prove that 15+17root3 be an irrational …

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

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