Logic and Mathematical Reasoningan introduction to proof writing

Each of the following statements are equivalent.

Since \(a,b\) are relatively prime, we have \(\gcd(a,b) = 1\text{,}\) and by a theorem from class there are integers \(x,y\) for which \(ax+by = 1\text{.}\) Multiplying by \(c\) we find \(acx+bcy = c\text{.}\) Now since \(a,b\) both divide \(c\text{,}\) there exist integers \(j,k\) so that \(c = aj\) and \(c = bk\text{.}\) Applying these to the above equation yields

Thus \(ab\) divides \(c\) since \(kx+jy\) is a integer.

\(\neg (\forall x \in \mathbb{R}, \exists y \in \mathbb{R}, y^2 = x) \text{.}\)

Since \(a \mid b\) and \(b \mid c\) there are integers \(m,n\) so that \(b = ma\) and \(c = nb\text{.}\) Therefore \(c = nb = n(ma) = (nm)a\text{.}\) Since \(nm\) is an integer, \(a \mid c\text{.}\)

This shows that the divides relation on \(\mathbb{N}\) is transitive.

The minimal elements of the divides relation on \(N\) are precisely the prime numbers. Indeed, suppose \(p \in \mathbb{N}\) is prime. Then its only divisors are \(1,p\text{,}\) but \(1 \notin N\text{.}\) Therefore, for \(a \in N\text{,}\) \(a \mid p\) if and only if \(a = p\text{,}\) hence \(p\) is minimal.

Similarly, if \(b \in N\) is not prime, then there is some divisor \(a\) with \(1 \lt a \lt b\text{,}\) so \(a \in N\) and \(a \mid b\text{.}\) Therefore, \(b\) is not minimal.

Suppose \(n\) is odd so that there is an integer \(m\) satisfying \(n = 2m+1\text{.}\) Then

Since \((m+1)(m+2)\) is the product of two consecutive integers, it must be even (we proved this elsewhere), and hence there is some integer \(k\) so that \((m+1)(m+2) = 2k\text{.}\) Hence \(n^2+4n+3 = 8k\text{.}\)

Given an odd positive integer \(n\text{,}\) we can write it as \(2m+1\) for some nonnegative integer \(m\text{.}\) Our proof will proceed by induction on \(m\text{.}\)

Base case: \(m=0\). Notice that when \(m=0\text{,}\) then \(n=1\text{,}\) and \(n^2+4n+3 = 1+4+3 = 8\) which is obviously divisible by \(8\text{.}\)

Inductive step. Suppose that \(m\) is a nonnegative integer and for \(n = 2m+1\text{,}\) the quantity \(n^2+4n+3\) is divisible by \(8\) so that there is some integer \(k\) for which \(n^2+4n+3 = 8k\text{.}\) Then consider \(r = 2(m+1)+1 = 2m+3 = n+2\text{.}\) Then

which is divisible by \(8\text{.}\)

By induction, we have proven the desired result.

Supose \(\sqrt{p}\) were rational, so that \(\sqrt{p} = \frac{a}{b}\) for some integers \(a,b\text{.}\) We may assume that the fraction is in lowest terms, i.e., \(\gcd(a,b) = 1\text{.}\) Then squaring our equation and multiplying both sides by \(b^2\text{,}\) we find \(b^2 p = a^2\text{.}\) In particular, this means \(p\) divides \(a^2\text{.}\) Since \(p\) is prime, by Euclid's lemma we know \(p\) divides \(a\text{.}\) Thus \(a = pk\) for some integer \(k\text{.}\) Hence \(b^2 p = a^2 = (pk)^2 = p^2 k^2\text{.}\) Cancelling one of the factors of \(p\text{,}\) we obtain \(b^2 = p k^2\text{,}\) and hence \(p\) divides \(b\text{.}\) This implies \(\gcd(a,b) \ge p \gt 1\text{,}\) contradicting the fact that \(\frac{a}{b}\) is in lowest terms. Therefore \(\sqrt{p}\) is irrational.

Notice that \(\gcd(a,b)\) divides \(f(x,y) = ax+by\) for any \(x,y\) since it divides each of \(a,b\text{.}\) Therefore, if \(\gcd(a,b) \gt 1\text{,}\) then \(f(x,y) \not= 1\) for any \(x,y \in \mathbb{Z}\text{,}\) hence \(f\) is not surjective.

Now suppose \(\gcd(a,b) = 1\text{.}\) Then by a theorem from class, there exist some integers \(r,s \in \mathbb{Z}\) so that \(ar+bs = 1\text{.}\) Now let \(c \in \mathbb{Z}\) be arbitrary. Notice that \(arc + bsc = (ar+bs)c = c\text{,}\) and so letting \(x = rc \in \mathbb{Z}\text{,}\) \(y = sc \mathbb{Z}\text{,}\) we have \(f(x,y) = c\text{.}\) Since \(c \in \mathbb{Z}\) was arbitrary, \(f\) is surjective.

