Prove 3+√(5 is irrational

Let us assume, to the contrary, that is {tex}3 + \sqrt { 5 }{/tex} rational.
That is, we can find coprime integers a and b {tex}( b \neq 0 ){/tex} such that
{tex}3 + \sqrt { 5 } = \frac { a } { b } \text { Therefore, } \frac { a } { b } - 3 = \sqrt { 5 }{/tex}
{tex}\Rightarrow \frac { a - 3 b } { b } = \sqrt { 5 }{/tex}
{tex}\Rightarrow \frac { a - 3 b } { b } = \sqrt { 5 } \Rightarrow \frac { a } { b } - \frac { 3 } { 2 }{/tex}
Since a and b are integers,
We get {tex}\frac { a } { b } - \frac { 3 } { 2 }{/tex} is rational, also so {tex}\sqrt { 5 }{/tex} is rational.
But this contradicts the fact that {tex}\sqrt { 5 }{/tex} is irrational.
This contradiction arose because of our incorrect 
assumption that {tex}3 + \sqrt { 5 }{/tex} is rational.
So, we conclude that {tex}3 + \sqrt { 5 }{/tex} is irrational.