Mathematics 220 Workshop 2
1. The proofs below contain errors. Please identify these errors, and prove or disprove the original
statement.
(a) Statement: ∃x ∈ R such that for any y ∈ R, y ≥ 0, we have xy = 2x.
Proof: Let x ∈ R, and y = 2. Then, y ≥ 0, and xy = 2x, as we wanted.
√
(b) Statement: For all real numbers x, we have that |x| > x2 − 1.
√
√
√
Proof. Let x ∈ R. Then 0 > −1 =⇒ x2 > x2 − 1 =⇒ x2 > x2 − 1 =⇒ |x| > x2 − 1
Solution:
(a) The proof is proving the statement
∀x ∈ R, ∃y ∈ R, y ≥ 0 such that xy = 2x,
which is not an equivalent statement to what was given. The original statement is
true, and here is a proof of it.
Proof: Let x = 0. Then, let y be any positive real number. Then, xy = 0 = 2x, as
desired.
(b) The error in the proof is at the step that takes square root of both sides of the
inequality. To be able to do so, one needs to ensure that both sides of the inequality
are non-negative, which x2 − 1 is not necessarily if x ∈ R.
The original statement is false: for x = 0, the inequality is not even defined, since we
cannot take square root of negative numbers.
2. Let f : A → B be a function. Let C ⊆ A, D ⊆ B.
(a) Write down the definition of f is injective.
(b) Write down the definition of f is surjective.
(c) Write down the definition of f (C)
(d) Write down the definition of f −1 (D).
3. Prove that the function g : R − 12 → R − {2} defined by g(x) =
4x+2
2x−1
is bijective.
Solution:
We have to show that the function is both injective and surjective.
• Injective. Let x, z ∈ R −
1
2
and assume that g(x) = g(z). Then
4x + 2
4z + 2
=
2x − 1
2z − 1
(2z − 1)(4x + 2) = (4z + 2)(2x − 1)
8xz + 4z − 4x − 2 = 8xz − 4z + 4x − 2
8z = 8x
and so x = z. Thus g is injective.
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1
2
expand
since x, z 6=
clean up
Mathematics 220 Workshop 2
• Surjective. Let y ∈ R − {2} and pick x =
and so x is in the domain of g. Now
y+2
g(x) = g
2y − 4
y+2
+2
4 2y−4
= y+2
2 2y−4
−1
y+2
2y−4
=
1
2
+
4
2y−4 .
Hence x ∈ R and x 6=
1
2
multiply num/denom by 2y − 4
4(y + 2) + 2(2y − 4)
2(y + 2) − (2y − 4)
8y
=
=y
8
=
clean up
Hence for any y in the codomain there is some x in the domain so that g(x) = y and
so g is surjective.
4. Let A be a non-empty set and let h : A → A be a function. Prove that if h ◦ h is bijective then
h is bijective.
Solution:
For convenience write h ◦ h = H and assume it is bijective. We must then show that h is
both injective and surjective.
• Injective. Let x, z ∈ A and assume that h(x) = h(z) = y. Then h(y) = H(x) = H(z).
Since H = h ◦ h is injective, it follows that x = z. Hence h is injective.
• Surjective. Let y ∈ A. Since H is surjective, there exists some z ∈ A so that H(z) = y.
Set x = h(z). Then h(x) = h(h(z)) = H(z) = y. Thus h is surjective.
5. Find an example of functions f : A → B and g : B → C so that f and g ◦ f are injective but g
is not injective.
Solution: Let A = {1} , B = {2, 3} , C = {4}. Then define f and g by
f (1) = 2
g(2) = g(3) = 4
and so
g(f (1)) = g(2) = 4
Then f and g ◦ f are both injective, but since g(2) = g(3), g is not injective.
6. Find an example of functions f : A → B and g : B → C so that g and g ◦ f are surjective but
f is not surjective.
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Mathematics 220 Workshop 2
Solution: Let A = {1} , B = {2, 3} , C = {4}. Then define f and g by
f (1) = 2
g(2) = g(3) = 4
and so
g(f (1)) = g(2) = 4
Then g and g ◦ f are both surjective, but since there is no a ∈ A so that f (a) = 3 ∈ B, it
follows that f is not surjective.
7. Let g : A → C and h : B → C are functions so that h is bijective. Prove that there exists a
function f : A → B so that g = h ◦ f .
Solution:
Since h is bijective, its inverse function h−1 : C → B exists. Now let f = h−1 ◦ g, which is
a function that maps A → B (since it maps A → C → B). Now
h ◦ f = h ◦ (h−1 ◦ g)
= (h ◦ h−1 ) ◦ g
since composition is associative
= iC ◦ g
inverse functions
=g
as required.
