Axioms and models: reinterpreting and solving a problem.

Question

I found the following problem on a "problem solving techniques" book:

Suppose $a,b,c,d \in \mathbb{R}$ are such that $a^2+b^2=1$, $c^2+d^2=1$ and $ac+bd=0$. Find the value of $ab+dc$.

Here is the solution posted there:

We can interpret $a^2+b^2=1$ and $c^2+d^2=1$ as saying that the points $(a,b)$ and $(c,d)$ belong to the circle with radius $1$ centered at the origin in $\mathbb{R}^2$. The equation $ac+bd=0$ says that the dot product of the vectors $(a,b)$ and $(c,d)$ is zero, so the vectors are orthogonal, so we must have $(c,d) = (b,-a)$ or $(c,d)=(-b,a)$. In the first case we have $ab+cd = (-d)c + cd =0$, in the second case we have $ab+cd = d(-c) + cd = 0$. Since these are the only two possibilities we have that $ab+cd=0$.

Here is my question:

The solution beings by interpreting the equations $a^2+b^2=1$, $c^2+d^2=1$ and $ac+bd=0$ in a geometric setting. So in our model of the problem (the geometry of the circle in $\mathbb{R}^2$) we conclude that "$(c,d) = (b,-a)$ or $(c,d)=(-b,a)$". Why can we conclude that" $(c,d) = (b,-a)$ or $(c,d)=(-b,a)$" holds in our "original" problem?

I mean, the circle is one model of our problem, how do we know that the conclusion "$(c,d) = (b,-a)$ or $(c,d)=(-b,a)$" is not specific to this model?

For example, consider the axioms for a group. We can interpret the axioms for a group in $\mathbb{Z}$ with respect to the ordinary addition operation, in this model the group is commutative (this is a property of our specific model), but surely this is not so in every model of the group axioms, since there are non abelian groups.

Can you see the parallel? In both cases we are considering a specific model, in the first case we are using a specific property of our model to solve our question. But in the second case it's clear that we can't assume that every group is commutative just by the verification of this condition in our model.

In the first case, why can we assume that "$(c,d) = (b,-a)$ or $(c,d)=(-b,a)$"? How do we know that this holds in "every model" (i.e., it follows just from $a^2+b^2=1$, $c^2+d^2=1$ and $ac+bd=0$) by the verification in this specific model?

Answer 1

In this problem, $a,b,c,d$ are real numbers. $a^2+b^2= 1$ means exactly the same thing as "$(a,b)$ is on the unit circle." Similarly for the dot product. So, it's just saying the same thing in different words. The algebra problem and the geometry problem are identical. It's easier to reason about the geometric problem though (at least for some people.)

You're correct in saying that this reasoning isn't sufficient, so far as I can see, to conclude that $ab+cd=0$ if $a,b,c,d$ are elements of a ring, for example.

Answer 2

You are reading way too much into the word "interpret". We're not "interpreting in a model" or anything like that. There aren't any "other models"; we're literally just talking about real numbers. So, it's not just that we can interpret $(a,b)$ as a point on the unit circle in $\mathbb{R}^2$; it literally is a point on the unit circle in $\mathbb{R}^2$. By definition, a point on the unit circle is a pair of real numbers such that $a^2+b^2=1$, and that's exactly what $(a,b)$ is.

Answer 3

I think it's the other way round: the theory is (for example) Hilbert's axioms for plane geometry, and the model is $\mathbb{R}^2$ together with the Cartesian-coordinate interpretations of Hilbert's primitives and proofs that Hilbert's axioms hold for these interpretations.

We've implicitly done a lot of work to show that, e.g., $\ulcorner P\urcorner \in\{(x,y)\mid x^2+y^2=1\}$ is the interpretation of $OP\cong OX$ where $\ulcorner O \urcorner=(0,0)$, $\ulcorner X\urcorner=(1,0)$. But once we've shown that the loci we're interested in are interpretations in our model of some formulae in our theory, then theorems involving those formulae that are proved using the theory are automatically true (since, presumably, somebody has shown that Hilbert's axioms are sound).