\documentclass[12pt]{article}
\usepackage{xcolor}
\textwidth=35pc
\textheight=54pc
\topmargin=-3pc
\linespread{1.3} % gives about one-and-one-half line spacing
\begin{document}
\begin{center}
\large
\bf
Scholarship Skills 2020\\[1ex] % an ex is the height of a letter x, and thus depends on the current font size
\Large
Revise Mathematics
\normalsize
\vspace{1ex}
Exercise due Monday, 3$^{\textrm{\small rd}}$ February 2020
\\[2pc] % pc is the abbreviation for a pica, 1/6 inch
\end{center}
\noindent
{\color{red}
Apply what you've learned about writing mathematics to rewrite this proof.
Don't be afraid to \emph{rewrite} it, rather than tinker about with it in small ways.
The \LaTeX{} source for this proof is on the web site, so you can edit it to create your own version.
}
\vspace{15pt} % pt is the abbreviation for a point, 12 pt = 1 pc
\begin{center}
\large
\textbf{The Largest Prime}
\end{center}
\linespread{2.5} % gives about double line spacing
\normalsize
\noindent
Suppose that there were a largest prime number $p_i$.
Then consider the product $\prod_{j=0}^{p_i-1} \; p_i-j$.
Then $\left( \prod_{j=0}^{p_i-1} \; p_i - j \right) \,+\,1$
cannot be divided evenly by any of the numbers
up to $p_i$, $2,3,4,\ldots,p_i$ because each of these divides the left
factor evenly, but not the right factor, hence not their sum.
(Recall that if $a_1$ divides $a_2$ and $a_2=a_3+a_4$ then if $a_1$
divides $a_3$, it will also divide $a_4$.)
Since we are assuming
$p_i$ is the largest prime, $\left(\prod_{j=0}^{p_i-1}
\;p_i-j\right)\,+\,1$ can have no prime factors greater than $p_i$,
hence $\left(\prod_{j=0}^{p_i-1} \;p_i-j\right)\,+\,1$ is a prime, and
it is greater than $p_i$, since $\prod_{j=0}^{p_i-1}\;p_i-j \ge p_i$.
This contradicts the maximality of $p_i$.
Hence the assumption that
$p_i$ is the largest prime must be false, and so there is no largest
prime.
\end{document}