گام به گام تمرین صفحه 33 درس آشنایی با مبانی ریاضیات آمار و احتمال

\(\begin{array}{*{20}{l}}{A = \{ - 1\,,\,0\,,\,1\} }\\{}\\{C = \{ 0\,,\,1\,,\,2\} }\\{}\\{E = \{ 0\,,\,1\,,\,2\} }\end{array}\,\,\,\,\,\,\,\,\,\begin{array}{*{20}{l}}{B = \{ - 1\,,\,0\,,\,1\} }\\{}\\{D = \{ - 1\,,\,0\,,\,1\} }\\{}\\{}\end{array}\)

بنابراین نتیجه می شود که A=B=D و C=E

گیریم مجموعه A دارای n عضو باشد در نتیجه \({2^n} - 384 = {2^{n - 2}}\) ؛ بنابراین به حل این معادله می پردازیم:

\(\begin{array}{l}{2^n} - 384 = {2^{n - 2}} \Rightarrow {2^n} - 384 = \frac{{{2^n}}}{4} \Rightarrow {2^n} - \frac{{{2^n}}}{4} = 384\\\\ \Rightarrow \frac{3}{4} \times {2^n} = 384 \Rightarrow {2^n} = 384 \times \frac{4}{3} = 512 \Rightarrow {2^n} = {2^9}\\\\ \Rightarrow n = 9\end{array}\)

مجموعه دارای 9 عضو است.

\(\left\{ \begin{array}{l}x - y = 4\\\\x + 2y = 5\end{array} \right.\,\,\,\,\, \Rightarrow \,\,\,\,\,\left\{ \begin{array}{l}x = 3\\\\y = 1\end{array} \right.\)

\(\forall x;[x \in A - B] \Rightarrow x \in A\,,\,x \notin B \Rightarrow x \in A \Rightarrow A - B \subseteq A\)

می دانیم که \(A \subseteq B\) ؛ بنابراین:

\(\begin{array}{l}\forall x;(x \in A \cup C) \Rightarrow (x \in A \vee x \in C) \Rightarrow (x \in B \vee x \in C)\\\\ \Rightarrow (x \in B \cup C) \Rightarrow A \cup C \subseteq B \cup C\end{array}\)

می دانیم که \(A \subseteq B\)؛ بنابراین:

\(\begin{array}{l}\forall x;(x \in A \cap C) \Rightarrow (x \in A \wedge x \in C) \Rightarrow (x \in B \wedge x \in C)\\\\ \Rightarrow (x \in B \cap C) \Rightarrow A \cap C \subseteq B \cap C\end{array}\)

به این دلیل که \(A \subseteq B\) و \(C \subseteq D\)، داریم:

\(\begin{array}{l}\forall x;(x \in A \cap C) \Rightarrow (x \in A \wedge x \in C) \Rightarrow (x \in B \wedge x \in D)\\\\ \Rightarrow (x \in B \cap D) \Rightarrow A \cap C \subseteq B \cap D\end{array}\)

به این دلیل که \(A \subseteq B\) و \(C \subseteq D\)، داریم:

\(\begin{array}{l}\forall x;(x \in A \cap C) \Rightarrow (x \in A \wedge x \in C) \Rightarrow (x \in B \wedge x \in D)\\\\ \Rightarrow (x \in B \vee x \in D) \Rightarrow (x \in B \cup D) \Rightarrow A \cap C \subseteq B \cup D\end{array}\)

می دانیم تهی، زیرمجموعه هر مجموعه است. بنابراین:

\(\left\{ \begin{array}{l}\emptyset \subseteq A\\\\A \subseteq \emptyset \end{array} \right. \Rightarrow A = \emptyset \)

می دانیم هر مجموعه، زیرمجموعه مجموعۀ مرجع است. بنابراین:

\(\left\{ \begin{array}{l}A \subseteq U\\\\U \subseteq A\end{array} \right. \Rightarrow A = U\)

