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Tommy Ratliff \\
Homework \#0 \\
August 28, 2013
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\textit{I have abided  by the Wheaton College Honor Code in this work. } 
-- \textbf{TCR}

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\textcolor{Maroon}{\underline{1.3.23:}  \textit{Prove that the sum of any positive real number and its reciprocal must be at least 2.}}

(I worked with Rachelle DeCoste, Leonard Euler, and Carl Friedrich Gauss on this problem.)
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Let $r\in\mathbb{R}$ be a real number with $r>0$.  We will proceed with a proof by contradiction.  

Suppose that 
$$ r + \frac{1}{r}<2$$
Since $r\ne 0$, we can multiply both sides of the inequality by $r$ and  the statement remains valid.  Further, since $r>0$, the inequality does not switch directions.
\begin{align*}
 r\cdot \left( r + \frac{1}{r} \right) & < 2 \cdot r \\
\Rightarrow r^2 + 1 - 2r &< 0\\
\Rightarrow  (r-1)^2 &< 0 
\end{align*}
Since the square of every real number is greater than or equal to zero, this is a contradiction.  

Therefore, $r+\frac{1}{r}\ge 2$.  $\Box$

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\textcolor{Maroon}{\underline{1.5.8:}  \textit{Prove that for every pair of integers $n$ and $m$, if $mn$ is odd, then $n+m$ is even.}}

(I worked alone on this problem.)
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From Example 3 in Section 1.3 on pg 21, we know that both $m$ and $n$ are odd.  By the definition of odd integers, there exist integers $k_1$ and $k_2$ such that $m=2k_1 + 1$ and $n=2k_2 + 1$.  

Therefore, 

$$ n+m = (2k_2+1) + (2k_1+1) = 2(k_2+k_1 + 1) $$
and $n+m$ is even by definition.  $\Box$


	
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