DEPARTMENT OF MATHEMATICS AND STATISTICS
MATH 1020 Mathematical Foundations
Practice Exam Questions
Winter 2023
1. Write the symbolic form of the proposition “At least one subset of the real numbers has a smallest
element.”
Solution.
∃S ∈ P(R), ∃a ∈ S, ∀x ∈ S, a ≤ x.
2. Write the symbolic form of the proposition “There is no largest natural number.”
Solution.
∀x ∈ N, ∃y ∈ N, x < y.
3. Write the symbolic form of the proposition “Between any two distinct real numbers there is a rational
number.”
Solution.
∀x, y ∈ R, if x 6= y, then ∃z ∈ Q, x < z < y or y < z < x.
4. Write the symbolic form of the proposition “The cube of every even integer is even.”
Solution.
∀x ∈ Z, if ∃a ∈ Z, x = 2a, then ∃b ∈ Z, x3 = 2b.
5. Write the symbolic form of the proposition “The sum of a rational number and an irrational number
is always irrational.”
Solution.
∀x, y ∈ R, if x is rational and y is irrational, then x + y is irrational.
6. Write the negation of the proposition “The sum of a rational number and an irrational number is
always irrational.”
Solution.
∃x, y ∈ R, x is rational and y is irrational and x + y is rational.
7. Write the negation of the proposition ∀a, b ∈ R, if ∀x ∈ (a, ∞), b ≤ x, then b ≤ a.
Solution.
∃a, b ∈ R, ∀x ∈ (a, ∞), b ≤ x, and a < b.
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8. Write the negation of the proposition ∀a ∈ R, if (∀x ∈ R, if x > 0, then a ≤ x), then a ≤ 0.
Solution.
∃a ∈ R, (∀x ∈ R, if x > 0, then a ≤ x), and 0 < a.
9. Write the negation of the proposition ∀y ∈ R, ∃x ∈ R, x > y 2 + 1.
Solution.
∃y ∈ R, ∀x ∈ R, x ≤ y 2 + 1.
10. Write the negation of the proposition ∀x ∈ N, if ∃n ∈ N, x2 = n3 , then ∃m ∈ N, x = m3 .
Solution.
∃x ∈ N, ∃n ∈ N, x2 = n3 and ∀m ∈ N, x 6= m3 .
11. Let a ∈ R. Write the contrapositive form of the following implication:
if ∃x ∈ R, a + x = x, then ∀y ∈ R, a + y = y.
Solution.
if ∃y ∈ R, a + y 6= y, then ∀x ∈ R, a + x 6= x.
12. Prove, using a proof by contraposition: ∀a ∈ R, if (∀x ∈ R, if x > 0, then a ≤ x), then a ≤ 0.
Proof.
Let a ∈ R.
Assume 0 < a.
Put x = a2 .
Since 0 < a, we have 0 < a2 ; hence x > 0.
Also, since 1 < 2 and 0 < a, we have a < 2a; hence a2 < a.
We now have x > 0 and x < a.
Therefore, ∃x ∈ R, x > 0 and x < a.
Therefore, if 0 < a, then ∃x ∈ R, x > 0 and x < a.
Therefore, if (∀x ∈ R, if x > 0, then a ≤ x), then a ≤ 0.
Therefore, ∀a ∈ R, if (∀x ∈ R, if x > 0, then a ≤ x), then a ≤ 0.
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13. Prove, using only the axioms of the real numbers, ∀x, y, z ∈ R, if x ≤ y and y < z, then x < z.
Proof.
Let x, y, z ∈ R.
Assume x ≤ y and y < z.
Since x ≤ y means x < y or x = y, we consider two cases.
Case 1: x < y.
In this case, since x < y and y < z, we have x < z.
Case 2: x = y.
In this case, since y < z, we have x < z.
Therefore, if x ≤ y and y < z, then x < z.
Therefore, ∀x, y, z ∈ R, if x ≤ y and y < z, then x < z.
14. Prove ∀a ∈ R, if ∀x ∈ R, ax ≤ x, then a = 1.
Proof.
Let a ∈ R.
Assume ∀x ∈ R, ax ≤ x.
Since 1 ∈ R, we have a(1) ≤ 1; hence a ≤ 1.
Since −1 ∈ R, we have a(−1) ≤ −1; hence 1 ≤ a.
Now, since a ≤ 1 and 1 ≤ a, we have a = 1.
Therefore, if ∀x ∈ R, ax ≤ x, then a = 1.
Therefore, ∀a ∈ R, if ∀x ∈ R, ax ≤ x, then a = 1.
15. Prove ∀a ∈ R, if ∀x ∈ R, ax ≤ 0, then a = 0.
Proof.
Let a ∈ R.
Assume ∀x ∈ R, ax ≤ 0.
Since 1 ∈ R, we have a(1) ≤ 0; hence a ≤ 0.
Since −1 ∈ R, we have a(−1) ≤ 0; hence 0 ≤ a.
Since a ≤ 0 and 0 ≤ a, we have a = 0.
Therefore, if ∀x ∈ R, ax ≤ 0, then a = 0.
