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\lhead{MATH 311-02}
\chead{Problem Set \#4 Solutions}
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\rfoot{March 4, 2011}
\begin{document}
\begin{enumerate}
\item \textbf{(12 points)} \emph{Below are three existence statements
which are either true or false. For each of them, either prove
them true (by either an example or a nonconstructive proof) or
prove them false (by a disproof of existence).}
\begin{enumerate}
\item \emph{There is a real number $x$ such that $x^6+x^4+1=2x^2$.}
This statement is false. The disproof follows.
\begin{proof}
We may rephrase the above equation as $x^6+x^4-2x^2+1=0$; the
last three terms invite a factorization into the form
$x^6+(x^2-1)^2$. We know that the square $x^6$ is greater than
or equal to $0$ for all real $x$, achieving equality only when
$x=0$; likewise the square $(x^2-1)^2$ is greater than or equal
to zero for all real $x$, achieving equality only when $x^2-1=0$
(i.e. at the values $x=-1$ and $x=1$). Thus, for all real $x$
except $0$, $1$, and $-1$, it is clear that $x^6>0$ and
$(x^2-1)^2>0$; thus for all real $x$ except $0$, $1$, and $-1$,
$x^6+(x^2-1)^2>0$. Also, we may specifically compute that
$0^6+(0^2-1)^2>0$, $1^6+(1^2-1)^2>0$, and $(-1)^6+((-1)^2-1)>0$,
so for all real $x$, $x^6+(x^2-1)^2>0$, so there is no real
solution to the equation $x^6+x^4-2x^2+1=0$.
\end{proof}
\item \emph{There is an integer $n$ such that $4\mid n^2+2$.}
This statement is false. The disproof follows.
\begin{proof}
The most effective approach is casewise on the parity of $n$:
\textbf{Case I: $n$ is even.} Then $n=2k$ for some integer $k$,
and thus $n^2+2=(2k)^2+2=4k^2+2$. By the definition of
divisibility $4\mid 4k^2$, but $4\nmid 2$, so $4\nmid 4k^2+2$.
\textbf{Case II: $n$ is odd.} Then $n=2k+1$ for some integer $k$,
and thus $n^2+2=(2k+1)^2+2=4k^2+4k+3$. By the definition of
divisibility $4\mid 4k^2+4k$, but $4\nmid 3$, so $4\nmid 4k^2+4k+3$.
\end{proof}
\item \emph{There is an integer $n$ such that $n^3\equiv 6\pmod 7$.}
This statement is true, and we demonstrate the existence of such
an $n$ with the specific value $n=3$. $3^3=27$, and since
$7\mid(27-6)$, it follows by definition that $3^3\equiv 6\pmod
6$. Other specific values may also work.
\end{enumerate}
\item \textbf{(6 points)} \emph{Prove that there is exactly one
solution to the equation $x=\cos x$.}
\begin{proof}
The solutions to the given equation are identically the solutions
to $x-\cos x=0$, so if we let $f(x)=0$, the above assertion is
identical to the assertion that $f(x)$ has exactly one zero.
We prove that $f(x)$ has at least one zero with the Intermediate
Value Theorem. $f(0)=0-\cos 0=-1$, while $f(\frac\pi 2)=\frac\pi
2-\cos\frac\pi2=\frac\pi2$. Since $f(0)$ is negative and
$f(\frac\pi2)$ is positive, and $f(x)$ is a continuous function,
it follows from the Intermediate Value Theorem that $f(c)=0$ for
some value of $c\in[0,\frac\pi2]$.
To prove that this zero is unique, we may appeal to the fact that
$f(x)$ is increasing throughout, which is easily proven with
calculus: $f'(x)=1-\sin x$. Since $\sin x\leq 1$, $f'(x)\geq
1-1=0$, so $f(x)$ is increasing. Thus, if we had two zeroes $a$
and $b$ of $f(x)$ with $a