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CmSc 180 – Discrete mathematics
Problems on Functions
1. Give an example of a function that is one-to-one but not onto
2. Give an example of a function that is onto but not one-to-one.
3. Give an example of a function that is neither one-to-one nor onto
4. Give an example of a function that is both one-to-one and onto
5. How many functions are there from A = {1,2} to B = {a, b}? Write them as sets
of ordered pairs. Which are one-to-one? Which are onto?
6. Let X = {1, 2, 3, 4}, Y = {a, b, c, d}. For each of the following subsets of X x Y
determine whether it is a function or not. If it is a function, determine whether it is
one-to-one, onto, or both. If it is a bijection, determine its inverse function as a set
of ordered pairs.
A1 = {(1,a), (2,a), (3,c), (4, b)}
A2 = {(1, c), (2, a), (3, b), (4, c), (2, d)}
A3 = {(1, c), (2, d), (3, a), (4, b)}
A4 = {(1, d), (2, d), (4, a)}
A5 = {(1, b), (2, b), (3, b), (4, b)}
7. Do the following sets define functions? If so, give their domain and range:
F1 = {(1, (2,3)), (2, (3,4)), (3, (1,4)), (4, (2,4))}
F2 = {((1,2), 3), ((2,3), 4), ((3,3), 2)}
F3 = {(1, (2,3)), (2, (3,4)), (1, (2,4))}
F4 = {(1, (2,3)), (2, (2,3)), (3, (2,3))}
8. Let N be the set of all non-negative integers. Determine which of the following
functions are one-to-one, which are onto, and which are one-to-one and onto:
a. f: N N
f(n) = n2 + 2
b. f: N N
f(n) = n(mod 3)
c. f: N N
f(n) = 1 if n is odd, 0 if n is even
d. f: N {0, 1} f(n) = 1 if n is odd, 0 if n is even
9. Let X and Y be finite sets. Find a necessary condition for the existence of one-toone mappings from X to Y. Show that if the condition is not present, there are no
one-to-one mappings from X to Y.
10. Let A be a finite set. Show that any function from A to A that is one-to-one must
also be onto and conversely.
11. Show that there exists a one-to-one function from A x B to B x A. Is it also onto?
12. Let g = {(1, a), (2, c), (3, a)} be a function from X = {1, 2, 3} to Y = {a, b, c, d},
and f = {(a, x), (b, x), (c, z), (d, w)} be a function from Y to Z = {w, x, y, z}.
Write g  f as a set of ordered pairs.
13. Let f and g be functions from N to N defined by the equations:
f(n) = 2n + 1, g(n) = 3n -1
Find the compositions f  f , g  g, g  f , and f  g
Hint: f  g = g(f(x))
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14. Let f and g be functions from N to N defined by the equations:
f(n) = n2 , g(n) = 2n
Find the compositions f  f , g  g, g  f , and f  g
Hint: f  g = g(f(x))
15. Represent each function below as a composition of simpler functions
Example:
Let f(x) = 2sin(x + 1)
Consider the functions:
g(x) = x + 1
h(x) = sin(x)
k(x) = 2x
f(x) is the composition (g  h)  k = k(h(g(x))
a. f(x) = log 2 (x2 + 2)
b. f(x) = sin(2x)
c. f(x) = (3 + sin (x) ) 4
d. f(x) = 2sin(x)
e. f(x) = 1 / (cos(x))3
16. Let f = {(a, b), (b, a), (c, b)} be a function from X = {a, b, c} to X. Write f  f and
ff f
17. Let f = {(1, c), (2, a), (3, b), (4, c)}, g = {(a,2), (b,1), (c,4)}, and h = {(1,A), (2,B),
(3, D), (4,D)}. Find g  f,
18. Let f be a function from X to Y, and g be a function from Y to Z. consider f  g –
the function from X to Z (f  g = g(f(x)). Prove or disprove the following
statements:
a. If f is one-to-one then f  g is one-to-one
Not true
Disprove by counterexample:
Let X = {1,2,3}, Y = {a,b,c}, Z = {A,B,C}
f = {(1,a), (2,b), (3,c)}
g = {(a,A), (b,A), (c,A)}
f  g = {(1,A), (2,A), (3,A)} not one-to-one
b. If g is one-to-one then f  g is one-to-one
Not true
Disprove by counterexample:
Let X = {1,2,3,4}, Y = {a,b,c}, Z = {A,B,C}
f = {(1,a), (2,b), (3,c), (4,c)}
g = {(a,A), (b,B), (c,C)}
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f  g = {(1,A), (2,B), (3,C), (4,C)} not one-to-one
c. If f and g are one-to-one then f  g is one-to-one
True:
Proof by contradiction:
Assume that f  g is not one-to-one.
Then there exist x1 and x2 in X, x1  x2, such that g(f(x1)) = g(f(x2))
f is one-to-one, therefore f(x1)  f(x2)
g is one-to-one. Therefore for any y1,y2, y1 y2, g(y1) g(y2).
Therefore g(f(x1))  g(f(x2))
This contradicts the assumption that g(f(x1)) = g(f(x2))
d. If f is not one-to-one then f  g is not one-to-one
True
Proof by contradiction
Assume f  g is one to one.
Then g(f(x1))  g(f(x2)) for all x1 and x2, such that x1  x2
However, f is not one-to one. Hence there is at least one pair x1 and x2,
x1  x2, such that f(x1) = f(x2)
Therefore g(f(x1)) = g(f(x2) (because g is a function)
This contradicts the assumption that g(f(x1))  g(f(x2))
e. If g is not one-to-one then f  g is not one-to-one
Not True
Disproof by counterexample
:
Let X = {1,2,3}, Y = {a,b,c,d}, Z = {A,B,C}
f = {(1,a), (2,b), (3,c)}
g = {(a,A), (b,B), (c,C), (d,C)}
f  g = {(1,A), (2,B), (3,C)} one-to-one
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f. If f  g is one-to-one then f is one-to-one
True
Proof by contradiction.
g.
h.
i.
j.
k.
l.
m.
n.
If f  g is one-to-one then g is one-to-one
If f and g are onto then f  g is onto
If f is onto then f  g is onto
If g is onto then f  g is onto
If f  g is onto then f is onto
If f  g is onto then g is onto
If f  g is not onto then f is not onto
If f  g is not onto then g is not onto
Fill your findings in a table:
f
g
one-toonto
one-toonto
one
one
Y
Y
Y
Y
Y
N
Y
Y
N
Y
Y
Y
N
N
Y
Y
Y
Y
Y
N
Y
N
Y
N
N
Y
Y
N
N
N
Y
N
Y
Y
N
Y
Y
N
N
Y
N
Y
N
Y
N
N
N
Y
Y
Y
N
N
Y
N
N
N
N
Y
N
N
N
N
N
N
f g
one-toonto
one
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