Therefore, ∀a ∈ R, if ∀x ∈ R, ax ≤ 0, then a = 0.
16. Prove ∀x ∈ Z, if 10 divides x and 4 divides x, then 20 divides x.
Proof.
Let x ∈ Z.
Assume 10 divides x and 4 divides x.
Choose a, b ∈ Z with x = 10a and x = 4b.
Put c = b − 2a.
x = 5x − 4x = 5(4b) − 4(10a) = 20b − 40a = 20(b − 2a) = 20c.
Therefore, ∃c ∈ Z, x = 20c.
This means 20 divides x.
Therefore, if 10 divides x and 4 divides x, then 20 divides x.
Therefore, ∀x ∈ Z, if 10 divides x and 4 divides x, then 20 divides x.
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17. Using a proof by contradiction, prove ∀x, y ∈ Z, if x is odd and xy is even, then y is even.
Proof.
Let x, y ∈ Z.
Assume x is odd and xy is even and y is not even.
Since y is not even, y is odd.
Choose a, b, c ∈ Z with x = 2a + 1 and xy = 2b and y = 2c + 1.
Then (2a + 1)(2c + 1) = 2b; 4ac + 2a + 2c + 1 = 2b; 1 = 2(b − 2ac − a − c).
Since 0 < 1 and 0 < 2, we must have 0 < b − 2ac − a − c. It follows that 1 ≤ b − 2ac − a − c.
We then have 2 ≤ 2(b − 2ac − a − c); hence 2 ≤ 1. This is a contradiction, since 1 < 2.
Therefore, if x is odd and xy is even, then y is even.
Therefore, ∀x, y ∈ Z, if x is odd and xy is even, then y is even.
18. Prove ∀x, y ∈ R, if min(x, y) < |x − y|, then 2 min(x, y) < max(x, y).
Proof.
Let x, y ∈ R.
Assume min(x, y) < |x − y|.
Case 1: x < y.
In this case, min(x, y) = x and max(x, y) = y.
Also, in this case x − y < 0, so |x − y| = y − x.
By assumption, we then have x < y−x, which gives us 2x < y. Thus, 2 min(x, y) < max(x, y).
Case 2: y < x.
In this case, min(x, y) = y and max(x, y) = s.
Also, in this case 0 < x − y, so |x − y| = x − y.
Our assumption then becomes y < x − y, which implies 2y < x. That is, 2 min(x, y) <
max(x, y).
Case 3: x = y.
In this case, min(x, y) = x and max(x, y) = x.
Also, |x − y| = |0| = 0.
Our assumption then gives us x < 0.
Adding x to both sides gives 2x < x, which means 2 min(x, y) < max(x, y).
Therefore, if min(x, y) < |x − y|, then 2 min(x, y) < max(x, y).
Therefore, ∀x, y ∈ R, if min(x, y) < |x − y|, then 2 min(x, y) < max(x, y).
19. Prove ∀x, y ∈ R, min(x, y) = 12 (x + y − |x − y|).
Proof.
Let x, y ∈ R.
Case 1: x ≤ y.
In this case, min(x, y) = x.
Also, in this case x − y ≤ 0, so |x − y| = y − x.
Now, min(x, y) = x = 21 (x + x) = 12 (x + y − (y − x)) = 21 (x + y − |x − y|).
Case 2: y < x.
In this case, min(x, y) = y.
Also, in this case 0 < x − y, so |x − y| = x − y.
min(x, y) = y = 21 (y + y) = 12 (x + y − (x − y)) = 21 (x + y − |x − y|).
Therefore, min(x, y) = 21 (x + y − |x − y|).
Therefore, ∀x, y ∈ R, min(x, y) = 12 (x + y − |x − y|).
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20. Prove ∀a, x, y ∈ R, if a < max(x, y), then a < x or a < y.
Proof.
Let x, y, a ∈ R.
Assume a < max(x, y).
Case 1: x ≥ y.
In this case, max(x, y) = x, and hence a < x.
Therefore, a < x or a < y.
Case 2: x < y.
In this case, max(x, y) = y, so a < y.
Again, a < x or a < y is true.
Therefore, if a < max(x, y), then a < x or a < y.
Therefore, ∀a, x, y ∈ R, if a < max(x, y), then a < x or a < y.
21. Prove ∀x, y ∈ B, if x ∧ y = F , then ¬x ∨ y = ¬x.
Proof.
Let x, y ∈ B.
Assume x ∧ y = F .
Then ¬x ∨ (x ∧ y) = ¬x ∨ F .
This gives us (¬x ∨ x) ∧ (¬x ∨ y) = ¬x. So, T ∧ (¬x ∨ y) = ¬x.
Thus, ¬x ∨ y = ¬x.
Therefore, if x ∧ y = F , then ¬x ∨ y = ¬x.
Q.E.D.
22. Prove ∀x, y, z ∈ B, if x ⇒ y and ¬z ⇒ x, then y ∨ z = T .
Proof.
Let x, y, z ∈ B.
Assume x ⇒ y and ¬z ⇒ x.
