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\begin{center}
{\bf \Large Math 112 Homework 1 Solutions}\medskip
\end{center}

{\bf Problem 1.}

By contradiction: take $r$ rational, $x$ irrational, and assume $y=r+x\in\Q$.
Since $y\in\Q$, $r\in\Q$ and $\Q$ is a field, $x=y-r\in\Q$; contradiction,
hence $y$ is irrational.

Similarly, take $r\neq 0$ rational, $x$ irrational, and assume $y=rx\in\Q$.
Since $y\in\Q$, $r\in\Q$, $r$ is non-zero and $\Q$ is a field, $x=y/r\in\Q$;
contradiction, hence $y$ is irrational.
\bigskip

{\bf Problem 2.}

Since $A$ is bounded below, $-A$ is bounded above (because if $x$ is
a lower bound of $A$, i.e.\ $x\le y$ $\forall y\in A$, then $-x\ge -y$
$\forall y\in A$, so $-x$ is an upper bound of $-A$). Since $A$ is not
empty, $-A$ is not empty either. Therefore, by the least upper bound
property of $\R$, the set $-A$ admits a least upper bound $\alpha=\sup(-A)$.
We must show that $\inf(A)=-\alpha$.

First we show that $-\alpha$ is a lower bound of $A$. Let $x$ be any element
of $A$: then $-x\in -A$, so $-x\le \alpha$ ($\alpha$ is an upper bound of
$-A$). Multiplying by $-1$ we get $x\ge -\alpha$; since this holds for any
$x\in A$, we get that $-\alpha$ is a lower bound of $A$.

Next we show that $-\alpha$ is the greatest lower bound of $A$. Let $y$
be any lower bound of $A$: then $\forall x\in A$, $x\ge y$, so
$-x\le -y$. Since all elements of $-A$ are of the form $-x$ where $x\in A$,
we get that $-y$ is an upper bound of $-A$. Therefore, $-y\ge \alpha$
(because $\alpha$ is the least upper bound). Multiplying by $-1$ again we get
$y\le -\alpha$, so $-\alpha$ is the greatest lower bound of $A$.
\bigskip

{\bf Problem 3.}

We must check that the two axioms of an order relation (\S 1.5 of Rudin)
hold:

(i) Let $z=a+bi$, $w=c+di\in \C$. We must show that exactly one of the
three properties $z<w$, $z=w$ and $w<z$ holds. 
There are three cases to consider: if $a<c$, then $z<w$ (while $z\neq w$
and $w\not<z$); if $a>c$, then $w<z$ (while $z\neq w$ and $z\not<w$); 
the last case is $a=c$. When $a=c$, there are again three subcases:
if $b<d$ then $z<w$ (while $z\neq w$ and $w\not<z$); if $b>d$ then $w<z$
(while $z\neq w$ and $z\not<w$); if $b=d$ then $w=z$ (while $z\not<w$ and
$w\not<z$).

(ii) Let $z=a+bi$, $w=c+di$, $u=e+fi\in\C$. Assume that $z<w$ and $w<u$.
We must show that $z<u$. We know that $a\le c$ and $c\le e$, therefore
$a\le e$. If $a<e$ then by definition $z<u$. The remaining case to consider
is when $a=e$, where $c$ is also necessarily equal to $a$ and $e$; then
we must have $b<d$ and $d<f$, so $b<f$, and therefore $z<u$.

This ordered set does not have the least-upper-bound property: for example
consider $A=\{a+bi,\ a<0\}$: then $c+di$ is an upper bound of $A$ if and
only if $c\ge 0$. However, given any upper bound $w=c+di$ of $A$, then
$w'=c+(d-1)i$ is also an upper bound of $A$ (since $c\ge 0$), and $w'<w$
(since $d-1<d$). So there is no least upper bound of $A$.
\bigskip

{\bf Problem 4.}

(a) Since $\Q(\sqrt{2})$ is a subset of $\R$, the usual commutativity,
associativity and distributivity properties are clearly satisfied. Moreover
it is obvious that $0$ and $1$ belong to $\Q(\sqrt{2})$; therefore it
is enough to check that the usual operations are well-defined in
$\Q(\sqrt{2})$ (axioms (A1), (A5), (M1), (M5) of Rudin \S 1.12).

Let $x=a+b\sqrt{2}$ and $y=c+d\sqrt{2}$ be two elements of
$\mathbb{Q}(\sqrt{2})$. Then
$$x+y=(a+c)+(b+d)\sqrt{2},\qquad xy=(ac+2bd)+(ad+bc)\sqrt{2},\qquad
-x=(-a)+(-b)\sqrt{2}$$
are clearly elements of $\Q(\sqrt{2})$. 

Moreover, if $x\neq 0$, i.e.\ if $a$ and $b$ are not simultaneously equal to
$0$, then $a^2-2b^2\neq 0$ because there is no rational number $r\in\Q$ with
the property that $r^2=2$; therefore
$$x^{-1}=\frac{1}{a+b\sqrt{2}}=\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}
\in\Q(\sqrt{2}).$$

Therefore $\Q(\sqrt{2})$ with the usual operations is a subfield of $\R$.

