View source for Math 360, Fall 2021, Assignment 3

__NOTOC__
''We admit, in geometry, not only infinite magnitudes, that is to say, magnitudes greater than any assignable magnitude, but infinite magnitudes infinitely greater, the one than the other.  This astonishes our dimension of brains, which is only about six inches long, five broad, and six in depth, in the largest heads.''

: - Voltaire

==Read:==

# Section 2.

==Carefully define the following terms, then give one example and one non-example of each:==

# Injective (function; a.k.a. ''one-to-one'' function).
# Surjective (function; a.k.a. ''onto'' function).
# Bijective (function).
# Equinumerous (sets).
# Countable (set).
# Uncountable (set).
# Binary operation (on a set $S$).
# Binary structure.
# Commutative (binary structure).
# Associative (binary structure).
# Left identity element (in a binary structure).
# Right identity element (in a binary structure).
# Identity element (in a binary structure).
# Invertible element (in a binary structure with identity).
# Inverse (of an element of a binary structure with identity).

==Carefully state the following theorems (you do not need to prove them):==

# Cantor's Theorem.
# Theorem bounding the number of two-sided identity elements in one binary structure.

==Solve the following problems:==

# Section 2, problems 1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12, 13, 17, 18, 19, 20, 21, 22, and 23.
# Give an example of a binary structure $(S,\triangle)$ which has two distict left identity elements.  ''(Hint: let $S$ be a set with a very small number of elements, e.g. $S=\{a,b,c\}$.  Define your very own binary operation $\triangle$ on $S$ by means of a table, as discussed on page 24 of the text.  Remember that you are the author of the table and may complete it however you wish, as long as the result is a legitimate binary structure.  Within a short time, you are likely to see several ways of completing such a table which lead to multiple left identity elements.)''
# Using the method of the previous exercise, give an example of a binary structure with two distinct right identity elements.
# Now try to use the method of the last two exercises to produce a binary structure with two distinct (two-sided) identity elements.  What goes wrong?
# Show that the closed interval $[0,1]$ of the real line is equinumerous with the closed interval $[0,2]$, by constructing an explicit bijection between these two sets.  Then formally verify that your map is a bijection.
# Consult a calculus book for a graph of the function $f(x)=\tan^{-1}(x)$.  Assuming that the graph is not misleading, explain why the whole of the real number system $\mathbb{R}$ is equinumerous with the open interval $\left(\frac{-\pi}{2},\frac{\pi}{2}\right)$.  ''(Hint: you may need to look up the ''horizontal line test'' for injectivity of functions from $\mathbb{R}$ to $\mathbb{R}$, and you may also need to think about how to determine the image of such a function by inspecting its graph.)''

<center><big>'''--------------------End of assignment--------------------'''</big></center>

==Questions:==

==Solutions:==
==Definitions:==
# Injective: $f(a) = f(b) \iff a = b$
# Surjective: $im(f) = codom(f)$
# Injective $\wedge$ Surjective
# $\exists$ a bijection from A to B
# A set {{mvar|S}} is ''countable'' if its cardinality |S| is less than or equal to <math>\aleph_0</math> ([[aleph-null]]), the cardinality of the set of [[natural numbers]] {{math|1='''N'''}}.
# not countable
# Let $S$ be a set. A binary operation on $S$ is a function $\triangle S \ times S \rightarrow S$
# $(S, \triangle)$
# $a * b = b * a$
# $a * (b * c) = (a * b) * c$
# $e * a = a$
# $a * e = a$
# $e * a = a$ and $a * e = a$
# has an inverse
# $a * a^{-1} = e$

==Theorems:==
# Cantor's Theorem:  the cardinality of a set is strictly less than the cardinality of its power set,  Even for infinite sets, there are size with greater cardinality than this infinite size, such as the power set of this infinite set, and the power set of this power set will have greater cardinality than this powers...These increasing cardinality are called transfinite numbers. And $\mathbb Z_{\geq 0} \text{and} \mathbb R$ are not equinumerous is possible because they have different transfinite numbers as their cardinality. 
# A binary structure has at most one two-sided identity element.
----