Then ¬z ⇒ y by transitivity. This means ¬z ∧ y = ¬z. Then (¬z ∧ y) ∨ z = ¬z ∨ z.
Expanding, we have (¬z ∨ z) ∧ (y ∨ z) = T , and so T ∧ (y ∨ z) = T .
Thus, y ∨ z = T .
Therefore, if x ⇒ y and ¬z ⇒ x, then y ∨ z = T .
Q.E.D.
Alternate Proof.
Let x, y, z ∈ B.
Assume x ⇒ y and ¬z ⇒ x.
Since ¬z ⇒ x, we have ¬x ⇒ z by contraposition.
Now, x ∧ y = x and ¬x ∧ z = ¬x.
Then (x ∧ y) ∨ (¬x ∧ z) = x ∨ ¬x. Thus, (x ∧ y) ∨ (¬x ∧ z) = T .
This gives us y ∨ (x ∧ y) ∨ (¬x ∧ z) = y ∨ T .
By absorption, we then have y ∨ (¬x ∧ z) = T .
Now, y ∨ (¬x ∧ z) ∨ z = T ∨ z.
By absorption again, we have y ∨ z = T .
Therefore, if x ⇒ y and ¬z ⇒ x, then y ∨ z = T .
Q.E.D.
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23. Prove ∀a ∈ R, if 1 ≤ a, then ∀n ∈ N, a ≤ an .
Proof.
Let a ∈ R.
Assume 1 ≤ a
Let A = {n ∈ N | a ≤ an }.
Since a = a1 , we have a ≤ a1 . Thus, 1 ∈ A.
Let n ∈ N, and assume n ∈ A.
Then a ≤ an .
Since 0 < 1 and 1 ≤ a and a ≤ an , we have 0 ≤ an .
Now, since 1 ≤ a and 0 ≤ an , we have an (1) ≤ an (a). That is, an ≤ an+1 .
Finally, since a ≤ an and an ≤ an+1 , we have a ≤ an+1 .
Thus, n + 1 ∈ A.
Therefore, ∀n ∈ N, if n ∈ A, then n + 1 ∈ A.
By the Principle of Mathematical Induction N ⊆ A.
Therefore, ∀n ∈ N, a ≤ an .
Therefore, if 1 ≤ a, then ∀n ∈ N, a ≤ an .
Therefore, ∀a ∈ R, if 1 ≤ a, then ∀n ∈ N, a ≤ an .
24. Prove ∀x ∈ N, 7 divides 8x − 1.
Proof.
Let A = {x ∈ N | 7 divides 8x − 1}.
Letting q = 1 gives us 81 − 1 = 7 = 7q.
Therefore, ∃q ∈ Z, 81 − 1 = 7q. That is, 7 divides 81 − 1, and hence 1 ∈ A.
Let n ∈ A. Then 7 divides 8n − 1.
Accordingly, let p ∈ Z with 8n − 1 = 7p.
Let k = 8n + p
8n+1 − 1 = 8(8n ) − 1
= (7 + 1)(8n ) − 1
= 7(8n ) + 8n − 1
= 7(8n ) + 7p
= 7(8n + p)
= 5k
Therefore, ∃k ∈ Z, 8n+1 − 1 = 7k, and so n + 1 ∈ A.
Therefore, ∀n ∈ N, if n ∈ A, then n + 1 ∈ A.
By the Principle of Mathematical Induction N ⊆ A.
Therefore, ∀x ∈ N, 7 divides 8x − 1.
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25. Prove ∀x ∈ Z, if x is odd then ∀n ∈ N, xn is odd.
Proof.
Let x ∈ Z.
Assume x is odd.
Choose a ∈ Z with x = 2a + 1.
Let A = {n ∈ N | xn is odd}.
Since x1 = x and x is odd, we have 1 ∈ A.
Let n ∈ N.
Assume n ∈ A.
Hence xn is odd.
Choose b ∈ Z with xn = 2b + 1.
Then xn+1 = xn x = (2b + 1)(2a + 1) = 4ab + 2a + 2b + 1
Putting c = 2ab + a + b then gives us xn+1 = 2c + 1.
Therefore, xn+1 is odd.
Hence n + 1 ∈ A.
Therefore, if n ∈ A then n + 1 ∈ A.
Therefore, by the Principle of Mathematical Induction, N ⊆ A.
Therefore, ∀n ∈ N, xn is odd.
Therefore, if x is odd, then ∀n ∈ N, xn is odd.
Therefore, ∀x ∈ Z, if x is odd, then ∀n ∈ N, xn is odd.
26. Prove ∀a, r ∈ R, if r 6= 1 and r 6= 0, then ∀n ∈ N,
n
X
ark =
k=0
Proof.
Let a, r ∈ R, and assume r 6= 1 and r 6= 0.
Pn
a(1−r n+1 )
k
Let A = {n ∈ N |
}.
k=0 ar =
1−r
P1
a(1+r)(1−r)
k
0
1
=
k=0 ar = ar + ar = a(1 + r) =
1−r
Pn
a(1−r n+1 )
k
Let n ∈ A. Then k=0 ar =
.
