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chapter_7/exercises.md
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@ -2389,7 +2389,7 @@ $$ -x_1 = -x_2 $$
$$ x_1 = x_2 $$
Thus it has been shown that $(x_1, y_1) = (x_2) = y_2$.
Thus it has been shown that $(x_1, y_1) = (x_2, y_2)$.
By the definition of one-to-one, it can be concluded that $G$ is one-to-one.
@ -2425,7 +2425,7 @@ $$ x = -w $$
Now, note that $\dfrac{t}{2} \in \mathbb{R}$, and $-w \in \mathbb{R}$. It
follows that $\left(\dfrac{t}{2}, -w\right) \in \mathbb{R} \times \mathbb{R}$.
Now, evaluating for $G\left(-w, \dfrac{t}{2}\right)$:
Now, evaluating for $G(x, y)$, which is $G\left(-w, \dfrac{t}{2}\right)$:
$$ G\left(-w, \frac{t}{2}\right) = \left(2\left(\frac{t}{2}, -(-w)\right)\right) $$
@ -2447,7 +2447,7 @@ a. Is $H$ one-to-one? Prove or give a counterexample.
$H$ is one-to-one.
**Proof:
**Proof:**
Suppose $(x_1, y_1), (x_2, y_2) \in \mathbb{R} \times \mathbb{R}$ such that
$H(x_1, y_1) = H(x_2, y_2)$.
@ -2558,7 +2558,73 @@ Omitted.
a. Is $\log_{8}27 = \log_{2}3$? Why or why not?
a. Is $\log_{16}9 = \log_{4}3$? Why or why not?
Let $x = \log_{8}27$, and let $y = \log_{2}3$. By definition of logarithms:
$$ 8^x = 27 \text{ and } 2^y = 3 $$
Now, $8 = 2^3$, so:
$$ 8^x = (2^3)^x = 2^{3x} $$
Also, $27 = 3^3, so:$
$$ 27 = 3^3 = (2^y)^3 = 2^{3y}$$
Hence, since $8^x = 27$:
$$ 8^x = 2^{3x} = 27 = 2^{3y} $$
Since:
$$ 2^{3x} = 2^{3y} $$
By the laws of exponents:
$$ 3x = 3y $$
Then, by algebra:
$$ x = y $$
Now, we back-substitute our original definitions of $x$ and $y$, and find that:
$$ \log_{8}27 = \log_{2}3 $$
It can therefore be concluded that the answer to the query is yes.
b. Is $\log_{16}9 = \log_{4}3$? Why or why not?
Let $x = \log_{16}9$ and $y = \log_{4}3$. Then by definition of log:
$$ 16^x = 9 \text{ and } 4^y = 3 $$
Note that $16 = 4^2$, so:
$$ 9 = (4^2)^x = 4^{2x} $$
Note that $9 = 3^2$, so:
$$ 9 = 3^2 = (4^y)^2 = 4^{2y} $$
So, by the laws of equivalency:
$$ 4^{2x} = 9 = 4^{2y} $$
$$ 4^{2x} = 4^{2y} $$
By the laws of exponents then:
$$ 2x = 2y $$
Then, by algebra:
$$ x = y $$
Back-substituting in the definitions for $x$ and $y$:
$$ \log_{16}9 = \log_{4}3 $$
Therefore the answer to the given question is yes.
The properties of logarithm established in 33-35 are used in Sections 11.4 and
11.5.
@ -2567,15 +2633,112 @@ The properties of logarithm established in 33-35 are used in Sections 11.4 and
$$ \log_{b}\left(\frac{x}{y}\right) = \log_{b}x - \log_{b}y $$
**Proof:**
Suppose that $b$, $x$, and $y$ are any positive real numbers with $b \neq 1$.
Let $u = \log_{b}x$ and $v = \log_{b}y$. By definition of logarithm then:
$$ b^u = x \text{ and } b^v = y $$
By substitution:
$$ \frac{x}{y} = \frac{b^u}{b^v} $$
By the laws of exponents:
$$ = b^{u - v} $$
Taking the logarithm base $b$ of both sides now gives:
$$ \log_{b}\left(\frac{x}{y}\right) = \log_{b}(b^{u - v}) $$
$$ = u - v $$
Back-substituting the definitions of $u$ and $v$ yields:
$$ = \log_{b}x - \log_{b}y $$
This is what was to be shown.
Q.E.D.
34. Prove that for all positive real numbers $b$, $x$, and $y$ with $b \neq 1$,
$$ \log_{b}(xy) = \log_{b}x + \log_{b}y $$
**Proof:**
Suppose $b$, $x$, and $y$ are any positive real numbers with $b \neq 1$.
Let $u = \log_{b}x$, and $v = \log_{b}y$.
