Question

State true or false:

If a, b, c, d are rationals, $b>0,d>0,$ and $\sqrt{b},\sqrt{d}$ are surds and $a+\sqrt{b}=c+\sqrt{d}$, then $a=c,b=d.$

• A. True
• B. False

Answer 1

The correct option is A True

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

Case (i): Let $a=c$, then

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

Therefore, $b=d$

Case (ii): Let $a\ne c$

Let us take $a=c+k$ where $k$ is a rational number not equal to zero. Then we have,

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

Let us now square on both the sides,

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

Notice that the RHS is a rational number.

Hence $\sqrt{b}$ is a rational number

This is possible only when $b$ is square of a rational number.

Hence, $d$ is also square of a rational number as $k+\sqrt{b}=\sqrt{d}$.