Logic and Mathematical Reasoningan introduction to proof writing

1. \((\neg P) \wedge (\neg Q)\) (or equivalently, \(\neg (P \vee Q)\) where \(P\) is “I am a good basketball player,” and \(Q\) is “I am a good baseball player.”
2. \(\neg (P \wedge Q)\) where \(P\) is “I am american,” and \(Q\) is “I am Japanese.”
3. \((P \wedge Q) \implies (R \vee S)\) where \(P\) is “\(17\) is prime,” \(Q\) is “\(17\) divides \(476\text{,}\)”, \(R\) is “\(17\) divides \(68\text{,}\)” and \(S\) is “\(17\) divides \(7\text{.}\)”
4. Not a proposition because it is a question and so has no truth value.
5. \(P \implies (Q \vee R)\) where \(P\) is “I owe you a dollar,” \(Q\) is “I am in debt,” and \(R\) is “you owe me more than a dollar.”
6. Not a proposition because \(x\) is an unquantified variable; this is an open sentence.
7. \(P \iff Q\) where \(P\) is \(2 \lt 5\) and \(Q\) is \(4 \lt 25\text{.}\)
8. \((\neg P) \wedge Q\) where \(P\) is “I am a quick reader,” and \(Q\) is “I can do mathematics easily.”
9. Not a proposition. If this statement were true, then by its meaning it would be false, which is a contradiction. If this statement were false, then by its meaning it would be true, which is a contradiction. Thus this statement can be neither true nor false.
10. \(P \implies Q\) where \(P\) is “it is raining,” and \(Q\) is “there are clouds.”
11. \(P \iff \neg Q\) where \(P\) is \(3+2 = 5\text{,}\) and \(Q\) is \(3 \cdot 2 = 7\text{.}\)

For the quantified sentences, we have

1. \((\exists ! x \in \mathbb{N}) (\forall y \in \mathbb{N}) (x \le y)\)
2. \((\forall x \in \mathbb{R}) \big((x \lt 0) \vee (x \gt 0)\big)\)
3. \((\forall x \in \mathbb{Z}) (\exists y \in \mathbb{Z}) (x \gt y)\)
4. \((\exists x \in \mathbb{Z}) (\forall y \in \mathbb{Z}) \big((x \not= y) \implies (x>y) \big)\)
5. \((\forall x \in \mathbb{Q}) (x \in \mathbb{R})\)

For the first equivalence, we have the following chain:

For the second equivalence, we have the following chain:

For the last equivalence, we have the following chain:

9.

1. Every natural number is greater than or equal to one.
2. There exists a unique real number which is both greater than or equal to zero and less than or eqaul to zero.
3. For any prime natural number \(x\text{,}\) if \(x\) is not equal to \(2\text{,}\) then \(x\) is odd. Alternatively, you could say: every prime natural number other than \(2\) is odd.
4. There exists a unique real number \(x\) for which is the base-\(e\) logarithm of \(x\) is \(1\text{.}\)
5. There is no real number whose square is negative.
6. There exists a unique real number whose square is zero.
7. For every natural number \(x\text{,}\) if \(x\) is odd, then \(x^2\) is also odd. Alternatively, you could say: The square of any odd natural number is itself odd.

10. Here I provide the truth value as well as an English translation to help you see why it has that truth value.

1. True, “for every real number, there is some other number so that the sum of the two is zero.”
2. False, “there exists some real number so that when you add it to any other real number, the sum is zero.”
3. False, “there exist two real numbers \(x,y\) so that \(x^2 + y^2 = -1\text{.}\)”
4. False, “for every positive real number there exists some negative real number so that the product is positive.”
5. True, “for every real number \(y\) there is some real number \(x\) so that no matter which real number \(z\) is given, the products \(xy, xz\) are equal.”
6. False, “there exists a smallest real number.”
7. True, “every real number is at least as small as some real number.”
8. False, “there exists a unique negative real number which, becomes positive upon adding \(3\text{.}\)”
9. False, “there exists a unique number which is the square root of every real number.”
10. False, “every real number has a unique sqaure root.”
11. False, “there exist unique real numbers \(x,y\) whose difference is less than the square of every real number \(w\text{.}\)”

1(a). Suppose \((G,\ast)\) is a cyclic group, …, therefore \((G,\ast)\) is abelian.

