abstract algebra: a study guide for beginners - Northern Illinois ...

14 CHAPTER 2. FUNCTIONS
of a function of x if and only if no vertical line intersects the curve more than once.
Explain why this agrees with Definition 2.1.1.
22. The “Horizontal Line Test” from calculus says that a function is one-to-one if and
only if no horizontal line intersects its graph more than once. Explain why this agrees
with Definition 2.1.4.
23. In calculus the graph of an inverse function f −1 is obtained by reflecting the graph of
f about the line y = x. Explain why this agrees with Definition 2.1.6.
24. Show that the function f : R 2 → R 2 defined by f(x, y) = (x 3 +y, y), for all (x, y) ∈ R 2 ,
is a one-to-one correspondence.
25. Define f : R → R by f(x) = x 3 + 3x − 5, for all x ∈ R. Is f a one-to-one function?
Is f an onto function?
Hint: Use the derivative of f to show that f is a strictly increasing function.
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26. Does the following formula define a function from Q to Z? Set f = m, where
n
m, n are integers and n �= 0.
27. Define f : Z12 → Z8 by f([x]12) = [2x]8, for all [x]12 ∈ Z12, and define g : Z12 → Z8
by g([x]12) = [3x]8, for all [x]12 ∈ Z12. Show that f is a function, but g is not.
28. Let a be a fixed element of Z × 17 . Define the function θ : Z× 17 → Z× 17
by θ(x) = ax, for
all x ∈ Z × 17 . Is θ one-to-one? Is θ onto? If possible, find the inverse function θ−1 .
29. For integers m, n, b with n > 1, define f : Zn → Zn by f([x]n) = [mx + b]n.
(a) Show that f is a well-defined function.
(b) Prove that f is a one-to-one correspondence if and only if gcd(m, n) = 1.
(c) If gcd(m, n) = 1, find the inverse function f −1 ,
30. Let f : S → T be a function, and let A, B be subsets of S. Prove the following:
(a) If A ⊆ B, then f(A) ⊆ f(B).
(b) f(A ∪ B) = f(A) ∪ f(B)
(c) f(A ∩ B) ⊆ f(A) ∩ f(B)
31. Let f : S → T be a function. Prove that f is a one-to-one function if and only if
f(A ∩ B) = f(A) ∩ f(B) for all subsets A, B of S.
32. Let f : S → T be a function, and let X, Y be subsets of T . Prove the following:
(a) If X ⊆ Y , then f −1 (X) ⊆ f −1 (Y ).
(b) f −1 (X ∪ Y ) = f −1 (X) ∪ f −1 (Y )
(c) f −1 (X ∩ Y ) = f −1 (X) ∩ f −1 (Y )