By definition of logarithms, this means that:
$$ b^u = x \text{ and } b^v = y $$
By substitution, this means that:
$$ xy = b^u \cdot b^v $$
$$ = b^{u + v} $$
Taking the logarithm of base $b$ of both sides yields:
$$ \log_{b}(xy) = \log_{b}(b^{u + v}) $$
$$ = u + v $$
Back-substituting in the values for $u$ and $v$ shows:
$$ \log_{b}(xy) = \log_{b}x + \log_{b}y $$
This is what was to be shown.
Q.E.D.
35. Prove that for all real numbers $a$, $b$, and $x$ with $b$ and $x$ positive
and $b \neq 1$,
$$ \log_{b}(x^a) = a\log_{b}x $$
**Proof:**
Suppose $a$, $b$, and $x$ are any real numbers with $x$ and $b$ being positive
and $b \neq 1$.
Let $r = \log_{b}(x^a)$ and $s = \log_{b}x$.
By definition of logarithms, this means that:
$$ b^r = x^a \text{ and } b^s = x $$
Since $b^s = x$, by substitution:
$$ b^r = x^a = (b^s)^a = b^{sa} $$
So:
$$ x^a = b^{sa} $$
Now, applying $\log_{b}$ to both sides:
$$ \log_{b}(x^a) = \log_{b}(b^{sa}) $$
$$ = sa $$
Back-substituting in the definition for $s$, this yields:
$$ \log_{b}(x^a) = \log_{b}x \cdot a $$
Or:
$$ \log_{b}(x^a) = a\log_{b}x $$
This is what was to be shown.
Q.E.D.
Exercises 36 and 37 use the following definition: If
$f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ are functions,
then the function $(f + g): \mathbb{R} \to \mathbb{R}$ is defined by the formula
@ -2584,9 +2747,61 @@ $(f + g)(x) = f(x) + g(x)$ for every real number $x$.
36. If $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ are
both one-to-one, is $f + g$ also one-to-one? Justify your answer.
No.
**Disproof (by counterexample):**
Suppose $f$ and $g$ are functions such that $f: \mathbb{R} \to \mathbb{R}$ and
$g: \mathbb{R} \to \mathbb{R}$ and both $f$ and $g$ are one-to-one functions.
Furthermore, suppose $(f + g)$ is a function where
$(f + g): \mathbb{R} \to \mathbb{R}$ such that $(f + g)(x) = f(x) + g(x)$.
Consider $f(x) = x$ and $g(x) = -x$. Note that $f$ and $g$ are one-to-one
functions still follow the definitions of $f$ and $g$ in the supposition.
Then, by definition of $(f + g)$, $(f + g)(x) = f(x) + g(x) = x + (-x) = 0$.
Then consider $x_1 = 1$, and $x_2 = 2$, then:
$$ f(x_1) = 1 \text{ and } g(x_1) = -1 \text{ and } (f + g)(x_1) = 1 + (-1) = 0 $$
$$ f(x_2) = 2 \text{ and } g(x_2) = -2 \text{ and } (f + g)(x_2) = 2 + (-2) = 0 $$
So $(f + g)(x_1) = (f + g)(x_2)$, but $x_1 \neq x_2$.
By the definition of one-to-one, it can therefore be concluded that $(f + g)$ is
not one-to-one.
Q.E.D.
37. If $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ are
both onto, is $f + g$ also onto? Justify your answer.
**Disproof (by counterexample):**
Suppose $f$ and $g$ are both functions where $f: \mathbb{R} \to \mathbb{R}$, and
$g: \mathbb{R} \to \mathbb{R}$. Furthermore, suppose
$(f + g): \mathbb{R} \to \mathbb{R}$ where $(f + g)(x) = f(x) + g(x)$ for some
$x \in \mathbb{R}$.
Consider $f(x) = x$ and $g(x) = -x$. Note that both $f$ and $g$ are still onto
based off the definition of onto as required by the supposition.
Then by definition of $(f + g)$:
$$ (f + g)(x) = x + (-x) = 0 $$
Since no matter what the value for $x$ will always output $0$, while
$0 \in \mathbb{R}$, by the definition of onto, every element in the co-domain of
$\mathbb{R}$ must have a corresponding input image.
Consider that $1 \in \mathbb{R}$, but there is no input image $x$ such that
$(f + g)(x) = 1$.
Therefore, by the definition of onto, $(f + g)$ is not onto.
Q.E.D.
Exercises 38 and 39 use the following definition: If
$f: \mathbb{R} \to \mathbb{R}$ and $c$ is a nonzero real number, the function
$(c \cdot f): \mathbb{R} \to \mathbb{R}$ is defined by the formula
@ -2596,16 +2811,284 @@ $(c \cdot f)(x) = c \cdot (f(x))$ for every real number $x$.
number. If $f$ is one-to-one, is $c \cdot f$ also one-to-one? Justify your
answer.
