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\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
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\vspace{1ex}
Exercise due Monday, 3$^{\textrm{\small rd}}$ February 2020
\\[2pc]  % pc is the abbreviation for a pica, 1/6 inch
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\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.
}

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\textbf{The Largest Prime}
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\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.


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