| Suppose the address of x is a1a2...an(0infinity). |
| Then x = a1/2 + a2/4 + ... + an/2n =
(2n-1a1 + 2n-2a2 + ... + an)/2n,
a fraction with denominator a power of 2. |
| Now suppose the address of x is a1a2...an(1infinity)
with an = 0. From our study of non-unique addresses in problem 5, we see an equivalent address for x
is a1a2...an-11(0infinity). This is the first case above. |