Solutions to Homework 2 Mathematics 503 Foundations of Mathematics Spring 2014

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1 Solutions to Homework 2 Mathematics 503 Foundations of Mathematics Spring : 1. (a) Here is a short table. P m is an even integer Hypothesis P 1 There is an integer q such that m = 2q Definition of even integer P 2 m + 1 = 2q + 1 Algebra Q1 There is an integer k such that m + 1 = 2k + 1 Let q = k Q m + 1 is an odd integer Definition of odd integer Proof. If m is an even integer, then there is an integer q such that m = 2q. If we add 1 to both sides of this equation, we obtain m + 1 = 2q + 1 This equation implies that m + 1 is an odd integer. (b) Here is a short table. P m is an odd integer Hypothesis P 1 There is an integer q such that m = 2q + 1 Definition of odd integer P 2 m + 1 = 2q + 2 Algebra P 3 m = 2(q + 1) Algebra P 4 q + 1 is an integer Closure property of the integers Q1 There is an integer k such that m + 1 = 2k Let k = q + 1 Q m + 1 is an odd integer Definition of even integer Proof. If m is an odd integer, then there is an integer q such that m = 2q + 1. When we add the number 1 to both sides of this equation and do some algebra, we obtain m + 1 = 2q + 2 = 2(q + 1). The preceding equation implies that m is even, for q + 1 is an integer.

2 2. (a) Here is a short table. P x and y are even integers Hypothesis P 1 There are integers p and q such that x = 2p and y = 2q Definition of even integer P 2 x + y = 2p + 2q Algebra P 3 x + y = 2(p + q) Algebra P 4 p + q is an integer Closure property of the integers Q1 There is an integer k such that x + y = 2k Let k = p + q Q x + y is an even integer Definition of even integer Proof. Suppose that x and y are even integers; then, there exist integers p and q such that x = 2p and y = 2q. When we add x and y, we obtain x + y = 2p + 2q = 2(p + q). This equation implies that x + y is an even integer, for p + q is an integer. (b) Here is a short table. P x is an even integer and y is an odd integer Hypothesis P 1 There are integers p and q such that x = 2p and y = 2q + 1 Definition of even, odd integers P 2 x + y = 2p + 2q + 1 Algebra P 3 x + y = 2(p + q) + 1 Algebra P 4 p + q is an integer Closure property of the integers Q1 There is an integer k such that x + y = 2k + 1 Let k = p + q Q x + y is an odd integer Definition of odd integer Proof. If x is an even integer and y is an odd integer, then there exist integers p and q such that x = 2p and y = 2q + 1. When we add x and y and do some algebra, we obtain x + y = 2p + 2q + 1 = 2(p + q) + 1. Now, p + q is an integer. It follows from the definition of odd integer that x + y is an odd integer.

3 (c) Here is a short table. P x and y are even integers Hypothesis P 1 There are integers p and q such that x = 2p + 1 and y = 2q + 1 Definition of odd integer P 2 x + y = 2p + 2q + 2 Algebra P 3 x + y = 2(p + q + 1) Algebra P 4 p + q + 1 is an integer Closure property of the integers Q1 There is an integer k such that x + y = 2k Let k = p + q + 1 Q x + y is an even integer Definition of even integer Proof. If x and y are both odd integers, then there exist integers p and q such that x = 2p + 1 and y = 2q + 1. When we add x with y and do some algebra, we obtain the following equation: x + y = 2p + 2q + 2 = 2(p + q + 1). Since (p + q + 1) is an integer, it follows that x + y is an even integer. 3. (b) Here is a short table. P n is an even integer Hypothesis P 1 There is an integer k such that n = 2k Definition of even integer P 2 n 2 = (2k) 2 Substitution P 3 n 2 = 4k 2 Algebra P 4 n 2 = 2(2k 2 ) Algebra P 5 2k 2 is an integer Closure Property of the Integers Q1 There is an integer q such that n 2 = 2q Let q = 2k 2 Q n 2 is an even integer Definition of even integer Proof. Suppose that n is an even integer; then, there is an integer k such that We square n, and note that n = 2k. n 2 = (2k) 2 = 4k 2. The second equation follows from some properties of the integers. Finally, we note that 4k 2 = 2(2k 2 ).

