Project I

% COMMENT: THIS TELLS US THE DOCUMENT TYPE - NAMELY, AN ARTICLE %

\documentclass{article}

% COMMENT: THESE ARE PACKAGES THAT WE WANT SHARELATEX TO INCLUDE, THAT WE WILL - OR MAY - USE %

\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{makeidx}
\usepackage{graphicx}
\usepackage{wrapfig}
\usepackage{color}
\usepackage{xcolor,colortbl}
\usepackage{framed}
\usepackage{mdframed}
\usepackage{hyperref}
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage[english]{babel}

\setlength{\parindent}{4em}
\setlength{\parskip}{1em}
\renewcommand{\baselinestretch}{1}
% COMMENT: THIS AUTOMATICALLY MAKES HYPERLINKS BLUE %

\hypersetup{colorlinks=true,linkcolor=blue}

% COMMENT: WE FILL IN THE TITLE, AUTHOR & DATE %

\title{Structured Proof }
\author{Richard Charette}
\date{September 16, 2016}

% COMMENT: ANYTHING UP TO NOW IS PREAMBLE. WE NOW BEGIN THE DOCUMENT %

\begin{document}
% COMMENT: WE MAKE THE TITLE (INCLUDES AUTHOR & DATE) %

\maketitle

% COMMENT: WE BEGAN A NEW PAGE, AFTER THE TABLE OF CONTENTS %

\newpage
% COMMENT: WE BEGAN A NEW NUMBERED SECTION. FOR AN UNNUMBERED SECTION USE "section*{}"* %

\section*{The Proof}

% COMMENT: WE BEGAN THE TEXT. NOTE THE FORMAT FOR HYPERLINKS %
The first thing is to assume that $\sqrt{7}=\frac{a}{b}$
where both a,b ${\in}$ ${Z^+}$, both ${a}$ and ${b}$ ${\ne}$ 0. Next, assume that ${a}$ and ${b}$ do not have any common divisors that are ${>}$ ${1}$. What this means is the fraction $\frac{a}{b}$ gets reduced to its lowest possible terms, this can be done to any fraction. Neither ${a}$ and ${b}$ are even, this has to do with the common divisor of ${7}$, which is based upon the fraction $\frac{a}{b}$ that cannot be in its lowest terms.

The equation $\sqrt{7}=\frac{a}{b}$ is rearranged into ${a^2} ={7b^2}$, this means that ${a^2}$ is ${Z^+}$ because any ${Z^+}$ is written as ${7k}$, where ${k}$ is any ${Z}$. For ${a^2}$ is a ${Z^+}$, what this means is  ${a}$ is also ${Z^+}$, this is due to the square of any even number is also an even number. The square of an odd number is also an odd number.

Since ${a}$ is even, it is written as ${a}$=${7k}$.
The next step is to substitute ${a}$=${2k}$ into the original equation ${a^2}$=${7b^2}$ and get ${49k^2}$ = ${7b^2}$. Next comes the simplification to obtain ${7k^2}$ =${b^2}$. Following the same logic for ${a}$ one can see that ${b^2}$ is an even number, from this ${b}$ is also an even number.

For both ${a,b}$ ${\in}$ ${Z^+}$, but there is a contradiction from this. At the very start of the proof there was an assumption that $\frac{a}{b}$ was simplified in the lowest terms. This means that both a and b have no common divisors. In this case they are both even, this would mean that ${7}$ would be a common divisor between a and b.

This is a contradiction of the original assumption that $\sqrt{7}$ is incorrect, thus $\sqrt{7}$ is not a rational number.


\end{document}