Yes, $(c \cdot f)$ is one-to-one.
**Proof:**
Suppose $f: \mathbb{R} \to \mathbb{R}$ is a one-to-one function, and that $c$ is
a nonzero real number such that $(c \cdot f): \mathbb{R} \to \mathbb{R}$ is
defined as $(c \cdot f)(x) = c \cdot (f(x))$ for any real number $x$.
To prove $(c \cdot f)$ is one-to-one, it must be shown that there are some
$x_1, x_2 \in \mathbb{R}$ such that if $(c \cdot f)(x_1) = (c \cdot f)(x_2)$,
then $x_1 = x_2$.
By definition of $(c \cdot f)$:
$$ (c \cdot f)(x_1) = c \cdot (f(x_1)) = c \cdot (f(x_2)) = (c \cdot f)(x_2) $$
$$ c \cdot (f(x_1)) = c \cdot (f(x_2)) $$
By arithmetic:
$$ f(x_1) = f(x_2) $$
By the supposition, $f$ is a one-to-one function, so therefore, by definition of
one-to-one:
$$ x_1 = x_2 $$
This is what was to be shown.
Q.E.D.
39. Let $f: \mathbb{R} \to \mathbb{R}$ be a function and $c$ a nonzero real
number. If $f$ is onto, is $c \cdot f$ also onto? Justify your answer.
$c \cdot f$ is onto.
**Proof:**
Suppose $f: \mathbb{R} \to \mathbb{R}$ such that $f$ is onto. Furthermore,
suppose $c$ is a nonzero real number, where
$(c \cdot f): \mathbb{R} \to \mathbb{R}$ is defined as
$(c \cdot f)(x) = c \cdot (f(x))$ for any real number $x$.
To prove that $(c \cdot f)(x)$ is onto, it must be shown that there exists some
$y \in \mathbb{R}$, such that $(c \cdot f)(x) = y$.
By definition for $c \cdot f$:
$$ (c \cdot f)(x) = c \cdot (f(x)) = y $$
$$ c \cdot (f(x)) = y $$
By algebra:
$$ f(x) = \frac{y}{c} $$
Since $f$ is onto (by the supposition), this means that there exists some
$z \in \mathbb{R}$ such that $f(z) = \dfrac{y}{c}$.
Let $x = z$, then:
$$ (c \cdot f)(x) = c \cdot (f(x)) $$
$$ = c \cdot (f(z)) $$
$$ = c \cdot \left(\frac{y}{c}\right) $$
$$ = y $$
This is what was to be shown. Therefore it can be concluded that $(c \cdot f)$
is onto.
Q.E.D.
40. Suppose $F: X \to Y$ is one-to-one.
a. Prove that for every subset $A \subseteq X$, $F^{-1}(F(A)) = A$.
**Proof:**
Suppose $A \subseteq X$.
To prove that $F^{-1}(F(A)) = A$, it must be shown that:
$$ F^{-1}(F(A)) \subseteq A $$
and also that:
$$ A \subseteq F^{-1}(F(A)) $$
_Proof ($F^{-1}(F(A)) \subseteq A$):_
Let $x \in F^{-1}(F(A))$.
By the definition of inverse image:
$$ F^{-1}(F(A)) = \{x \in X | F(x) \in F(A)\} $$
By the definition for $F(A)$, there exists $r \in A$ such that $F(r) = F(x)$.
Since $F(r) = F(x)$, and since $F$ is one-to-one, it follows that $x \in A$
Since $x \in F^{-1}(F(A))$ and $x \in A$, it can be concluded that
$F^{-1}(F(A)) \subseteq A$.
This is what was to be shown.
_Proof ($A \subseteq F^{-1}(F(A))$):_
Let $x \in A$.
Since $x \in A$, then $F(x) \in F(A)$, by the definition of $F(A)$.
By the definition of inverse image:
$$ x \in F^{-1}(F(A)) $$
Since $x \in A$ and $x \in F^{-1}(F(A))$, it can be concluded that
$A \subseteq F^{-1}(F(A))$.
This is what was to be shown.
_Conclusion:_
Since both subset definitions have been shown, it can be concluded that
$F^{-1}(F(A)) = A$.
Q.E.D.
b. Prove that for all subsets $A_1$ and $A_2$ in $X$,
$F(A_1 \cap A_2) = F(A_1) \cap F(A_2)$.
**Proof:**
Suppose $A_1, A_2 \in X$.
