CMSC250, Spring 2004 Homework 12 Answers

Due Wednesday, April 28 at the beginning of your discussion section.

You must write the solutions to the problems single-sided on your own lined paper, with all sheets stapled together, and with all answers written in sequential order or you will lose points.

1. Prove or disprove the following:
   1. $f : (0, \infty) \to {\bf {R}}$ defined by $f(x) = \ln(x)$ is 1-1.
      • Answer: TRUE. Proof:
        Let $x_1$ and $x_2$ be arbitrary numbers in $(0, \infty)$ such that $f(x_1) = f(x_2)$.
        By the definition of $f$, $f(x_1) = \ln(x_1)$ and $f(x_2) = \ln(x_2)$.
        By substitution, $\ln(x_1) = \ln(x_2)$.
        By taking antilogs, $e^{\ln(x_1)} = e^{\ln(x_2)}$.
        By simplifying, $x_1 = x_2$.
        By closing the conditional world and generalizing from the generic particular, $\forall x_1, x_2 \in (0, \infty) \ [ f(x_1) = f(x_2) ] \to [x_1=x_2]$.
        Therefore, $f$ is 1-1 by the definition of 1-1.

   2. $f : (0, \infty) \to {\bf {R}}$ defined by $f(x) = \ln(x)$ is onto.
      • Answer: TRUE. Proof:
        Let $y$ be arbitrary in ${\bf {R}}$.
        Define $x$ to be $e^y$.
        Because $e > 0$, we know that $x = e^y$ is greater than zero, and therefore is in $(0, \infty)$.
        By the definition of $f$, $f(x) = f\left(e^y \right) = \ln(e^y) = y\cdot \ln (e) = y$ by substitution, algebra, and properties of logarithms.
        By generalizing from the generic particular, $\forall y \in {\bf {R}}\ \exists x \in (0, \infty) \ f(x)=y$.
        Therefore, $f$ is onto by the definition of onto.

2. Let $A$ and $B$ be sets, and $f : A \to B$ be the function $f(x) = x^2$. For each of the four following questions, define the sets $A$ and $B$ such that $f$ is
   1. 1-1 and onto.
      • Answer: Let $A = B = {\bf {R}}^{\geq 0}$. Others are possible, too.
   2. 1-1 but not onto.
      • Answer: Let $A = {\bf {R}}^{\geq 0}$ and $B = {\bf {R}}$. Others are possible, too.
   3. onto but not 1-1.
      • Answer: Let $A = {\bf {R}}$ and $B = {\bf {R}}^{\geq 0}$. Others are possible, too.
   4. neither 1-1 nor onto.
      • Answer: Let $A = B = {\bf {R}}$. Others are possible, too.

3. Let $f$ be an arbitrary bijective function from ${\bf {R}}$ to ${\bf {R}}$.
   1. Define a function $g_1 : {\bf {R}}\to {\bf {R}}$ in terms of $f$ so that $g_1$ is 1-1 but not onto. Prove that $g_1$ satisfies this criteria.
      • Answer: There are many answers. Here's one.
        Let $g_1(x) = e^{f(x)}$.

        
        

        First, we prove that $g_1$ is 1-1:
        Let $a$ and $b$ be arbitrary real numbers and assume $g_1(a) = g_1(b)$.
        By definition of $g_1$, $g_1(a) = e^f(a)$ and $g_1(b) = e^f(b)$.
        By substitution, $e^f(a) = e^f(b)$.
        By taking the natural log of both sides, we get $f(a) = f(b)$.
        Since $f$ is 1-1, we know $a = b$. By closing the conditional world and generalizing from the generic particular,
        $\forall a, b \in {\bf {R}}\ \left[ g_1(a) = g_1(b) \right] \to \left[ a = b \right]$.
        Then $g_1$ is 1-1 by the definition of 1-1.

        
        

        Now we show $g_1$ is not onto:
        Let $y = 0$. Zero is a real number, but since $e > 0$, raising $e$ to any power will never give you zero. Therefore, $g_1$ is not onto.

