Logic and Mathematical Reasoningan introduction to proof writing

Call the left-hand set \(A\) and the right-hand set \(B\text{.}\)

(\(\subseteq\)). Suppose that \(x \in A\text{.}\) Then \(x = 5m+2\) for some \(m \in \mathbb{Z}\text{.}\) Thus \(x = 5m+2+3-3 = 5(m+1)-3\text{.}\) Since \(m+1 \in \mathbb{Z}\text{,}\) we find \(x \in B\text{.}\)

(\(\supseteq\)). Suppose that \(x \in B\text{.}\) Then \(x = 5n-3\) for some \(n \in \mathbb{Z}\text{.}\) Thus \(x = 5n-3-2+2 = 5(n-1)+2\text{.}\) Since \(n-1 \in \mathbb{Z}\text{,}\) we find \(x \in A\text{.}\)

Let \(m,n \in \mathbb{Z}\) be arbitrary.

(\(\Rightarrow\)). Suppose that \(m\mathbb{Z} \subseteq n\mathbb{Z}\text{.}\) Since \(m = m\cdot 1 \in m\mathbb{Z}\text{,}\) we know that \(m \in n\mathbb{Z}\text{.}\) Therefore \(m = nk\) for some \(k \in \mathbb{Z}\text{.}\) Thus \(n \mid m\text{.}\)

(\(\Leftarrow\)). Suppose that \(n \mid m\text{.}\) Then \(m = nk\) for some \(k \in \mathbb{Z}\text{.}\) Now take any element \(x \in m\mathbb{Z}\text{.}\) By definition of \(m\mathbb{Z}\text{,}\) we find \(x = mj\) for some integer \(j\text{.}\) Hence \(x = mj = (nk)j = n(kj)\text{.}\) Since \(kj\) is an integer, we find \(x \in n\mathbb{Z}\text{.}\)

#6.

1. \(A = \{1,2\}, B = \{2\}, C = \{1\}\)
2. \(A = B = C = \{1\}\)
3. \(A = B = \{1\}, C = \{1,2\}\)
4. \(A = \{1,3\}, B = \{2,3\}, C = \{3\}\)
5. \(A = \{1\}, B = C = \{1,2\}\)
6. \(A = B = C = \{1\}\)

#9.

