The 0.bar(9)=10.¯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)x=0.¯9
=> 10x = 9.bar(9)⇒10x=9.¯9
=> 10x - x = 9.bar(9) - 0.bar(9)⇒10x−x=9.¯9−0.¯9
=> 9x = 9⇒9x=9
=> x = 1⇒x=1
Using the equality 0.bar(3) = 1/30.¯3=13,
0.bar(9) = 0.bar(3)+0.bar(3)+0.bar(3)0.¯9=0.¯3+0.¯3+0.¯3
=1/3+1/3+1/3=13+13+13
=1=1
Suppose 0.bar(9)!=10.¯9≠1. Then x = 1-0.bar(9) > 0x=1−0.¯9>0. Because 10^(-n)->010−n→0 as n->oon→∞, 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.
