MATH327 – HOMEWORK SOLUTIONS HOMEWORK #1

Section 0: Problems 1,4,5,6,7,8,10,13

Krzysztof Galicki Problem 0.1.

a) True, because 4 = 2 + 2 is true and 7 < \sqrt{50} is true.

b) False, because 4 ≠ 2 + 2 is false.

c) True, because 4 = 2 + 2 is true and 7 < \sqrt{50} is true.

d) True, because 4 ≠ 2 + 2 is false (the hypothesis).

e) False, because the hypothesis 4 = 2 + 2 is true but the conclusion 7 > \sqrt{50} is false.

f) False, because the area of the circle is not 2πr and its circumference is not πr^{2}.

g) False, because the hypothesis 2 + 3 = 5 is true but the conclusion 5 + 6 = 10 is false.

Problem 0.4.

a) Converse: If a/c is an integer then a/b and b/c are integers. Contrapositive: If a/c is not an integer then either a/b or b/c is not an integer.

b) Converse: If x ± 1 then x^{2} = 1. Contrapositive: If x ≠ ±1 then x^{2} ≠ 1.

c) Converse: If graph G is connected than G is Euclidean. Contrapositive: If graph G is not connected then it is not Euclidean.

d) Converse: If a = 0 or b = 0 then ab = 0. Contrapositive: If a ≠ 0 and b ≠ 0 then ab ≠ 0.

e) Converse: If A is a four-sided figure then A is a square. Contrapositive: If A is not a four-sided figure than A is not a square.

f) Converse: If a^{2} = b^{2} + c^{2} then ΔBAC is a right triangle. Contrapositive: If a^{2} ≠ b^{2} + c^{2} then ΔBAC is not a right triangle.

Problem 0.5.

a) There exists a continuous function which is not differentiable.

b) For all x ∈ ℝ, 2^{x} ≥ 0.

c) For all x ∈ ℝ, there exists y ∈ ℝ such that y > x.

d) For each prime number p, there exists a prime number q such that q > p.

e) For each positive x ∈ ℝ, x is a product of primes.

f) For each positive x ∈ ℝ, \sqrt{x} is real.

Problem 0.6. Theorem A: For all x ∈ ℝ, there exists y ∈ ℝ such that y > x.

Proof: Let x ∈ ℝ be arbitrary. Consider y = x + 1. Then y > x.

Theorem B: There exists y ∈ ℝ such that y > x for every x ∈ ℝ.

Proof: (This statement is false. We will disprove it assuming that it is true via the “proof by contradiction”.) Suppose Theorem B is true. Let y_{max} be such a number. Set x = y_{max} + 100. Then x > y_{max}. The contradiction shows that the statement is false. Problem 0.7. Theorem A: If a ∈ ℤ then either a or a + 1 is even.

Proof: Case 1: Suppose a is even. Then the statement is true. Case 2: Suppose a is odd. Then a = 2k + 1 for some integer k. But then a + 1 = 2k + 2 = 2(k + 1) and, hence, a + 1 is even. Again, the statement is true in this case. Theorem B: If n ∈ ℤ then n^{2} + n is even.

Proof: Case 1: Let n = 2k be even. Then n^{2} + n = 4k^{2} + 2k = 2(2k^{2} + k) is even. Case 2: Let n = 2k+1 be odd. Then n^{2} +n = (2k+1)^{2} +(2k+1) = 4k^{2} +4k+1+2k+1 = 4k^{2} + 6k + 2 = 2(2k^{2} + 3k + 1) is even. Problem 0.8. Theorem: If n ∈ ℤ then n^{2} − n + 5 is odd.

Proof: Case 1: Let n = 2k be even. Then n^{2} − n + 5 = 4k^{2} − 2k + 5 = 2(2k^{2} − k + 2) + 1 is odd. Case 2: Let n = 2k +1 be odd. Then n^{2} −n+5 = (2k +1)^{2} −(2k +1)+5 = 4k^{2} +2k +5 = 2(2k^{2} + k + 2) + 1 is odd. Problem 0.10. Theorem: a^{2} − b^{2} is odd if and only if a and b have opposite parity.

Proof: Case 1: If a = 2k and b = 2l then a^{2} − b^{2} = (2k)^{2} − (2l)^{2} = 2(2k^{2} − 2l^{2}) is even. Case 2: If a = 2k+1 and b = 2l+1 then a^{2}−b^{2} = (2k+1)^{2}−(2l+1)^{2} = 2(2k^{2}+2k−2l^{2}−2l) is even. Case 3: If a = 2k + 1 and b = 2l then a^{2} − b^{2} = (2k + 1)^{2} − (2l)^{2} = 2(2k^{2} + 2k − 2l^{2}) + 1 is odd. Case 4: If a = 2k and b = 2l + 1 then a^{2} − b^{2} = (2k)^{2} − (2l + 1)^{2} = 2(2k^{2} − 2l^{2} − 2l − 1) + 1 is odd. Problem 0.13. Theorem: There exists no smallest positive real number.

Proof: Suppose such a number does exist and let us denote it by x_{min}. Consider y = \frac{1}{2}x_{min}. Then y < x and y is positive. But this contradicts the fact that x_{min} was the smallest real number.