Theoretical Mechanics IPSP

Jürgen Vollmer, Universität Leipzig

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book:chap2:2.2_sets [2021/10/25 23:24] – jvbook:chap2:2.2_sets [2024/10/26 13:02] (current) – jv
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 +[[forcestorques|2. Balancing Forces and Torques]]
 +  * [[  2.1 Motivation and Outline| 2.1 Motivation and outline: forces are vectors ]]
 +  * ** 2.2 Sets **
 +  * [[  2.3 Groups| 2.3 Groups ]]
 +  * [[  2.4 Fields| 2.4 Fields ]]
 +  * [[  2.5 Vector spaces| 2.5 Vector spaces ]]
 +  * [[  2.6 Physics application balancing forces| 2.6.  Physics application: balancing forces]]
 +  * [[  2.7 The inner product | 2.7 The inner product]]
 +  * [[  2.8 Cartesian coordinates | 2.8 Cartesian coordinates]]
 +  * [[  2.9 Cross products --- torques| 2.9 Cross products — torques ]]
 +  * [[ 2.10 Worked example Calder's mobiles| 2.10 Worked example: Calder's mobiles ]]
 +  * [[ 2.11 Problems| 2.11 Problems ]]
 +  * [[ 2.12 Further reading| 2.12 Further reading ]]
 +
 +----
 +
 ===== 2.2 Sets ===== ===== 2.2 Sets =====
  
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-<WRAP box round>**Example 2.2** <wrap em>Sets</wrap> +<WRAP box round #bsp_sets >**Example 2.2** <wrap em>Sets</wrap> 
 \\  \\ 
   *  Set of capitals of German states: \\   *  Set of capitals of German states: \\
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 The cardinalities of these sets are \\ The cardinalities of these sets are \\
-\centering $|A_C| = 16$, $|A_L = 30|$, and $|A_M| = 12$.+<wrap center> $|A_C| = 16$, $|A_L = 30|$, and $|A_M| = 12$. </wrap>
 </WRAP> </WRAP>
  
-<WRAP box round>**Example 2.3** <wrap em>Sets of sets</wrap> \\ +<WRAP box round #bsp_setsofsets >**Example 2.3** <wrap em>Sets of sets</wrap> \\ 
 A set can be an element of a set. A set can be an element of a set.
 For instance the set For instance the set
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 In obvious cases we use ellipses such as In obvious cases we use ellipses such as
 $A_L = \{$a, b, c, \dots, z, ä, ö, ü, ß$\}$ $A_L = \{$a, b, c, \dots, z, ä, ö, ü, ß$\}$
-for the set given in \Example{Sets}.+for the set given in  [[#bsp_Sets|Example 2.2]].
 Alternatively, one can provide a set $M$ by specifying the properties $A(x)$ Alternatively, one can provide a set $M$ by specifying the properties $A(x)$
 of its elements $x$ in the following form of its elements $x$ in the following form
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 |$ A \Leftrightarrow B $  |  1  |  0  |  0  |  1  | $A$ is equivalent to $B$ | |$ A \Leftrightarrow B $  |  1  |  0  |  0  |  1  | $A$ is equivalent to $B$ |
 |$ A \lor \lnot B      $  |  1  |  0  |  1  |  1  | $A$ or not $B$ | |$ A \lor \lnot B      $  |  1  |  0  |  1  |  1  | $A$ or not $B$ |
-|$ \lnot A \land B     $  |  0  |  1  |  0  |  0  | not $A$ or $B$ |+|$ \lnot A \land B     $  |  0  |  1  |  0  |  0  | not $A$ and $B$ |
 |$ A \land \lnot B     $  |  0  |  0  |  1  |  0  | $A$ and not $B$ | |$ A \land \lnot B     $  |  0  |  0  |  1  |  0  | $A$ and not $B$ |
 Table 2.1: List of the results of different junctors acting on two statements $A$ and $B$. Table 2.1: List of the results of different junctors acting on two statements $A$ and $B$.
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 In the rightmost column we state the contents of the expression in the left column in words. In the rightmost column we state the contents of the expression in the left column in words.
 The final three lines provide examples of more complicated expressions. The final three lines provide examples of more complicated expressions.
-<wrap hide>\label{table:Junktors}</wrap> 
 </WRAP> </WRAP>
  
