x is an integer such that 1≤x≤9, Prove that 0.(0x)recurring=x/99
r=0.0.x.
r=0.0x0x0x0x....
100r=x.0x0x0x (1)
10,000r=x0x.0x0x0x (2)
(2) - (1): 9,900r=x00
r=x00/9,990 r=x/99
EE
r=0.0.x.
r=0.0x0x0x0x....
100r=x.0x0x0x (1)
10,000r=x0x.0x0x0x (2)
(2) - (1): 9,900r=x00
r=x00/9,990 r=x/99