1(b). Suppose \(B\) is a nonsingular matrix, …, hence \(\det B \not= 0\text{.}\)

1(c). Suppose \(A,B\text{,}\) and \(C\) are sets and that \(A\) is a subset of \(B\) and that \(B\) is a subset of \(C\text{,}\) …, thus \(A\) is a subset of \(C\text{.}\)

5(a). Suppose \(x,y\) are even integers. By the definition of even, we have \(x = 2m\) and \(y = 2n\) for some integers \(m,n \in \mathbb{Z}\text{.}\) So \(x+y = 2m + 2n = 2(m+n)\text{.}\) Since the sum of integers is an integer, \(m + n \in \mathbb{Z}\) is an integer, and so by the definition of even, \(x+y\) is even.

5(c). Suppose \(x,y\) are even integers. By the definition of even, we have \(x = 2m\) and \(y = 2n\) for some integers \(m,n \in \mathbb{Z}\text{.}\) So \(xy = (2m)(2n) = 4(mn)\text{.}\) Since the product of integers is an integer, \(mn \in \mathbb{Z}\) is an integer, and so using the definition of divisibility, we see \(4\) divides \(xy\text{.}\)

7(j). Suppose \(a,b\) are positive integers and \(a\) divides \(b\) and \(b\) divides \(a\text{.}\) By the definition of divisibility, there are some integers \(m,n\) so that \(b = am\) and \(a = bn\text{.}\) Since, \(a,b\) are positive, \(m,n\) must be positive as well. Thus \(a = bn = (am)n = a(mn)\text{.}\) Since, \(a \not= 0\text{,}\) we may cancel it on both sides to obtain \(1 = mn\text{.}\) Because \(m,n\) are positive integers, \(m,n \ge 1\text{.}\) Therefore, \(1 \le m \le mn = 1\) and thus \(m = 1 = n\text{.}\) Finally, \(b = am = a\text{.}\)

An alternate proof proceeds as follows. Suppose \(a,b\) are positive integers and \(a\) divides \(b\) and \(b\) divides \(a\text{.}\) By a result proven in class, if a positive integer \(x\) divides another positive integer \(y\text{,}\) then \(x \le y\text{.}\) Therefore, we find that \(a \le b\) and \(b \le a\text{,}\) therefore \(a = b\text{.}\)

7(k). Suppose \(a,b,c,d \in \mathbb{Z}\) and \(a\) divides \(b\) and \(c\) divides \(d\text{.}\) By the definition of divisibility, there are some integers \(m,n\) so that \(b = am\) and \(d = cn\text{.}\) Thus \(bd = (am)(cn) = (ac)(mn)\text{.}\) Since the product of integers is an integer, \(mn \in \mathbb{Z}\) is an integer, and so using the definition of divisibility, we see \(bd\) divides \(ac\text{.}\)

7(l). Suppose \(a,b,c \in \mathbb{Z}\) and \(ab\) divides \(c\text{.}\) By the definition of divisibility, there is some integer \(m\) so that \(c = (ab)m\text{.}\) Thus \(c = a(bm)\text{.}\) Since the product of integers is an integer, \(bm \in \mathbb{Z}\) is an integer, and so using the definition of divisibility, we see \(a\) divides \(c\text{.}\)

Use proof by cases.

Suppose that \(k \in \mathbb{Z}\text{.}\)

Case 1: \(k\) is even. Then \(k = 2n\) for some integer \(n \in \mathbb{Z}\text{,}\) so \(k(k+1) = (2n)(k+1) = 2(n(k+1))\text{,}\) which is even since \(n(k+1)\) is an integer.

Case 2: \(k\) is odd. Then \(k = 2n + 1\) for some integer \(n \in \mathbb{Z}\text{,}\) so \(k+1 = 2n+2 = 2(n+1). Then \)\(k(k+1) = k(2(n+1)) = 2(k(n+1))\text{,}\) which is even since \(k(n+1)\) is an integer.

Note that the contrapositive of this statement is that if both \(x,y\) are not even, then \(xy\) is not even. This is equivalent to saying that if both \(x,y\) are odd, then \(xy\) is odd. This is the statement we will prove.