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 <WRAP box round>**Example 2.5** <wrap em>Subsets</wrap> \\  <WRAP box round>**Example 2.5** <wrap em>Subsets</wrap> \\ 
  
-  *  The set of month with names that end with 'ber' is a subset of the set $A_M$ of \Example{Sets}+  *  The set of month with names that end with 'ber' is a subset of the set $A_M$ of [[#bsp_Sets|Example 2.2]]:
 \begin{align*} \begin{align*}
       \{ \text{September, October, November, December} \} \subseteq A_M       \{ \text{September, October, November, December} \} \subseteq A_M
 \end{align*} \end{align*}
  
-  *  For the set  $M$  of  \Example{SetsOfSets}  one has+  *  For the set  $M$  of  [[#bsp_SetsOfSets|Example 2.3]]  one has
 \begin{align*} \begin{align*}
       \{ 1 \} \subseteq M \, , \quad        \{ 1 \} \subseteq M \, , \quad 
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 Two sets are the same when they are subsets of each other. Two sets are the same when they are subsets of each other.
-<WRAP box round>**Theorem 2.1 <wrap hi>Äquivalenz von Mengen</wrap>** \\ +<WRAP box round #Thm_SetEquivalence >**Theorem 2.1 <wrap em> Equivalence of Sets</wrap>** \\ 
 Two sets  $A$ and $B$   are //equal// or //equivalent//, iff Two sets  $A$ and $B$   are //equal// or //equivalent//, iff
 \begin{align*} \begin{align*}
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 </WRAP> </WRAP>
  
-<wrap lo>**Anmerkung.** [iff] +<wrap lo>**Remark.** In mathematics "iff" indicates that something holds "if and only if".
-In mathematics ``iff'' indicates that something holds +
-``if and only if''.+
 Observe its use in the following two statements: Observe its use in the following two statements:
 A number is an even number if it is the product of two even numbers. A number is an even number if it is the product of two even numbers.
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 </wrap> </wrap>
  
-<wrap lo #rem_LogicalPrecedence >**Anmerkung.** [precedence of operations in logical expressions.] +<wrap lo #rem_LogicalPrecedence >**Remark.** In logical expressions we first evaluate $\in$, $\not\in$ and other set operations that are used to build logical expressions. 
-In logical expressions we first evaluate $\in$, $\not\in$ and other set operations +
-that are used to build logical expressions. +
 Then we evaluate the junctor $\lnot$ that is acting on a a single logical expression. Then we evaluate the junctor $\lnot$ that is acting on a a single logical expression.
 Finally the other junctors $\land$, $\lor$, $\Rightarrow$, and $\Leftrightarrow$ are evaluated. Finally the other junctors $\land$, $\lor$, $\Rightarrow$, and $\Leftrightarrow$ are evaluated.
-Hence, the brackets are not required in \Thm{SetEquivalence}.+Hence, the brackets are not required in [[#Thm_SetEquivalence|Theorem 2.1]].
 </wrap> </wrap>
  
-=== Proof of \Thm{SetEquivalence} ===+=== Proof of Theorem 2.1 ===
 $A \subseteq B$ implies that $a \in A \Rightarrow a \in B$.  $A \subseteq B$ implies that $a \in A \Rightarrow a \in B$. 
 $B \subseteq A$ implies $b \in B \Rightarrow b \in A$.  $B \subseteq A$ implies $b \in B \Rightarrow b \in A$. 
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 $\rule{3mm}{3mm}$ $\rule{3mm}{3mm}$
  
-The description of sets by properties of its members, \Example{SetOfDigits},+The description of sets by properties of its members, [[#bsp_SetOfDigits|Example 2.4]],
 suggests that one will often be interested in operations on sets. suggests that one will often be interested in operations on sets.
 For instance the odd and even numbers are subsets of the natural numbers. For instance the odd and even numbers are subsets of the natural numbers.
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 <WRAP 200px right> <WRAP 200px right>
-<wrap #fig_setIntersection >+<wrap #fig_setIntersection > 
 {{ book:chap2:mengen_schnitt.png?direct&200 |}} {{ book:chap2:mengen_schnitt.png?direct&200 |}}
  
