Download Homework 4 - Solutions

Some problems on: Relations, Partial Orders and Functions
Problem 1
Identify the minimal and maximal elements of S with respect to the given partial order R.
S is the set of all real numbers x such that 0<= x < 1 and x R y iff x <= y.
SOLUTION:
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.
Problem 2
Find a subset Y of the set of the real numbers such that g: Y -> Y defined
by g(x) = -1/x is a one-to-one correspondence. Then compute the
function inverse of g.
SOLUTION:
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)
Problem 3
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Let X be the set of real numbers. Compute the inverse of the following function f: X -> X
if it exists: f(x) = x3 + 5
SOLUTION:
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)
Problem 4
Show that the following relation R is an equivalence relation on S. Then describe
the distinct equivalence classes of R.
S = R - {0}
For all x,y in S:
xRy means xy>0
SOLUTION:
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.
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