1. Note that \(A \subseteq B\) if and only if \(\forall x, x \in A \implies x \in B\) if and only if \(\forall x, \neg( x \in A \wedge x \notin B)\) if and only if \(\forall x, \neg( x\in A \setminus B)\) if and only if \(\forall x, x \notin A \setminus B\) if and only if \(A \setminus B = \varnothing\text{.}\) Note that the second “if and only if” uses the equivalence \((P \implies Q) \iff \neg(P \wedge \neg Q)\text{,}\) and the last “if and only if” uses the definition of the empty set.
2. Suppose \(A \subseteq B \cup C\) and \(A \cap B = \varnothing\text{.}\) Take an \(x \in A\text{.}\) Then since \(A \subseteq B \cup C\text{,}\) \(x \in B \cup C\text{,}\) so either \(x \in B\) or \(x \in C\text{.}\) But since \(A \cap B = \varnothing\text{,}\) it cannot be the case that \(x \in B\) (for otherwise \(x \in A \cap B\)), and therefore \(x \in C\text{.}\) Hence \(A \subseteq C\text{.}\)
3. (\(\Rightarrow\)) Suppose that \(C \subseteq A \cap B\text{.}\) Since \(A \cap B \subseteq A\) and \(A \cap B \subseteq B\text{,}\) we can use the small lemma we proved in class to conclude \(C \subseteq A\) and \(C \subseteq B\text{.}\) (\(\Leftarrow\)) Suppose that \(C \subseteq A\) and \(C \subseteq B\text{.}\) Now take any \(x \in C\text{.}\) By the first condition, \(x \in A\text{,}\) and by the second condition, \(x \in B\text{.}\) Therefore \(x \in A \cap B\text{.}\) Thus \(C \subseteq A \cap B\text{.}\)
4. Suppose \(A \subseteq B\text{.}\) Now take any \(x \in A \setminus C\text{.}\) Then \(x \in A\) and \(x \notin C\text{.}\) Since \(A \subseteq B\text{,}\) we have \(x \in B\text{.}\) Thus \(x \in B \setminus C\text{.}\) Therefore \(A \setminus C \subseteq B \setminus C\text{.}\)
5. Notice that for any \(x\text{,}\) \begin{align*} x \in (A \setminus B) \setminus C &\iff x \in A \setminus B \wedge x \notin C\\ &\iff x \in A \wedge x \notin B \wedge x \notin C\\ &\iff (x \in A \wedge x \notin C \wedge x \notin B) \vee (x \in A \wedge x \notin C \wedge x \in C)\\ &\iff (x \in A \wedge x \notin C) \wedge (x \notin B \vee x \in C)\\ &\iff (x \in A \wedge x \notin C) \wedge \neg(x \in B \wedge x \notin C)\\ &\iff x \in (A \setminus C) \wedge x \notin (B \setminus C)\\ &\iff x \in (A \setminus C) \setminus (B \setminus C) \end{align*}
6. Suppose \(A \subseteq C\) and \(B \subseteq C\text{.}\) Take any \(x \in A \cup B\text{.}\) Then either \(x \in A\) or \(x \in B\text{.}\) If \(x \in A\text{,}\) then \(x \in C\) since \(A \subseteq C\text{.}\) If \(x \in B\text{,}\) then \(x \in C\) since \(B \subseteq C\text{.}\) Either way, \(x \in C\text{.}\) Thus \(A \cup B \subseteq C\text{.}\)
7. Take any \(x \in (A \cup B) \cap C\text{.}\) Then \(x \in A \cup B\) and \(x \in C\text{.}\) So, either \(x \in A\) or \(x \in B\text{.}\) Case 1: \(x \in A\). Then certainly \(x \in A \cup (B \cap C)\text{.}\) Case 2: \(x \in B\). Then since \(x \in C\) also, we have \(x \in B \cap C\text{.}\) Hence \(x \in A \cup (B \cap C)\text{.}\)
8. To show \((A \setminus B) \cap B = \varnothing\text{,}\) we will use a proof by contradiction. That is, suppose this set is nonempty. Then there is some \(x \in (A \setminus B) \cap B\text{,}\) so \(x \in A \setminus B\) and \(x \in B\text{.}\) But the first statement means \(x \in A\) and \(x \notin B\text{,}\) contradicting the fact that \(x \in B\text{.}\) Hence our assumption (that \((A \setminus B) \cap B \not= \varnothing\)), was false. Therefore \((A \setminus B) \cap B = \varnothing\text{.}\)

#10. Let \(A,B,C,D\) be sets.

1. Suppose \(C \subseteq A\) and \(D \subseteq B\text{.}\) Take any \(x \in C \cap D\text{.}\) Then \(x \in C\) and \(x \in D\text{.}\) Since \(C \subseteq A\) we have \(x \in A\text{.}\) Since \(D \subseteq B\) we have \(x \in B\text{.}\) Thus \(x \in A \cap B\text{.}\)
2. Suppose \(C \subseteq A\) and \(D \subseteq B\text{.}\) Take any \(x \in C \cup D\text{.}\) Then \(x \in C\) or \(x \in D\text{.}\) Case 1: suppose \(x \in C\text{.}\) Since \(C \subseteq A\) we have \(x \in A\text{.}\) Case 2: suppose \(x \in D\text{.}\) Since \(D \subseteq B\) we have \(x \in B\text{.}\) So \(x \in A\) or \(x \in B\text{.}\) Thus \(x \in A \cup B\text{.}\)
3. Suppose \(C \subseteq A\) and \(D \subseteq B\text{,}\) and \(A,B\) are disjoint. Then \(A \cap B = \varnothing\text{,}\) and by #10(a), \(C \cap D \subseteq A \cap B = \varnothing\text{.}\) But trivially \(\varnothing \subseteq C \cap D\text{,}\) so they must be equal.
4. Suppose \(C \subseteq A\) and \(D \subseteq B\text{.}\) Take any \(x \in D \setminus A\text{.}\) Then \(x \in D\) and \(x \notin A\text{.}\) Since \(D \subseteq B\) we have \(x \in B\text{.}\) Since \(C \subseteq A\text{,}\) by an earlier problem, we have \(x \notin C\text{.}\) Thus \(x \in B \setminus C\text{.}\)
5. Suppose \(A \cup B \subseteq C \cup D\text{,}\) \(A \cap B = \varnothing\text{,}\) and \(C \subseteq A\text{.}\) Take \(x \in B\text{,}\) then \(x \in A \cup B\text{,}\) and hence \(x \in C \cup D\text{.}\) Since \(A \cap B = \varnothing\text{,}\) \(x \notin A\text{,}\) and hence \(x \notin C\text{.}\) But because \(x \in C \cup D\) and \(x \notin C\text{,}\) we must have \(x \in D\text{.}\) Therefore \(B \subseteq D\text{.}\)

