From 4ccc1f3750bf38bf6a210f9f051ceb10033c1f1f Mon Sep 17 00:00:00 2001 From: tomit4 Date: Fri, 22 May 2026 14:24:50 -0700 Subject: [PATCH] :construction: Fin 1 --- chapter_1/1_5/additional_exercises.md | 191 ++++++++++++++++++++++++++ 1 file changed, 191 insertions(+) diff --git a/chapter_1/1_5/additional_exercises.md b/chapter_1/1_5/additional_exercises.md index 2d8f788..fab8210 100644 --- a/chapter_1/1_5/additional_exercises.md +++ b/chapter_1/1_5/additional_exercises.md @@ -374,6 +374,98 @@ $f(f^{-1}(D)) \neq D$. A: +(A) For $f: \mathbb{N} \to \mathbb{N}$ defined by $f(n) = n^2$, what are each of +the following sets? + +(a) $f^{-1}(\{1, 4, 9\})$ + +$$ f^{-1}(\{1, 4, 9\}) = \{1, 2, 3\} $$ + +(b) $f^{-1}(\{2, 3, 5, 7\})$ + +$$ f^{-1}(\{2, 3, 5, 7\}) = \emptyset $$ + +There are no natural numbers, $n$ such that $n^2 \in \{2, 3, 5, 7\}$ since none +of these numbers are perfect squares. + +\(c\) $f^{-1}(\{1, 2, \dots, 10\})$ + +$$ f^{-1}(\{1, 2, \dots, 10\}) = \{1, 2, 3\} $$ + +(B) Prove that for any set $C \subseteq X, C \subseteq f^{-1}(f(C))$. + +- Let $x$ be some element $x \in C$. + +- If we then apply the function $f$ to $x$ as $f(x)$, since $C \subseteq X$, it + follows that $f(x) \in f(C)$ by the definition of the image of a set. + +- Since $f(x) \in f(C)$, it then follows that $x \in f^{-1}(f(C))$ by definition + of an inverse image of a set. + +- Therefore $C \subseteq f^{-1}(f(C))$. + +\(C\) Give an example of a function $f$ and a set $C$ such that +$C \neq f^{-1}*(f(C))$. + +- Let $X$ be the set $\{1, 2\}$, and let $C$ bu a subset of $X$, $C \subseteq X$ + such that $C = \{1\}$. + +- Let $f$ be a function defined as: + +$$ f: X \to Y \quad f(x) = 0 \forall x \in X $$ + +- Therefore $f(C) = \{0\}$ + +- By the definition of an inverse image of a set + $f^{-1}(f(C)) = \{x \in X: f(x)\in f(C)\}$, so therefore + $f^{-1}(f(C)) = \{x \in X: f(x) = 0\}$ + +- Since $X = \{1, 2\}$, then $f(1) = 0$ and $f(2) = 0$, so + $f^{-1}(C) = \{1, 2\}$. + +- Therefore $C = \{1\}$ and $f^{-1}(f(C)) = \{1, 2\}$, which means that + $C \neq f^{-1}(f(C))$. + +(D) Prove that for any set $D \subseteq Y, f(f^{-1}(D)) \subseteq D$. + +- By definition of an inverse image: + +$$ f^{-1}(D) = \{x \in X: f(x) \in D\} $$ + +- By applying $f$ to the inverse, we get: + +$$ f(f^{-1}(D)) = \{f(x): x \in X, f(x) \in D\} $$ + +- Since $x \in f^{-1}(D)$, we have $f(x) \in D$, so every element of + $f(f^{-1}(D))$ lies in $D$. + +- Therefore, $f(f^{-1}(D)) \subseteq D$. + +(E) Give an example of a function $f$ and a set $D$ such that +$f(f^{-1}(D)) \neq D$. + +- Let the following be: + +$$ X = \{1, 2} $$ + +$$ Y = \{0, 1\} $$ + +$$ f(1) = 0, f(2) = 0 $$ + +$$ D = \{0, 1\} $$ + +- The inverse image is: + +$$ f^{-1}(D) = \{1, 2\} $$ + +- Apply $f$ to the inverse: + +$$ f(f^{-1}(D)) = f(\{1, 2\}) = \{0\} $$ + +- Comparing: + +$$ f(f^{-1}(D)) = \{0\} \neq \{0, 1\} = D $$ + 10. Q: Let $f: X \to Y$ be a function, and let $A$ and $B$ be subsets of $Y$. Prove @@ -381,6 +473,37 @@ that $f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B)$. A: +- Let $x \in f^{-1}(A \cap B)$. + +- By definition of an inverse image, this means that $f(x) \in A \cap B$ + +- This means that $f(x) \in A \wedge f(x) \in B$ by definition of an + intersection. + +- It follows then that $x \in f^{-1}(A) \cap f^{-1}(B)$ + +- Therefore $f^{-1}(A \cap B) \subseteq f^{-1}(A) \cap f^{-1}(B)$ + +To prove equality, we must also show the reverse inclusion. + +- Let $x \in f^{-1}(A) \cap f^{-1}(B)$. + +- This means that $x \in f^{-1}(A) \wedge x \in f^{-1}(B)$ by definition of an + intersection. + +- It follows then that $f(x) \in A \wedge f(x) \in B$ by definition