1−r
n+1
X
k=0
ark =
n
X
k=0
ark + arn+1 =
a(1 − rn+1 )
.
1−r
a(1−r 2 )
1−r .
Thus, 1 ∈ A.
a(1 − rn+1 )
a(1 − rn+1 + rn+1 (1 − r))
a(1 − rn+2 )
+ arn+1 =
=
1−r
1−r
1−r
Thus, n + 1 ∈ A.
Therefore, A is inductive. By the PMI, ∀n ∈ N, n ∈ A.
n+1
Pn
)
Therefore, ∀n ∈ N, k=0 ark = a(1−r
.
1−r
Pn
Therefore, ∀a, r ∈ R, if r 6= 1 and r 6= 0, then ∀n ∈ N, k=0 ark =
7
a(1−r n+1 )
.
1−r
27. Let A ∈ P(U ). Prove if U \ A ⊆ A, then A = U .
Proof.
Assume U \ A ⊆ A and assume A 6= U .
Since A ⊆ U , it must be the case that U * A.
That is, ∃x ∈ U , x ∈
/ A. Choose such an x.
Then x ∈ U and x ∈
/ A, which means x ∈ U \ A.
Since U \ A ⊆ A, we then have x ∈ A.
We now have the contradiction x ∈ A and x ∈
/ A.
Therefore, if U \ A ⊆ A, then A = U .
28. Let A, B, C, and D be sets. Prove if A ⊆ C and B ⊆ D, then A ∪ B ⊆ C ∪ D.
Proof.
Assume A ⊆ C and B ⊆ D.
Let x ∈ U and assume x ∈ A ∪ B.
Then x ∈ A or x ∈ B.
Case 1: x ∈ A.
Since A ⊆ C, we then have x ∈ C.
This proves x ∈ C or x ∈ D, which means x ∈ C ∪ D.
Case 1: x ∈ B.
Since B ⊆ D, we then have x ∈ D.
This proves again x ∈ C or x ∈ D, so x ∈ C ∪ D.
Therefore, ∀x ∈ U , if x ∈ A ∪ B, then x ∈ C ∪ D.
Thus, A ∪ B ⊆ C ∪ D.
Therefore, if A ⊆ C and B ⊆ D, then A ∪ B ⊆ C ∪ D.
Alternate Proof.
Assume A ⊆ C and B ⊆ D.
Since A ⊆ C, we have A ∪ B ⊆ C ∪ B.
Since B ⊆ D, we have C ∪ B ⊆ C ∪ D.
By transitivity, we then have A ∪ B ⊆ C ∪ D.
Therefore, if A ⊆ C and B ⊆ D, then A ∪ B ⊆ C ∪ D.
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29. Prove ∀a, b ∈ R, if a < b, then (−∞, b) \ (−∞, a) = [a, b).
Proof.
Let a, b ∈ R.
Assume a < b.
Let x ∈ R and assume x ∈ (−∞, b) \ (−inf ty, a).
Then x ∈ (−∞, b), so x < b, and x ∈
/ (−∞, a), so a ≤ x.
Since a ≤ x and x < b, we have x ∈ [a, b).
Therefore, ∀x ∈ R, if x ∈ (−∞, b) \ (−inf ty, a), then x ∈ [a, b).
Thus, (−∞, b) \ (−∞, a) ⊆ [a, b).
Conversely, let x ∈ R and assume x ∈ [a, b).
Then a ≤ x and x < b.
Since a ≤ x, we have x ∈
/ (−∞, a).
Since x < b, we have x ∈ (−∞, b).
Now, x ∈ (−∞, b) and x ∈
/ (−∞, a), so x ∈ (−∞, b) \ (−∞, a).
Therefore, [a, b) ⊆ (−∞, b) \ (−∞, a).
Therefore, (−∞, b) \ (−∞, a) = [a, b).
Therefore, if a < b, then (−∞, b) \ (−∞, a) = [a, b).
Therefore, ∀a, b ∈ R, if a < b, then (−∞, b) \ (−∞, a) = [a, b).
30. Prove (−∞, 3] \ (1, 2) = (−∞, 1] ∪ [2, 3].
Proof.
Let x ∈ R and assume x ∈ (−∞, 3] \ (1, 2).
Then x ∈ (−∞, 3], so x ≤ 3, and x ∈
/ (1, 2), so either x ≤ 1 or 2 ≤ x.
Case 1: x ≤ 1.
In this case, we have x ∈ (−∞, 1], so x ∈ (−∞, 1] ∪ [2, 3].
Case 2: 2 ≤ x.
Now, 2 ≤ x and x ≤ 3, so x ∈ [2, 3].
This implies that x ∈ (−∞, 1] ∪ [2, 3].
Therefore, (−∞, 3] \ (1, 2) ⊆ (−∞, 1] ∪ [2, 3].
Conversely, let x ∈ R and assume x ∈ (−∞, 1] ∪ [2, 3].
Then x ∈ (−∞, 1] or x ∈ [2, 3].
Case 1: x ∈ (−∞, 1].
In this case, x ≤ 1, so x ∈
/ (1, 2).
