Question

Prove that, if $a,b,c$ and $d$ be positive rationals such that, $a+\sqrt{b}=c+\sqrt{d},$ then either $a=c$ and $b=d$ or $b$ and $d$ are squares of rationals.

Answer 1

If $a=c$, then $a+\sqrt{b}=c+\sqrt{d}\Rightarrow \sqrt{b}=\sqrt{d}\Rightarrow b=d$

So, let $a\ne c$. Then, there exists a positive rational number $x$ such that $a=c+x$.

Now,

$\Rightarrow a+\sqrt{b}=c+\sqrt{d}$

$\Rightarrow c+x+\sqrt{b}=c+\sqrt{d}$ $[\because a=c+x]$

$\Rightarrow x+\sqrt{b}=\sqrt{d}$

${\Rightarrow (x+\sqrt{b})}^2={\left(\sqrt{d}\right)}^2$

${\Rightarrow x}^2+2\sqrt{b}x+b=d$

$\Rightarrow \sqrt{b}=\frac{d-x^2-b}{2x}$

$\Rightarrow \sqrt{b}$ is rational $[\therefore d,x,barerationals\therefore \frac{d-x^2-b^2}{2x}isrational]$

$\Rightarrow$ $b$ is the square of a rational number.

From$\left(i\right)$, we have

$\sqrt{d}=x+\sqrt{b}$

$\Rightarrow$ $\sqrt{d}$ is rational

$\Rightarrow$ $d$ is the square of a rational number.

Hence, either $a=c$ and $b=d$ or $b$ and $d$ are the squares of rationals.