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{{:Real analysis/Section 1 Exercises}}
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#With the view of getting a contradiction, assume <math>\sqrt{p}</math> is rational. Then <math>\sqrt{p}=r/s</math> for some integers <math>s\ </math> and <math>r\ </math> such that they have no common factor other than one, (that is, <math>s\ </math> and <math>r\ </math> are in lowest terms). Squaring both sides and rearranging terms gives <math>s^2p=r^2\ </math>. Since <math>p\ </math> is prime, <math>r\ </math> must be divisible by <math>p\ </math>, say <math>r=pt\ </math> for some integer <math>t\ </math>. By substitution, <math>s^2p=p^2t^2\ </math>, so that <math>s^2=pt^2\ </math>, and thus <math>s\ </math> must also be divisible by <math>p\ </math>, a contradiction.
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[[Category:Real analysis]]