Does the Pythagorean formula $a^2+b^2=c^2$ hold in the plane $\mathbb{Q} \times \mathbb{Q}$?

Question

Does the Pythagorean formula $a^2+b^2=c^2$ hold in the plane $\mathbb{Q} \times \mathbb{Q}$ ?

For example,

The triangle with vertices $(0,0), \ (1,0), \ (0,1) \in \mathbb{Q} \times \mathbb{Q}$ and

$c^2=\sqrt{(0-1)^2+(1-0)^2}=2 \\ \implies c=\sqrt2 $.

So can I say that changing the field $\mathbb{R}$ to $\mathbb{Q}$, Pythagorean formula does not hold ?

All I want to know that in $\mathbb{R}^2$ the pythagorean formula holds .

My question is -

when or in which field pythagorean formula does not hold?

Let me confess that this my confusion.

Can someone help me?

Answer 1

Your question is not really about the Pythagorean formula. The problem is that in your "plane over the field $\mathbb Q$", distance itself is less well-behaved than over $\mathbb R$. On the "plane" $\mathbb Q \times \mathbb Q$, the Euclidean distance between points $(0, 0)$ and $(1, 1)$ is not an element of $\mathbb Q$ -- it is, however, an element of $\mathbb R$. Whether or not the Pythagorean theorem still holds does not even factor into that.

Certainly if you have three points $A, B, C$ in $\mathbb Q \times \mathbb Q$, and they form a right angle at $B$, and their pairwise distances are all rational numbers, then the Pythagorean theorem "still" holds in the sense that $\overline{AB}^2 + \overline{BC}^2 = \overline{AC}^2$, as this is just a special case of the Pythagorean theorem in the ordinary plane $\mathbb R^2$.