# Existence and uniqueness of function satisfying intuitive properties of distance in $\mathbb{R}^2$?

This question is probably a fixed up version of the unclear closed question Non-geometric Proof of Pythagorean Theorem. I have two questions. My first question is

Is $$d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$$ the unique binary function from $$\mathbb{R}^2$$ to $$\mathbb{R}$$, or unique function from $$(\mathbb{R}^2)^2$$ to $$\mathbb{R}$$ that satisfies the following properties?

1. For any points $$(x, y)$$ and $$(z, w)$$ in $$\mathbb{R}^2$$, $$d((0, 0), (z, w)) = d((x, y), (x + z, y + w))$$
2. For any point $$(x, y)$$ in $$\mathbb{R}^2$$, $$d((0, 0), (x, y))$$ is nonnegative
3. For any nonnegative real number $$x$$, $$d((0, 0), (x, 0)) = x$$
4. For any point $$(x, y)$$ in $$\mathbb{R}^2$$, $$d((0, 0), (x, -y)) = d((0, 0), (x, y))$$
5. For any points $$(x, y)$$ and $$(z, w)$$ in $$\mathbb{R}^2$$, $$d((0, 0), (xz - yw, xw + yz)) = d((0, 0), (x, y))d((0, 0), (z, w))$$

My second question is

If the answer to my first question is yes, does that function also satisfy the additional properties using my definition of $$\cos$$ and $$\sin$$?

1. The area of any square in $$\mathbb{R}^2$$ is the square of the length of its edges
2. $$\forall x \in \mathbb{R}d((0, 0), (\cos(x) ,\sin(x))) = 1$$

I define $$\cos$$ and $$\sin$$ by the following differential equations.

• $$\cos(0) = 1$$
• $$\sin(0) = 0$$
• $$\sin' = \cos$$
• $$\cos' = -\sin$$

If there's exactly one binary function from $$\mathbb{R}^2$$ to $$\mathbb{R}$$ satisfying all 7 properties, that doesn't necessary mean there's exactly one binary function from $$\mathbb{R}^2$$ to $$\mathbb{R}$$ satisfying the first 5 properties. That's why I specifically asked if there exists a unique function satisfying the first 5 properties.

• Could you please clarify what the actual question is? Feb 6, 2019 at 19:46
• @Micapps I'm not sure if that's possible since distance was treated as an undefined concept that we assumed satisfies certain properties, so I had to specify that the task was to pick a definition that satisfies those properties if you can. I suppose I could have asked a clearer question that is a totally different less good question about whether you can prove the Pythaogren theorem when distance wasn't defined and we cannot assume it satisfies those property. Then the answer would be a very simple answer of no which doesn't help mathematics. With no assumptions about the distance formula, Feb 6, 2019 at 19:52
• "How do you prove that a way of defining distance that satisfies the 5 stated properties exists at all?" By simply doing the math. $\sqrt{(x-w)^2 + (y-z)^2} \ge 0$. $\sqrt{(x-w)^2 + (y-z)^2} = \sqrt{(w-x)^2 + (z-y)^2}$ and $\sqrt{(x-w)^2 + (y-z)^2} = 0 \iff (x,y) = (w,z)$ and $\sqrt{(x-k)^2+(y-j)^2}\le\sqrt{(x-w)^2 + (y-z)^2} +\sqrt{(w-j)^2 + (z-k)^2}$. It's just math. Feb 6, 2019 at 20:34
• @Martin-Blas How would you define a rotation? If you define it by, say, $\{\{\cos\theta,-\sin\theta\},\{\sin\theta,\cos\theta\}\}$, then that would be as strong as axiom (5); why not, for example $\{\{\cos\theta,-\alpha\sin\theta\},\{\sin\theta/\alpha,\cos\theta\}\}$? Feb 14, 2019 at 10:44
• If you want to get useful answers it would help to first write a much, much more focused question. This one is far too wordy. I have one suggestion for doing this: separate the question about the existence of a distance function satisfying nice properties from all the stuff about measuring volumes, which is only tangentially related. Then if you like, post a separate question about volumes that mentions this one (if you think it is good motivation). Feb 18, 2019 at 18:16

The answer to both questions is yes. It can be shown that there is exactly one binary function from $$\mathbb{R}^2$$ to $$\mathbb{R}$$ satisfying the first 5 properties and that function also satisfies properties 6 and 7. The existence and uniqueness of a function satisfying the first 5 properties can be proven as follows. First I will show uniqueness of that function.