8. Define a relation on N by aRb iff a2 ≡ b2 (mod 5) . Show R is an equivalence relation. Show
that [2] = [3].
Solution:
• Reflexive: a2 ≡ a2 (mod 5) as a2 − a2 = 0 and 5|0.
• Symmetric: If a2 ≡ b2 (mod 5), then a2 − b2 = 5k for some k ∈ Z, so b2 − a2 = 5(−k).
So b2 ≡ a2 (mod 5).
• Transitive: If a2 ≡ b2 (mod 5) and b2 ≡ c2 (mod 5), then a2 − b2 = 5k1 , b2 − c2 = 5k2
for some k1 , k2 ∈ Z. So a2 − c2 = 5(k1 + k2 ). So a2 ≡ c2 (mod 5).
Therefore, R is an equivalence relation on N.
To show [2] ⊆ [3]: For any x ∈ [2], xR2 and so x2 − 22 = 5k for some k ∈ Z. Now
x2 − 32 = x2 − 22 − 5 = 5(k − 1). So 5|(x2 − 32 ). Hence xR3. Thus x ∈ [3].
To show [3] ⊆ [2]: For any x ∈ [3], xR3 and so x2 − 9 = 5k for some k ∈ Z. Then
x2 − 4 = 5k + 5 = 5(k + 1). So 5|(x2 − 4) and so xR2. Thus x ∈ [2]. Hence [2] = [3] as
required.
9. (a) Let f : A → B be a function and let D ⊆ B. Prove that
f −1 (B \ D) ⊆ A \ f −1 (D)
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Mathematics 220 Workshop 2
Solution:
Let x ∈ f −1 (B \ D), and hence x ∈ A. Then f (x) ∈ B \ D. Thus f (x) ∈ B and
f (x) 6∈ D. Since f (x) 6∈ D it follows that x 6∈ f −1 (D). Finally since x ∈ A and
x 6∈ f −1 (D) we have x ∈ A \ f −1 (D).
(b) Let f : A → B be an injective function and let C ⊆ A. Prove that
f −1 (f (C)) ⊆ C
Hence ”=” holds as the other direction holds without assuming f is injective, as shown in
class.
Solution: Let f : A → B be injective and let x ∈ f −1 (f (C)). Thus f (x) ∈ f (C).
Since f (x) ∈ f (C), there is some z ∈ C such that f (z) = f (x). Because f is injective,
it follows that x = z. Thus x ∈ C.
10. Prove that 13 + · · · + n3 = (1 + 2 + · · · + n)2 for all n ∈ N.
Solution: n = 1: LHS = 13 = 1, RHS = 12 = 1. True. Assume 13 +· · ·+k 3 = (1+· · ·+k)2 .
For k + 1:
LHS = (1 + · · · + k)2 + (k + 1)3 (by induction hypothesis)
k(k + 1) 2
=
+ (k + 1)3
2
2
k
= (k + 1)2
+k+1
4
1
(k + 1)2 (k + 2)2
=
4
(k + 1)(k + 2) 2
=
2
= (1 + 2 + · · · + k + (k + 1))2
= RHS
By mathematical induction, we are done.
11. Use induction to prove that 81(10n+1 − 9n − 10) for every non-negative integer n.
Solution: We prove the statement by mathematical induction.
• Base case. When n = 0 we have (10 − 0 − 10) = 0 and so 81|(101 − 9 · 0 − 10) as
required.
• Inductive step. Assume that 81|(10k+1 − 9k − 10). Hence there is some q ∈ Z so that
10k+1 − 9k − 10 = 81q
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Mathematics 220 Workshop 2
Now we must show the case n = k + 1 and to do so we consider 810q.
810q = 10k+2 − 90k − 100
and now consider
10k+2 − 9(k + 1) − 10 = (810q + 90k + 100) − 9k − 9 − 10
= 810q + 81k − 81 = 81(10q + k − 1)
Since 10q + k − 1 ∈ Z, we have that 81|(10k+2 − 9(k + 1) − 10) as required.
By mathematical induction, the statement is true for all non-negative integers.
12. Prove (5)1/3 is irrational.
Solution: Assume (5)1/3 is rational. There are m, n ∈ N s.t. m, n have no common factors
other than 1 and (5)1/3 = m/n. Then 5 = (m/n)3 , and so 5n3 = m3 . Hence 5|m3 , and then
5|m, and so m = 5k for some k ∈ N. It follows: 5n3 = (5k)3 = 53 k 3 , and then n3 = 52 k 3 .
Thus, 5|n3 . So 5|n. Therefore, 5 is a common factor of m and n, which contradicts the
choice of m, n. We now conclude (5)1/3 is irrational.
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