Claim: \(\bigcup_{n=1}^{\infty} A_n = B \cup C\) where \(B := \{5,6,10,11,12,15,16,17,18\}\) and \(C := \{ n \in \mathbb{N} \mid n \ge 20\}\text{.}\)

“\(\subseteq\)” Suppose \(x \in \bigcup_{n=1}^{\infty} A_n\text{.}\) Then there is some \(k \in \mathbb{N}\) so that \(x \in A_k\text{.}\)

Case 1: \(k=1,2,3\text{.}\) Then \(x \in A_k \subseteq A_1 \cup A_2 \cup A_3 = B\text{,}\) and hence \(x \in B \cup C\text{.}\)

Case 2: \(k \ge 4\text{.}\) Then \(20 = 5 \cdot 4 \le 5k \le x \le 6k\text{,}\) and hence \(x \in C\text{,}\) so \(x \in B \cup C\text{.}\)

“\(\supseteq\)” Suppose that \(x \in B \cup C\text{.}\) If \(x \in B = A_1 \cup A_2 \cup A_3\text{,}\) then clearly \(x \in \bigcup_{n=1}^{\infty} A_k\text{.}\)

If \(x \in C\text{,}\) then by the division algorithm we can write \(x = 5m + r\) with \(m,r \in \mathbb{Z}\) and \(0 \le r \lt 5\text{.}\) Moreover, since \(20 = 5 \cdot 4\) and \(x \ge 20\) we know \(m \ge 4\text{.}\) Furthermore, \(x = 5m+r \le 5m + 4 \le 5m + m = 6m\text{.}\) Thus \(5m \le x \le 6m\text{.}\) And therefore \(x \in A_m\text{,}\) so \(x \in \bigcup_{n=1}^{\infty} A_n\text{.}\)

Simple example: set \(A_n = \{\frac{1}{n+1}\}\text{.}\) Clearly these are pairwise disjoint because each only contains one element and none of those are the same one. Moreover, \(0 \lt \frac{1}{n+1} \lt 1\text{,}\) so \(A_n \subseteq (0,1)\text{.}\)

Consider \(A_n = (0,\frac{1}{n})\text{.}\) Clearly, if \(n \lt m\text{,}\) then \(\frac{1}{m} \lt \frac{1}{n}\text{,}\) so \(A_m \subseteq A_n\) and thus \(A_m \cap A_n = A_m \not= \emptyset\text{.}\)

We now prove the intersection over all the sets is empty. Let \(x \in \mathbb{R}\text{.}\) If \(x \le 0\text{,}\) then \(x \notin A_1\text{,}\) so \(x\) is not in the intersection either. Now suppose \(x \gt 0\text{.}\) Then by the Archimedean Principle, there is some \(n \in \mathbb{N}\) so that \(\frac{1}{n} \lt x\text{.}\) Therefore \(x \notin A_n\) and hence \(x\) is not in the intersection either. This proves that every \(x\) is not in the intersection, so the intersection is empty.

Subset based solution.

