Does there exists a solution for $x + y\sqrt2 = p + q\sqrt{2}$ such that $x,y,p,q \in \mathbb{Q}$ and that $x \neq p$ and $y \neq q$?

Question

Does there exists a solution for $x + y\sqrt2 = p + q\sqrt{2}$ such that $x,y,p,q \in \mathbb{Q}$ and that $x \neq p$ and $y \neq q$?

Apologies for the lack of coherent attempt. Im not too sure if this is the proper way to do it.

My Attempt.

Suppose there exists a solution to the equation above, which satisfies the condition. Then:

$$y \neq q \Rightarrow y\sqrt{2} \neq q\sqrt{2}$$.

Note that the difference between $y\sqrt{2}$ and $q\sqrt{2}$ can be written in the form of $(y-q)\sqrt{2}$, that is, the difference is an irrational number.

Now, consider the following:

$$x + y\sqrt2 = p + q\sqrt{2} \Rightarrow x + (y-p)\sqrt2 + y\sqrt2 -(y-p)\sqrt2 = p + q\sqrt{2}$$

$$\Rightarrow (x+(y-p)\sqrt2) + (y-y+q)\sqrt2) = p+q\sqrt{2} $$

$$\Rightarrow(x+(y-p)\sqrt2) + q\sqrt2 = p + q\sqrt2 \Rightarrow (x+(y-p)\sqrt2) = p $$

Note that $x + (y-p)\sqrt2$ is irrational, but we defined that $p$ is rational. This is a contradiction. So no such solutions exists.

I was wondering if this proof is correct?

Any inputs?

Answer 1

Your approach is fine but is can be shortened. $x+y\sqrt{2}=p+q\sqrt{2}$ with the given constraints implies $$\frac{x-p}{q-y}=\sqrt{2}$$ where the LHS$\in\mathbb{Q}$ but the RHS$\not\in\mathbb{Q}$, contradiction.

Answer 2

The set $\lbrace 1, \sqrt{2}\rbrace$ is a $\Bbb Q$-basis for the vector space $\Bbb Q(\sqrt{2})$. If $x, y, p, q\in \Bbb Q$ then

$$x + y\sqrt{2} = p + q\sqrt{2}$$ implies $$(x - p) + (y - q)\sqrt{2} = 0.$$

Since $1$ and $\sqrt{2}$ are linearly independent, this implies $x - p = 0$ and $y - q = 0$, or equivalently that $x = p$ and $y = q$.