1.xn = a0.a1a2 ... an has no ellipsis, so it is limited. The part before n is the same as the part before X=a0.a 1a2a3 ... One ... except the ellipsis, which is smaller than the ellipsis of x ... so Xn is an n-bit underapproximation of X.

2. In fact, insufficient approximation means that all the digits on the decimal point become 0 from n+ 1, so when n increases, Xn must increase because the digits on n+ 1 are always greater than or equal to 0. The approximation of the remainder is to add 1 to the nth digit, which is actually equivalent to replacing all the digits of n+ 1 with 9. When n increases, because the number of n+ 1 bits is at most 9, it must decrease!