“\(\subseteq\)” Suppose that \(x \in B \cap \bigcup_{\alpha \in I} A_{\alpha}\text{,}\) then \(x \in B\) and \(x \in \bigcup_{\alpha \in I} A_{\alpha}\text{.}\) Thus there is some \(\beta \in I\) so that \(x \in A_{\beta}\text{.}\) Therefore \(x \in B \cap A_{\beta}\text{.}\) Hence \(x \in \bigcup_{\alpha \in I} (B \cap A_{\alpha})\text{.}\)

“\(\supseteq\)” Suppose that \(x \in \bigcup_{\alpha \in I} (B \cap A_{\alpha})\text{,}\) then there is some \(\beta \in I\) so that \(x \in B \cap A_{\beta}\text{.}\) Thus \(x \in B\) and \(x \in B \cap A_{\beta}\text{.}\) Hence \(x \in \bigcup_{\alpha \in I} A_{\alpha}\text{.}\) Therefore \(x \in B \cap \bigcup_{\alpha \in I} A_{\alpha}\text{.}\)

The equality \(B \cup \bigcap_{\alpha \in I} A_{\alpha} = \bigcap_{\alpha \in I} (B \cup A_{\alpha})\) follows from a similar proof.

Iff based solution.

Note that

and

We claim that \(\bigcap_{n=1}^\infty A_n = \emptyset\text{.}\)

Since our universe is \(\mathbb{N}\text{,}\) it suffices to show that every \(k \in \mathbb{N}\) is not in this intersection. So, let \(k \in \mathbb{N}\) be arbitrary. Notice that \(k \lt k+1\text{,}\) so \(k \not\ge k+1\text{.}\) Therefore \(k \notin A_{k+1}\text{.}\) Hence \(k \notin \bigcap_{n=1}^{\infty} A_n\text{.}\)

• \(\{n \mid n = 2^k\ \text{for some}\ k \in \mathbb{N} \} := \{a_n \mid n \in \mathbb{N}\}\text{,}\) where \(a_1 := 2\) and \(a_{n+1} = 2 a_n\text{.}\)
• \(\{a,a+d,a+2d,a+3d,\ldots\} := \{a_n \mid n \in \mathbb{N}\}\text{,}\) where \(a_1 := a\) and \(a_{n+1} = a_n + d\text{.}\)
• \(\{a,ar,ar^2,ar^3,\ldots\} := \{a_n \mid n \in \mathbb{N}\}\text{,}\) where \(a_1 := a\) and \(a_{n+1} = a_n r\text{.}\)
• \(\bigcup_{i=1}^n A_i\text{,}\) for some indexed family \(\{ A_i \mid i \in \mathbb{N} \}\text{.}\) \(\bigcup_{i=1}^1 A_i := A_1\text{,}\) and \(\bigcup_{i=1}^{n+1} A_i := \left( \bigcup_{i=1}^n A_i \right) \cup A_{n+1}\text{.}\)

For the base case, notice that

Now let \(k \in \mathbb{N}\) be arbitrary and assume \(\sum_{j=1}^k j \cdot j! = (k+1)! - 1\text{.}\) Then we find

For the base case, notice that

Now let \(k \in \mathbb{N}\) be arbitrary and assume \(\sum_{j=1}^k \frac{1}{j(j+1)} = \frac{k}{k+1}\text{.}\) Then we find

For the base case, notice that

Now let \(k \in \mathbb{N}\) be arbitrary and assume \(\prod_{j=1}^k \left( 1 - \frac{1}{j+1} \right) = \frac{1}{n+1}\text{.}\) Then we find

For the base case, notice that

Now let \(k \in \mathbb{N}\) be arbitrary and assume \(\prod_{j=1}^k (2j-1) = \frac{(2k)!}{k!2^k}\text{.}\) Then we find

For the base case, notice that

Now let \(k \in \mathbb{N}\) be arbitrary and assume \(\sum_{j=0}^{k-1} ar^j = \frac{a(r^k-1)}{r-1}\text{.}\) Then we find

For the base case, notice that \(8 = 1 \cdot 8\text{,}\) so \(8 \mid 8 = 9^1 -1\text{.}\) Now let \(k \in \mathbb{N}\) be arbitrary and assume \(8 \mid 9^k-1\text{.}\) Then there is some \(j \in \mathbb{Z}\) for which \(9^k-1 = 8j\text{.}\) Thus

Since \(9j+1 \in \mathbb{Z}\text{,}\) we have shown \(8 \mid 9^{k+1}-1\text{.}\) By induction we have proven the result.

