Consider the whole number 127. Hence it can be converted into Binary into following steps 127 i)‌127=2(63)+1→1 ii)‌63=2(31)+1→1 iii)‌31=2(15)+1→1 iv)‌15=2(7)+1→1 v)‌7=2(3)+1→1 vi)‌3=2(1)+1→1 vii)‌1=2(0)+1→1 (127)10→(1111111)2 Hence, the binary of 127 is seven times 1 or (1111111). Now for the binary of the number after the point, we get 0.25×2=0.5→0 0.5×2=1.0→1 ∴(0.25)10=(.01)2 ∴(127.25)−10=(1111111.01)2