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\headline{\vtop{\hbox to \hsize{\strut Math 2106 - Dr. Gupta\hfil Due Thursday
2022-01-27 at 11:59 pm}\hrule height .5pt}}

\centerline{\bigbf Homework 3 - Holden Rohrer}
\bigskip

\noindent{\bf Collaborators:} None

\section{Hammack 2.7: 2, 9, 10}

\item{2.} Write the following as an English sentence:
$\forall x\in\bb R, \exists n\in \bb N, x^n\geq 0.$

\answer
For all real numbers $x,$ there is a natural number $n$ such that $x^n$
is nonnegative.
This statement is true because, for all real numbers, $x^2 \geq 0$ and
$2\in\bb N.$
\endanswer

\item{9.} Write the following as an English sentence:
$\forall n\in\bb Z, \exists m\in\bb Z, m = n+5.$

\answer
For all integers $n,$ there is an integer $m$ which is 5 greater than
$n.$
This statement is true because the integers are closed under addition.
\endanswer

\item{10.} Write the following as an English sentence:
$\exists m\in\bb Z, \forall n\in\bb Z, m = n + 5.$

\answer
There is an integer $m$ such that for all integers $n,$ $m$ is 5 greater
than $n.$
This statement is false because $m$ cannot equal $0+5$ and $1+5$ at the
same time.
\endanswer

\section{Hammack 2.9: 1, 7, 10}

\item{1.} Translate the following sentence into symbolic logic: ``If $f$
is a polynomial and its degree is greater than 2, then $f'$ is not
constant.
\answer
Where $P$ is the set of polynomials, and $\mathop{\rm degree}(p)$ is the
degree of a polynomial $p,$
$$\forall p\in P, \left(\mathop{\rm degree}(p) > 2\right) \impl \exists
a,b\in\bb R, f'(a) \neq f'(b).$$
\endanswer

\item{7.} Translate the following sentence into symbolic logic: ``There
exists a real number $a$ for which $a+x = x$ for every real number $x.$
\answer
$$\exists a\in\bb R, \forall x\in\bb R, a+x = x.$$
\endanswer

\item{10.} Translate the following sentence into symbolic logic: ``If
$\sin(x) < 0,$ then it is not the case that $0\leq x\leq\pi.$
\answer
$$\forall x\in\bb R, \sin(x) < 0 \impl \lnot(0\leq x\leq\pi).$$
\endanswer

\section{Hammack 2.10: 2, 5, 10}
\item{2.} Negate the following sentence: ``If $x$ is prime, then $\sqrt
x$ is not a rational number.''

\answer
There is a prime number $x$ such that $\sqrt x$ is a rational number.
\endanswer

\item{5.} Negate the following sentence: ``For every positive number
$\epsilon,$ there is a positive number $M$ for which $|f(x)-b|<\epsilon$
whenever $x > M.$

\answer
There is a positive number $\epsilon$ such that for all $M$ there is an
$x > M$ such that $|f(x)-b|>\epsilon$
\endanswer

\item{10.} If $f$ is a polynomial and its degree is greater than 2, then
$f'$ is not constant.

\answer
There is a polynomial with degree greater than 2 such that $f'$ is
constant.
\endanswer

\section{Hammack 4: 4, 12, 20}

\item{4.} Prove ``Suppose $x,y\in\bb Z.$ If $x$ and $y$ are odd, then
$xy$ is odd'' with direct proof.
\answer
Suppose $x,y\in\bb Z$ and that $x$ and $y$ are odd.
Since $x$ is odd, there exists $j\in\bb Z$ such that $x = 2j+1.$
Since $y$ is odd, there exists $k\in\bb Z$ such that $y = 2k+1.$
$xy = (2j+1)(2k+1) = 4jk + 2j + 2k + 1 = 2(2jk + j + k) + 1.$
Because $2jk + j + k$ is an integer, $xy$ is odd because it is one more
than two times an integer.
\endanswer

\item{12.} Prove ``If $x\in\bb R$ and $0<x<4,$ then ${4\over x(4-x)}\geq
1.$'' with direct proof.
\answer
Let $x\in\bb R$ and $0<x<4.$
$$(x-2)^2 \geq 0 \to 4 - (x-2)^2\leq 4\to 4x-x^2 = x(4-x) \leq 4 \to
{4\over x(4-x)}\geq {4\over 4} = 1.$$
\endanswer

\item{20.} Prove ``If $a$ is an integer, and $a^2|a,$ then
$a\in\{-1,0,1\}.$'' with direct proof.

\answer
$a^2|a$ requires $a^2 \leq |a|.$ For $|a| > 1,$ $a^2 = |a|^2 > |a|,$ so
we will check $a^2|a$ for the remaining cases $\{-1,0,1\}.$

$n|m$ iff there is a $k\in\bb Z,$ $k\neq 0,$ $m = nk.$
For $0,$  $0^2 = 1(0),$ so $0^2|0.$
For $1,$  $1^2 = 1(1),$ so $1^2|1.$
For $-1,$ $(-1)^2 = -1(-1),$ so $(-1)^2|1.$
\endanswer

\section{Problem not from the textbok}

\item{1.} Prove that for all positive real numbers $x,$ the sum of $x$ and its
reciprocal is greater than or equal to 2.

\answer
Let $x$ be a positive real number.
For all real numbers $y,$ $y^2 \geq 0,$ so $(x-1)^2 \geq 0.$
This is equal to
$$x^2 - 2x + 1 \geq 0 \to x^2 + 1 \geq 2x \to x + 1/x \geq 2,$$
since dividing by $x > 0$ is a valid algebraic operation.
\endanswer

\bye