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\fancyhead[L]{\Large \textbf{Math 157 Fall 2013}}
\fancyhead[R]{\large \textbf{Homework 4 - selected answers}}



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\bp [Section I 3.5 Exercise 5]  First, let me record what I will use.
By applying Theorem 1.22 with $a=0$, we know that 
if $b>0$ and $c<0$,
then $bc<0$.  This proves that if $a>0$ and $b>0$, then the quotient
$\frac{a}{b}>0$ (otherwise, if $c=\frac{a}{b}<0$, the product
$cb=a<0$).  Also, by exercise 4, we know that if
$a>0$, then $\frac{1}{a}>0$. 


Now suppose that $0<a<b$.  This means that
$a>0$, $b>0$ and $0<b-a$.  Since $0<a$ and $0<b$ we know $0<ab$.
Since $0<b-a$ and $0<ab$, we have 
$$0<\frac{b-a}{ab}=\frac{1}{a}-\frac{1}{b}.$$
This proves that $\frac{1}{b}<\frac{1}{a}$.  Also, since $0<b$, we
have $0<\frac{1}{b}$, so we get $0<\frac{1}{b}<\frac{1}{a}.$
\ep

\bp [Section I 3.12 Exercise 1] Suppose $x<y$.    Let
$z=\frac{x+y}{2}$.   Then $x<z<y$.  To see this, first note that $$x<y\Rightarrow 
\frac{x}{2}<\frac{y}{2}.$$  
Then, $$x=\frac{x}{2}+\frac{x}{2}<\frac{x}{2}+\frac{y}{2}=z$$  and
$$z=\frac{x}{2}+\frac{y}{2}< \frac{y}{2}+\frac{y}{2}=y.$$  
\ep

\bp [Section I 3.12 Exercise 3]   Let $x>0$.  Since $\N$ is unbounded,
there exists $n\in \N$ with $\frac{1}{x}<n.$  Then, $\frac{1}{n}<x$.
\ep

\bp [Section I 3.12, part of Exercise 7] I will prove that if $x$ is rational and $y$ is irrational, then
$x+y$ is irrational.  

First, I'll prove that $x,y\in \Q \Rightarrow x-y\in \Q$.  To prove this, suppose
$x=\frac{a}{b}$ for some $a,b\in \Z$ with $b\neq 0$ and $y=\frac{c}{d}$ for some
$c,d\in \Z$ with $d\neq 0$.  Then $x-y=\frac{ad-bc}{bd}$.  Since $a,b,c,d \in \Z$, so
is $ad-bc$ and $bd$.  Since $b,d\neq 0$, the product $bd\neq 0.$  So,
$x-y=\frac{ad-bc}{bd} \in \Q$.

Now if $x+y$ is rational and $x$ is
rational,  then the difference $x+y-y=x$ is rational also.  Therefore,
if $x$ is rational and $y$ is irrational, it's impossible for $x+y$ to
be rational.
\ep

\bp $\sqrt{3}$ is irrational.  So is $-\sqrt{3}$.  But the sum
$\sqrt{3}-\sqrt{3}=0$ is rational.
\ep

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