For the base case, notice that

Now let \(k \in \mathbb{N}\) be arbitrary and assume \(\sum_{j=1}^k \frac{1}{j^2} \le 2-\frac{1}{k}\text{.}\) Before we continue, recall the following elementary fact: for any \(m \in \mathbb{N}\text{,}\) \(\frac{1}{m(m+1)} = \frac{1}{m}-\frac{1}{m+1}\text{.}\) Then we find

By induction we have proven the result.

For the base case, notice that \((1+x)^1 = 1+x = 1+1x\text{.}\) Now let \(k \in \mathbb{N}\) be arbitrary and assume \((1+x)^k \ge 1+kx\text{.}\) Then we find (be careful! the first inequality only works because \(1+x>0\) since \(x>0\text{.}\) When you start introducing other variables, you have to pay attention to such things!)

where the last inequality follows because \(k \in \mathbb{N}\) and so \(kx^2>0\text{.}\) By induction we have proven the result.

The base case is one of our assumptions, so we are done. Now let \(k \in \mathbb{N}\) be arbitrary and assume \(\frac{d}{dx}(x^k) = kx^{k-1}\text{.}\) Then notice that \(x^{k+1} = x^k \cdot x\text{,}\) so applying the product rule to the functions \(f(x) = x^k\) and \(g(x) = x\text{,}\) we find

By induction we have proven the result.

Use a standard fact from geometry.

The base case, when \(n=3\) is just the standard geometry fact that the sum of the interior angle measures of a triangle is always \(\pi\text{.}\)

Now suppose \(k \in \mathbb{N}\) and \(k \gt 2\) (i.e., \(k \ge 3\)) and the sum of the interior angle measures of any \(k\)-sided polygon is \((k-2) \pi\text{.}\) Then consider any \((k+1)\)-sided polygon. Pick a vertex \(v\) with an interior angle less than \(\pi\text{.}\) Since \(k+1 \ge 4\text{,}\) we can divide our polygon into a triangle and a \(k\)-sided polygon by drawing a line between the vertices adjacent to \(v\text{.}\) Moreover, the sum of the interior angles of the original \((k+1)\)-gon is the same as the sum of the interior angles of the \(k\)-gon (which has total angle measure \((k-2) \pi\)) added to the sum of the interior angles of the triangle (which has total angle measure \(\pi\text{.}\) Therefore, the total angle measure for the \((k+1)\)-gon is \((k-2) \pi + \pi = ((k+1)-2) \pi\text{,}\) as desired.

For each relation, determine if it is reflexive, symmetric or transitive.

1. #1(f). The relation is \(\not=\) on \(\mathbb{N}\text{.}\) This is not reflexive since \(x = x\) for any \(x \in \mathbb{N}\text{.}\) This relation is symmetric since \(x \not= y\) if and only if \(y \not= x\text{.}\) This relation is not transitive on \(\mathbb{N}\) since \(0 \not= 1\) and \(1 \not= 0\text{,}\) but \(0 = 0\text{.}\)
2. #1(h). The relation is \(R := \{ (x,y) \in \mathbb{Z}^2 \mid x+y = 10 \}\) on \(\mathbb{Z}\text{.}\) This relation is not reflexive since \(0+0 = 0 \not= 10\) and hence \((0,0) \notin R\text{.}\) This relation is symmetric since \((x,y) \in R\) if and only if \(x+y = 10\) if and only if \(y+x = 10\) if and only if \((y,x) \in R\text{.}\) This relation is not transitive since \((4,6) \in R\) and \((6,4) \in R\text{,}\) but \((4,4) \notin R\text{.}\)
3. #1(k). The relation is \(R\) on \(\mathbb{R}^2\) where \((x,y)\ R\ (z,w)\) if and only if \(x+z \le y+w\text{.}\) This relation is not reflexive since \(1+1 \not\le 0+0\text{,}\) so \((1,0)\ \not R\ (1,0)\text{.}\) This relation is symmetric since \((x,y)\ R\ (z,w)\) if and only if \(x+z \le y+w\) if and only if \(z+x \le w+y\) if and only if \((z,w)\ R\ (x,y)\text{.}\) This relation is not transitive since \((1,0)\ R\ (0,1)\) and \((0,1)\ R\ (1,0)\text{,}\) but \((1,0)\ \not R\ (1,0)\text{.}\)
4. #1(m). The relation is \(T\) on \(\mathbb{R}^2\) where \((x,y)\ R\ (z,w)\) if and only if \(x+y \le z+w\text{.}\) This relation is reflexive since for any \((x,y) \in \mathbb{R}^2\text{,}\) \(x+y \le x+y\) and so \((x,y)\ T\ (x,y)\text{.}\) This relation is not symmetric since \(0+0 \le 1+1\) but \(1+1 \not\le 0+0\text{,}\) so \((0,0)\ T\ (1,1)\) but \((1,1)\ \not T\ (0,0)\text{.}\) This relation is transitive. Indeed, if \((x,y)\ T\ (z,w)\) and \((z,w)\ T\ (a,b)\text{,}\) then \(x+y \le z+w\) and \(z+w \le a+b\text{,}\) and therefore \(x+y \le a+b\text{.}\) Thus \((x,y)\ T\ (a,b)\) and so \(T\) is transitive.

We will prove that the relation \(R\) on \(\mathbb{R}\) given by \(x\ R\ y\) if and only if \(x - y \in \mathbb{Q}\) is an equivalence relation. Note that this relation is reflexive since for any \(x \in \mathbb{R}\text{,}\) \(x - x = 0 \in \mathbb{Q}\text{,}\) and hence \(x\ R\ x\text{.}\) This relation is also symmetric since for any \(x,y \in \mathbb{R}\text{,}\) we have \(x-y \in \mathbb{Q}\) if and only if \(y-x = -(x-y) \in \mathbb{Q}\text{.}\) This relation is also transitive since for any \(x,y,z \in \mathbb{R}\text{,}\) if \(x\ R\ y\) and \(y\ R\ z\text{,}\) then \(x-y, y-z \in \mathbb{Q}\text{,}\) and hence \(x-z = (x-y)+(y-z) \in \mathbb{Q}\text{.}\)

We give the equivalence classes of \(0,1,\sqrt{2}\) below.

We will prove that the relation \(V\) on \(\mathbb{R}\) given by \(x\ V\ y\) if and only if \(x = y\) or \(xy = 1\) is an equivalence relation.

This relation is reflexive since for any \(x \in \mathbb{R}\text{,}\) \(x = x\) and hence \(x\ V\ x\text{.}\) This relation is symmetric since \(x = y\) or \(xy = 1\) if and only if \(y = x\) or \(yx = 1\text{.}\) This relation is also transitive, for if \(x,y,z \in \mathbb{R}\) and \(x\ V\ y\) and \(y\ V\ z\text{,}\) then if either \(x = y\) or \(y = z\text{,}\) the conclusion \(x\ V\ z\) is trivial. So suppose \(x \not= y \not= z\text{.}\) Thus \(xy = 1 = yz\text{,}\) and so none of \(x,y,z\) are zero. Dividing by \(z\) we obtain \(x = z\text{.}\) Thus \(x\ V\ z\text{.}\)

We give the equivalence classes of \(3,-\frac{2}{3},0\) below.

Consider the relation \(R\) on the set of differentiable functions defined by \(f\ R\ g\) if and only if \(f\) and \(g\) have the same first derivative. This relation is obviously an equivalence relation because it is a notion of “sameness.” By the Fundamental Theorem of Calculus, the elements of any equivalence class are things of the form \(f + C\) where \(C\) is an arbitrary constant.

Notice that \(a_1 = 2 = 2^1\) and \(a_2 = 4 = 2^2\text{.}\) Now let \(m \ge 3\) and suppose that for all \(k \lt m\text{,}\) \(a_k = 2^k\text{.}\) Then by the definition of \(a_m\) and the strong inductive hypothesis, we find

Notice that \(5\) obviously divides \(5 = 8-3 = 8^1-3^1\text{.}\) Now suppose \(m \in \mathbb{N}\) and assume \(5\) divides \(8^m-3^m\) so that \(8^m-3^m =5k\) for some integer \(k\text{.}\) Then

Suppose to the contrary that not every natural number greater than \(1\) has a prime divisor. Then by the Well-Ordering Principle, there is a smallest such natural number, which we will call \(n\text{.}\) Note that \(n\) cannot be prime, for otherwise it would be its own prime divisor. Therefore, since \(n \gt 1\text{,}\) it must be composite, and thus \(n = mk\) for \(1 \lt m,k \lt n\text{.}\) By the minimality of \(n\text{,}\) \(k\) must have a prime divisor, which we call \(p\text{.}\) Thus \(k=xp\) for some integer \(x\text{.}\) Finally, \(n=mk=mxp\) and hence \(p\) is a prime divisor of \(n\text{,}\) which is a contradiction. Therefore, every natural number greater than \(1\) has a prime divisor.

1. \(A = \{0,1\}\text{,}\) \(R = \{(0,0)\}\text{.}\) \(R\) is not reflexive because \(1\,\not R\,1\text{,}\) and it is not irreflexive because \(0\,R\,0\text{.}\)
2. \(A = \{0,1,2,\}\text{,}\) \(R = \{(0,0),(1,1),(2,2),(0,1),(1,0),(1,2),(2,1)\}\text{.}\) This is clearly reflexive and symmetric, but it is not transitive because \(0\,R\,1\) and \(1\,R\,2\) but \(0\,\not R\,2\text{.}\)
3. \(A = \{0,1,2,\}\text{,}\) \(R = \{(0,0),(1,1),(2,2),(0,1),(1,2)\}\text{.}\) This is clearly reflexive and antisymmetric, but it is not transitive because \(0\,R\,1\) and \(1\,R\,2\) but \(0\,\not R\,2\text{.}\)
4. \(A = \{0,1,2,\}\text{,}\) \(R = \{(0,0),(1,1),(2,2),(0,1),(1,2),(0,2)\}\text{.}\) This is reflexive and transitive, but it is not symmetric because \(0\,R\,1\) but \(1\,\not R\,0\text{.}\) As an alternative example you could take \(\le\) on \(\mathbb{R}\text{.}\)

1. The equivalence classes consist of sets of points whose two nearest neighbors are exactly distance 1 apart. One of these equivalence class is just \(\mathbb{Z}\) itself, and the others are just shifted copies of \(\mathbb{Z}\text{.}\)
2. Notice that \(\sin x = \sin y\) if and only if either \(y = x + 2\pi k\) for some integer \(k\) or if \(y = (\pi - x) +2\pi k\) for some integer \(k\text{.}\) So the equivalence classes are the union of two sets of points each having an equal spacing of \(2\pi\text{.}\) Alternatively, you can view the equivalence classes as the set of points of intersection of a horizontal line (between \(-1\) and \(1\) on the vertical axis) with the graph of \(\sin(x)\text{.}\)
3. The equivalence classes from this equivalence relation are just the union of a point with its negative, e.g. \(\{2,-2\}\text{.}\) These are all two point sets except the line \(\{0\}\) which is a singleton. As in the previous example, you can also think of this as the set of points of intersection of a horizontal line with the graph of the parabola \(x^2\text{.}\)
4. Given two points \((x,y), (u,v)\text{,}\) \(xy = 0\) if and only if either \(x=0\) or \(y =0\text{,}\) that is, if \((x,y)\) lies on one of the axes. So, two points are similar if either \(xy = uv = 0\) (i.e. they both lie on the axes) or \(xyuv \gt 0\) (which can only happen if both points are not on the axes). Thus the axes are one of the partitions. Now, how can \(xyuv \gt 0\text{?}\) Consider the different quadrants: if \((x,y)\) is in: quadrant 1, then \(xy \gt 0\text{;}\) quadrant 2, then \(xy \lt 0\text{;}\) quadrant 3, then \(xy \gt 0\text{;}\) quadrant 4, then \(xy \lt 0\text{.}\) So, the only way for \(xyuv \gt 0\) is if both \((x,y),(u,v)\) are in quadrants 1,3 or both are in quadrants 2,4. Thus the partition has 3 sets, the axes, the union of quadrants 1 and 3, and the union of quadrants 2 and 4.

