The #0.bar(9)=1# equality is well known and often confusing for students encountering it for the first time. There are many ways to show that the equality holds. For example:
Let #x = 0.bar(9)#
#=> 10x = 9.bar(9)#
#=> 10x - x = 9.bar(9) - 0.bar(9)#
#=> 9x = 9#
#=> x = 1#
Using the equality #0.bar(3) = 1/3#,
#0.bar(9) = 0.bar(3)+0.bar(3)+0.bar(3)#
#=1/3+1/3+1/3#
#=1#
Suppose #0.bar(9)!=1#. Then #x = 1-0.bar(9) > 0#. Because #10^(-n)->0# as #n->oo#, there must be some #n in ZZ# such that #10^(-n) < x#. Then, for such an #n#, #0.bar(9) + 10^(-n) < 0.bar(9)+x = 1#. However, #0.bar(9)+10^(-n) = 1.000...000999... = 1+(0.bar(9))*10^(-(n+1)) > 1#, a contradiction. Thus the initial assumption was false, meaning #0.bar(9) = 1#.
Using the geometric series formula #sum_(k=0)^oo r^(k)=1/(1-r)# for #|r| < 1#,
#0.bar(9) = 0.9 + 0.09 + 0.009 + ...#
#=sum_(k=1)^oo 9(1/10)^k#
#=-9 + sum_(k=0)^oo9(1/10)^k#
#=-9 + 9sum_(k=0)^oo(1/10)^k#
#=-9+9(1/(1-1/10))#
#= -9 + 9(10/9)#
#=-9 + 10#
#=1#
And so on. Note that the first technique of multiplying by #10# (or #10^n# for longer repeating sequences) and the subtracting is quite useful for figuring out the fractional representation of any repeating decimal.