Also, since x ≤ 1 and 1 ≤ 3, we have x ≤ 3, so x ∈ (−∞, 3].
Now, x ∈ (−∞, 3] and x ∈
/ (1, 2), so x ∈ (−∞, 3] \ (1, 2).
Case 2: x ∈ [2, 3].
In this case, 2 ≤ x, so x ∈
/ (1, 2), and x ≤ 3, so x ∈ (−∞, 3].
Again, we have x ∈ (−∞, 3] and x ∈
/ (1, 2), so x ∈ (−∞, 3] \ (1, 2).
Therefore, (−∞, 1] ∪ [2, 3] ⊆ (−∞, 3] \ (1, 2).
Therefore, (−∞, 3] \ (1, 2) = (−∞, 1] ∪ [2, 3].
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31. Let A = (−7, 1) and B = [−2, 17). Find A ∪ B. Prove your result.
Solution.
A ∪ B = (−7, 17).
Proof.
Let x ∈ R and assume x ∈ A ∪ B.
Then x ∈ A or x ∈ B.
Case 1: x ∈ A.
In this case, we have x ∈ (−7, 1), so −7 < x and x < 1.
Since x < 1 and 1 < 17, we have x < 17.
Now, −7 < x and x < 17, so x ∈ (−7, 17).
Case 2: x ∈ B.
Then x ∈ [−2, 17), so 2 ≤ x and x < 17.
Since −7 < 2 and 2 ≤ x, we have −7 < x.
Again, −7 < x and x < 17, so x ∈ (−7, 17).
Therefore, A ∪ B ⊆ (−7, 17).
Conversely, let x ∈ R and assume x ∈ (−7, 17).
Then −7 < x and x < 17. We consider two cases: x < 1 and 1 ≤ x.
Case 1: x < 1.
In this case, −7 < x and x < 1, so x ∈ A. Thus, x ∈ A ∪ B.
Case 2: 1 ≤ x.
Since −2 ≤ 1 and 1 ≤ x, we have −2 ≤ x.
Now, −2 ≤ x and x < 17, so x ∈ B. Thus, x ∈ A ∪ B.
Therefore, (−7, 17) ⊆ A ∪ B.
Therefore, A ∪ B = (−7, 17).
32. Prove h5i ∩ h4i = h20i.
Proof.
Let x ∈ Z and assume x ∈ h5i ∩ h4i.
Then x ∈ h5i and x ∈ h4i.
Choose a, b ∈ Z with x = 5a and x = 4b.
Let c = b − a.
x = 5x − 4x = 5(4b) − 4(5a) = 20b − 20a = 20c.
Therefore, ∃c ∈ Z, x = 20c. Thus, x ∈ h20i.
Therefore, h5i ∩ h4i ⊆ h20i
Conversely, let x ∈ Z and assume x ∈ h20i.
Accordingly, choose q ∈ Z with x = 20q.
Let s = 4q.
Then x = 20q = 5(4q) = 5s.
Therefore, ∃s ∈ Z, x = 5s, which means x ∈ h5i.
Let t = 5q.
x = 20q = 4(5q) = 4t.
Therefore, ∃t ∈ Z, x = 4t. Thus, x ∈ h4i.
We now have x ∈ h5i and x ∈ h4i, which means x ∈ h5i ∩ h4i.
Therefore, h20i ⊆ h5i ∩ h4i.
Therefore, h5i ∩ h4i = h20i.
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33. Prove
[
[1, 3n) = [1, ∞).
n∈N
Proof.
S
Let x ∈ R and assume x ∈ n∈N [1, 3n).
This means ∃n ∈ N, x ∈ [1, 3n). Choose such an n.
Since x ∈ [1, 3n), we have 1 ≤ x and x < 3n.
Since 1 S
≤ x, we have x ∈ [1, ∞).
Therefore, n∈N [1, 3n) ⊆ [1, ∞).
Conversely, let x ∈ R and assume x ∈ [1, ∞).
Then 1 ≤ x.
By the Archimedean property, since 0 < 3, ∃n ∈ N, x < 3n. Choose such an n.
Now, 1 ≤ x and x < 3n, so x ∈ [1, 3n).
Therefore, ∃n ∈ S
N, x ∈ [1, 3n).
This means x ∈ Sn∈N [1, 3n).
Therefore, [1,
S ∞) ⊆ n∈N [1, 3n).
Therefore, n∈N [1, 3n) = [1, ∞).
34. Prove
\
[1, n + 1) = [1, 2).
n∈N
Proof.
T
Let x ∈ R and assume x ∈ n∈N [1, n + 1).
This means ∀n ∈ N, x ∈ [1, n + 1).
Since 1 T
∈ N, we have x ∈ [1, 1 + 1), which means x ∈ [1, 2).
Therefore, n∈N [1, n + 1) ⊆ [1, 2).
Conversely, let x ∈ R and assume x ∈ [1, 2).
This means 1 ≤ x and x < 2.
Let n ∈ N.
Then 1 ≤ n, so 2 ≤ n + 1.
Since x < 2 and 2 ≤ n + 1, we have x < n + 1.
