COT 3100 Discrete Mathematics
HW #4 Solutions
Section 1.7
3. Proof: If x≥y, then max(x,y)=x, min(x,y)=y. Thus max(x,y)+min(x,y)=x+y.
If x<y, then max(x,y)=y, min (x,y)=x. Thus max(x,y)+min(x,y)=x+y.
In either case, max(x,y)+min(x,y)=x+y holds.
4.Proof: If a≤b and a≤c, b≤c , min(a,min(b,c))= min(a,b)=a.
min(min(a,b),c)=min(a,c)=a.
If a≤b and a≤c, b>c , min(a,min(b,c))= min(a,c)=a.
min(min(a,b),c)=min(a,c)=a.
If b≤a and b≤c, a≤c , min(a,min(b,c))= min(a,b)=b.
min(min(a,b),c)=min(b,c)=b.
If b≤a and b≤c, a>c , min(a,min(b,c))= min(a,b)=b.
min(min(a,b),c)=min(b,c)=b.
If c≤a and c≤b, a≤b , min(a,min(b,c))= min(a,c)=c.
min(min(a,b),c)=min(a,c)=c.
If c≤a and c≤b, a>b , min(a,min(b,c))= min(a,c)=c.
min(min(a,b),c)=min(b,c)=c.
5.Proof: Since | || |
we have | |
Meanwhile | |
| |
| |≥|
|
|
,
| || |
| |
| |
|
| .
,
| must hold.
6. Proof: Since 1+2=3, 3 satisfied the requirement, which indictes that the statment is true.
This is a constructive proof.
11. Proof: If
√
is irrational, then we can set
and
√ .
Suppose √ is rational, then we set x=
irrational.
15. Proof: Suppose there is a c satisfied |
Hence
Since
=2
,
|=|
√
and
√ /4. We have
|, then |
|
|
√ which is
| .
.
2 must holds. Then =
.
Also, since a and b are odd, c is an integer.
Hence such c is unique.
40. Proof: This is easily done, by laying the dominoes horizontally, three in the first and last rows and
four in each of the other six rows.
Section 2.1
2．a) {3n|n=0, 1, 2, 3, 4}
b) {x|-3≤x≤3, x is integer}
c) {x| x is a letter of the word monopoly other than l or y}.
3．a) Yes. Order and repetition do not matter.
b) No. The first set has one element and the second has two.
c) No. The first set has none element and the second has one.
4. Each of the sets is a subset of itself. Aside from that, the only relations are
8. a) True .
b) True.
c) False.
d) True.
e) True.
f) True.
g) False.
and
.
13.
A
B
C
U
18. a) 0.
b) 1.
c) 2.
d) 3.
23. a) {(a, y), (a, z), (b, y), (b, z), (c, y), (c, z), (d, y), (d, z)}.
b) {(y, a), (y, b), (y, c), (y, d), (z, a), (z, b), (z, c), (z, d)}.
36. a) Truth set is {
|
b) Truth set is .
c) Truth set is {
|
Bouns
48. We will use a coloring of the
showing that no such tiling exists.
board with four colors as the basis for a proof by contradicion
Assume that 25 straight tetrominoes can cover the board. Some will be placed horizontally and some
vertically. Because there is an odd number of tiles, the number placed horizontally and the number
placed vertically can not both be odd, so assume without loss of generality that an even number of
tiles are placed horizontally.
Color the squares in order using the colors red, blue, green, yellow in that order repeatedly, starting in
the upper left corner and preceeding row by row, from left to right in each row. Then it is clear that
every horizontally placed tile covers one square of each color and each vertically placed tile covers
either zero or two squares of each color. It follows that in this tiling an even number of squares of
each color are covered. But this contradicts the fact that there are 25 squares of each color. Therefore
no such coloring exists.
38. a) If
S, then by the defining condition for S we conclude that
S, a contradiction.
b) If
S, then by the defining condition for S we conclude that it is not the case that
contradiction.
S, again