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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions Midterm Exam 1 -- October 4, 2004}
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II.  Consider the following definition:  A function $f$
is periodic if and only if there exists a real number
$P > 0$ such that for all real numbers $x$, $f(x + P) = f(x)$.
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\item{B)} (10) Give the negation of the statement ``$f$ is periodic''.
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{\it Solution:}  The negation of ``there exists a real number
$P > 0$ such that for all real numbers $x$, $f(x + P) = f(x)$''
is:  ``for all real numbers $P > 0$, there exists a real number
$x$ such that $f(x+P) \ne f(x)$.''  Note:  You do {\it not}
negate the side condition $P > 0$ when you apply the general
rule 
$$\sim (\exists P \ni q(P)) \Leftrightarrow \forall P, \sim q(P)$$ 
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III.  Let $A = \{x \in {\bf R} : 1 < x^2 < 9\}$, let $B$ be the
interval $[-2,4)$, and let $C = \{1,2,3\}$.
\item{A)}  (10) What is the set $A\cap B$?
\item{B)}  (10) What is the set $(A\cup B)\ \backslash\ C$?
\item{C)}  (10) Let $f: {\bf R} \to {\bf R}$ be defined
by $f(x) = -x + 3$.  What is the set $f^{-1}(A)$?
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{\it Solution:}  Since all three of these parts deal with
the set $A$, let's start by determining exactly which 
real numbers are in that set!  The inequalities $1 < x^2 < 9$
are satisfied if and only if either $1 < x < 3$,  or $-3 < x < -1$.
Hence $A$ is the union of two open intervals: $A = (-3,-1) \cup (1,3)$. 
Then for part A of the problem,
$$A\cap B = \{x \in {\bf R} : x \in A\ {\rm and}\ x \in B\} = [-2,-1) \cup (1,3)$$
(that is the set of $x$ with either $-2 \le x < 1$, or $1 < x < 3$).
For part B of the problem, $A\cup B = (-3,4)$, so 
$$(A \cup B)\ \backslash\ C = \{x \in {\bf R} : -3 < x < 4,\ x \ne 1,2,3\}
=(-3,1) \cup (1,2) \cup (2,3) \cup (3,4)$$
For part C), the function $f(x)$ is defined by $f(x) = -x + 3$.
We want $f^{-1}(A) = \{x \in {\bf R} : f(x) \in A\}$.  This means either 
$-3 < -x + 3 < 1$ or $1 < -x + 3 < 3$ from the above identification
of $A$ as the union $(-3,-1) \cup (1,3)$.  
The first pair of inequalities
say $-6 < -x < -4 \Rightarrow 4 < x < 6$.  The second pair says
$-2 < -x < 0$, so $0 < x < 2$.  Hence $f^{-1}(A) = (0,2) \cup (4,6)$.
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\item{D)}  (10) Is the function $f$ from part C injective?  Is it 
surjective (onto ${\bf R}$)?  Justify your answers.
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{\it Solution:} Even though this part was listed under question III, 
note that the question does not make reference to 
the sets $A,B,C$ -- they are not relevant in any way here.  You must
read questions carefully before trying to answer them!!  The function 
$f$ is $f(x) = -x + 3$.  We want to think of this as a function 
$f : {\bf R} \to {\bf R}$.  It is injective because
$$\eqalign{f(x_1) &= f(x_2) \cr
       \Rightarrow -x_1+3 &= -x_2 + 3 \cr
       \Rightarrow -x_1 &= -x_2 \cr
       \Rightarrow x_1 &= x_2\cr}$$
The function $f$ is also surjective, because given any $y \in {\bf R}$,
we have 
$$y = f(x) = -x + 3 \Leftrightarrow x = -(y-3) = -y + 3$$
So $y = f(x)$ for $x = -y+3$.
(In other words, $f$ is actually its own inverse function(!))
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IV. (20) Show that the open interval $(0,1)$ in the real numbers
is not a denumerable set.
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See the class notes and Theorem 8.12 in the book for this.
I was seeing a lot of memorization without much understanding
on this question(!)
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V.  (15) Prove that for all $n \in {\bf N}$, 
$$1^3 + 2^3 + \cdots + n^3 = {1\over 4}n^2(n+1)^2.$$
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{\it Solution:}  We use mathematical induction.  The 
base case $n = 1$ is true because $1^3 = 1 = {1\over 4}1^2 (1+1)^2$.
For the induction step, assume the result is true for $n = k$, and 
consider the sum of the cubes of the first $k+1$ integers.  By
the induction hypothesis, we have
$$\eqalign{1^3 + 2^3 + \cdots + k^3 + (k+1)^3
&= {1\over 4}k^2(k+1)^2 + (k+1)^3\cr
&= (k+1)^2\left({1\over 4}k^2 + k+1\right)\cr
&= {1\over 4} (k+1)^2 (k^2 + 4k+ 4)\cr
&= {1\over 4} (k+1)^2 (k+2)^2\cr
&= {1\over 4} (k+1)^2 ((k+1)+1)^2\cr}$$
which is what we wanted to show.  Hence the formula 
for the sum of the cubes works for all $n \in {\bf N}$.
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{\it Extra Credit}
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(10)  Let $x$ be a non-negative real number.  Show that if
$x \le \varepsilon$ for all $\varepsilon > 0$, then $x = 0$.
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{\it Solution:} We will prove the contrapositive form:
If $x \ne 0$, then there exists a $\varepsilon > 0$
such that $x > \varepsilon$.  We are still working under
the hypothesis that $x$ is non-negative.  So if
$x \ne 0$, then by the trichotomy property of 
the order relation (O1 in the book), we have $x > 0$ (strictly 
positive).  But then there certainly is a positive real number
$\varepsilon$ such that $0 < \varepsilon < x$ (take
$\varepsilon = x/2$ for instance).
\bye
