Try the Free Math Solver or Scroll down to Tutorials!












Please use this form if you would like
to have this math solver on your website,
free of charge.

Worksheet 11 - Hints and Selected Solution

1. For which sets A is it the case that the only bijection from A to A is the identity?

Answer. For sets A consisting of exactly one element, as well as for the empty set.

2. Let A be the set of the days of the week. Let f assign to each day of the week the number of letters
in its name in English. Is f an injection?

Answer. No. f(Monday) = f(Sunday) but Monday ≠ Sunday.

3. Let A be the set of subsets of that have even size, and let B be the set of subsets of
that have odd size. Establish a bijection from A to B. ()

Proof. Let A be the collection of even-sized subsets of , and let B be the collection of
odd-sized subsets. For each x ∈ A, define f(x) as follows:

By this definition, #(x) and #(f(x)) differ by one, so f(x) is a set of odd size and f maps A into

We claim first that f is one-to-one. Consider distinct x, y ∈ A. If both contain or both omit n,
then f(x) and f(y) agree on whether they contain n but differ outside {n}. If exactly one of {x, y }
contains n, then exactly one of {f(x), f(y)} contains n. Thus f(x) ≠ f(y), and f is one-to-one.
We next claim that f is onto. If z ∈ B, then flipping whether n is present in z yields a subset x
such that f(x) = z, so f is also onto, and hence f is a bijection.

4. Verify that defines a bijection from the interval [0, 1] to R. (Hint: In the proof that
f is surjective, use the quadratic formula.) ()

Proof. f is injective Suppose that f(x) = f(y). Then we obtain that

(2x - 1)2y(1 - y) = (2y - 1)2x(1 - x),

which simplifies further to 2(y2 - x2) - 2(y - x) = 4xy(y - x). If y ≠ x, then we can divide
by 2(y - x) to obtain y + x - 1 = 2xy. Rewriting this as -xy = (x - 1)(y - 1) makes it clear
that there is no solution when x, y ∈[ 0, 1], since the left side is negative and the right side is
positive. Therefore we conclude that x = y, and so f is injective.

f is surjective Suppose that f(x) = b. We solve for x to obtain x ∈[0, 1] such that f(x) = b.
Observe that b = 0 is achieved by , so we may assume that b ≠ 0. Clearing fractions
leads to , or . The quadratic formula yields

The magnitude of the square root is larger than lbl. Therefore, choosing the negative sign in
the numerator yields a negative x, which is not in the domain of f. We therefore choose the
positive sign.

If b > 0, then the square root is less than b + 1, and we obtain . Also, the
square root is bigger than 1, so x > 0. If b < 0, then let b' = -b. The formula for x becomes
, where b' > 0. the square root is strictly between 1 and b' + 1, so x is strictly
between 1/2 and 0. In each case, we have found x in the domain [0, 1] such that f(x) = b, and
we have thus established that f is onto.

5. Let f and g be surjections from R to R, and let h = fg be their product. Must h also be surjective?
Give a proof or a counterexample.

Counterexample. Let f, g : R → R be defined by f(x) = g(x) = x. Then both f and g are
surjective, but (f ยท g)(x) = x2 is not surjective.

6. Determine which formulas below define surjections from onto :

(a) f(a, b) = a + b.

Answer. Not surjective. (Why?)

(b) f(a, b) = ab.

Answer. Surjective. (Why?)

(c) f(a, b) = ab(b + 1)/2.

Answer. Surjective. (Why?)

(d) f(a, b) = (a + 1)b(b + 1)/2.

Answer. Not surjective. (Why?)

(e) f(a, b) = ab(a + b)/2.

Answer. Not surjective (Why?)

7. Given f : R → R, suppose that there are positive constants c, α  such that for all x, y ∈ R,
. Prove that f is injective.

Proof. Suppose that f is not injective. Then there exist distinct numbers x and y such that
f(x) = f(y). Since , this contradicts the hypothesis.

8. If f : A → B and g : B → C are bijections, prove that .

Proof. We use exercise 11. Since f and g are bijections, both are invertible. Note that


9. Given f : A → B and g : B → C, let h = g o f. Determine which of the following statements are
true. Give proofs for the true statements and counterexamples for the false statements.

(a) If h is injective, then f is injective.

Answer. True. (Why?)

(b) If h is injective, then g is injective.

Answer. False. (Why?)

(c) If h is surjective, then f is surjective.

Proof. False. (Why?)

(d) If h is surjective, then g is surjective.

Proof. True. (Why?)

10. Consider f : A → B and g : B → A. Answer each question below by providing a proof or a

(a) If f(g(y)) = y for all y ∈ B, does it follow that f is a bijection?
Proof. No. (Why?)

(b) If g(f(x)) = x for all x ∈ A, does it follow that f(g(y)) = y for all y ∈ B?
Proof. No. (Why?)

11. Consider f : A → B and g : B → A. Prove that if f o g and g o f are identity functions, then f is
a bijection. In particular, prove that

(a) If f o g is the identity function on B, then f is surjective.

Proof. If y ∈ B then f(g(y)) = y. Hence there is an element of A mapped to y by f, namely,
the element g(y). This shows that f satisfies the definition of a surjection.

b) If g o f is the identity function on A, then f is injective.

Proof. If x ∈ A, we have that g(f(x)) = x. If f is not injective, then there exist distinct
elements ∈ A such that . If we apply g to both sides of the equality,
we obtain , which contradicts our choice of distinct elements.
Hence our assumption that f is not injective must be wrong.

12. Consider f : A → A. Prove that if f o f is injective, then f is injective.

Proof. Suppose that f(x) = f(y). Then f(f(x)) = f(f(y)). By the definition of composition, this
yields (f o f)(x) = (f o f)(y). The hypothesis that f o f is injective yields now that x = y. We
have proved that f(x) = f(y) implies x = y, and so f is injective.

13. Construct an explicit bijection from the closed interval [0, 1] onto the open interval [0, 1]. ()

Hint. This is a very nice problem. In short, no hint!

14. Consider the function g : R → R defined by

where for any x ∈ R, [x ] denotes the largest natural number less than or equal to x. (For example,
, and .) Let f : N → Q be the function defined by the following

Then prove that f is a bijection between N and , where denotes the set .
(At least convince yourself that this is true!)

Remark. This last exercise should be a final project.