prove that 15+17root3 be an irrational …

Suppose $\sqrt { 3 } = \frac { a } { b }$, where a and b are co-prime integers, $b\ne  0$
Squaring both sides, 

$\Rightarrow 3 =$ $\frac { a ^ { 2 } } { b ^ { 2 } }$
Multiplying with $b$ on both sides, 
$\Rightarrow 3b =$ $\frac { a ^ { 2 } } { b }$
LHS = $3\times  b$ $= Integer$
RHS = $\frac { a ^ { 2 } } { b } = \frac { \text { Integer } } { \text { Integer } }$$= Rational\ Number$
$\Rightarrow \mathrm { LHS } \neq \mathrm { RHS }$
$\therefore$ Our supposition is wrong.
$\Rightarrow \sqrt { 3 }$ is irrational.
Suppose $15 + 17$$\sqrt 3$ is a rational number.
$\therefore 15 + 17 \sqrt { 3 } = \frac { a } { b }$, where $a\ and\ b$ are co-prime, $b\ne0$
$\Rightarrow \quad 17 \sqrt { 3 } = \frac { a } { b } - 15$
$\sqrt { 3 } = \frac { a - 15 b } { 17 b }$
$\frac { a - 15 b } { 17 b }$ is rational number,
$\sqrt 3$ is irrational.
$\therefore \quad \sqrt { 3 } \neq \frac { a - 15 b } { 17 b }$
$\therefore$ Our supposition is wrong.
$\Rightarrow \quad 15 + 17 \sqrt { 3 }$ is irrational.