Show that $\sqrt{3}$ is not an element of $\mathbb{Q}(\sqrt{2})$

Question

Show that $\sqrt{3}$ is not an element of $\mathbb{Q}(\sqrt{2})$

I know that $\mathbb{Q}(\sqrt{2})$ is $a+b \sqrt{2}$ where $a,b \in \mathbb{Q}$ so I need to show that $a+b \sqrt{2}\not =\sqrt{3}$ it seems obvious that there is no way to get $\sqrt{3}$ yet I don't know how to explicitly show that.

Answer 1

If there exist rationals such that $$a+b\sqrt {2}=\sqrt {3} $$ then $$(a+b\sqrt {2})^2=3$$

and $$\sqrt {2}=\frac {3-a^2-2b^2}{2ab} $$

but $\sqrt {2}\notin \Bbb Q $.

Answer 2

Guide:

Squaring both sides

$$a^2+2\sqrt{2}ab+b^2=3$$

Now use what you know about $\sqrt{2}$ and $\sqrt{3}$ to deduce the result.

Answer 3

If $\sqrt 3=a+b\sqrt 2$ with $a,b\in\Bbb Q$ then $$3=a^2+2b^2+2ab\sqrt 2$$

If $ab\neq 0$ then $$\sqrt 2=\frac{3-a^2-2b^2}{2ab}\in\Bbb Q$$ which is a contradiction.

If $a$ and/or $b$ are $0$ is even easier to get a contradiction. Can you finish?

Answer 4

Suppose that there is some $c,d,e,f\in\mathbb{Z}$ such that $$a+b \sqrt{2}=\sqrt{3}\implies \frac cd+\frac ef\sqrt2=\sqrt3\implies cf+de\sqrt2=df\sqrt3$$ where $a=\dfrac cd$ and $b=\dfrac ef$. Then $$c^2f^2+2cdef\sqrt2+2d^2e^2=3d^2f^2\implies\sqrt2=\frac{3d^2f^2-2d^2e^2-c^2f^2}{2cdef}\in\mathbb{Q}$$ which is a contradiction since $\sqrt2$ is irrational. Hence result.