The_root_of_all_evil said:
maninahat said:
I'm not sure what you are saying as I am not familiar with most of these terms. Could you please explain in layman's terms?
Layman's terms: X times 10 is X0. If the last recurring digit isn't a zero, then it's not following the basic rules of maths - it's following the nearest possible because you're trying to use a finite operation on infinity.
Though I lack the knowledge to properly contend what you are saying, I think you're wrong and that you are somehow misapplying your knowledge). Recurring numbers are workable, despite being infinite in length. x can represent any number, and for the purpose of the proof, it represents 0.(9). It is perfectly possible to use operatives on infinitely long decimal numbers both on paper and practically. In fact, you could just do it counting beans. You have three beans. One of them is one third the total number of beans. In otherwords, 0.(3). I can take another bean along with that one. That makes two thirds of the total, or 0.(6). I add the final bean to the pile. 100% the beans, despite each individual bean representing 0.(3). How could this be possible? How could I be able to do this to the beans, when the fractions they represent are infinitely recurring numbers? How do you explain that 3/3rds makes a 1, when 1/3 makes 0.(3)?
Equally, 10 times infinity is still infinity. 10 divided by infinity ≡ 0.
I'm not asking anyone to multiply infinity. I am asking them to multiply a
real number that can be displayed to an infinite number of decimal places. That is perfectly straightfoward, and it is something that mathematicians do even with normal numbers like 5 or 17. That is because it these numbers aren't just "5" or "17", they are "5.(0)" and "17.(0)". For convenience sake, we terminate them at the first zero, but that doesn't change the fact that there are an infinite number of zeros behind any of these numbers and that these infinite zeros don't make a blind bit of difference when I want to apply operatives to them.
If I want to double 3.333... I can. The answer is 6.666. I'm not doubling infinity itself.
Any number can be represented by x, or any symbol I choose to give. And I can perform whatever operatives I like on the number. I recall there being a particulary famous one that represents an infinitely long irrational number.
Pi, I think they call it.
As someone so apparently savvy with mathematics, I don't understand why you are having this much trouble dealing with an issue that has long been accepted by mathematicians for decades. [http://en.wikipedia.org/wiki/0.999...]