go to subtract the original x = 0.99999… from the not multiplied x value, what you have is 10*x has one less nine after the decimal than the original nine.
10 * x = 9.99999...
- x = 0.99999...
9 * x = 9.00000...9
x = 1.00000...1
Of course none of this makes sense because I am assuming that even though the number of nines after the decimals is infinite, it is a fixed infinite… The problem is that infinity is a concept and 0.99999…,1 are numbers. In which case we either consider both of our arguments valid or invalid. Because I do not understand how we can be sure that 0.99999… cleanly cancels out with the decimal portion of 9.99999…!
1/3 = 0.33333...
2/3 = 0.66666...
3/3 = 0.99999...
1 = 0.99999...
Now consider 2.99999…/3, which interestingly enough equals 0.99999… . So what does that mean? If 0.99999… is assumed to be equal to 1, then 2.99999…/3 = 3/3. That in turn means that any number which has an infinite number of trailing nines after the decimal can in fact be always rounded up? In other words, every number has a infinite decimal representation…
The last proof worth mentioning is that if 0.99999… to be not equal to 1, then we need to find a number which is in between the two. What is between 0.99999… and 1? The answer is seems to be obviously 0.99999… Because you can always find a number that is greater than 0.99999… but less than 1. For example:
0.9 < x < 1, x is 0.99
0.99 < x < 1, x is 0.999
0.999 < x < 1, x is 0.9999
And so on…so what is greater than 0.99999… and less than 1? Well 0.99999… of course.
In reality of course this has nothing to do with the reality. In reality 0.99999… is NOT equal to 1. However 0.99999… behaves like 1 when current math rules and math assumptions are applied.
Now going a step further, if 0.99999… is interchangeable with 1.0 when doing mathematical calculations using Real numbers, then what is the number that can be interchanged with 0.333333… or 0.66666… or 0. What I'm saying is how come non-repeating numbers have substitutes of repeating numbers but not vice-versa.
I find that a post by Michael VanDeMar on a continuation of the original thread follows my logic well.
Saying that .999… is infinitely close to one is indeed saying that it is at the same time discreet from it. This is not just semantics. If you were to follow your argument through, you would then say that below .999… there is also a number infinitely close to it, which is equal therefore equal to it, and continued on infinitely, you would then end up with an infinite series proving 1 equals 0.
Next thing that bothers me, is 0.99999… equal to 0.99999…? They are as concepts but not as numbers. Because how can we possibly have a finite definition of infinity, meaning that 0.99999… cannot directly equal 0.99999… since infinity does not equal infinity and we don't know the exact number of trailing nines, there are infinite… Or maybe I'm far off the field to make such statements.
Anyways my brain already feels twisted up enough, so here are some links of interest: