Hints to exercises
\(x_1 x_2 x_3 <_{\mbox{lex}} y_1 y_2 y_3\) if and only if \(x_1\) comes before \(y_1\) in alphabetical order, or \(x_1 = y_1\) and \(x_2\) comes before \(y_2\) in alphabetical order, or \(x_1 = y_1,\) \(x_2 = y_2,\) and \(x_3\) comes before \(y_3\) in alphabetical order.
\(\{ (a, a) {\,:\,}a \in {\mathbb R}\}.\)
\(\mathsf{S}^{-1} = \{ (y,x) {\,:\,}y = \sin x \}.\)
We show \((\mathsf{S}\circ \mathsf{R})^{-1} \subseteq \mathsf{R}^{-1} \circ \mathsf{S}^{-1}.\) The reverse inclusion is left as an exercise.
Suppose that \(\mathsf{R}\) is a relation from \(A\) to \(B\) and \(\mathsf{S}\) is a relation from \(B\) to \(C\). Let \((c,a) \in (\mathsf{S}\circ \mathsf{R})^{-1}.\) Then by definition of the inverse relation, \((a,c) \in \mathsf{S}\circ \mathsf{R}.\) By the definition of composition, there exists \(b \in B\) such that \((a,b) \in \mathsf{R}\) and \((b,c) \in \mathsf{S}.\) Hence, \((b,a) \in \mathsf{R}^{-1}\) and \((c,b) \in \mathsf{S}^{-1},\) implying that \((c,a) \in \mathsf{R}^{-1} \circ \mathsf{S}^{1}.\)
Anti-symmetric because if \(x\) is smarter than \(y,\) then \(y\) cannot be smarter than \(x.\)
Symmetric because if \(x\) has the same astrological sign as \(y,\) then \(y\) has the same astrological sign as \(x.\)
If \(x\) is smarter than \(y\) and \(y\) is smarter than \(z,\) then \(x\) is smarter than \(z.\)
If \(x\) has the same astrological sign as \(y\) and \(y\) has the same astrological sign as \(z,\) then \(x,\) \(y,\) and \(z\) all have the astrological sign and so \(x\) has the same astrological sign as \(z.\)
All relations in the following table are over \({\mathbb R}.\)
Reflexive \(\{ (a,a) \}\) Not reflexive \(\{ (a,b) {\,:\,}a > b\}\) Irreflexive \(\{ (a,b) {\,:\,}a \neq b \}\) Not irreflexive \(\{ (a,b) {\,:\,}a \geq b\}\) Symmetric \(\{ (a,b) {\,:\,}a^2+b^2=1\}\) Not symmetric \(\{ (a,b) {\,:\,}a > b\}\) Anti-symmetric \(\{ (a,b) {\,:\,}a = b+1 \}\) Not anti-symmetric \(\{ (a,b) {\,:\,}a^2+b^2=1\}\) Transitive \(\{ (a,b) {\,:\,}a > b \}\) Not transitive \(\{ (a,b) {\,:\,}a=b+1\}\) \(1 \mid 2\) but it is not true that \(2 \mid 1.\)
Let \(a\) be a positive integer. Then \(a \mid a\) since \(a = 1 \cdot a.\) Thus, divisibility over positive integers is reflexive.
Let \(a\) and \(b\) be positive integers such that \(a \mid b.\) If \(b \neq a,\) then \(b > a.\) Thus, we cannot have \(b \mid a,\) showing that divisibility over positive integers is anti-symmetric.
If \(a \mid b\) and \(b \mid c,\) then there exist integers \(j\) and \(k\) such that \(b = ja\) and \(c = kb.\) Hence, \(c = k(ja) = (kj)a,\) giving \(a \mid c\) since \(kj\) is a product of two integers and is thus an integer. Thus, divisibility over positive integers is transitive.
We show that \(\mathsf{A}\) reflexive, symmetric, and transitive. It is reflexive because every \(x\) has the same astrological sign as \(x\); symmetric because if \(x\) has the same astrological sign as \(y,\) then \(y\) has the same astrological sign as \(x\); transitive because if \(x\) has the same astrological sign as \(y\) and \(y\) has the same astrological sign as \(z,\) then \(x,\) \(y,\) and \(z\) all have the astrological sign and so \(x\) has the same astrological sign as \(z.\)
It is not difficult to show that \(\square\) is reflexive, symmetric, and transitive.
\(0 / \square = \{0\}\) \(1 / \square = \{-1,1\}\) \(2 / \square = \{-2,2\}\) \(3 / \square = \{-3,3\}\) \(4 / \square = \{-4,4\}\) \(5 / \square = \{-5,5\}\) The solution becomes apparent by making the observation that two words are anagrams of each other if and only if they contain the same letters and matching in the number of occurrences.
Label the outer nodes clockwise starting at the top in the following order: A, B, C, H, J, E. See if you can label the inner nodes so that connections between labelled vertices are preserved.
Since \(ab = ba,\) we have \((a,b)\mathsf{Q}(a,b).\) Hence, \(\mathsf{Q}\) is reflexive.
Suppose that \((a,b)\mathsf{Q}(c,d).\) Then \(ad = bc,\) which is equivalent to \(cb = da,\) giving \((c,d)\mathsf{Q}(a,b).\) Hence, \(\mathsf{Q}\) is symmetric.
Suppose that \((a,b)\mathsf{Q}(c,d)\) and \((c,d)\mathsf{Q}(e,f).\) So \(ad = bc \land cf = de,\) giving \(adcf = bcde.\) Since \(c\) and \(d\) are nonzero, we obtain \(af = be,\) giving \((a,b)\mathsf{Q}(e,f).\) Hence, \(\mathsf{Q}\) is transitive.
This can be seen from that the equivalence class \((a,b)/\mathsf{Q}\) contains all \((c,d)\) such that \(\frac{c}{d} = \frac{a}{b}.\)
We needed non-zero numbers in proving that \(\mathsf{Q}\) is transitive.
\(\{ \mbox{Cow}, \mbox{Duck}, \mbox{Robin}, \mbox{Goose} \}\)
1: g, 2: f, 3: c, 4: d, 5: h, 6: b, 7: e, 8: a
Draw two 3-dimensional cubes and connect make some appropriate connections between the two copies.
Since \(210 = 2\cdot 3\cdot 5 \cdot7\), there are 16 node labels which can be divided into those that are even and those that are odd. The Hasse diagram for the odd labels is a 3-dimensional cube. So is the Hasse diagram for the even labels.
This exercise is similar to the previous one. Separate the labels into those that contain \(a\) and those that do not.
Note that \(11025 = 3^2 \cdot 5^2 \cdot 7^2.\) How many nodes are there in the Hasse diagram for the poset of divisors of \(11025\)?
The following collection works: \[\begin{gather*} \{a, a', b, b', c, c'\}, \\ \{a', b, b', c, c'\}, \{a, a', b, c, c'\}, \{a, a', b, b', c'\}, \\ \{a', b', c, c'\}, \{a', b, b', c'\}, \{a, a', b', c'\}, \{b, b', c, c'\}, \{a, a', c, c'\}, \{a, a', b, b'\}, \\ \{b', c, c'\}, \{b, b', c'\}, \{a', c, c'\}, \{a, a', c'\}, \{a, c, c'\}, \{a, a', b', c, c'\}, \{a', b', c'\}, \\ \{a,b\}, \{a,c\}, \{b,c\}, \{a,a'\}, \{b,b'\}, \{c, c'\}, \\ \{a\}, \{b\}, \{c\}, \\ \{\} \end{gather*}\]
Note that any set containing the range can serve as the codomain.
Domain: \({\mathbb R}\). Range: \([-1,1]\).
Domain: \({\mathbb R}\). Range: \((0,\infty)\).
Domain: \({\mathbb R}\). Range: \([0,\infty)\).
Domain: \({\mathbb R}\setminus\{-1,1\}\). Range: \((-\infty,-1]\cup (1,\infty)\).
Domain: \({\mathbb R}\). Range: \({\mathbb Z}\).
Let \(f(n) = \frac{\sqrt{n}-1}{2}.\) It is not difficult to see that \(f\) is injective. To see that \(f\) is surjective, let \(y \in \{0,1,\ldots\}.\) Then \(x = (2y+1)^2\) is the square of an odd integer and \(f(x) = \frac{\sqrt{(2y+1)^2}-1}{2}=\frac{2y+1-1}{2}= y.\)
Note that \(f^{-1}(n) = (2n+1)^2.\)
Let \(y > x > 0.\) Then \[\ln(y)-\ln(x) = \int_{t=x}^y \frac{1}{t} \mbox{d}t > 0\] since \(\frac{1}{t}\) is positive over the interval \([x,y].\) Hence, \(\ln(x)\) is strictly increasing on \((0,\infty).\)
Let \(y\) be a positive real number. Since \(\ln(2)\) is a fixed real number, setting \(b = y/\ln(2),\) we get that \(\ln(2^b) = y.\)
\(f(25) = 4/5,\) \(f(29) = 7/3,\) \(f^{-1}(3/4) = 15.\)
We can restrict to \([-0.5, \infty)\) for example.
\(T^{-1}(x) = \frac{-1+\sqrt{1+8x}}{2}\) with domain \([-\frac{1}{8},\infty).\)
\(\sin(t).\)
\(f^{-1}(x) = x\) for all \(x \in [0,\infty).\)
\(\pi_{12}(H)\) is the set of points on the unit circle on the \(x\)–\(y\) coordinate plane. The other two are the sine and cosine curves.
\(\{ (1,1), (2,1), (3,1), (4,0), (5,0), \ldots, (10,0)\}.\)
\(1_S \cdot 1_T = 1_{S \cap T}.\)
\(\frac{1}{2}+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}=\frac{247}{210}.\)