Why Pythagorean theorem is all about 2?

Question

We all know $a^2 + b^2 = c^2$ on a right angle triangle. Yes. It works. It can be proven using area of square and everything. But my question is: why?

What makes the number 2 so special, that it defines the 2-norm, gets involved everywhere in vector space, and even statistics when we talk about independent random variables are almost like perpendicular vectors and their variance can simply. (The last part might be a bit of a stretch on my part.)

I know this feels almost like asking why gravity is 9.8. It just is. It is not because 9.8 is a perfect number in number theory or anything. But our insistence on using the square has to come from somewhere. To me, the only natural connection of 2 or square of anything must be linked to the area of square on a flat surface. Why isn't it $a^4 + b^4 = c^4$? Is that a worse world to live in? In other words, what is the first principle reason that the square law comes out so often if the only first principle meaning is something multiply by itself?

This question comes from my long history of not understanding why statistics often use the standard deviation and not the average distance. Some people say because we want to emphasize the outliers so variance is the root cause (but why?) and some people say standard deviation has nice properties (precisely because of the 2 in the square law). In a way, I feel that it all comes down to the Pythagorean theorem. But why?

In the end, I can only comfortably establish that square is something multiplied by itself, or area of square on a flat surface. But what does length of sides on a right-angle triangle has to do with them? Any explanation other than: it just works? Why Pythagorean theorem works, that no matter the dimension, if we are interested in the distance, we still square them?

Answer 1

This is too long for a comment but points out the special role of $2$.

The usual distance formula states the distance between two points $x = (x_1,x_2)$ and $y = (y_1,y_2)$ is $$d(x,y) = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}$$

For any $p \ge 1$ you can similarly define a $p$-distance $$d_p(x,y) = \sqrt[p]{|x_1 - x_2|^p + |y_1 - y_2|^p}$$ and also $$d_\infty(x,y) = \max\{|x_1 - x_2|, |y_1 - y_2|\}.$$ It takes a bit of work, but you can show that each $d_p$ is in fact a metric, so that it satisfies the triangle inequality $$d_p(x,y) \le d_p(x,z) + d_p(y,z)$$ for any three points $x,y,z$.

If $C$ is a smooth curve in the plane you can use $d_p$ to define an arclength $\ell_p(C)$. Let $C_p$ denote the unit circle in the $p$-metric: $$C_p = \{(x_1,x_2) : |x_1|^p + |x_2|^p = 1\}.$$ The perimeter of $C_p$ is $\ell_p(C_p)$, and the diameter of $C_p$ can be calculated using $d_p$. This gives a definition of $\pi$ that depends on $p$: $$\pi_p = \frac{\ell_p(C_p)}{\mathrm{diam}\ C_p}$$

What is so special about $p=2$? Try to show that $$\pi_2 = \min_{1 \le p \le \infty} \pi_p$$ and $p \not= 2$ implies $\pi_p > \pi_2$. That is, $\pi_2$ is the unique minimum value of $\pi_p$.

Answer 2

There's a lot of ways of looking at it. One answer is that it comes from inner products (aka dot products, aka scalar products). The Pythagorean Theorem holds true in any space where inner products exist. It works in the plane, in space, in four-dimensional hyperspace, even in weird constructs like Hilbert spaces. All you need is an inner product.

Let $ \mathbf a, \mathbf b$ and $\mathbf c$ be vectors with $\mathbf a + \mathbf b = \mathbf c$. Then we have:

$$ \mathbf a + \mathbf b = \mathbf c $$ $$ (\mathbf a + \mathbf b) \cdot (\mathbf a + \mathbf b) = \mathbf c \cdot \mathbf c $$ $$ \mathbf a \cdot (\mathbf a + \mathbf b) + \mathbf b \cdot (\mathbf a + \mathbf b) = \mathbf c \cdot \mathbf c $$ $$ \mathbf a \cdot \mathbf a + 2 (\mathbf a \cdot \mathbf b) + \mathbf b \cdot \mathbf b = \mathbf c \cdot \mathbf c $$

When $\mathbf a$ and $\mathbf b$ are perpendicular, we have $ \mathbf a \cdot \mathbf b = 0$, and this becomes just: $$ \mathbf a \cdot \mathbf a + \mathbf b \cdot \mathbf b = \mathbf c \cdot \mathbf c $$

and we are done.

