Let a, b, c, d be rational numbers...

Question

Let $a, b, c, d$ be rational numbers, where $\sqrt{b}$ and $\sqrt{d}$ exist and are irrational. If $a + \sqrt{b} = c + \sqrt{d}$, prove that $a=c$ and $b=d$.

Answer 1

Set $t=a-c$. Then you have $t+\sqrt{b}=\sqrt{d}$. Therefore $$ t^2+2t\sqrt{b}+b=d. $$ If $t\ne 0$, this means…

Answer 2

Hint $\,\ 0\ne \alpha = \sqrt b\! -\!\sqrt d\in\Bbb Q\,\Rightarrow\ \alpha' = \sqrt b\!+\!\sqrt d = \dfrac{\ b-d}{\sqrt b \!-\! \sqrt d}\in\Bbb Q\,\Rightarrow\,\alpha\!+\!\alpha' = 2\sqrt b\in \Bbb Q\,\Rightarrow\Leftarrow$