   2. Define a function $g_2 : {\bf {R}}\to {\bf {R}}$ in terms of $f$ so that $g_2$ is onto but not 1-1. Prove that $g_2$ satisfies this criteria.
      • Answer: Again, there are many answers. Here's one.
        Let $g_2(x) = f(\ln \vert x\vert)$ if $x \neq 0$ and 0 otherwise.

        
        

        First, we prove that $g_2$ is onto:
        Let $y$ be arbitrary in ${\bf {R}}$.
        Define $x$ to be $e^{f^{-1}(y)}$.
        Then $g_2(x) = g_2(e^{f^{-1}(y)}) = f(\ln \vert e^{f^{-1}(y)}\vert) = f(f^{-1}(y)) = y$.
        We can do this because $f^{-1}$ exists since $f$ is a bijection, and since $e > 0$, $e^{f^{-1}(y)}$ will never be zero.
        By generalizing from the generic particular, $\forall y \in {\bf {R}}\ \exists x \in {\bf {R}}\ g_2(x) = y$.
        Therefore, $g_2$ is onto by the definition of onto.

        
        

        Now we show $g_2$ is not 1-1:
        Let $x_1 = -1$ and $x_2 = 1$.
        Then $f(x_1) = \ln\vert-1\vert = \ln\vert 1\vert = f(x_2)$, but obviously $x_1 \neq x_2$.

4. Let $D$ denote the set of odd integers.
   That is, $D = \{n \in {\bf {Z}}\mid n=2k+1$    for some integer $k\}$.
   1. Define a bijective function $h_1 : D \to {\bf {Z}}$. Prove that $h_1$ is a bijection.
      • Answer: Let $h_1(x) = (x-1)/2$.

        
        

        First we prove $h_1$ is 1-1:
        Let $a$ and $b$ be arbitrary odd numbers and assume $h_1(a) = h_1(b)$.
        By definition of $h_1$, $h_1(a) = (a-1)/2$ and $h_1(b) = (b-1)/2$.
        By substitution, $(a-1)/2 = (b-1)/2$.
        By multiplying both sides by 2 and adding 1, $a = b$.
        By closing the conditional world and generalizing from the generic particular,
        $\forall a, b \in D \ \left[ h_1(a) = h_1(b) \right] \to \left[ a = b \right]$.
        Then $h_1$ is 1-1 by the definition of 1-1.

        
        

        Now we show $h_1$ is onto:
        Let $y$ be arbitrary in ${\bf {Z}}$.
        Define $x$ to be $2y+1$. We know $2y+1 \in D$ by the definition of odd.
        Then $h_1(x) = ((2y+1)-1)/2 = (2y)/2 = y$.
        By generalizing from the generic particular, $\forall y \in {\bf {Z}}\ \exists x \in D \ h_1(x) = y$.
        Therefore, $h_1$ is onto by the definition of onto.

   2. Define a bijective function $h_2 : {\bf {Z}}\to D$. Prove that $h_2$ is a bijection.
      • Answer: Let $h_2(x) = 2x+1$.

        
        

        First we prove $h_1$ is 1-1:
        Let $a$ and $b$ be arbitrary integers and assume $h_2(a) = h_2(b)$.
        By definition of $h_2$, $h_2(a) = 2a+1$ and $h_2(b) = 2b+1$.
        By substitution, $2a+1=2b+1$.
        By subtracting 1 from both sides and dividing by 2, $a = b$.
        By closing the conditional world and generalizing from the generic particular,
        $\forall a, b \in {\bf {Z}}\ \left[ h_2(a) = h_2(b) \right] \to \left[ a = b \right]$.
        Then $h_2$ is 1-1 by the definition of 1-1.

        
        

        Now we show $h_2$ is onto:
        Let $y$ be arbitrary in $D$.
        Define $x$ to be $(y-1)/2$.
        Since $y$ is odd, there exists some integer $k$ such that $y=2k+1$. Therefore, $x = (y-1)/2 = ((2k+1)-1)/2 = k$, which means $x$ is an integer.
        Then $h_2(x) = h_2((y-1)/2) = 2((y-1)/2)+1 = y-1+1 = y$.
        By generalizing from the generic particular, $\forall y \in D \ \exists x \in {\bf {Z}}\ h_2(x) = y$.
        Therefore, $h_2$ is onto by the definition of onto.

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