Let c=2.

10c=20.

10c-c=18.

9c/9=2

c=c

2=2

Wrong again, once you multiply c *10, theoretically at the end of the infinite 9's there would be a zero.  Which is why 9.... or 9-bar is not .9999999 infinite, it's a symbol for a number that can't be expressed in decimal terms.

You people are crazy...even if it's not how you denote a repeating decimal, it's definitely widely accepted enough that .3333...... meant .3-bar or however you type that on a keyboard.

He made like 10 decimals for crying out loud.

I don't think they're equal... .3 repeating is as close as you could ever get to 1/3...like infinitely close, but I don't think it technically equals 1/3. I don't know, I haven't taken math theory.

You're essentially asking whether the limit of the sequence 0.9 * (0.1^n) is equal to 1 with n starting at 0 and going to infinity ... the answer is yes, but in order to prove it you must first take calculus

The sequence  0.5 * (0.5^n) is also equal to 1 with n starting at 0 and going to infinity

wait a minute, why don't we use apples and pears?

a apple=/= a pear

Well, who cares really, unless someone is a mathmatician.

And when you get into number theory you'll find out why that is an oversimplification that we use to express numbers.

YEs it is... Next question

What is the difference between a black cat and a black dog???

Theoretically, if infinity has no end, how is there zero at the end of it?