\(R\) is not antisymmetric since \((0,1),(1,0) \in R\text{.}\)

\(S\) is a partial order because it is reflexive (contains \((0,0),(1,1),(2,2)\)), antisymmetric (contains \((0,1),(1,2),(0,2)\) but not any of \((1,0),(2,1),(2,0)\)), and transitive (it contains \((0,1),(1,2)\) and also \((0,2)\)). Note that \(S\) is just the partial order \(\le\) on \(A\text{.}\)

\(T\) is not reflexive since \((2,2) \notin T\text{.}\)

\(U\) is not transitive since \((0,1),(1,2) \in U\) but \((0,2) \notin U\text{.}\)

We first prove that \(\prec\) is reflexive. Note that for any \((x,y) \in \mathbb{R}^2\text{,}\) we have \(x \le x\) and \(y \le y\text{,}\) and therefore \((x,y) \prec (x,y)\text{.}\)

To see that \(\prec\) is antisymmetric, note that for any \((x,y),(a,b) \in \mathbb{R}^2\text{,}\) if \((x,y) \prec (a,b)\) and \((a,b) \prec (x,y)\text{,}\) then \(x \le a\) amd \(y \le b\text{,}\) and also \(a \le x\) and \(b \le y\text{.}\) Therefore, by the antisymmetry of \(\le\text{,}\) we find \(x = a\) and \(y = b\text{,}\) therefore \((x,y) = (a,b)\text{.}\)

To see that \(\prec\) is transitive, note that for any \((x,y), (a,b), (w,z) \in \mathbb{R}^2\text{,}\) if \((x,y) \prec (a,b)\) and \((a,b) \prec (w,z)\text{,}\) then \(x \le a\) and \(y \le b\text{,}\) and also \(a \le w\) and \(b \le z\text{.}\) Therefore, by the transitivity of \(\le\text{,}\) we find \(x \le w\) and \(y \le z\text{.}\) Hence \((x,y) \prec (w,z)\text{.}\)

We have shown that \(\prec\) is a partial order on \(\mathbb{R}^2\) because it is reflexive, antisymmetric and transitive, but it is not a total order because there are incomparable elements. In particular, \((1,2) \not\prec (0,3)\) since \(1 \not\le 0\text{,}\) but also \((0,3) \not\prec (1,2)\) since \(3 \not\le 2\text{.}\)

The relation \(R\) is not a partial order on \(\mathbb{C}\) because it is not antisymmetric. In particular, \(1\ R\ i\) and \(i\ R\ 1\) since \(1^2 + 0^2 = 0^2 + 1^2\text{,}\) but \(1 \not= i\text{.}\) This completes the solution to this problem.

Note however that this relation is reflexive and transitive. A relation like this is called a pre-order. If \(S\) is any pre-order (reflexive, transitive) on a set \(A\text{,}\) then we can create an equivalence relation \(\sim\) on \(A\) by \(x \sim y\) if and only if \(x\ S\ y\) and \(y\ S\ x\text{.}\) This is easily seen to be symmetric, and the reflexivity and transitivity follow from those same properties of \(S\text{.}\) Then there is a natural partial order \(S/\sim\) on the set of equivalence classes \(A/\sim\) defined by \([x] S/\sim [y]\) if and only if \(x\ S\ y\text{.}\)

If we were to apply the procedure in the last paragraph to the relation \(R\) given in this problem, then the equivalence relation on \(\mathbb{C}\) we would obtain is the one in which \(x \sim y\) if and only if \(x,y\) lie on the same circle centered at \(0\text{.}\) So the equivalence classes are circles centered at \(0\text{.}\) Then the partial order \(R/\sim\) on the set of these circles is precisely the one that says one circle is less than or equal to another circle if and only if its radius is less than or equal to the radius of the other.