0.00...=0

[Meitan]

Hello.

I'm not sure if this is the right forum, but I hope it's close enough. Feel free to move if it isn't.

But yeah, I have heard many different proofs that when something is infinitely close to some number, it's equal that. However, I'm still not convinced.
Could someone explain what is wrong with my reasoning in this:
1/?=0.00...=0
x*0=0
x/y*y=x when y!=0
Let's put x=1, y=?
1/?*?= Now this is where the problem lies. By x/y*y=x, it should be 1. However, if 1/?=0, then it would be 0. But then the rule that y!=0 would need to be extended that y!=0 and y!=?.

Also,
Assuming 1-(1/?)=0.99..=1
then 1-(1/?)-(1/?)-(1/?)-(1/?)-....=0 or 1?

Thanks in advance

 

[tkhunny]

"when y!=0"

Why would you write it and then ignore it?

 

[daon]

If a is a real number greater than 0 and is infinitely close to it, then it is smaller than every positive number, right? But it isn't smaller than 1/2*a. That is one proof. Another proof is that, considering real numbers as defined by sequences of digits, when 2 real numbers are distinct they differ in at least one place. Hence either x>y or y>x. If each were "infinitely close" to 0, BOTH inequalities would have to be satisfied, which is nonsense. The conclusion then is that either both are zero, or at least one is not a real number.

You may want to look at the Surreal Number system. These infinitesimals do not belong to the reals or extended reals.

As for your reasoning, Infinity is not a real number. It does not satisfy many of the axioms of the real number system. For instance infinity^2=infinity would mean that x(x-1) has 3 roots, one of which is infinity. But, plugging it in yields that infinity^2=0. So you cannot write "1/infinity" and treat it as a real number.