==Book Solutions:==
# 1. c; b; a
# 2. a; a; This example is associative. But We have to compute all to see. This is not enough
# 3. a; c; No
# 4. Yes. If we look at line (a,a)(b,b),(c,c)..., it is a line of symmetric, the table is symmetric related to this line. It means that all $x, y \in$ table, $x*y == y*x$
# 5. From left to right, up to down, c, a, b, d.
# 6. Left to right d, d, d, anything. But it should not be possible, because $c*b =d, b*c = c$.
# 8. commutative, not associative.
# 9. both yes.
# 10. commutative, not associative.
# 11. both not.
# 12. $n * n * n \cdots n$, in total $n^{2}$ ns, because it is an $n*n$ table. $n^{n^{2}}$.
# 13. There are $n^{n}$ ways of filling $(x, x), \text{for any} x \in set$. Because of symmetry of tow triangles, we can only fill one triangle, therefore, $\frac{n*n-n}{2}$ blocks to fill, each $n$ ways, $n^{\frac{n*n-n}{2}}$, and in total $n^{n}*n^{\frac{n*n-n}{2}} = n^{\frac{n*n+n}{2}}$.
# 17. Violating 2. $1-3 = -2 \not\in \mathbb Z^{+}$
# 18. yes.
# 19. yes.
# 20. yes.
# 21. yes.
# 22. Violating 2. $1*1 = 0 \not\in \mathbb Z^{+}$
# 23. a. $(a1+a2)\in \mathbb R, (-b1-b2)\in \mathbb R, (b1+b2)\in \mathbb R$, under addition it is closed. b. $(a1a2+ b2(-b1))\in \mathbb R, (a1(-b2)+a2(-b1))\in \mathbb R, (a2b1+b2a1)\in \mathbb R, ((-b2)b1+a2a1)\in \mathbb R$, closed under multiplication

----

==Solve Problems:==
# Give an example of a binary structure with 2 left identities$(S,\triangle)$: $S = \{a, b, c\},\triangle: S \times S \rightarrow S.$ $\triangle: a \triangle a = a, a \triangle b = b, a \triangle c = c,$ $b \triangle a = a, b \triangle b = b, b \triangle c = c,$ $c \triangle a = b, c \triangle b = c, c \triangle c = a$. $a, b$ are two distinct left identity elements.
#  Give an example of a binary structure with 2 right identities $(S,\triangle)$: $S = \{a, b, c\},\triangle: S \times S \rightarrow S.$ $\triangle: a \triangle a = a, b \triangle a = b, c \triangle a = c,$ $a \triangle b = a, b \triangle b = b, c \triangle b = c,$ $a \triangle c = b, b \triangle c = c, c \triangle c = a$. $a, b$ are two distinct right identity elements.
# Assume there are 2 distinct two-sided identities $e1, e2$ for $(S,\triangle)$. Then for $e1$ as identity, $e1 \triangle e2 = e2, e1 \triangle e2 = e2$; for $e2$ as identity, $e2 \triangle e1 = e1, e1 \triangle e2 = e1$, solve the equations we will get $e1 = e2$, which contract out assumption that e1 and e2 are two distinct elements.
# Show that the closed interval $[0,1]$ of the real line is equinumerous with the closed interval $[0,2]$. $A = \{[0,1]\}, B = \{[0, 2\}$, $f:A \rightarrow B, f(a) = 2a = b, (2a = b)  a \in A \wedge b \in B$. $\forall b \in [0,2] \in \mathbb R, \exists a \in [0,1] \in \mathbb R$ that $2a = b$. Therefore, it is surjective. And $\forall a1, a2 \in [0,1]$, and $f(a1) = b1, f(a2) = b2, b1, b2 \in [0,2]$, if $ b1 = b2$, then $\frac{b1}{2} = \frac{b2}{2}. f(2a) = b, a = \frac{b}{2}$ Therefore, $a1 = \frac{b1}{2} = \frac{b2}{2} = a2$. $f(a1) = f(a2) \iff a1 = a2$, it is injective. It is both injective and surjective $\rightarrow$ it is surjective.
#$f(x) = \tan^{-1}(x)$, $g(y) = \tan(y)$. As we know about $\tan$ that $\tan(y) = x, then \tan^{-1}(x) = y$. Therefore, domain of $g(y)$ is the codomain of $f(x)$. $\tan(y) = \frac{\sin(y)}{\cos(y)}$, when $\cos(y) \rightarrow 0, \tan(y) \rightarrow \infty$, which means that for $\tan^{-1}(x) = y, x \rightarrow \pm \infty$. $\cos(y) \rightarrow$ when $y1 = \frac{\pi}{2}, y2 = -\frac{\pi}{2}$. Which means that for $\tan^{-1}(x) = y$, when $y \rightarrow \frac{\pi}{2} \vee y \rightarrow -\frac{\pi}{2}, x \rightarrow \infty \vee -\infty$. Also, for $\tan^{-1}(x) = y$, it is a function when $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ because all x maps to exactly one element in codomain. $\forall y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, there is a corresponding $x \in [-\infty, \infty]$, it is surjective. And $\forall x1, x2 \in [-\infty, infty], \text{if} \tan^{-1}(x1) = \tan^{-1}(x2), x1 = x2$ because if $y1 = y2, x1 = \tan(y1)= \tan(y2) = x2$, it is also injective. Therefore it is also bijective. There exists a bijective operation $\mathbb R \rightarrow [-\frac{\pi}{2}, \frac{\pi}{2}]$. It is equinumerous.
<center><big>'''--------------------End of Solutions--------------------'''</big></center>