Download Solutions to Eighteen Short Proofs

Fundamentals of Mathematics
December 9, 2010
Solutions to Eighteen Short Proofs
1. Let x, y ∈ R. If xy > 0 and x + y > 0, then x > 0 and y > 0.
Proof. (Contrapositive) Assume that x, y ∈ R and either x ≤ 0 or y ≤ 0. If x ≤ 0 and
y > 0 (or if y ≤ 0 and x > 0), then xy ≤ 0. If x ≤ 0 and y ≤ 0, then x + y ≤ 0. In either
case, xy ≤ 0 or x + y ≤ 0.
2. ere are no integers m and n such that 9n − 6 = 27m.
Proof. Assume, to the contrary, that there exist integers m and n such that 9n − 6 = 27m.
en 9n − 27m = 6, so 9(n − 3m) = 6. Since n − 3m is an integer, this implies that 6 | 9,
which is a contradiction.
3. Let A and B be sets. If B ⊆ A, then A ∪ B = A.
Proof. Assume B ⊆ A.
(⊆) Let x ∈ A ∪ B. en x ∈ A or x ∈ B. Since B ⊆ A, in either case x ∈ A. erefore
A ∪ B ⊆ A.
(⊇) Now let x ∈ A. By the definition of union, x ∈ A ∪ B, so A ⊆ A ∪ B. Finally, we
conclude that A ∪ B = A.
4. Let R be the relation from R to R × R defined by xR(y, z) if x = y. en
(a) R is not a function.
(b) R−1 is a function.
Proof. (a) Counterexample: 1R(1, 1) and 1R(1, 2) which violates the definition of
function.
(b) Note that R−1 is the relation defined from R × R to R by (y, z)R−1 x if x = y. Let
(a, b) ∈ R × R. en there exists c ∈ R such that (a, b)R−1 c. Namely, c = a.
Moreover, c is unique: if (a, b)R−1 c and (a, b)R−1 d, then a = c and a = d, so c = d.
5. ere exists no largest negative real number.
Proof. Assume, to the contrary, that there exists a largest negative real number. Let r be
that number. en r < r/2 < 0, so r/2 is a negative real number that is larger than r. is
is a contradiction.
6. e function f : Z → N given by f (n) = |n| + 1 is not a bijection.
Proof. e counterexample f (2) = 3 = f (−2) shows that f is not one-to-one, so f is not a
bijection.
7. e open interval (3, 6) is uncountable. (You may assume that (0, 1) is uncountable.)
Proof. Let f : (0, 1) → (3, 6) be the function f (x) = 3x + 3. We claim that f is a bijection.
Observe f (x) = f (y) ⇒ 3x + 3 = 3y + 3 ⇒ x = y, so f is one-to-one. Let y ∈ (3, 6).
en x = (y − 3)/3 ∈ (0, 1) and f (x) = 3(y − 3)/3 + 3 = y, so f is onto. erefore (0, 1)
and (3, 6) are numerically equivalent; since (0, 1) is uncountable, (3, 6) is uncountable.
8. e sum of the smallest n odd positive numbers equals n2 .
Proof. (By induction) e smallest odd positive number is 1 and 12 = 1, so the result is true
for n = 1. Assume that the sum of the smallest k odd positive numbers equals k 2 , for
k ≥ 1. at is,
1 + 3 + 5 + · · · + (2k − 1) = k 2
Add 2k + 1 to both sides of the equation:
1 + 3 + 5 + · · · + (2k − 1) + (2k + 1) = k 2 + (2k + 1) = (k + 1)2 .
erefore, the sum of the smallest k + 1 odd positive numbers equals (k + 1)2 .
9. Let x, y ∈ Z. en xy is even if and only if x is even or y is even.
Proof. (⇐) Suppose x is even or y is even. If x is even, then x = 2k for some k ∈ Z, so
xy = 2ky. Since ky ∈ Z, xy is even. If y is even, the argument is similar.
(⇒) (Contrapositive) Suppose x is odd and y is odd. en there exist k, ℓ ∈ Z such that
x = 2k + 1 and y = 2ℓ + 1. en
xy = (2k + 1)(2ℓ + 1) = 4kℓ + 2k + 2ℓ + 1 = 2(2kℓ + k + ℓ) + 1. Since 2kℓ + k + ℓ is an
integer, xy is odd.
10. Let R be the relation defined on Z × Z by (a, b)R(c, d) if a = c. en R is an equivalence
relation. (What are its equivalence classes?)
