Rationals

$a + \sqrt[]{b} = c + \sqrt[]{d} \\ \text{Let }a = c + x; x \in R \\ \text{case (i) : x = 0} \implies a = c\\ \implies c + \sqrt[]{b} = c + \sqrt[]{d} \\ \implies \sqrt[]{b} = \sqrt[]{d} \\ \implies b = d \text{ ... } \fbox{1}\\ \text{case (ii) : }x \ne 0 \implies a = c + x \\ \implies c + x + \sqrt[]{b} = c + \sqrt[]{d} \\ \implies x + \sqrt[]{b} = \sqrt[]{d} \\ \implies (x + \sqrt[]{b})^2 = (\sqrt[]{d})^2 \\ \implies x^2 + b + 2x\sqrt[]{b} = d \\ \implies 2x\sqrt[]{b} = d - x^2 - b\\ \implies \sqrt[]{b} = \dfrac{d - b - x^2}{2x} \\ \text{Left side of this equation contains of rational number, and is rational} \\ \text{So, }\sqrt[]{b} \text{ must be a rational number. So, b is a square no.} \\ \implies \sqrt[]{b}\text{ is rational.} \\ x + \sqrt[]{b} = \sqrt[]{d} \\ \text{x, }\sqrt[]{b} \text{ is both rational, so d is a square no.} \\ \text{Let b = d + x} \\ \text{Let x = 0, b = d} \\ a + \sqrt[]{b} = c + \sqrt[]{b} \\ \implies a = c \\ \text{Let b = d + x} \\ a + \sqrt[]{d + x} = c + \sqrt[]{d} \\ a^2 + 2a\sqrt[]{d+x} + d + x = c^2 + 2c\sqrt[]{d} + d \\ a^2 + 2a\sqrt[]{d+x} + x = c^2 + 2c\sqrt[]{d} \\ \implies a^2 - c^2 = 2c\sqrt[]{d} - 2a\sqrt[]{d+x} \\ \implies (a+c)(a-c) = 2c(\sqrt[]{d} - \sqrt[]{d+x}) \\ \text{On comparing both sides, you get d and d are square of rational numbers.}$