Apply the equivalence from the first question in this review.

No, \(f\) is not well-defined because \(f(\frac{1}{2}) = \frac{1}{2}\text{,}\) and \(f(\frac{2}{4}) = \frac{1}{4}\text{,}\) but \(\frac{1}{2} \not= \frac{1}{4}\)

Let \(f : A \to \mathbb{Z}\) be defined as follows: for \(n \in A\text{,}\) let \(f(n)\) be the quotient (from the Division Algorithm) of \(n\) divided by \(5\text{.}\) This function is well-defined because the quotient from the Division Algorithm always exists, and is a unique integer.

Now consider the function \(g : \mathbb{Z} \to A\) defined by \(g(k) = 5k+2\text{.}\) Clearly, \(g(k) \in A\) since it has remainder \(2\) when divided by \(5\text{,}\) and so \(A\) is a valid codomain.

Now consider \(g \circ f\text{.}\) Let \(n \in A\text{.}\) By the Division Algorithm, there exist unique integers \(q,r\) with \(0 \le r \lt 5\) so that \(n = 5q + r\text{.}\) Since \(n \in A\text{,}\) \(r = 2\text{.}\) Moreover, by definition \(f(n) = q\text{.}\) Therefore \((g \circ f)(n) = g(f(n)) = g(q) = 5q+2 = n\text{.}\) Hence \(g \circ f = Id_A\text{.}\)

Now consider \(f \circ g\text{.}\) Let \(n \in \mathbb{Z}\text{.}\) Notice that \(g(n) = 5n+2\text{.}\) Since the quotient from the division algorithm is unique, it is clear that \((f \circ g)(n) = f(5n+2) = n\) so that \(f \circ g = Id_{\mathbb{Z}}\text{.}\)

By a result proven previously, since \(f \circ g\) and \(g \circ f\) are the identity on their respective domains, we know that \(f,g\) are bijective and in fact \(f^{-1} = g\text{,}\) thereby proving the result.

Notice that the inclusion map \(f : [0,\infty) \to \mathbb{R}\) givne by \(f(x) = x\) is obviously injective. Moreover, consider the function \(g : \mathbb{R} \to [0,\infty)\) given by \(g(x) = 2^x\text{.}\) The codomain \([0,\infty)\) is valid since \(2^x \gt 0\) for all \(x \in \mathbb{R}\text{.}\) Moreover, \(g\) is strictly increasing (using, say, the first derivative test from calculus), and therefore injective. Indeed, if \(x \not= y\text{,}\) then \(x \lt y\) (or \(y \lt x\)). Thus, \(g(x) \lt g(y)\) (or \(g(y) \lt g(x)\)), and hence \(g(x) \not= g(y)\text{.}\) Since we have injective functions in both directions, by the Cantor--Schröder--Bernstein theorem, these sets have the same cardinality.

Let \(2\mathbb{Z}\) be the even integers and \(2\mathbb{Z}+1\) be the odd integers. Clearly these sets are disjoint since no integer is both even and odd. Moreover, any integer is either even or odd, so the union of these two sets is all of \(\mathbb{Z}\text{.}\) Therefore this set is a partition.

The equivalence relation which generates this partition is congurence moduluo \(2\text{.}\)

Consider the family of singletons (i.e., sets with only one element) \(\mathcal{A} := \{ \{x\} \mid x \in \mathbb{Z} \}\text{.}\) Clearly, any two sets from this family are either the same or disjoint since they each only contain a single element. Moreover, the union is all of \(\mathbb{Z}\) since for any \(x \in \mathbb{Z}\text{,}\) \(x \in \{x\} \subset \bigcup_{A \in \mathcal{A}} A\text{.}\)

The equivalence relation which generates this partition is equality.

This appeared on a previous review sheet.

For any prime \(p\text{,}\) we know that \(p \gt 2\) and therefore \(p \mathbb{N} \subseteq \mathbb{N} \setminus \{1\}\text{.}\) Taking the union over all \(p\) prime, we find

We now prove the other inclusion by using the well-ordering principle. Suppose the difference \(D := (\mathbb{N} \setminus \{1\}) \setminus \Big( \bigcup_{p\ \text{prime}} p\mathbb{N} \Big)\) is nonempty. Then since \(\mathbb{N}\) is well-ordered, \(D\) has a minimum element which we will call \(n\text{.}\)

Since \(n \gt 1\text{,}\) are two possibilities, either \(n\) is prime, in which case \(n \in n\mathbb{N}\) contradicting the fact that \(n \in D\text{;}\) or \(n\) is composite. In this case, \(n = ab\) for some \(1 \lt a,b \lt n\text{.}\) Since \(1 \lt a \lt n\text{,}\) \(a \notin D\text{,}\) and therefore \(a \in p\mathbb{N}\) for some prime \(p\text{.}\) Hence, \(a = pk\) for some \(k \in \mathbb{N}\text{.}\) Finally, \(n = ab = p(kb) \in p\mathbb{N}\text{,}\) contradicting the fact that \(n \in D\text{.}\) Therefore our assumption that \(D\) is nonempty is false. This proves the other inclusion.

