x = 1 - dx bar
10x = 10 - 10dx bar
10x - x = 9x, which is the same as (10 - 10dx) bar - (1 - 1dx) bar, which is (9 - 9dx) bar
9x = 9 - 9dx
x = 1 - 1dx
Your friend is assuming 1 - dx bar = .999..., which is what he's trying to prove. Your friend may think he's clever to assume what he's trying to prove in the beggining, because then all the work is done for him.
If he understands calculus, confront him with this proof:
If we let S(n) = .9999...9 (with n 9s)
S(n) = sigmar = 1 to n 9 * 10-r (From the definition of a base 10 number)
10S(n) = sigmar = 0 to n-1 9 * 10-r (Multiply each term of the sum by 10)
9S(n) = 10S(n) - S(n) = sigmar = 0 to n-1 9 * 10-r - sigmar = 1 to n 9 * 10-r = 9 - 9 * 10-n
S(n) = 9S(n)/9 = 1 - 10-n
limn goes to infinity S(n) = 1 - limn goes to infinity 10-n = 1 - 0 = 1
i'm going to give YOU the benifit of the doubt and assume you are telling me something hilariously not funny.
(I can't believe ppl can't put this through their heads)
if x= .99999ect... and y= 1 , then x equals point nine freaking nine and y equals 1.
If a = b, and a = c, then b = c
* let x = 0.9999... (a = b)
* 10x = 9.9999...
* 10x - x = 9
* (10 - 1)x = 9
* 9x = 9
* x = 1 (a = c)
hence c = b
2/2 looks different from 1, but it's still equal to 1.
That's the thing with math, you can't rely on your first instincts.