There is at most one way to represent a number as $a+b\sqrt 2$ with rational $a,b$

Question

If $a,b,c,d\in\mathbb Q$ and $a+b\sqrt 2= c + d\sqrt 2$, then prove $a=c$ and $b=d$ ?

I don't have any idea to solve this , it's freaking me out.

Answer 1

Hint: $$a+b\sqrt2=c+d\sqrt2\implies\sqrt2=\dfrac{c-a}{b-d}.\tag{$a,b,c,d\in\Bbb Q$}$$ Suppose that $b\neq d$ and $c\neq a$, can this be possible ?

Answer 2

Multiply the equation with common divisor and get integer equation $k+l\sqrt 2=m+n\sqrt 2$. Then $(l-n)\sqrt 2=m-k$, why $l=n$ and $m=k$.

Answer 3

$$a+b\sqrt{2}=c+d\sqrt{2}.$$

Rearranging the equation gives

$$(b-d)\sqrt{2}=c-a,\tag{1}$$ which we can also write $$\sqrt{2}=\frac{c-a}{b-d}\tag{2}.$$ But equation (2) is absurd since $\sqrt{2}$ cannot be expressed as a rational number, i.e. $\sqrt{2}\not\in\mathbb{Q}$. Hence, the only possible result, from equation (1), is that $b-d=0$ and $c-a=0$, or written differently, $b=d$ and $c=a$.

Hopefully you will stop freaking out now ;-)