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   *  //Union//:  $M_1 \bigcup M_2 = \left\{ m   \; \vert \;  m \in M_1 \lor m \in M_2 \right\}$,   *  //Union//:  $M_1 \bigcup M_2 = \left\{ m   \; \vert \;  m \in M_1 \lor m \in M_2 \right\}$,
   *  //Difference//:  $M_1 \backslash M_2 = \left\{ m   \; \vert \;  m \in M_1 \land m \notin M_2 \right\}$,   *  //Difference//:  $M_1 \backslash M_2 = \left\{ m   \; \vert \;  m \in M_1 \land m \notin M_2 \right\}$,
-  *  The //complement// of a set $M$ in a //universe// $U$ is defined for subsets $M \subseteq U$ +  *  The //complement// of a set $M$ in a //universe// $U$ is defined for subsets $M \subseteq U$ as $M^{C} = \left\{ m \in U   \; \vert \;  m \notin M \right\} = U \backslash M$. 
-as   $M^{C} = \left\{ m \in U   \; \vert \;  m \notin M \right\} = U \backslash M$. +  *  The //Cartesian product of two sets $M_1$ and $M_2$// is defined as the set of ordered pairs $(a, \, b)$ of elements $a \in M_1$ and \\ $b \in M_2$:  $\displaystyle M_1 \times M_2 = \left\{ (a, \, b)   \; \vert \;  a \in M_1, \, b \in M_2 \right\} $.
-  *  The //Cartesian product of two sets $M_1$ and $M_2$// is defined as the set of ordered pairs +
-$(a, \, b)$ of elements $a \in M_1$ and \\ $b \in M_2$:  $\displaystyle M_1 \times M_2 = \left\{ (a, \, b)   \; \vert \;  a \in M_1, \, b \in M_2 \right\} $.+
  
 A graphical illustration of the operations is provided in A graphical illustration of the operations is provided in
-\cref{figure:setIntersection,figure:setUnion,figure:setDifference,figure:setComplement}.+[[#fig_setIntersection|Figure 2.2]], [[#fig_setUnion|Figure 2.3]], [[#fig_setDifference|Figure 2.4]], [[#fig_setComplement|Figure 2.5]].
 </WRAP> </WRAP>
  
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 The set of non-female participants is  $P \backslash F$. The set of non-female participants is  $P \backslash F$.
 The set of heterosexual couples in the class is a subset of the Cartesian product $F \times M$.  The set of heterosexual couples in the class is a subset of the Cartesian product $F \times M$. 
-Furthermore, the union $F \bigcup M$ is a proper subset of$P$,  +Furthermore, the union $F \bigcup M$ is a proper subset of $P$, when there is a participant who is neither female nor male.
-when there is a participant who is neither female nor male.+
 </WRAP> </WRAP>
  
