离散部分作业答案

P125 21.
a)N={1,3} Φ(4)=2
b)N={1,3,7,9}Φ(10)=4
c)N={1,2,3,…,12}Φ(13)=12
P149 25
b) 5^2003 mod 1001=983
P183 4
“the sailing race will be held and the lifesaving demonstration will go on”: this sentence should be split into two propositions ,one is the “sailing race will be held “, the other is “the lifesaving demonstration will go on.”
P202 64
N is odd and not divisible by 3 , so we can express n=6k+1 or 6k+5(k>=1) Basis step: It is easy to prove the correction of the proposition when n=7 and n=11.
Inductive step: assume that if n=6k+1, the proposition is true,
and if n=6(k+1)+1=(6k+1)+6, as shown in the figure below :
the square is split into 4 parts as
a)(6k+1)*(6k+1), b) 7*7-1 ( please
allow me to write this way), c)6k*6 and
d)6*6k . it is clear that a) is fit to the
proposition , b) can be filled with
L-shaped pieces and c),d) can be filled with a rectangle consist of 2 L-shaped pieces.
If n=6(k+1)+5 , the proposition can be also proved as above.
So , we can believe that the proposition is true.
P259 49
a)We need to find something to count so that the left-hand side of the
equation counts it in one way and the right-hand side counts it in different way. After much thought , we might try the following. We will count the number of bit strings of length n+r+1containing exactly r 0’s and n+1 1’s . There are C(n+r+1,r) such strings , since
a string is completely specified by deciding which r of the bits are
to be the 0’s . To see that the left-hand side of the identity counts the same thing,let l+1 be the position of the last 1 in the string . Since there are n+1 1’s , we know that l cannot be less than n . thus there are disjoint cases for each l from n to n+r . For each such l, we completely determine the string by deciding which of the l positions
I the string before the last 1 are to be 0’s . Since there are n 1’s in
this range, there are l-n 0’s. Thus there are C(l,l-n) .ways to choose the positions of the 0’s . Now by the sum rule , the total number of bit strings will be ∑(l=n to n+r)C(l,l-n) . By making the change of variable k=l-n,this transforms into the left-hand side, and we are
finished.
P243 33
a)there can clearly be no one-to-one function from {1,2,…n} to {0,1}
if n>2. if n=1 ,then there are 2 such functions , the one that sends 1 to 0, and the one that sends 1 to 1. if n=2 , then there are again 2 such functions since once it is determined where 1 is sent , the function must send 2 to the other value in the codomain.
b)If the function assigns 0 to both 1 and n , then there are n-2 function
values free to be chosen. Each can be chosen in 2 ways. Therefore by product rule, there are 2^n-2 such functions ,as long as n>1. if n=1 ,then clearly there is just one such function.
c)If n=1, then there are no such functions.
If n>1, we will decide which of the numbers from 1 to n-1 , inclusive,will get sent to 1. there are n-1 ways to make this choice.finally , we are free to specify the value if the function at n,and this may be done in 2ways ,hence by the product rule ,the answer is 2(n-1).
P383 25
b)for each unordered pair {a.b} of distinct elements of A, we have a
3-way choice----either include (a,b) only, include (b,a) only , or
include neither .For each element of A we have a 2-way choice .
Therefore the answer is (3^C(n,2))*(2^n) .
c)as in part (b) we have a 3-way choice for a!=b . there is no choice
about including (a,a) in the relation——the definition prohibits it.
Therefore the answer is 3^C(n,2)。