Question 113395
{{{-sqrt(81)=-9}}} because {{{ 9*9=81}}}.  9 is clearly a rational number because it can be expressed as the ratio of two integers: {{{9/1}}} for example.


67 is not a perfect square, because {{{8^2=64}}} and {{{9^2=81}}}.  So, {{{sqrt(67)}}} is some number between 8 and 9.  But, is it rational or irrational?


Let's suppose the square root of 67 is a rational number. Then it must be 
of the form a/b, where a and b are whole numbers. Furthermore, we can 
assume that a and b don't have any common factor (otherwise we could 
just divide the numerator and denominator by the common factor).


So we have {{{sqrt(67)=a/b}}}, so 


    {{{67=a^2/b^2}}}
    {{{a^2=67b^2}}}


This means that a squared is a multiple of 67. Now you can easily see 
that to get a number that is a multiple of 67 by squaring a, a itself 
must be a multiple of 67, so {{{a = 67c}}} for some whole number c. Putting 
this back into the equation, we get


   {{{67^2c^2 = 67b^2}}}

Now we divide both sides of the equation by 67, getting

   {{{67c^2= b^2}}}

So {{{b^2}}} is a multiple of 67, and b must be a multiple of 67.

But now we are in trouble, because we have shown that both a and b are
multiples of 67, but we know that a and b have no common factor, since we chose {{{a/b}}} at the start to be reduced to lowest terms. So 
our original assumption that {{{sqrt(67)}}} was rational must 
have been wrong.  Therefore, {{{sqrt(67)}}} is irrational.


Hope this helps,
John