Now, 1 ≤ x and x < n + 1, so x ∈ [1, n + 1).
Therefore, ∀n ∈ T
N, x ∈ [1, n + 1).
This means x ∈T n∈N [1, n + 1).
Therefore, [1,
T 2) ⊆ n∈N [1, n + 1).
Therefore, n∈N [1, n + 1) = [1, 2).
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35. For n ∈ N, let ≡2n be the relation on Z given by: ∀x, y ∈ Z, x ≡2n y means (x − y)(x + y) ∈ hni.
(a) Prove ≡2n is an equivalence relation.
Proof.
Let x ∈ Z.
Then (x − x)(x + x) = 0(2x) = 0 = 0n. Thus, (x − x)(x + x) ∈ hni.
Therefore, x ≡2n x.
Therefore, ≡2n is reflexive.
Let x, y ∈ Z.
Assume x ≡2n y. Then (x − y)(x + y) ∈ hni.
Let q ∈ Z with (x − y)(x + y) = nq.
Then (y − x)(y + x) = −(x − y)(x + y) = −nq = n(−q).
Thus, (y − x)(y + x) ∈ hni, meaning y ≡2n x.
Therefore, if x ≡2n y, then y ≡2n x.
Therefore, ≡2n is symmetric.
Let x, y, z ∈ Z.
Assume x ≡2n y and y ≡2n z. Then (x − y)(x + y) ∈ hni and (y − z)(y + z) ∈ hni.
Let a, b ∈ Z with (x − y)(x + y) = an and (y − z)(y + z) = bn.
Then x2 − y 2 = an and y 2 − z 2 = bn.
Adding these equations gives x2 − z 2 = an + bn = (a + b)n.
Thus, (x − z)(x + z) ∈ hni, and so x ≡2n z.
Therefore, if x ≡2n y and y ≡2n z, then x ≡2n z.
Therefore, ≡2n is transitive.
Q.E.D.
(b) Prove ∀a ∈ Z, [a] = [−a].
Proof.
Let a ∈ Z.
Let x ∈ [a].
Then x ≡2n a, meaning (x − a)(x + a) ∈ hni.
Let q ∈ Z with (x − a)(x + a) = qn.
Now, (x − (−a))(x + (−a)) = (x + a)(x − a) = qn.
Thus, (x − (−a))(x + (−a)) ∈ hni.
Therefore, x ≡2n −a, and so x ∈ [−a].
Therefore, [a] ⊆ [−a].
Conversely, let x ∈ [−a].
Then x ≡2n −a, and so (x − (−a))(x + (−a)) ∈ hni.
As before, this gives us (x + a)(x − a) ∈ hni.
Thus, x ≡2n a. Hence, x ∈ [a].
Therefore, [−a] ⊆ [a].
Therefore, [a] = [−a].
Q.E.D.
Alternate Proof.
Let a ∈ Z.
(a − (−a))(a + (−a)) = 2a(0) = 0 = 0n.
Thus, (a − (−a))(a + (−a)) ∈ hni.
This shows a ≡2n −a, and thus [a] = [−a].
Q.E.D.
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36. Let R be an equivalence relation on a set U .
Prove ∀a, b ∈ U , if [a]R ∩ [b]R 6= ∅, then aRb.
Proof.
Let a, b ∈ U .
Assume [a]R ∩ [b]R 6= ∅.
Then ∃x ∈ U , x ∈ [a]R ∩ [b]R . Choose such an x.
Since x ∈ [a]R ∩ [b]R , we have x ∈ [a]R and x ∈ [b]R .
This means xRa and xRb.
By symmetry, since xRa, we have aRx.
Now, by transitivity, since aRx and xRb, we have aRb.
Therefore, if [a]R ∩ [b]R 6= ∅, then aRb.
Therefore, ∀a, b ∈ U , if [a]R ∩ [b]R 6= ∅, then aRb.
37. Let R be an equivalence relation on a set U . Prove ∀a, b ∈ U , if [a]R ∩ [b]R 6= ∅, then [a]R = [b]R .
Proof.
Let a, b ∈ U .
Assume [a]R ∩ [b]R 6= ∅.
Then ∃c ∈ U , c ∈ [a]R ∩ [b]R . Choose such a c.
Then c ∈ [a]R and c ∈ [b]R . This means cRa and cRb.
Let x ∈ U and assume x ∈ [a]R . This means xRa.
By symmetry, since cRa, we have aRc.
By transitivity, since xRa and aRc, we have xRc.
Again, by transitivity, since xRc and cRb, we have xRb. Therefore, x ∈ [b]R .
Thus, [a]R ⊆ [b]R .
Conversely, let x ∈ U and assume x ∈ [b]R . This means xRb.
By symmetry, since cRb, we have bRc.
By transitivity, since xRb and bRc, we have xRc.
Again, by transitivity, since xRc and cRa, we have xRa. Therefore, x ∈ [a]R .
Therefore, [b]R ⊆ [a]R , and hence [a]R = [b]R .
Therefore, if [a]R ∩ [b]R 6= ∅, then [a]R = [b]R .