It is clear that if \(A=B\) then \(A \setminus B = B \setminus A = A \setminus A = \emptyset\text{.}\)

For the other direction, suppose that \(A \setminus B = \emptyset = B \setminus A\text{.}\) Let \(x \in A\text{.}\) Then since \(A \setminus B = \emptyset\text{,}\) we know that \(x \in B\) (for otherwise \(x \in A \setminus B\)). Therefore \(A \subseteq B\text{.}\) A symmetric argument proves \(B \subseteq A\text{.}\)

This result says precisely that the divides relation is antisymmetric on \(\mathbb{N}\text{.}\) No, it does not hold on \(\mathbb{Z}\text{.}\) Indeed, if \(m \in \mathbb{Z}\) and \(m \not= 0\text{,}\) then \(m = (-1)(-m)\text{,}\) and therefore \(m \mid -m\text{.}\) Similarly \(-m = (-1)m\) so \(-m \mid m\text{.}\) But \(m \not= -m\) since \(m \not= 0\text{.}\)

Base case. For \(n = 1\text{,}\) the quantity on the left is \(\sum_{k=1}^1 (-1)^{1-1} (1-1) = 0 = \frac{-1+1}{2}\text{.}\)

Inductive step. Now suppose \(n \in \mathbb{N}\) and the formula holds for \(n\text{.}\)

Case 1: \(n\) even. In this case, \((-1)^{n+1} = -1\) since \(n+1\) is odd. By the inductive hypothesis, we have

as desired.

Case 1: \(n\) odd. In this case, \((-1)^{n+1} = 1\) since \(n+1\) is even. By the inductive hypothesis, we have

as desired.

Don't look now, but if we use this formula to define a function \(f : \mathbb{N} \to \mathbb{Z}\text{,}\) by \(f(n) = \sum_{k=1}^n (-1)^n (n-1)\text{,}\) then \(f\) is just our usual bijection between \(\mathbb{N}\) and \(\mathbb{Z}\text{.}\)

We need to show that the resulting equivalence class under addition is independent of the choice of representatives from the original equivalence classes. So, suppose \((x',y') \in [(x,y)]\) and \((a',b') \in [(a,b)]\text{.}\) Then \(x - y = x' - y'\) and \(a - b = a' - b'\text{.}\) Therefore,

and hence \([(x+a,y+b)] = [(x'+a',y'+b')]\text{.}\) Thus \(+\) is well-defined.

Clearly \(\sim\) is reflexive since for any \(x \in \mathbb{R}\text{,}\) \(x-x=0 \in \mathbb{Q}\) and so \(x \sim x\text{.}\)

Now suppose \(x,y \in \mathbb{R}\) and \(x \sim y\text{.}\) Then \(x-y \in \mathbb{Q}\) and hence \(y-x = -(x-y) \in \mathbb{Q}\text{.}\) Therefore \(\sim\) is symmetric.

For transitivity it will be helpful to remember that \(\mathbb{Q}\) is closed under addition since \(\frac{p}{q} + \frac{a}{b} = \frac{pb+aq}{qb}\text{.}\) Suppose that \(x \sim y\) and \(\sim z\text{.}\) Then \(x-y \in \mathbb{Q}\) and \(y-z \in \mathbb{Q}\text{.}\) Therefore \(x-z = (x-y)+(y-z) \in \mathbb{Q}\) and hence \(x \sim z\text{.}\) Thus \(\sim\) is transitive.

Since \(\sim\) is reflexive, symmetric and transitive, it is an equivalence relation.

Suppose \(n\) is even. Then \(n = 2k\) for some integer \(k\text{.}\) Thus \(n^2 = (2k)^2 = 4k^2 = 2(2k^2)\) which is even since \(2k^2 \in \mathbb{Z}\text{.}\)

For the other direction we use contraposition. Suppose \(n\) is not even. Then by parity \(n\) is odd and hence \(n = 2k+1\) for some integer \(k\text{.}\) Thus

which is odd since \(2k^2 +2k \in \mathbb{Z}\text{.}\) Thus \(n^2\) is not even. By contraposition, if \(n^2\) is even then \(n\) is even.

Euclid's lemma.

If \(n\) is a multiple of \(p\text{,}\) then \(n = pk\) for some integer \(k\text{.}\) Thus \(n^2 = (pk)^2 = p(pk^2)\) is a multiple of \(p\) as well.

For the oher direction, if \(n^2\) is a multiple of \(p\) then \(n^2 = pk\) for some integer \(p\text{.}\) This means that \(p\) divides \(n^2\text{.}\) By Euclid's lemma \(p\) divides \(n\text{.}\)

Consider \(p=4\) and \(n=2\text{.}\) Then \(n^2 = 4 = p\) is a multiple of \(p\text{,}\) but \(n\) is not a multiple of \(p\) since it is less than \(p\text{.}\)