Question

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

Answer 1

Hint: Consider two cases to solve the question. In case (i), assume that a = c and prove that b = d. In case (ii), assume that \[a\ne c\] but a = c + x where x is any rational number. Form relation between x, b, and d by squaring the terms on both the sides and prove that \[\sqrt{b}\] is rational to get the result.

Complete step by step answer:
Here, we have been provided with four rational terms a, b, c, and d which are positive. We have been given the expression: \[a+\sqrt{b}=c+\sqrt{d}\] and we have to prove either a = c and b = d or b and d are square of the rationales.
Now, let us consider the following two cases:
Case (I): Here, we are assuming a = c.
Now, since a = c, therefore, we have,
\[\Rightarrow a+\sqrt{b}=c+\sqrt{d}\]
\[\Rightarrow a+\sqrt{b}=a+\sqrt{d}\]
\[\Rightarrow \sqrt{b}=\sqrt{d}\]
On squaring both the sides, we get,
\[\Rightarrow b=d\]
Hence, the result for part (i) is obtained.
Case (II): Here, we are assuming \[a\ne c.\]
Now, since \[a\ne c,\] therefore we must have a rational number x such that a = c + x. So, we have,
\[\Rightarrow a+\sqrt{b}=c+\sqrt{d}\]
\[\Rightarrow c+x+\sqrt{b}=c+\sqrt{d}\]
\[\Rightarrow x+\sqrt{b}=\sqrt{d}......\left( i \right)\]
On squaring both the sides, we get,
\[\Rightarrow {{\left( x+\sqrt{b} \right)}^{2}}={{\left( \sqrt{d} \right)}^{2}}\]
\[\Rightarrow {{x}^{2}}+b+2x\sqrt{b}=d\]
\[\Rightarrow 2x\sqrt{b}=d-{{x}^{2}}-b\]
\[\Rightarrow \sqrt{b}=\dfrac{d-{{x}^{2}}-b}{2x}\]
Now, we know that d, b and x are the rationales. So, the right-hand side in the above expression contains a rational term. So, we have,
\[\Rightarrow \dfrac{d-{{x}^{2}}-b}{2x}=\text{ rational}\]
\[\Rightarrow \sqrt{b}=\text{ rational}\]
On squaring both the sides, we get,
\[\Rightarrow b={{\left( \text{rational} \right)}^{2}}\]
Therefore, we can say that b is a square of a rational. Now, using equation (i), we have,
\[\Rightarrow \sqrt{d}=x+\sqrt{b}\]
Since, \[\sqrt{b}\] is proved to be rational. So, we have,
\[\Rightarrow \sqrt{d}=\text{ rational}\]
On squaring both the sides, we get,
\[\Rightarrow d={{\left( \text{rational} \right)}^{2}}\]
Therefore, we can say that d is a square of a rational.
Hence, the result for part (ii) is proved.

Note:
One may note that we have assumed a = c + x in case (ii). You may also assume c = a + x as it will not change the conclusion. But remember that x must be assumed as rational. You must know the properties of rationals and irrationals to solve the question. Remember that in part (i) we can also assume b = d at the initial step and then prove that a = c to get the answer.