Therefore, ∀a, b ∈ U , if [a]R ∩ [b]R 6= ∅, then [a]R = [b]R .
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38. Let | be the relation on N given by: For x, y ∈ N, x|y means ∃q ∈ N, y = xq. Prove | is a partial
ordering.
Proof.
Let x ∈ N.
Putting t = 1 gives us x = xt; hence x|x.
Therefore, | is reflexive.
Let x, y ∈ N.
Assume x|y and y|x.
Choose s, t ∈ N such that y = xs and x = yt.
Then x = xst. Since x 6= 0, we then have st = 1.
Since t divides 1, we have t ≤ 1; hence t = 1 since t ∈ N.
Therefore, x = y(1), which means x = y.
Therefore, if x|y and y|x, then x = y.
Hence, | is antisymmetric.
Let x, y, z ∈ N.
Assume x|y and y|z.
Choose s, t ∈ N such that y = xs and z = yt.
Then z = x(st); hence x|z.
Therefore, if x|y and y|z, then x|z.
Therefore, | is transitive.
Since | is reflexive, symmetric, and transitive, | is a partial ordering.
39. Let R be the relation R = {(x, y) ∈ R × R | ∃n ∈ Z≥0 , y = 2n x}. Prove R is antisymmetric.
Proof.
Let x, y ∈ R.
Assume xRy and yRx.
This means ∃n ∈ Z≥0 , y = 2n x and ∃n ∈ Z≥0 , x = 2n y.
Choose n ∈ Z≥0 with y = 2n x, and choose m ∈ Z≥0 with x = 2m y.
Then x = 2m+n x; hence either x = 0 or 1 = 2m+n .
Case 1: x = 0.
In this case, y = 2n x = 2n (0) = 0; thus y = x.
Case 2: 1 = 2m+n .
Since m + n < 2m+n (Proposition 1.2.20), we have m + n < 1; hence m + n ≤ 0.
Since 0 ≤ n, we have m ≤ m + n ≤ 0, and so m ≤ 0. Since we also have 0 ≤ m, this means
m = 0.
Now, x = 2m y = 20 y = (1)y = y.
Therefore, if xRy and yRx, then x = y.
Therefore, ∀x, y ∈ R, if xRy and yRx, then x = y.
Thus, R is antisymmetric.
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40. Let 2 be the relation on N given by: For x, y ∈ N, x 2 y means ∃n ∈ Z≥0 , y ∈ h2n i and x ∈
2n + h2n+1 i.
(a) Prove 2 is transitive.
Proof.
Let x, y, z ∈ N.
Assume x 2 y and y 2 z.
Let m, n ∈ Z≥0 with x ∈ h2m i and y ∈ 2m + h2m+1 i and y ∈ h2n i and z ∈ 2n + h2n+1 i.
Let a, b, c, d ∈ Z with x = 2m a and y = 2m + 2m+1 b and y = 2n c and z = 2n + 2n+1 d.
Then 2m + 2m+1 b = 2n c, and so 2m (1 + 2b) = 2n c.
Suppose m < n. Then 0 < n − m.
We then have 1 + 2b = 2n−m c with 0 < n − m.
This is impossible, since the left-hand-side is odd, but the right-hand-side is even.
This proves n ≤ m, and so 0 ≤ m − n.
Now, x = 2m a = 2n (2m−n a) ∈ h2n i.
We now have x ∈ h2n i and z ∈ 2n + h2n+1 i. Thus, x 2 z.
Therefore, if x 2 y and y 2 z, then x 2 z.
(b) Using complete induction on x, prove 2 is reflexive.
Proof.
Let S = {x ∈ N | x 2 x}.
Note that 1 ∈ h1i and 1 = 1 + (0)2 ∈ 1 + h2i.
This means 1 ∈ h20 i and 1 ∈ 20 + h21 i. Thus, 1 2 1. Hence, 1 ∈ S.
Let x ∈ N and assume {1, . . . x} ⊆ S.
Case 1: x + 1 is odd.
Let a ∈ Z with x + 1 = 2a + 1.
Then x + 1 = 20 + 21 a, and so x + 1 ∈ 20 + h21 i.
Also, since x + 1 = (x + 1)(1) = (x + 1)20 , we have x + 1 ∈ h20 i.
Therefore, x + 1 2 x + 1, and so x + 1 ∈ S.
Case 2: x + 1 is even.
Let y ∈ N with x + 1 = 2y.
Then y < x + 1, giving us y ∈ {1, . . . , x}, and hence y ∈ S.
Choose n ∈ N with y ∈ h2n i and y ∈ 2n + h2n+1 i.
Let a, b ∈ Z with y = 2n a and y = 2n + 2n+1 b.
Then x + 1 = 2y = 2n+1 and x + 1 = 2y = 2n+1 + 2n+2 b.
This gives us x + 1 ∈ h2n+1 i and x + 1 ∈ 2n+1 + h2n+2 i.
Therefore, x + 1 2 x + 1. Hence, x + 1 ∈ S.
Therefore, ∀x ∈ N, if {1, . . . , x} ⊆ S, then x + 1 ∈ S.