To prove $F(A_1 \cap A_2) = F(A_1) \cap F(A_2)$, it must be shown that:
$$ F(A_1 \cap A_2) \subseteq F(A_1) \cap F(A_2) $$
and that:
$$ F(A_1) \cap F(A_2) \subseteq F(A_1 \cap A_2) $$
_Proof ($F(A_1 \cap A_2) \subseteq F(A_1) \cap F(A_2)$):_
Suppose $y \in F(A_1 \cap A_2)$.
It must be shown that $y \in F(A_1) \cap F(A_2)$.
By the definition of $F(A_1 \cap A_2)$, there exists some $x \in A_1 \cap A_2$
such that $F(x) = y$.
By the definition of intersection:
$$ x \in A_1 \text{ and } x \in A_2 $$
Since $x \in A_1$ and $x \in A_2$, it follows that:
$$ y \in F(A_1) \text{ and } y \in F(A_2) $$
By the definition of intersection, this means that:
$$ y \in F(A_1) \cap F(A_2) $$
Since $y \in F(A_1 \cap A_2)$ and $y \in F(A_1) \cap F(A_2)$, it can be
concluded that $F(A_1 \cap A_2) \subseteq F(A_1) \cap F(A_2)$.
This is what was to be shown.
_Proof ($F(A_1) \cap F(A_2) \subseteq F(A_1 \cap A_2)$):_
Suppose $y \in F(A_1) \cap F(A_2)$.
It must be shown that $y \in F(A_1 \cap A_2)$.
By the definition of intersection:
$$ y \in F(A_1) \text{ and } y \in F(A_2) $$
By the definition of $F(A_1)$, there exists some $x_1 \in A_1$ such that:
$$ F(x_1) = y $$
Similarly, by definition of $F(A_2)$, there exists some $x_2 \in A_2$ such that:
$$ F(x_2) = y $$
Since $F$ is one-to-one (by the supposition), and since $F(x_1) = y = F(x_2)$,
or $F(x_1) = F(x_2)$, this means that:
$$ x_1 = x_2 $$
By the definition of intersection:
$$ x_1 \in A_1 \cap A_2 $$
It follows then that since $y = F(x_1)$, that:
$$ y \in F(A_1) \cap F(A_2) $$
Since $y \in F(A_1) \cap F(A_2)$ and $y \in F(A_1) \cap F(A_2)$, it can be
concluded that $F(A_1) \cap F(A_2) \subseteq F(A_1 \cap A_2)$.
This is what was to be shown.
_Conclusion:_
Since both subset relations have been shown, it can be concluded that
$F(A_1 \cap A_2) = F(A_1) \cap F(A_2)$.
Q.E.D.
41. Suppose $F: X \to Y$ is onto. Prove that for every subset $B \subseteq Y$,
$F(F^{-1}(B)) = B$.
**Proof:**
Suppose $F: X \to Y$ such that $F$ is onto.
Let $B \subseteq Y$.
To prove that $F(F^{-1}(B)) = B$, it must be shown that:
$$ F(F^{-1}(B)) \subseteq B $$
and that:
$$ B \subseteq F(F^{-1}(B)) $$
_Proof ($F(F^{-1}(B)) \subseteq B$):_
Suppose $y \in F(F^{-1}(B))$.
It must be shown that $y \in B$.
By definition of $F$, there exists some $x \in F^{-1}(B)$ such that $F(x) = y$.
By definition of inverse image, since $x \in F^{-1}(B)$, this means that:
$$ F(x) \in B $$
Since $F(x) = y$, it follows then that:
$$ y \in B $$
Since $y \in F(F^{-1}(B))$ and $y \in B$, it can be concluded that
$F(F^{-1}(B)) \subseteq B$.
_Proof ($B \subseteq F(F^{-1}(B))$):_
Suppose $y \in B$.
It must be shown that $y \in F(F^{-1}(B))$.
Since $y \in B$, and since $B \subseteq Y$, it follows that $y \in Y$.
By the supposition, $F$ is onto. It follows that since $y \in Y$, there exists
some $x \in X$ such that $F(x) = y$.
Since $F(x) = y$ and $y \in B$, by the definition of inverse function:
$$ x \in F^{-1}(B) $$
It follows then that:
$$ y \in F(F^{-1}(B)) $$
Since $y \in B$ and $y \in F(F^{-1}(B))$, it can be concluded that
$B \subseteq F(F^{-1}(B))$.
_Conclusion:_
Since both subset relations have been shown, it can be concluded that
$F(F^{-1}(B)) = B$.
Q.E.D.
Let $X = \{a, b, c, d, e\}$ and $Y = \{s, t, u, v, w\}$. In each of 42 and 43 a
one-to-one correspondence $F: X \to Y$ is defined by an arrow diagram. In each
case draw an arrow diagram for $F^{-1}$.