Question
Prove that $3+2\sqrt { 5 }$ is irrational.

Answer

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

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