\(\begin{array}{l}\forall x;x \in B - A \Rightarrow (x \in B \wedge x \notin A) \Rightarrow x \in B \Rightarrow B - A \subseteq B\\\\\left\{ \begin{array}{l}\forall x;x \in B\\\\A \cap B = \emptyset \end{array} \right. \Rightarrow (x \in B \wedge x \notin A) \Rightarrow x \in B - A \Rightarrow B \subseteq B - A\\\\ \Rightarrow B - A = B\end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}\forall x;x \in B\\\\A \cap B = \emptyset \end{array} \right. \Rightarrow (x \in B \wedge x \notin A) \Rightarrow x \notin A \Rightarrow x \in A'\\\\ \Rightarrow B \subseteq A'\end{array}\)

\(\begin{array}{l}A \cap B = \{ x \in U|x \in A \wedge x \in B\} = \\\\\{ x \in U|x \in B \wedge x \in A\} = B \cap A\end{array}\)

\(\begin{array}{l}A \cap (B \cap C) = \{ x \in U|x \in A \wedge x \in B \cap C\} = \\\\\{ x \in U|x \in A \wedge (x \in B \wedge x \in C)\} = \\\\\{ x \in U|(x \in A \wedge x \in B) \wedge x \in C\} = \\\\\{ x \in U|x \in A \cap B \wedge x \in C\} = (A \cap B) \cap C\end{array}\)

\(\begin{array}{l}x \in A \cap (B \cup C) \Rightarrow x \in A \wedge x \in B \cup C\\\\ \Rightarrow x \in A \wedge (x \in B \vee x \in C)\\\\ \Rightarrow (x \in A \wedge x \in B) \vee (x \in A \wedge x \in C)\\\\ \Rightarrow (x \in A \cap B) \vee (x \in A \cap C) \Rightarrow x \in (A \cap B) \cup (A \cap C)\\\\ \Rightarrow A \cap (B \cup C) \subseteq (A \cap B) \cup (A \cap C)\end{array}\)

به طور مشابه ثابت می کنیم \((A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)\)؛ داریم:

\(\begin{array}{l}x \in (A \cap B) \cup (A \cap C) \Rightarrow (x \in A \cap B) \vee (x \in A \cap C)\\\\ \Rightarrow (x \in A \wedge x \in B) \vee (x \in A \wedge x \in C)\\\\ \Rightarrow x \in A \wedge (x \in B \vee x \in C)\\\\ \Rightarrow x \in A \wedge x \in B \cup C \Rightarrow x \in A \cap (B \cup C)\\\\ \Rightarrow (A \cap B) \cup (A \cap C) \subseteq A \cap (B \cup C)\end{array}\)

در نهایت داریم:

\(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\)

\((A \cap B) \cup (B' \cap A) = A \cap (B \cup B') = A \cap U = A\)

\(\begin{array}{l}(A' \cap B') \cap A = (B' \cap A') \cap A = B' \cap (A' \cap A) = \\\\B' \cap \emptyset = \emptyset \end{array}\)

\(A \cap (B \cap C) = (A \cap A) \cap (B \cap C) = (A \cap B) \cap (A \cap C)\)

\(A \cup (B \cup C) = (A \cup A) \cup (B \cup C) = (A \cup B) \cup (A \cup C)\)

\(\begin{array}{l}(A' \cap B) \cup ([(B \cap A) - B'] \cap (B \cup A)) = \\\\(A' \cap B) \cup ([(B \cap A) \cap B] \cap (B \cup A)) = \\\\(A' \cap B) \cup ([(B \cap B) \cap A] \cap (B \cup A)) = \\\\(A' \cap B) \cup ((B \cap A) \cap (B \cup A)) = \\\\(A' \cap B) \cup ([B \cap (B \cup A)] \cap [A \cap (B \cup A)]) = \\\\(A' \cap B) \cup (B \cap A) = (A' \cap B) \cup (A \cap B) = \\\\(A' \cup A) \cap B = U \cap B = B\end{array}\)