Proof. (Reflexive) For all (a, b) ∈ Z × Z, (a, b)R(a, b) because a = a.
(Symmetric) Suppose (a, b)R(c, d). en a = c, so c = a and therefore (c, d)R(a, b).
(Transitive) Suppose (a, b)R(c, d) and (c, d)R(e, f ), so a = c and c = e. us a = e and
therefore (a, b)R(e, f ).
e equivalence class of (a, b) is
[(a, b)] = {(x, y) ∈ Z × Z : (a, b)R(x, y)} = {(a, y) : y ∈ Z}.
ere is one equivalence class for every integer a. It consists of all the pairs of integers
where the first coordinate equals a. For example, [(−2, 4)] = {(−2, y) : y ∈ Z}.
11. Let f : A → A be a function. en f ◦ iA = iA ◦ f = f , where iA is the identity function
on A.
Proof. For all x ∈ A, (f ◦ iA )(x) = f (iA (x)) = f (x) and
(iA ◦ f )(x) = iA (f (x)) = f (x).
12. Let x, y, n ∈ Z and let n ≥ 2. If x ≡ y (mod n), then for all k ∈ N, kx ≡ ky (mod nk).
Proof. Assume x ≡ y (mod n). en x − y = ℓn for some integer ℓ. erefore,
kx − ky = k(x − y) = kℓn = (nk)ℓ. Since ℓ is an integer, nk divides kx − ky, so kx ≡ ky
(mod nk).
13. Let A and B be sets. If A and B are denumerable, then A and B are numerically equivalent.
Proof. Assume A and B are denumerable. en there exist bijections f : N → A and
g : N → B. Since f is a bijection, f −1 is a bijection from A to N, and g ◦ f −1 is a bijection
from A to B. erefore A and B are numerically equivalent.
14. For all n ∈ N, 6 | (7n − 1).
Proof. (Induction) Since 71 − 1 = 6, the result is true for n = 1. Assume that 6 | (7k − 1)
for some integer k ∈ N. en 7k − 1 = 6ℓ for some ℓ ∈ Z, so 7k = 6ℓ + 1. Now,
7k+1 − 1 = 7(6ℓ + 1) − 1 = 42ℓ + 6 = 6(7ℓ + 1). Since 7ℓ + 1 ∈ Z, 6 | (7k+1 − 1).
15. ere exists no irrational number whose square root is rational.
√
Proof. Assume,√to the contrary, that there exists an irrational number
r
such
that
r is
√ 2
2
2
rational. en r = p/q for some p, q ∈ Z, where q ̸= 0, so r = ( r) = p /q , which
contradicts the assumption that r is irrational.
16. Every finite nonempty set of real numbers has a largest element. (Hint: Use induction on
the size of the set.)
Proof. Since any set containing one real number has a largest element, the result is true for
sets of size n = 1. Assume that every finite nonempty set of k real numbers has a largest
element, for some k ≥ 1. Let A be a set containing k + 1 real numbers, and let a ∈ A.
en A − {a} is a set with k elements where k ≥ 1. By our inductive hypothesis, A − {a}
has a largest element; call it b. If a > b, then a is the largest element of A and if a < b then
b is the largest element of A. (Note that a = b is impossible because a set A cannot contain
repeated elements.)
17. Let A be a nonempty set and let f : A → P(A) be the function f (x) = {x}. en f is an
injection but not a surjection.
Proof. (Injection) Suppose f (x) = f (y). en {x} = {y}, so x = y.
(Not a surjection) Counterexample: since ∅ ⊆ A, ∅ ∈ P(A). However, there does not exist
x such that f (x) = ∅, since {x} is nonempty.
18. e function f : R − {1} → R − {2} defined by f (x) =
2x−1
x−1
is a bijection.
Proof. (One-to-one) Suppose there exist x, z ∈ R − {1} such that f (x) = f (z). en
2x−1
= 2z−1
⇒ 2xz − z − 2x + 1 = 2xz − x − 2z + 1 ⇒ x = z.
x−1
z−1
. en x ̸= 1. (To prove this, assume, to the
(Onto) For each y ∈ R − {2} let x = y−1
y−2
y−1
contrary, that x = 1. en y−2 = 1 ⇒ y − 1 = y − 2 ⇒ −1 = −2. is is a contradiction,
y−1
so x = y−2
∈ R − {1}.) Moreover,
(
f
y−1
y−2
)
=
2 y−1
−1
y−2
y−1
y−2
−1
=
2y − 2 − y + 2
= y.
y−1−y+2