2.2 28
在有四個 sets 的 Venn Diagram 中，應該用以下兩種圖表示，才能夠表示出所有
可能的狀況。
若以以下這種圖表示的話，則無法包含只有[紅、綠]、以及只有[藍、黃]這兩個
Set 的交集(即減少兩個可能的 Dimensions)。
參考資料：http://en.wikipedia.org/wiki/Venn_diagram
2.2 52
a) No elements are included, so this is the empty set.
b) All elements are included, so this is the universal set.
2.3
54
Since a byte is eight bits, all we are asking for in each case is n / 8 , where n is the
number of bits. Therefore,
a) 4 / 8  1
2.3
b) 10 / 8  2
c) 500 / 8  63
d) 3000 / 8  375
70
a) This is true. Since  x  is already an integer,   x     x  .
b) A little experimentation shows that this is not always true. To disprove it, we need
only produce a counterexample, such as x  y  3 / 4 . In this case the left-hand
side is 3 / 2  1, while the right-hand side is 0 + 0 = 0.
c) A little trial and error fails to produce a counterexample, so maybe this is true. We
look for a proof. Since we are dividing by 4, let us write x = 4n + k, where
0  k  4 . In other words, write x in terms of how much it exceeds the largest
multiple of 4 not exceeding it. There are three cases. If k = 0, then x is
n  1/ 2  n  1 . Of course the right-hand side is n + 1 as well, so again the two
side agree. Finally, suppose that 2  k  4 . Then  x / 2  2n  2 , and the
left-hanc side is n  1  n  1 ; of course the right-hand side is still n + 1, as well.
Since we proved that the two sides are equal in all cases, the proof is complete.
d) For x = 8.5, the left-hand side is 3, whereas the right-hand side is 2.
e) This is true. Write x  n   and y  m   , where n and m are integers and ε and δ
are nonnegative real numbers less than 1. The left-hand side is n + m + (n + m + 1),
the latter occurring if and only if     1. The right-hand side is the sum of two
quantities. The first is either 2n ( if   1/ 2 ) or 2n + 1 ( if   1/ 2 ). The only way,
then, for the left-hand side to exceed the right-hand side is to have the left-hand
side be 2n + 2m + 1 and the right-hand side be 2n + 2m. This can occur only if
    1 while   1/ 2 and   1/ 2 . But that is an impossibility, since the sum of
two numbers less than 1/2 cannot be as large as 1. Therefore, the right-hand side is
always at leat as large as the left-hand side.