\(\begin{array}{l}(A \cup B) - B = (A \cup B) \cap B' = (A \cap B') \cup (B \cap B') = \\\\(A \cap B') \cup \emptyset = A \cap B' = A - B\end{array}\)

\([(A \cup B) - A] \cup (A \cap B) = [(A \cup B) \cap A'] \cup (A \cap B) = \)

\([(A \cap A') \cup (B \cap A')] \cup (A \cap B) = \)

\(\begin{array}{l}[\emptyset \cup (B \cap A')] \cup (A \cap B) = (B \cap A') \cup (A \cap B) = \\\\(A' \cap B) \cup (A \cap B) = (A' \cup A) \cap B = U \cap B = B\end{array}\)

\((A \subseteq X) \wedge (A' \subseteq X) = (A \cup A') \subseteq X \Rightarrow U \subseteq X\)

از طرفی می دانیم همواره \(X \subseteq U\)، بنابراین X=U

\(\begin{array}{l}(A - B) \cup (A \cap B) = (A \cap B') \cup (A \cap B) = \\\\A \cup (B' \cap B) = A \cup \emptyset = A\end{array}\)

\(\begin{array}{l}(A \cap B) - C = (A \cap B) \cap C' = (A \cap B) \cap (C' \cap C') = \\\\(A \cap C') \cap (B \cap C') = (A - C) \cap (B - C)\end{array}\)

\((A - B) \cup (B - A) = (A \cap B') \cup (B \cap A') = \)

\([A \cup (B \cap A')] \cap [B' \cup (B \cap A')] = \)

\([(A \cup B) \cap (A \cup A')] \cap [(B' \cup B) \cap (B' \cup A')] = \)

\([(A \cup B) \cap U] \cap [U \cap (B' \cup A')] = (A \cup B) \cap (A' \cup B') = \)

\((A \cup B) - (A' \cup B')' = (A \cup B) - (A \cap B)\)

\(\begin{array}{l}(A \cup B) \cap (A' \cap B') = (A \cup B) - (A' \cap B')' = \\\\(A \cup B) - (A \cup B) = \emptyset \end{array}\)

\(\begin{array}{l}\left\{ \begin{array}{l}(A \cup B) = (A \cup C)\,\,\,\,\,\,\,\,\,\,(i)\\\\(A \cap B) = (A \cap C)\,\,\,\,\,\,\,\,\,\,(ii)\end{array} \right.\\\\B = B \cup (A \cap B)\mathop = \limits^{(iI)} B \cup (A \cap C) = (B \cup A) \cap (B \cup C) = \\\\(A \cup B) \cap (B \cup C)\mathop = \limits^I (A \cup C) \cap (B \cup C) = (A \cap B) \cup C\mathop = \limits^{(ii)} \\\\(A \cap C) \cup C = C\\\\ \Rightarrow B = C\end{array}\)

از \(A \times B = B \times A\) نتیجه می شود A=B؛ بنابراین:

{y+2 , 5 , z}={x+1 , 4 , -2}

واضح است که 5 فقط می تواند با x+1 برابر باشد لذا x=4 است. اما در موارد دیگر دو حالت داریم :

\(\begin{array}{l}[(y + 2 = 4) \wedge (z = - 2)] \vee [(y + 2 = - 2) \wedge (z = 4)]\\\\ \Rightarrow [(y = 2) \wedge (z = - 2)] \vee [(y = - 4) \wedge (z = 4)] \Rightarrow y + z = 0\end{array}\)

در نتیجه x+y+z=4 خواهد بود.

\(\begin{array}{l}A \times B = \{ (2\,,\,2)\,,\,(2\,,\,3)\,,\,(2\,,\,4)\,,\,(3\,,\,2)\,,\,(3\,,\,3)\,,\,(3\,,\,4)\} \\\\B \times A = \{ (2\,,\,2)\,,\,(2\,,\,3)\,,\,(3\,,\,2)\,,\,(3\,,\,3)\,,\,(4\,,\,2)\,,\,(4\,,\,3)\} \end{array}\)