of an image + of a set. + +- Then $f(x) \in A \cap B$ by definition of an intersection. + +- So $x \in f^{-1}(A \cap B)$ by definition of an inverse image of a set. + +- Therefore $f^{-1}(A) \cap f^{-1}(B) \subseteq f^{-1}(A \cap B)$ + +Since both inclusions hold, this proves that: + +$$ f^{-1}(A \cap B) = f^{-1}(A) \cap f^{-1}(B) $$ + 11. Q: Let $f: X \to Y$ be a function, and let $A$ and $B$ be subsets of $Y$. Prove @@ -388,6 +511,37 @@ that $f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B)$. A: +- Let $x \in f^{-1}(A \cup B)$ + +- Then $f(x) \in A \cup B$ by definition of an inverse image. + +- This means that $f(x) \in A \vee f(x) \in B$ by definition of a union. + +- It follows then that $x \in f^{-1}(A) \vee x \in f^{-1}(B)$ by definition of + an inverse image. + +- So $x \in f^{-1}(A) \cup f^{-1}(B)$ by definition of a union. + +- Therefore $f^{-1}(A \cup B) \subseteq f^{-1}(A) \cup f^{-1}(B)$ + +Now reverse the inclusion. + +- Let $x \in f^{-1}(A) \cup f^{-1}(B)$ + +- Then $x \in f^{-1}(A) \vee x \in f^{-1}(B)$ by definition of a union. + +- So $f(x) \in A \vee f(x) \in B$ by definition of an inverse image. + +- It then follows that $f(x) \in A \cup B$ by definition of a union. + +- So $x \in f^{-1}(A \cup B)$ by definition of an inverse image. + +- Therefore $f^{-1}(A) \cup f^{-1}(B) \subseteq f^{-1}(A \cup B)$ + +Since both inclusions hold, we have proved that: + +$$ f^{-1}(A \cup B) = f^{-1}(A) \cup f^{-1}(B) $$ + 12. Q: For each relation below, determine whether it is transitive. If it is, prove @@ -396,11 +550,48 @@ it. If it is not, give a counterexample. (a) The relation "$|$" (divides) on $\mathbb{Z}$ defined by $a|b$ provided $b$ is a multiple of $a$. +$a|b$ is transitive on $\mathbb{Z}$ provided $b$ is a multiple of $a$. + +Let $k$ be some multiple of $a$, this means that: + +$$ b = a \cdot k $$ + +Let $c$ then be a multiple of $b$, $c|b$. Then let $m$ be some multiple of $c$. +This would then mean that: + +$$ c = b \cdot m $$ + +$$ c = (a \cdot k) \cdot m = a(km) $$ + +And this means then that $c$ is a multiple of $b$ and also $a$ and that $a|b|c$. +Therefore the property is transitive since this then means that $a|c$. + (b) The relation "$\leq$" (less than or equal to) on $\mathbb{R}$. +The relation $\leq$ is transitive. + +If we have two numbers, $b$, and $c$ such that $b \leq c$, and then we also have +a number $a$, such that $a \leq b$ , then it is true that $a \leq c$, which +demonstrates the transitive property. + \(c\) The relation "$\perp$" (is perpendicular to) on the set of lines in the plane. +The relation $\perp$ is not transitive. Consider one horizontal plane $a$ and a +vertical plane $b$ such that $a \perp b$. Then say you have another horizontal +plane $c$ such that $b \perp c$. But this does NOT mean that $a \perp c$ since +this means that $a$ and $c$ are actually horizontal to each other, i.e. +$\overline{ac}$. + (d) The relation "$\sim$" (is similar to) on the set of triangles in the plane (two triangles are similar if they have the same angles, bu are not necessarily the same size). + +The relation $\sim$ is transitive. Consider two triangles $A$ and $B$ such that +$A \sim B$. This means that both $A$ and $B$ have the same angles (but not +necessarily the same leg dimensions). If we then have a third triangle, $C$, +such that $B \sim C$, then this means that both $B$ and $C$ have the same angles +(but again, not necessarily the same leg dimensions). Since $A$ and $B$ have the +same angles (are $\sim$), and $B$ and $C$ have the same angles (are $\sim$), +then we can conclude that $A \sim C$ and we have shown that the relation $\sim$ +is transitive.