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 <WRAP box round>**Example 2.7** <wrap em>Logical quantors and properties of set elements</wrap> \\  <WRAP box round>**Example 2.7** <wrap em>Logical quantors and properties of set elements</wrap> \\ 
-Let $|m|$ denote the number of days in a month $a \in A_M$  +Let $|m|$ denote the number of days in a month $a \in A_M$ (Refer to [[#bsp_Sets|Example 2.2]]).
-(\cf\Example{Sets}).+
 Then the following statements are true: Then the following statements are true:
 There is exactly one month that has exactly $28$ days:  There is exactly one month that has exactly $28$ days: 
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 Many sets of numbers that are of interest in physics have infinitely many elements. Many sets of numbers that are of interest in physics have infinitely many elements.
-We construct them in \cref{table:SetsOfNumbers} based on the natural numbers+We construct them in [[#tab_SetsOfNumbers|Table 2.2]] based on the natural numbers
 \begin{align*} \begin{align*}
   \mathbb N   \mathbb N
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 | negative numbers         | $-\mathbb{N}$  | $\left\{ -n   \; \vert \;  n \in \mathbb{N} \right\}$ | | negative numbers         | $-\mathbb{N}$  | $\left\{ -n   \; \vert \;  n \in \mathbb{N} \right\}$ |
 | even numbers             | $2 \mathbb{N}$ | $\left\{ 2\,n   \; \vert \;  n \in \mathbb{N} \right\}$ | | even numbers             | $2 \mathbb{N}$ | $\left\{ 2\,n   \; \vert \;  n \in \mathbb{N} \right\}$ |
-| odd numbers              | $2 \mathbb{N} - 1$   | $\left\{ 2\,n -1  \; \vert \;  n \in \mathbb{N} \right\} |+| odd numbers              | $2 \mathbb{N} - 1$   | $\left\{ 2\,n -1  \; \vert \;  n \in \mathbb{N}\right\}$ |
 | integer numbers        | $\mathbb{Z}$   | $\left( -\mathbb{N} \right) \bigcup \mathbb{N}_0$ | | integer numbers        | $\mathbb{Z}$   | $\left( -\mathbb{N} \right) \bigcup \mathbb{N}_0$ |
 | rational numbers       | $\mathbb{Q}$   | $\left\{ \frac{p}{q}   \; \vert \;  p \in \mathbb{Z}, \, q \in \mathbb{N} \right\}$ | | rational numbers       | $\mathbb{Q}$   | $\left\{ \frac{p}{q}   \; \vert \;  p \in \mathbb{Z}, \, q \in \mathbb{N} \right\}$ |
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 <WRAP group> <WRAP group>
 <WRAP half column> <WRAP half column>
-  * **a)** $\quad\lbrace A, B \rbrace \quad {{ \Box}} \quad \lbrace A, B, C \rbrace$,+  * **a)** $\quad\lbrace A, B \rbrace \quad {{ \Box}} \quad \lbrace A, B, C \rbrace$, 
   * **c)** $\quad\lbrace \emptyset \rbrace \quad  {\Box} \quad \emptyset$,   * **c)** $\quad\lbrace \emptyset \rbrace \quad  {\Box} \quad \emptyset$,
   * **e)** $\quad A  \quad { \Box} \quad \lbrace A, B, C \rbrace$,   * **e)** $\quad A  \quad { \Box} \quad \lbrace A, B, C \rbrace$,
   *  **g)** $\quad\lbrace A, C, D \rbrace \setminus \lbrace A, B \rbrace \;\Box\; \lbrace A, B, C \rbrace$,   *  **g)** $\quad\lbrace A, C, D \rbrace \setminus \lbrace A, B \rbrace \;\Box\; \lbrace A, B, C \rbrace$,
 </WRAP> </WRAP>
- 
 <WRAP half column> <WRAP half column>
   *  **b)** $\lbrace A \rbrace \quad { \Box} \quad B$,   *  **b)** $\lbrace A \rbrace \quad { \Box} \quad B$,
-  *  **d)** $\lbrace \lbrace A \rbrace \rbrace \quad  { \Box}  \lbrace \lbrace A \rbrace, \lbrace B \rbrace \rbrace$,+  *  **d)** $\lbrace \lbrace A \rbrace \rbrace \quad  { \Box}  \quad\lbrace \lbrace A \rbrace, \lbrace B \rbrace \rbrace$,
   *  **f)** $\lbrace A, C, D \rbrace \cap \lbrace A, B \rbrace \, {\Box} \, \lbrace A, B, C, D \rbrace$,   *  **f)** $\lbrace A, C, D \rbrace \cap \lbrace A, B \rbrace \, {\Box} \, \lbrace A, B, C, D \rbrace$,
   *  **h)** $\lbrace A, C, D \rbrace \cup \lbrace A, B \rbrace \quad { \Box} \quad  A$.   *  **h)** $\lbrace A, C, D \rbrace \cup \lbrace A, B \rbrace \quad { \Box} \quad  A$.
 </WRAP> </WRAP>
 </WRAP> </WRAP>
 +
 +++++ Solution: |
 +<WRAP group>
 +<WRAP half column>
 +  * **a)** $\quad \lbrace A, B \rbrace \quad \subset \quad \lbrace A, B, C \rbrace$, 
 +  * **c)** $\quad \lbrace \emptyset \rbrace \quad  {\ni} \quad \emptyset \, , \quad$ or  $\quad \lbrace \emptyset \rbrace \quad  {\supset} \quad \emptyset$
 +  * **e)** $\quad A  \quad { \in} \quad \lbrace A, B, C \rbrace$,
 +  *  **g)** $\quad\lbrace A, C, D \rbrace \setminus \lbrace A, B \rbrace \;\not\subset\; \lbrace A, B, C \rbrace$,
 +</WRAP>
 +<WRAP half column>
 +  *  **b)** $\lbrace A \rbrace \quad { \not\ni} \quad B$,
 +  *  **d)** $\lbrace \lbrace A \rbrace \rbrace \quad  { \subset }  \lbrace \lbrace A \rbrace, \lbrace B \rbrace \rbrace$,
 +  *  **f)** $\lbrace A, C, D \rbrace \cap \lbrace A, B \rbrace \, {\subset} \, \lbrace A, B, C, D \rbrace$,
 +  *  **h)** $\lbrace A, C, D \rbrace \cup \lbrace A, B \rbrace \quad { \ni} \quad  A$.
 +</WRAP>
 +</WRAP>
 +++++
 </WRAP> </WRAP>
  