4 This implies that n 2 = 2(2k 2 ). This last equation implies that n 2 is even. 5. (a) Here is a short table: P m is an even integer Hypothesis P 1 There is an integer k such that m = 2k Definition of even integer P 2 3m 2 + 2m + 3 = 3(2k) 2 + 2(2k) + 3 Substitution P 3 3(2k) 2 + 2(2k) + 3 = 2(6k 2 + 2k + 1) + 1 Algebra P 4 6k 2 + 2k + 1 is an integer Closure Properties of the Integers Q1 There is an integer q such that 3m 2 + 2m + 3 = 2q + 1 Let q = (6k 2 + 2k + 1) Q 3m 2 + 2m + 3 is an odd integer Definition of odd integer Proof. Suppose that m is an even integer; then, there is a k Z such that This means that It follows that m = 2k. 3m 2 + 2m + 3 = 3(2k) 2 + 2(2k) m 2 + 2m + 3 = 2(6k 2 + 2k + 1) + 1. Since 6k 2 + 2k + 1 is an integer, it follows that 3m 2 + 2m + 3 is odd. 6. (a) There are many possible answers. For instance, let x and y be the real numbers under consideration. If, for each real number ɛ > 0, we find that we can conclude that x y = 0. ɛ < x y < ɛ, (b) There are many possible answers. For instance: let x be the real number under consideration. If we can show that, for every real ɛ > 0, that then we can conclude that x = 0. ɛ < x < ɛ, (c) There are many possible answers. One might come from coordinate geometry. In this situation, the two lines are defined by two linear equations: ax + by = c and dx + ey = f. The two lines would be parallel if b = e = 0, or if a b = d e. One could also think of constructing vectors u and v parallel to the given lines, and examing the dot product of the two vectors as a means of determining the cosine of the angle between them. If one thinks of classical geometry, two lines l 1 and l 2 are parallel provided that when the two lines are intersected with a transversal, the corresponding angles at the intersections are equal.

5 (d) There are many possible answers. One could show that the given triangle possessed at least two sides of equal length. 7. Proposition. If m is a real number and m, m + 1, and m + 2 are the lengths of the three sides of a right triangle, then m = 3. Proof. Let m be a real number. Suppose that m, m + 1, and m + 2 are the lengths of the three sides of a right triangle. Since m + 2 > m + 1 > m, the side of length m + 2 must be the hypotenuse of the triangle. Thus, by the Pythagorean Theorem, m 2 + (m + 1) 2 = (m + 2) 2. Expanding the powers of the binomials, we obtain Collecting all of the terms, we obtain Factoring this quadratic expression gives m 2 + m 2 + 2m + 1 = m 2 + 4m + 4. m 2 2m 3 = 0. (m 3)(m + 1) = 0. It follows that either m 3 = 0 or m + 1 = 0, i.e., m = 3 or m = 1. Since m is the length of one of the sides of the triangle, m > 0. It follows that m 1, so m = (a) 1 = , 4 = , 2 = 3 ( 1) + 1, and 5 = 3 ( 2) + 1 are all type 1 integers. (b) 2 = , 5 = , 1 = 3 ( 1) + 2, and 4 = 3 ( 2) + 2 are all type 2 integers. (c) Yes, it does appear that the product of two type 1 integers is of type 1. In fact, if a = 3n + 1 and b = 3m + 1 for some integers m and n, then ab = (3n + 1)(3m + 1) = 9mn + 3n + 3m + 1 = 3(3mn + n + m) + 1, so ab must be of type 1 if a and b are. 10. (a) The proposition in question is: Proposition. If a and b are type 1 integers, then a + b is a type 2 integer. Proof. If a and b are type 1 integers, then there exist integers m and n such that a = 3m + 1 and b = 3n + 1.

6 If we add a and b, we obtain a + b = (3m + 1) + (3n + 1). After some algebra is performed, we get a + b = 3(m + n) : 1. Daisy s statement is true. The hypothesis in this case is false, which means that the conditional statement she made is true. 3. Note that the conditional P Q is false when P is true and Q is false. (a) True (b) True (c) True 4. If Q is false and P Q is true, then it must be that P is false. This means that P is true. (a) True (b) True (c) Since P is true, P R will be true when R is true, and false when R is false. (d) The conditional R P will be true when R is false, and false when R is true, since P is false. 5. (a) (b) (c) P Q P Q T T T T F F F T T F F T P Q Q P T T T T F T F T F F F T P Q P Q P Q T T F F T T F F T T F T T F F F F T T T

7 (d) P Q P Q Q P T T F F T T F F T F F T T F T F F T T T Note that the truth tables for (a) and (d) are the same. The truth tables for (b) and (c) are the same.