Why couldn't that 1 just be another 0...The 0 goes on forever...You can't add 1 to an infinity... Can you? Can you add 1 to an infinity...NO...Infinity goes on forever...So do the 0s...

You mean its true value is not 0.3 repeating? What is it then? Is a 4 gonna show up? Is a 2 gonna show up? Or are you saying the methods to do division on paper are just wrong...Guess what, they aren't...3 goes into 10 3 times, do that over and over, you keep getting 0.3333333333333333...FOREVER...Thats why its called...0.3 repeating...

So now you are gonna disagree with 0.9 repeating minus 0.9 repeating is 0? Because thats what you did up there...9 + 0.9 repeating - 0.9 repeating = 0...That is basic math...Come on...

Are seriously gonna tell me 0.9r - 0.9r don't equal 0?

Whats between 3.9r and 4? Nothing...Because 3.9r DOES equal 4...Haven't you figured out that 0.9 repating = 1?

So once again...What is in between 0.9 repeating and 1? There has to be something, if they aren't equal

The problem with this my simple minded friend is that you are basing your whole logic on conclusions that are false. You base everything on the assumption that 0.3repeating and 1/3 is equal when they arent. Think about it, you can NEVER stop righting the 3s hence the infinite part and because of that, you can never use that number in any calculation with 100% accuracy. BTW, you said I dont use a system of equations and then used a system of equations but youre still doing it wrong because it doesnt apply to a single variable equation. Youre right about the difference between 1 and 0.9 repeating when you ask why couldnt you just add another 0 where you put the 1. The PROBLEM with your logic is that no matter how many times you do it 10-9 will always leave a 1 so at some point you would have 0.0-and then 1. Never even with infinite 9s will you get 10-9=0 and thus they arent equivalent.

If I can't add something to an infinity, that also means you can't subtract from an infinity, because subtraction IS actually addition, negative addition.

You can't add a number to infinity...That goes against the concept of infinity

But you can subtract infinity from infinity which gives you 0...Bascially...The answer must still go through the rules of infinity, or still be an infinity...

Like you CAN subtract 0.6 repeating - 0.3 repeating...Why? Because the answer is still an infinity of 3s, nothing surprising...You are going to keep getting 3s no matter what...The answer to that is 0.3r

yep, it's a MADE number, to illustrate the value of 1/3. But it does not really exist. Neither does it have the same value

edit: actually, Jerm is right. IceAge is just typing the same nonsense für the whole 8 pages. He's like a mis-programmed robot that can't grasp that any information he has could be wrong. Just the same all over again, every post

The problem with this my simple minded friend is that you are basing your whole logic on conclusions that are false. You base everything on the assumption that 0.3repeating and 1/3 is equal when they arent. Think about it, you can NEVER stop righting the 3s hence the infinite part and because of that, you can never use that number in any calculation with 100% accuracy. BTW, you said I dont use a system of equations and then used a system of equations but youre still doing it wrong because it doesnt apply to a single variable equation. Youre right about the difference between 1 and 0.9 repeating when you ask why couldnt you just add another 0 where you put the 1. The PROBLEM with your logic is that no matter how many times you do it 10-9 will always leave a 1 so at some point you would have 0.0-and then 1. Never even with infinite 9s will you get 10-9=0 and thus they arent equivalent.

How does my equation not apply?

It works right here

x=8
10x=80
10-x=80-8
9x=72
x=8

Does the equation work or not? Its not really a system of equation solving for anything...It is just using basic math principles, that should work out equally...Why does it work on EVERY OTHER NUMBER, and not on the 0.9 repeating?

And strange, you didn't attempt to answer the question, whats between 0.9r and 1 if they aren't equal?

yep, it's a MADE number, to illustrate the value of 1/3. But it does not really exist. Neither does it have the same value

edit: actually, Jerm is right. IceAge is just typing the same nonsense für the whole 8 pages. He's like a mis-programmed robot that can't grasp that any information he has could be wrong. Just the same all over again, every post

I'm trying to improve myself. I'm the new, better Jerm

Join Date: Jun 2006

Location: Okokomaiko

Posts: 799

Quote:

Originally Posted by IceMan2

Another thing...

ENOUGH with all the posts saying its like an asymtope, "approaching" 1...

0.9 repeating is a NUMBER...1 NUMBER...It doesn't move, and go approaching 1, it has its own place...So none of that either...

Umm...It's very much like an asymptote and it isn't a single number. .99>.9 so .9 repeating is increasing and moving towards 1 but it will never get there. .9 repeating isn't and will never be equal to 1, this thread should have gone past page 1.

And what is between 0.9 repeating and 1 then? If they aren't equal, there must be something between them...

In order for youre 3rd statement to be true, there would have to be a 0 at the end of the 9.9 repeating once you multiply it by 10. IE If you multiply .9 with 10 9s, by 10 you would get 9.999999999-.9999999999which gives you 8.9999999991=9x. Then when you divide by 9 you would get .9 with 10 9s. That would work no matter how many nines you put on it but the point is that when you multiply by 10 you are moving the decimal point to the left 1 so the .9 r that you are subtracting would have one more 9 to the right of the decimal no matter what number of 9s you decided to use for the begining.