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INFS 501 Fall 2002 Solutions – HW4
Homework 4: Solutions
Ex. 20 Page 71
One minimal element: 0 (zero)
No maximal element, because for any real x, with 0x<1,
there exists a real y, with 0y<1, and x<y.
Therefore, for any real x, 0x<1, there exists a real y, with y  x and xRy.
Ex. 36 Page 72
(a) Proof:
Let s be an element of S.
If s=x, then xRs by reflexive property.
Let s x. Then there exists s1  S, s1 s, such that s1Rs, for otherwise s is minimal (that is against the fact that
x is a unique minimal element).
if s1=x, then xRs and we are done with the proof.
Let s1  x. Then there exists s2  S, s2  s1, such that s2Rs1, for otherwise s1 is minimal (that is against the fact
that x is a unique minimal element).
Note that s2  s.
Infact if s2 = s, then sRs1, s1Rs.
By antisymmetry, s=s1 (that is against our choice of s1).
Continuing in this manner, at the ith step, we consider an element si  S such that siRsi-1 and si sj for j=i-1, i2, ... , 1 and si  s.
Since S is a finite set, and since at each step we consider an element si  S not considered before, the
number of steps is bounded by the cardinality of S.
In particular, we stop when si = x, since x is the minimal element.
Then we have
xRsi-1
si-1Rsi-2
.
.
.
s1Rs
By the transitive property, we get xRs.
Since s is an arbitrary element of S, statement (a) is proved.
Ex. 40 Page 83
f(x)=3x and g(x) = 1/x
g f(x) = g(f(x)) = g(3x)=1/3x
f g(x) = f(g(x))= f(1/x)=3/x
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INFS 501 Fall 2002 Solutions – HW4
Ex. 42. Page 83
f(x) = 2x g(x) = 5x - x2
g f(x) = g(f(x)) = g(2x) = 5(2x) – (2x) 2
= 5 X 2x - 22x
f g(x) = f(g(x)) = f(5x-x2)=2 (5x-x^2)
Ex 62, Page 84
Y = {-1/2, -1, 1, 2}
g:YY
g(x) = -1/x
g(-1/2) = 2
g(-1) = 1
g(1) = -1
g(2) = -1/2
-1/2˙
˙ 2
-1 ˙
˙ 1
1 ˙
˙-1
2 ˙
˙-1/2
That is:
g = {(-1/2, 2), (-1, 1), (1, -1), (2, -1/2)}
Then,
g-1 = {(2, -1/2), (1, -1), (-1, 1), (-1/2, 2)}
that is:
g-1(x) = -1/x = g(x)
Ex 36 Page 104
f(x) = x3 + 5
f: R  R
f is one to one.
Infact : suppose f(x1) = f(x2).
Then, x13 + 5 = x23 + 5
x13 = x23
then x1 = x2
f is onto.
In fact: let y R and set x = (cube root of (y-5)) R.
Then, f(x) = f(cube root of (y-5)) = y-5+5 = y
Then f-1: X  X exists and f-1 = cube root of (x-5)
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INFS 501 Fall 2002 Solutions – HW4
Ex. 44 Page 104
S = R - {0}
xRy means xy>0
R is reflexive because for all x  S, x2 >0, and therefore xRx
Given x,y  S, if xRy, then xy>0. It follows that yx>0, therefore R is symmetric.
Given x,y,z  S, if xRy and yRz, then xy>0, yz>0.
From xy>0, we have that x and y are both positive or both negative.
If x and y are both positive, then z must also be positive since yz>0.
If x and y are both negative, then z must also be negative since yz>0.
In each case, xz >0.
Therefore xRz holds true.
Hence R is transitive.
If x is a negative real number:
[x] = {y S : y <0} = set of negative real numbers.
If x is a positive real number:
[x] = {y S : y>0} = set of positive real numbers.
In summary, we have two equivalence classes, they are
(i)
the set of negative real numbers and
(ii)
the set of positive real numbers.
Ex. 54 Page 104
S=set of people
given x,y  S, xRy means x=y or x is a descendant of y
For each x  S xRx since x=x, therefore R is reflexive.
Given x,y  S,
suppose xRy and yRx, then (x=y or x is a descendant of y) and (y=x or y is a descendant of x)
This means that, either x=y or
(x is a descedant of y) and (y is a descendant of x).
But (x is a descendant of y) and (y is a descendant of x) can hold true only when x and y are the same
person, i.e. when x=y. It follows that R is antisymmetric.
Given x,y,z  S,
suppose xRy, yRz, then (x=y or x is a descendant of y) and (y=z or y is a descendant of z)
We have four possibilities:
(1) x=y and y is a descendant of z. In this case, clearly x is a descendant of z, therefore xRz.
(2) x=y, y=z. In this case x=z, therefore xRz.
(3) x is a descendant of y and y=z. Clearly, x is a descendant of z, therefore xRz.
(4) x is a descendant of y and y is a descendant of z. It follows that x is a descendant of z.
We conclude that R is transitive.
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INFS 501 Fall 2002 Solutions – HW4
Exercise 1 (not in textbook)
X = {1,2,3}
Y = {1,2,3,4}
Z = {1,2}
(a) Function f: X  Y that is one to one, but not onto.
f = {(1,1), (2,2), (3,3)}
(b) Function g: X  Z that is onto, but not one to one.
g = {(1,1), (2,2), (3,2)}
(c) Function h: X  X that is neither one to one nor onto.
h = {(1,1), (2,2), (3,2)}
(d) Function k: X  X that is one to one and onto but is not the identity function on X.
k = {(1,2), (2,3), (3,1)}
Exercise 2 (not in textbook)
(a) f is not one to one.
Let A = {a} and B = {b}
Then,
f(A) = 1
f(B) = 1
but A  B
(b) f is not onto
The number 5  Z but f(A)  5 for any set A in S, because no subset of {a,b,c,d} has five elements.
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