By the PCI, ∀x ∈ N, x ∈ S.
Therefore, 2 is reflexive.
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41. For the partial order | on N, prove 6 is the least upper bound of the set S = {2, 3}.
Proof. Let a = 3.
Then 6 = 2a, and so 2|6.
Let b = 2.
Then 6 = 3b, and so 3|6.
Therefore, 6 is an upper bound of S.
Next, let b ∈ N and assume b is an upper bound of S.
This means 2|b and 3|b.
Let p, q ∈ N with b = 2p and b = 3q.
Let r = p − q.
b = 3b − 2b = 3(2p) − 2(3q) = 6(p − q) = 6r.
Thus, 6|b.
Therefore, ∀b ∈ N, if b is an upper bound of S, then 6|b.
Q.E.D.
42. For the partial order ⇒ on B, prove ∀x, y ∈ B, x ∧ y is the greatest lower bound of the set {x, y}.
Proof. By consistency, we have x ∧ y ⇒ x and x ∧ y ⇒ y.
This means x ∧ y is a lower bound of the set {x, y}.
Next, let b ∈ B and assume b is a lower bound of {x, y}.
This means b ⇒ x and b ⇒ y.
Then b ∧ x = b and b ∧ y = b.
This gives us b ∧ (x ∧ y) = (b ∧ x) ∧ y = b ∧ y = b.
Therefore, b ⇒ x ∧ y.
Therefore, ∀b ∈ B, if b is a lower bound of {x, y}, then b ⇒ x ∧ y.
Therefore, x ∧ y is the greatest lower bound of {x, y}.
43. For the partial order ≤ on R, prove 0 is the greatest lower bound of the set (0, ∞).
Proof. Let x ∈ (0, ∞).
then 0 < x, and hence 0 ≤ x.
Therefore, ∀x ∈ (0, ∞), 0 ≤ x.
Thus, 0 is a lower bound of (0, ∞).
Next, let b ∈ R and assume b is a lower bound of (0, ∞).
Suppose 0 < b.
Then 0 < 2b , which means 2b ∈ (0, ∞).
Since b is a lower bound of (0, ∞), we then have b ≤ 2b .
This gives us 2b ≤ b, and so b ≤ 0.
This is a contradiction to the assumption that 0 < b.
Therefore, b ≤ 0.
Therefore, ∀b ∈ R, if b is a lower bound of (0, ∞), then b ≤ 0.
Therefore, 0 is the greatest lower bound of (0, ∞).
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44. Let R be a relation on a set U that is both reflexive and transitive. Let S be the relation on U given
by: ∀x, y ∈ U , xSy provided xRy and yRx.
(a) Prove S is an equivalence relation.
Proof.
Let x ∈ U .
Since R is reflexive, we have xRx.
Then xRx and xRx is true, meaning xSx.
Therefore, S is reflexive.
Let x, y ∈ U .
Assume xSy. This means xRy and yRx.
Then yRx and xRy is true, meaning ySx.
Therefore, if xSy, then ySx.
Therefore, S is symmetric.
Let x, y, z ∈ U .
Assume xSy and ySz.
Then xRy and yRx and yRz and zRy.
Since xRy and yRz, we have xRz by transitivity of R.
Similarly, since zRy and yRx, we have zRx by transitivity of R.
Now, xRz and zRx, which means xSz.
Therefore, if xSy and ySz, then xSz.
Therefore, S is transitive.
(b) Prove ∀a, b, x, y ∈ U , if [a]S = [b]S and [x]S = [y]s and aRx, then bRy.
Proof.
Let a, b, x, y ∈ U .
Assume [a]S = [b]S and [x]S = [y]S and aRx.
Then aSb and xSy and aRx.
This means aRb and bRa and xRy and yRx and aRx.
Since bRa and aRx, we have bRx by transitivity.
Now, since bRx and xRy, we have bRy by transitivity.
Therefore, if [a]S = [b]S and [x]S = [y]s and aRx, then bRy.
Q.E.D.
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(c) Define the relation T on the set of equivalence classes U/S = {[x]S | x ∈ U } by:
For all [x]S , [y]S ∈ U/S, [x]S T [y]S provided xRy.
Prove T is a partial order.
Proof.
Let [x]s ∈ U/S.
Since R is reflexive, we have xRx. Thus, [x]S T [x]S .
Therefore, T is reflexive.
Let [x]S , [y]S ∈ U/S.
Assume [x]S T [y]S and [y]S T [x]S .
Then xRy and yRx. This means xSy.
Therefore, [x]S = [y]S .
Therefore, if [x]S T [y]S and [y]S T [x]S , then [x]S = [y]S .
Therefore, T is antisymmetric.
Let [x]S , [y]S , [z]S ∈ U/S.
Assume [x]S T [y]S and [y]S T [z]S .
This means xRy and yRz.
By transitivity of R, we have xRz.
Thus, [x]S T [z]S .
Therefore, if [x]S T [y]S and [y]S T [z]S , then [x]S T [z]S .
Therefore, T is transitive.
Q.E.D.
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