It's inaccurate to think of it all at once. In mathematics, the concept of infinity is used to show what happens if you continue on, never ending.
.333 repeating infinitely is the decimal value we give to 1/3 because it's simply the closest we can ever get to that actual number.
If I take the equation 1/8=x, I'm finding how many times x goes into 8. That's one way to define it.
.333 repeating will go into 1 three times, and it's the closest you can get because any number greater than .333 repeating will be greater than 1.
It's simple Calculus.
Prove to me that if you took .3333 repeating infinitely and added .33333 repeating infinitely, then did it again, you would get 1.
You don't; you get .99999 repeating infinitely.
There is never an actual point that .99 infinitely repeating is equivalent to 1. They are two separate numbers.
Again, the whole point of my post is that 1/3 is not .33333 repeating infinitely, but is the closest you can get. Not exact.
THANK YOU!! This is exactly what I was trying to say!
Infinitely close = Infinitely small seperation = Nothing = Zero
See, the problem here is that your turning something that's not equal to nothing, but saying it is. For something to be small, it has to have some quantity, and nothing doesn't have any quantity. Because of this, you're statement is just as ridiculous as this:
.333 repeating infinitely is the decimal value we give to 1/3 because it's simply the closest we can ever get to that actual number.
Like I said before, if it is not 1/3, then what number is it?
It's simple Calculus.
Which you clearly don't understand. 0.333... is a short-hand notation for the infinite series given by:
Sum[ 3 / 10^n for n from 1 to infinity ]
Which, by "simple calculus" is exactly 1/3, which is what Wolfram|Alpha will tell you.
Prove to me that if you took .3333 repeating infinitely and added .33333 repeating infinitely, then did it again, you would get 1.
You don't; you get .99999 repeating infinitely.
And since 0.333... is the decimal representation of 1/3, therefore 0.999... must represent the same real value as 1. Your argument here unfortunately is circular, you've assumed that 0.999... is not equal to 1.
As an aside, I ask you to please add 0.333... to 0.999... and tell me what you get. You get 1.333... which, by no accident, is also 1 + 0.333...
Unless, of course, you're going to tell me that it's somehow 1.333...32. But your very own argument would explain why that doesn't make sense. You insist that infinity has no end, so there will be no room for you to put anything at the end of an infinite sequence of digits. I've had to argue this point numerous times in this very topic, I'd rather not revisit it.
Again, the whole point of my post is that 1/3 is not .33333 repeating infinitely, but is the closest you can get. Not exact.
The computation it does is based on an algorithm that tests its convergence.
The series converges. The value it converges to is the value of the series. The series of partial sums 0.3, 0.33, 0.333, and so on converge to 1/3 and so we say that 0.333... is exactly 1/3. And the algorithm that it uses is actually more complicated than you make it sound, because it can evaluate all sorts of complicated infinite series. Wolfram|Alpha is actually doing algebra and calculus. What it's ultimately probably using is the following relationship, from calculus:
Try to do long division of 1 divided by 3. You'll never have an exact answer, because it goes on forever.
Yes, if I stop at any finite place, I won't have an exact answer. But the limit (there's that Calculus again) of the finite sequences is 1/3, therefore we say that the infinite repeating decimal is exactly 1/3.
Is this thread about rounding errors that computers have
This topic has nothing to do with rounding errors or truncated decimal expansions.
Yeah, all that crap you said despite us pointing out why it's not right is totally proof.
I've actually provided proofs throughout this topic and pointed out many misunderstandings that people have about the topic. All you've done is simply assert that the two decimals represent different numbers and have offered nothing in the way of argument.
If you wish to demonstrate that I'm wrong, I require that you provide an actual proof beginning with the construction of the real numbers. I will be suitably impressed if you can even explain to me the construction of the real numbers.
THANK YOU!! This is exactly what I was trying to say!
See, the problem here is that your turning something that's not equal to nothing, but saying it is. For something to be small, it has to have some quantity, and nothing doesn't have any quantity. Because of this, you're statement is just as ridiculous as this:
1/10 = 0.1 = 0.11 = 11/100
none the counter-claims provide anything but direct assertions to the contrary. Provide a counter-proof.
http://www.wolframalpha.com/input/?i=n+from+1+to+infinity+Summation+of+3%2F10%5En
http://pcpartpicker.com/user/SteevyT/saved/21PI
The computation it does is based on an algorithm that tests its convergence. That is what an infinite series test does.
Here's proof that 1/3 does not equal 1:
Try to do long division of 1 divided by 3. You'll never have an exact answer, because it goes on forever.
.3333 repeating infinitely is not an exact number.
Infinitely is used to show its convergence.
"Programmers never repeat themselves. They loop."
Yeah, all that crap you said despite us pointing out why it's not right is totally proof.
Like I said before, if it is not 1/3, then what number is it?
Which you clearly don't understand. 0.333... is a short-hand notation for the infinite series given by:
Sum[ 3 / 10^n for n from 1 to infinity ]
Which, by "simple calculus" is exactly 1/3, which is what Wolfram|Alpha will tell you.
And since 0.333... is the decimal representation of 1/3, therefore 0.999... must represent the same real value as 1. Your argument here unfortunately is circular, you've assumed that 0.999... is not equal to 1.
As an aside, I ask you to please add 0.333... to 0.999... and tell me what you get. You get 1.333... which, by no accident, is also 1 + 0.333...
Unless, of course, you're going to tell me that it's somehow 1.333...32. But your very own argument would explain why that doesn't make sense. You insist that infinity has no end, so there will be no room for you to put anything at the end of an infinite sequence of digits. I've had to argue this point numerous times in this very topic, I'd rather not revisit it.
Again, you're wrong.
The series converges. The value it converges to is the value of the series. The series of partial sums 0.3, 0.33, 0.333, and so on converge to 1/3 and so we say that 0.333... is exactly 1/3. And the algorithm that it uses is actually more complicated than you make it sound, because it can evaluate all sorts of complicated infinite series. Wolfram|Alpha is actually doing algebra and calculus. What it's ultimately probably using is the following relationship, from calculus:
Sum[ a r^i for i from 1 to infinity ] = a r / (1 - r) when |r| < 1
Yes, if I stop at any finite place, I won't have an exact answer. But the limit (there's that Calculus again) of the finite sequences is 1/3, therefore we say that the infinite repeating decimal is exactly 1/3.
This topic has nothing to do with rounding errors or truncated decimal expansions.
I've actually provided proofs throughout this topic and pointed out many misunderstandings that people have about the topic. All you've done is simply assert that the two decimals represent different numbers and have offered nothing in the way of argument.
If you wish to demonstrate that I'm wrong, I require that you provide an actual proof beginning with the construction of the real numbers. I will be suitably impressed if you can even explain to me the construction of the real numbers.