Suppose \(x,y\) are arbitrary integers. Assume that \(x\) is odd and \(y\) is odd. So there exist \(m,n \in \mathbb{Z}\) with \(x = 2m+1\) and \(y = 2n+1\text{.}\) Hence \(xy = (2m+1)(2n+1) = 4mn+2m+2n+1 = 2(2mn+m+n)+1\) is odd since \(2mn+m+n \in \mathbb{Z}\text{.}\)

1(d). To prove this statement (\(\neg (\exists m,n \in \mathbb{Z}) P(m,n)\)) we will prove the equivalent form (\((\forall m,n \in \mathbb{Z})(\neg P(m,n))\)). To this end, suppose \(m,n \in \mathbb{Z}\) are arbitrary. Then \(12m + 15n = 3\cdot 4m + 3 \cdot 5n = 3(4m+5n)\text{.}\) Since \(4m + 5n\) is an integer, we see that 3 divides \(12m + 15n\text{.}\) But 3 does not divide 1, and so \(12m + 15n \not= 1\text{.}\)

1(h). Let \(m\) be an arbitrary odd integer. Then there exists \(k \in \mathbb{Z}\) with \(m = 2k+1\text{,}\) and hence

By a previous problem, \(k(k+1)\) must be even, and so there is some \(j \in \mathbb{Z}\) so that \(k(k+1) = 2j\text{.}\) Hence,

2(c). Suppose \(a,b \in \mathbb{Z}\) are aribtrary and \(a\) divides \(b\text{.}\) Then \(b = ka\) for some \(k \in \mathbb{Z}\text{.}\) Let \(n \in \mathbb{N}\) be any natural number. Then \(b^n = (ka)^n = k^n a^n\text{.}\) Since \(k^n \in \mathbb{Z}\text{,}\) \(a^n\) divides \(b^n\text{.}\) Because \(n \in \mathbb{N}\) was arbitrary, this proves the result.

4(c). False. Counterexample: \(x = 2\text{,}\) \(y = \frac{1}{2}\text{,}\) then \(y^x = \left( \frac{1}{2} \right)^2 = \frac{1}{4} \le 2 = x\text{.}\)

4(d). False. Counterexample: \(a = 6\text{,}\) \(b = 2\text{,}\) \(c = 3\text{.}\) Certainly \(a = 6\) divides \(bc = 2 \cdot 3 = 6\) since \(6 = 6 \cdot 1\) and \(1 \in \mathbb{Z}\text{.}\) However, \(a = 6\) cannot divide either of \(b = 2\) or \(c = 3\) since these are less than \(6\text{.}\)

4(f). False. Counterexample: \(x = \frac{1}{2}\text{.}\) \(x^2 - x = \left( \frac{1}{2} \right)^2 - \frac{1}{2} = \frac{1}{4} - \frac{1}{2} = -\frac{1}{4} \lt 0\text{.}\) Alternative counterexample: Consider any \(0 \lt x \lt 1\text{.}\) Then \(x \gt 0\) but \(x-1 \lt 0\text{.}\) Hence \(x^2 - x = x(x-1) \lt 0\) since it is the product of a negative and a positive.

5(a). We will use a two-part proof.

Suppose \(x\) is any prime natural number. That means \(x \gt 1\) and \(x\) has no divisor \(y\) in the range \(1 \lt y \lt x\text{.}\) Since \(\sqrt{x} \lt x\) (because \(x \gt 1)\text{,}\) \(x\) has no divisor \(y\) in the range \(1 \lt y \le \sqrt{x}\text{.}\)

For the other direction, we will prove the statement using contraposition. Suppose \(x\) is a natural number which is not prime. There are two cases:

Case 1: \(x = 1\text{.}\) Then clearly \(x\) is not greater than one.

Case 2: \(x \gt 1\text{.}\) Since \(x\) is greater than one and not prime, it is composite (by definition). Thus \(x = ab\) for some positive integers \(1 \lt a,b \lt x\text{.}\) Either \(a \le b\) or \(b \le a\text{.}\) Without loss of generality, we may assume \(a \le b\text{.}\) Then multiplying both sides by \(a\text{,}\) we find

Hence \(a \le \sqrt{x}\) (by properties of square roots, in particular, because it is an increasing function). So, \(a\) is a divisor of \(x\) which is greater than one (because it is a prime) and less than or equal to the sqaure root of \(x\text{.}\) By contraposition, we have proven the desired result.

