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--- book:chap2:2.2_sets [2022/04/01 20:00] – jv
+++ book:chap2:2.2_sets [2024/10/26 13:02] (current) – jv
@@ Line 115: / Line 115: @@
 |$ A \Leftrightarrow B $  |  1  |  0  |  0  |  1  | $A$ is equivalent to $B$ |
 |$ 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 $B$ |
 Table 2.1: List of the results of different junctors acting on two statements $A$ and $B$.
@@ Line 337: / Line 337: @@
 <WRAP group>
 <WRAP half column>
   * **a)** $\quad\lbrace A, B \rbrace \quad {{ \Box}} \quad \lbrace A, B, C \rbrace$,
   * **c)** $\quad\lbrace \emptyset \rbrace \quad  {\Box} \quad \emptyset$,
   * **e)** $\quad A  \quad { \Box} \quad \lbrace A, B, C \rbrace$,
   *  **g)** $\quad\lbrace A, C, D \rbrace \setminus \lbrace A, B \rbrace \;\Box\; \lbrace A, B, C \rbrace$,
 </WRAP>
 <WRAP half column>
   *  **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)** $\lbrace A, C, D \rbrace \cap \lbrace A, B \rbrace \, {\Box} \, \lbrace A, B, C, D \rbrace$,
   *  **h)** $\lbrace A, C, D \rbrace \cup \lbrace A, B \rbrace \quad { \Box} \quad  A$.
 </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>
@@ Line 360: / Line 376: @@
   -  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>
+++++ Solutions: |
+  -   $\lbrack 1; 17 \rbrack \cap \rbrack 0; 5 \lbrack = [1; 5[$
+  -   $[-1; 4] \backslash [1; 2[ = [-1, 1[ \cup [2; 4]$
+++++
 ----
@@ Line 377: / Line 398: @@
 Example:  $T(2) = \{1,2\}$, $T(3) = \{1,3\}$, and $T(6) = \{1,2,3,6\}$.
 </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~~