11. Prove that for any prime positive integer

    \[\mathbf{p},~\surd p\]

is an irrational number.
11. Prove that for any prime positive integer

    \[\mathbf{p},~\surd p\]

is an irrational number.

Solution:

Consider

    \[\surd p\]

as a rational number

Assume

    \[\surd p\text{ }=\text{ }a/b\]

 where a and b are integers and

    \[b\text{ }\ne \text{ }0\]

By squaring on both sides

    \[p\text{ }=\text{ }{{a}^{2}}/{{b}^{2}}\]

    \[pb\text{ }=\text{ }{{a}^{2}}/b\]

                                                        

p and b are integers

    \[pb=\text{ }{{a}^{2}}/b\]

will also be an integer

But we know that

    \[{{a}^{2}}/b\]

is a rational number so our supposition is wrong

Therefore,

    \[\surd p\]

is an irrational number.