Theoretical Mechanics IPSP

Jürgen Vollmer, Universität Leipzig

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book:chap2:2.2_sets [2021/10/26 11:33] – abrilbook: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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 |$ 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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 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 #Thm_SetEquivalence >**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>**Remark.** [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 >**Remark.** [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.
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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
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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 [[#bsp_Sets|Example 2.2]]).+
 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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 <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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 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.1635240837.txt.gz · Last modified: 2021/10/26 11:33 by abril