Prove that, if a,b,c and d …

We have equation {tex}​​a+\sqrt b=c+\sqrt d{/tex}........(i).
let a = c, then putting this value in equation (i) we get,
{tex}\Rightarrow a+\sqrt b=a+\sqrt d{/tex} 
 {tex}\Rightarrow\sqrt b=\sqrt d{/tex} 
squaring both side we get.
{tex}\Rightarrow b=d{/tex}
again let possible  {tex} a \neq c{/tex}. Then, there exists a positive rational number x such that a = c + x.
Now,
{tex} a + \sqrt { b } = c + \sqrt { d }{/tex}
{tex} \Rightarrow \quad c + x + \sqrt { b } = c + \sqrt { d }{/tex} {tex} [ \because a = c + x ]{/tex}  ...(ii)
squaring both side we get
{tex} \Rightarrow \quad ( x + \sqrt { b } ) ^ { 2 } = ( \sqrt { d } ) ^ { 2 }{/tex}
{tex} \Rightarrow \quad x ^ { 2 } + 2 \sqrt { b } x + b = d{/tex}
{tex} \Rightarrow \quad d - x ^ { 2 } - b = 2 x \sqrt { b }{/tex}
{tex} \Rightarrow \quad \sqrt { b } = \frac { d - x ^ { 2 } - b } { 2 x }{/tex}
{tex} \Rightarrow \sqrt { b }{/tex} is rational {tex} \left[ \because d , x , b \text { are rationals } \therefore \frac { d - x ^ { 2 } - b ^ { 2 } } { 2 x } \text { is rational } \right]{/tex}
{tex} \Rightarrow{/tex} b is the square of a rational number.
From (i), we have
{tex} \sqrt { d } = x + \sqrt { b }{/tex}
{tex} \Rightarrow \quad \sqrt { d }{/tex} is rational {tex} [ \because \sqrt { b } \text { is rational } ]{/tex}
{tex} \Rightarrow{/tex} d is the square of a rational number.
Hence, either a = c and b = d or b and d are the squares of rationals.