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   -  Provide $\lbrack 1; 17 \rbrack \cap \rbrack 0; 5 \lbrack$ as a single interval.   -  Provide $\lbrack 1; 17 \rbrack \cap \rbrack 0; 5 \lbrack$ as a single interval.
-  -  Provide $[-1, 4] \backslash [1, 2[$ as union of two intervals.+  -  Provide $[-1; 4] \backslash [1; 2[$ as union of two intervals.
 </WRAP> </WRAP>
 +++++ Solutions: |
 +  -   $\lbrack 1; 17 \rbrack \cap \rbrack 0; 5 \lbrack = [1; 5[$ 
 +  -   $[-1; 4] \backslash [1; 2[ = [-1, 1[ \cup [2; 4]$
 +++++
 +
  
 ---- ----
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   -  $\lbrace 6 \cdot z \vert z \in \mathbb{Z} \rbrace \subset \lbrace 2 \cdot z \vert z \in \mathbb{Z} \rbrace$.   -  $\lbrace 6 \cdot z \vert z \in \mathbb{Z} \rbrace \subset \lbrace 2 \cdot z \vert z \in \mathbb{Z} \rbrace$.
   -  $\left\lbrace 2 \cdot z \vert z \in \mathbb{Z} \right\rbrace \cap \left\lbrace 3 \cdot z \vert z \in \mathbb{Z} \right\rbrace = \left\lbrace 6 \cdot z \vert z \in \mathbb{Z} \right\rbrace$.   -  $\left\lbrace 2 \cdot z \vert z \in \mathbb{Z} \right\rbrace \cap \left\lbrace 3 \cdot z \vert z \in \mathbb{Z} \right\rbrace = \left\lbrace 6 \cdot z \vert z \in \mathbb{Z} \right\rbrace$.
-  -  Let $T(a)$ be the set of numbers that divide $a$. Then +  -  Let $T(a)$ be the set of numbers that divide $a$. Then  $\quad\displaystyle  \forall a, b \in \mathbb{N}  \; \vert  \; T(a) \cup T(b) = T(a \cdot b) $. 
-\begin{align*} +
-      \forall a, b \in \mathbb{N}  \; \vert \;\quad T(a) \cup T(b) = T(a \cdot b)  +
-\end{align*}+
 Example:  $T(2) = \{1,2\}$, $T(3) = \{1,3\}$, and $T(6) = \{1,2,3,6\}$.  Example:  $T(2) = \{1,2\}$, $T(3) = \{1,3\}$, and $T(6) = \{1,2,3,6\}$. 
 </WRAP> </WRAP>
 +++++ Solutions: |
 +   - True: When a number can be divided by $2$, then it can also be divided by $6$. 
 +   - True: When a number can be divided by $2$ and by $3$, then it can also be divided by $6$. 
 +   - False: Assume that $a > 1$ and $b > 1$. Then $a\cdot b \in T(a\cdot b)$, while it can not be in $T(a)$ and $T(b)$ since $a\cdot b > a$ and $a\cdot b > b$, respectively.
 +++++ 
  
 ~~DISCUSSION|Questions, Remarks, and Suggestions~~ ~~DISCUSSION|Questions, Remarks, and Suggestions~~
book/chap2/2.2_sets.1635197086.txt.gz · Last modified: 2021/10/25 23:24 by jv