Can a = b when a and b are different integers?

[bugzpodder]

lol i didn't post that message. you got your names wrong Alphanos!

but exactly, as you yourself pointed out

then (1+0.9 repeated)/2 could not possibly be greater than 0.9 repeated!

so what is it equal to? it can't be greater than 1. it can't be less than 0.9 repeating. so its either 1, 0.9 repeating or undefined.

undefined won't make any sense.

so if its 1 or 0.9 repeating, it would indicate that the three numbers are the same, like:

if the average of 4 and x is 4, what is x? you would say 4 wouln't you?

the truth is, if there is indeed a difference between two numbers x and y, their average should be in between them. however as indicated, this would happen iff there isn't a difference between x and y, so x=y, or in other words, 0.9 repeating =1

i understand that conceptual numbers can be represented algebraically or can be used as an approximation, but i don't see how it is relevant to anything we've talked about. btw i am representing those conceptual numbers algebraically, not using an approximation, in case thats what u r saying.

Gandalf:

but that last digit would be at infinity making it infinitely less than .99 repeated, which is infinitely less than 1. so if .99 repeated is infinitely less than one and .999 .... 8 is infinitely less than .999 repeated then .999 repeated = .999 ... 8 so they both equal 1. If done an infinited amount of times 1 could equal 0. This is with your logic though. I agree that it isn't possible. Even though the difference is infitesimally small it adds up

you cannot have .999...8 where there is an infinate number of 9's. you can't just say at the end of infinity I will add a 8, because there is no end to infinity!!! That was what I am trying to say. sorry about the bold, i was just illustrating my point, or in NotLKH's words:IMHO, lol!