(b) By contradiction: assume that $\sqrt{3}\in\Q(\sqrt{2})$, i.e.\ there
exist $a,b\in\Q$ such that $\sqrt{3}=a+b\sqrt{2}$. Then $(a+b\sqrt{2})^2=
(a^2+2b^2)+2ab\sqrt{2}$, so one gets $$3-a^2-2b^2=2ab\sqrt{2}.$$
Since $\sqrt{2}\not\in\Q$ the only possibility is that $2ab=0$, which
implies that either $a=0$ or $b=0$. If $a=0$ then one gets $\sqrt{3}=
b\sqrt{2}$, i.e.\ $\sqrt{6}=2b\in\Q$, which is a contradiction. If $b=0$
then one gets $\sqrt{3}=a\in\Q$, which is again a contradiction.
Therefore $\sqrt{3}\not\in\Q(\sqrt{2})$.

(We omit the proof that $\sqrt3$ and $\sqrt6$ are irrational, which
is similar to that for $\sqrt2$ given in Rudin).%
\bigskip

{\bf Problem 5.}

Recall that $|z|^2=z\bar{z}$. Then:
\begin{eqnarray*}
&&|1+z_1|^2+|1+z_2|^2+\dots+|1+z_n|^2\\
&=&(1+z_1)(1+\bar{z}_1)+(1+z_2)(1+\bar{z}_2)+\dots+(1+z_n)(1+\bar{z}_n)\\
&=&(1+z_1+\bar{z}_1+|z_1|^2)+(1+z_2+\bar{z}_2+|z_2|^2)+\dots+(1+z_n+\bar{z}_n
+|z_n|^2)\\
&=&n+(|z_1|^2+|z_2|^2+\dots+|z_n|^2)+(z_1+z_2+\dots+z_n)+(\bar{z}_1+\bar{z}_2
+\dots+\bar{z}_n).
\end{eqnarray*}

Since $z_1+z_2+\dots+z_n=0$ one also has that $\bar{z}_1+\bar{z}_2+\dots+
\bar{z}_n=0$. In addition $|z_1|=|z_2|=\dots=|z_n|=1$, so the sum above is
equal to $2n$.
\bigskip

{\bf Problem 6.}

For $\mathbf{x}=0$ the statement clearly holds (any non-zero
$\mathbf{y}\in\R^k$ satisfies $\mathbf{x}\cdot\mathbf{y}=0$). So we
can restrict ourselves to the case where $\mathbf{x}\neq 0$, i.e.\ 
$\mathbf{x}=(x_1,\dots,x_k)$ where at least one of the $x_i$ is non-zero.
By permuting the components if necessary, we can assume without loss of 
generality that $x_1\neq 0$. Then let 
$\mathbf{y}=(-x_2,x_1,0,\dots,0)\in\R^k$: we have that $\mathbf{y}\neq 0$
(its second component is non-zero), and $\mathbf{x}\cdot\mathbf{y}=-x_1x_2+
x_2x_1=0$.

(Or more geometrically: for $\mathbf{x}\neq 0$, the set
$\{\mathbf{y}\in\R^k,\ \mathbf{x}\cdot\mathbf{y}=0\}$ is a hyperplane,
which always contains non-zero elements when $k\ge 2$).

For $k=1$ this is no longer true: if $x$ is non-zero, then the equation
$x\cdot y=0$ admits $y=0$ as only solution.
\bigskip

{\bf Problem 7.}

For every positive integer $n$, let $M_n$ be the set whose elements are all
the subsets of the finite set $\{-n,\dots,n\}$. The set $M_n$ is finite
(in fact it has $2^{2n+1}$ elements). However, every finite subset 
of $\Z$ is bounded and therefore contained in $\{-n,\dots,n\}$ for some 
integer $n$ (of course $n$ depends on the chosen subset). 
So every element of $M$ belongs to $M_n$ for some $n$, and therefore
$M=\bigcup_{n=1}^{\infty} M_n$. Since it is a countable union of
finite sets, $M$ is at most countable; since $M$ is clearly infinite, it is
countable.

{\it Alternative solution:} for every integer $n\ge 0$, let $A_n$ be the set of 
all subsets of $\Z$ containing exactly $n$ elements. The set $A_0$
admits the empty subset as its only element and is therefore finite. 
If $n\ge 1$, then to an element $\{x_1,\dots,x_n\}$ of $A_n$ we can associate 
the element $(x_1,\dots,x_n)$ of $\Z^n$ (the set of $n$-tuples of integers), 
where the $x_i$'s are ordered so that $x_1<x_2<\dots<x_n$. This defines a 
1-1 mapping of $A_n$ into $\Z^n$. However $\Z^n$ is countable (see Rudin 
\S 2.13), so $A_n$ which is equivalent to an infinite subset of $\Z^n$ is 
also countable. We conclude that $M=\bigcup_{n=0}^{\infty} A_n$ is also
countable.


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