Suppose $$d$$ is a binary function from $$\mathbb{R}^2$$ to $$\mathbb{R}^2$$ satisfying the first 5 properties. Using property 1, once you determine for every ordered pair $$(x, y)$$ in $$\mathbb{R}^2$$, $$d((0, 0), (x, y))$$, there is an obvious way to determine what the whole function is. Now all I have to do is determine $$d((0, 0), (x, y))$$ for every ordered pair $$(x, y)$$ in $$\mathbb{R}^2$$. $$\forall x \in \mathbb{R}\forall y \in \mathbb{R} (d((0, 0), (x, y)))^2 = d((0, 0), (x, y))d((0, 0), (x, y)) = d((0, 0), (x, y))d((0, 0), (x, -y)) = d((0, 0), (x^2 + y^2, 0)) = x^2 + y^2$$. Since $$\forall x \in \mathbb{R}\forall y \in \mathbb{R} (d((0, 0), (x, y)))^2 = x^2 + y^2$$, then also $$\forall x \in \mathbb{R}\forall y \in \mathbb{R} d((0, 0), (x, y)) = \sqrt{x^2 + y^2}$$, which shows the uniqueness of a function satisfying the first 5 properties. That function can be shown to be $$d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$$.

Now I'll show that that function does satisfy those properties. It's trivial to show that it satisfies the first 4 properties. It can also be shown to satisfy property 5 as follows. $$\forall x \in \mathbb{R}\forall y \in \mathbb{R}\forall z \in \mathbb{R}\forall w \in \mathbb{R} d((0, 0), (xz - yw, xw + yz)) = \sqrt{(xz - yw)^2 + (xw + yz)^2} = \sqrt{x^2z^2 - 2xyzw + y^2w^2 + x^2w^2 + 2xyzw + y^2z^2} = \sqrt{x^2z^2 + x^2w^2 + y^2z^2 + y^2w^2} = \sqrt{(x^2 + y^2)(z^2 + w^2)} = \sqrt{x^2 + y^2}\sqrt{z^2 + w^2} = d((0, 0), (x, y))d((0, 0), (z, w))$$

Now that I've shown that there exists a unique function satisfying the first 5 properties, I will from now on define $$d$$ to mean that function. I will also define $$d(x, y)$$ as shorthand for $$d((0, 0), (x, y))$$.

The binary function $$d$$, not the unary function $$d$$ that I also defined, can also be shown to satisfy properties 6 and 7. I'll derive some properties of the unary function $$d$$ to make the proof more concise, but it's only a proof that the binary function $$d$$ satisfies properties 6 and 7, not a proof that the unary function $$d$$ satisfies those properties.

The binary function $$d$$ can be proven to satisfy property 6 as follows. Take any square. The displacement along one of its edges going in the counterclockwise direction has both components nonnegative. Let's call the first component of that displacement $$x$$ and its second component $$y$$. Using property 1 of the binary function $$d$$, we can show that the length of that edge going in that direction is $$d(x, y)$$. In Calculus, the area of that square can be defined as the definite integral from $$-\infty$$ to $$\infty$$ of the function that assigns to each real number $$t$$ the length of the intersection of the square and the line of points in $$\mathbb{R}^2$$ with $$t$$ as the first component. This image shows that the area of the square is $$(x - y)^2 + 2xy = x^2 - 2xy + y^2 + 2xy = x^2 + y^2 = (\sqrt{x^2 + y^2})^2 = (d(x, y))^2$$. That proves that the binary function $$d$$ satisfies property 6.

The binary function $$d$$ can also be proven to satisfy property 7 as follows. $$\frac{d}{dx}(\cos^2(x) + \sin^2(x)) = \frac{d}{dx}(\cos^2(x)) + \frac{d}{dx}(\sin^2(x)) = 2\cos(x)(-\sin(x)) + 2\sin(x)\cos(x) = 0$$. That shows that the function $$\cos^2(x) + \sin^2(x)$$ is constant. Also $$\cos^2(0) + \sin^2(0) = 1$$. So $$\forall x \in \mathbb{R}\cos^2(x) + \sin^2(x) = 1$$. Therefore, $$\forall x \in \mathbb{R}d(\cos(x), \sin(x)) = \sqrt{\cos^2(x) + \sin^2(x)} = \sqrt{1} = 1$$. In conclusion, $$d((x, y), (z, w)) = \sqrt{(z - x)^2 + (w - y)^2}$$ thus is the unique binary function from $$\mathbb{R}^2$$ to $$\mathbb{R}$$ satisfying the first 5 properties and it also satisfies properties 6 and 7.