12(a). Grade: F. The student tried to prove the inverse (\((\neg P) \implies (\neg Q)\)) of the implication \(P \implies Q\) given in the problem. However, this is a logical error since the two are not equivalent.

12(b). Grade: A. The student correctly proved this statement by contraposition. There are no errors.

Suppose \(a\) is a nonzero rational number. Then there exist integers \(p,q \in \mathbb{Z}\) with \(q \not= 0\) so that \(a = \frac{p}{q}\text{.}\) Since \(a \not= 0\text{,}\) we know \(p \not= 0\text{,}\) and hence \(\frac{q}{p}\) is a rational number. Moreover, \(a \cdot \frac{q}{p} = \frac{p}{q} \cdot \frac{q}{p} = \frac{pq}{qp} = 1\text{.}\) Therefore, \(\frac{q}{p}\) is a multiplicative inverse for \(a\text{.}\) Thus every nonzero \(a \in \mathbb{Q}\) has a multiplcative inverse.

Now suppose \(b,c \in \mathbb{Q}\) are multiplicative inverses of \(a\text{.}\) Then \(ab = 1\) and \(ac = 1\text{,}\) and hence \(ab = ac\text{.}\) Since \(a \not= 0\text{,}\) we may cancel to obtain \(b = c\text{.}\) Thus the multiplicative inverse of \(a\) is unique.

Notice that without the word “unique,” this statement is precisely Bezout's identity. So, certainly \(x,y\) exist, the question is whether they are unique. We show below that they are not unique.

Counterexample: \(a = 2\) and \(b = 3\text{.}\) Two different solutions are \(x = -1\text{.}\) \(y = 1\) and \(x = 2\text{,}\) \(y = -1\text{.}\)

Comment: In fact, it is possible to list out precisely all the pairs that satisfy Bezout's identity by starting with a single pair \(x,y\text{.}\) I will state this without proof. Given nonzero integers \(a,b\) with \(d := \gcd(a,b) = ax+by\text{,}\) then all other solutions to Bezout's identity have the form:

Mimic the proof that \(\sqrt{2}\) is irrational that I provided in class, except use Euclid's Lemma for the key steps.

Assume to the contrary that \(x \in \mathbb{Q}\) and \(x^2 = p\) for some prime \(p\text{.}\) Then we may write \(x = \frac{a}{b}\) for \(a,b \in \mathbb{Q}\) with \(b \not= 0\text{.}\) Moreover, by <<Unresolved xref, reference "lem-rational-lowest-terms"; check spelling or use "provisional" attribute>>  we may even assume \(\gcd(a,b) = 1\text{.}\) Since \(p = x^2 = \frac{a^2}{b^2}\) we know \(pb^2 = a^2\text{,}\) which means that \(p\) divides \(a^2\) since \(b^2\) is an integer. By Euclid's Lemma we know that \(p\) must divide \(a\) as well. Thus \(a = pk\) for some \(k \in \mathbb{Z}\text{.}\) Therefore, \(pb^2 = a^2 = (pk)^2 = p^2 k^2\) and so \(b^2 = pk^2\text{.}\) Thus, \(p\) divides \(b^2\) since \(k^2 \in \mathbb{Z}\text{.}\) By Euclid's Lemma we know that \(p\) must divide \(b\) as well. Hence, \(p\) is a common divisor of \(a,b\text{,}\) which contradicts the fact that \(\gcd(a,b) = 1\text{.}\) Therefore our assumption was wrong and instead \(x \notin \mathbb{Q}\text{.}\)

Suppose that \(a,b\) are relatively prime integers and each divides \(c\text{.}\) Then \(c = am\) and \(c = an\) for some \(m,n \in \mathbb{Z}\text{.}\) Moreover, by Bezout's identity, \(1 = \gcd(a,b) = ax+by\) for some \(x,y \in \mathbb{Z}\text{.}\) Multiplying this last equation by \(c\text{,}\) we find

and therefore \(ab\) divides \(c\text{.}\)