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So it needs squares because inner products are the product of two vectors. It doesn't work with cubes, because there's no good way to multiply three vectors together - you can try, but with inner products you'll find that in general $(\mathbf a \cdot \mathbf a) \cdot \mathbf b \neq \mathbf a \cdot (\mathbf a \cdot \mathbf b)$, so the two sides of the equation don't cancel out.

Answer 3

For what it is worth: the exponent $2$ exhibits isotropy.

Consider the position vector $(x,y,z)$, and the locus of the points at equal distance of the origin, which form a spherical surface, of implicit equation

$$x^2+y^2+z^2=d^2.$$

The normal to this surface is given by the gradient $2(x,y,z)$, which is aligned with the position vector. This does not occur with other exponents.

Answer 4

Your "why this, not that" question thinks too small. A weaker version of the Pythagorean theorem states a function $f$ exists for which each right-angled triangle in a Euclidean plane has catheti $a,\,b$ and hypotenuse $c$ satisfying $f(a)+f(b)=f(c)$. That such an $f$ exists is already remarkable enough, but even if you accept it does the question shouldn't be "why can it be taken as $x^2$ (or $2x^2$ etc.) rather than $x^p$ for some $p\ne2$?", but "why can it be taken as $x^2$ as opposed to just about anything else?"

We'll come back to that. The greatest length $d$ of an $n$-dimensional hypercuboid of side lengths $x_i$ satisfies $\sum_{i=1}^nf(x_i)=f(d)$. To prove this by induction, note in the case $n=k+1$ this diagonal ia a hypotenuse of a triangle whose catheti are an edge and the longest diagonal of a hypercuboid orthogonal to that edge, in a subspace of dimension $k$. So the $2$-dimensional case is the only hard part.

Let's finally return to it. Drop a perpendicular from the hypotenuse to the opposite vertex. By angle-chasing, this splits the triangle in to two right-angled triangles similar to the original, of hypotenuses $a,\,b$. Since area is in $2$ dimensions, it scales as side squared. Since the original area has been split into two pieces, $f(x)=x^2$ can be taken.

So the short answer to your question is that $\sum_{i=1}^nf(x_i)+f(x_{n+1})$ specifies the inductive step in terms of the case $n=2$ (which is also the induction's base case), and in that case we can prove the additive quantity I've denoted $f$ is an area, and it's therefore quadratic.

Note that this proof of the planar Pythagorean theorem says "areas are additive, therefore squared sides are additive". Usually, people prove the second part however they like, then note the area implication. The above argument, which I've seen attributed to Einstein, reverses this, and finds a shockingly simple proof areas are additive, which is why it's among my favourite proofs of the theorem.

And if you still don't think these points are fundamental enough, the argument for the primacy of $n=2$ - that it makes the inductive step work - can be restated as addition being a binary operation we generalize to $n$ addends. This "addition has $2$ arguments, therefore use $p=2$" argument is very close to @OrangeMushroom's comment's point.

Answer 5

Partial answer.

The Pythagorean Theorem is a theorem about areas of squares. You can prove it without squaring anything:

The exponent $2$ appears because the areas of similar figures are proportional to the squares of any linear dimension. That's a fundamental property of area, not an arbitrary choice mathematicians made.

Here is Euclid's proof.

You may find this answer helpful: Is Pythagoras' Theorem a theorem?

(Image from https://www.nagwa.com/en/lessons/436176831673/.)

Answer 6

You question why three consecutive displacements made so as to lead back to the starting point, where the first two them are orthogonal to each other, have a relationship that is of power 2 and additive.

You wonder at why the spread measure definition commonly used in statistics uses a power 2 function of error rather than a power 1 function.

Need there be an underlying reason for these (and perhaps other) math facts that will meet your need for "mathematical integrity" or whatever ?

If so, why ?

What are you really looking for here ?

I see no strong connection between statistical variance and Pythagoras' theorem.

My own understanding of the definition of statistical variance was that, when defined thus, it handled spread on either side of the mean additively rather than having error on one side of the target cancel out error on the opposite side. One could achieve the same freedom from error cancellation by using the mean of the sum of absolute differences but squaring the differences also fitted in well with the mathematical form of various commonly encountered distributions, esp. the Normal/Gaussian distribution. This was essentially what Martin Gardner said in one of his books that I read as a 17 year old and what an odd professor may have intimated in class later.

I believe other definitions of spread can be deployed but do not lead to such compact expressions in their deductions.

Nobody wants to discourage mathematical intuition. But we must always be careful not to see more in something than is actually there. Wait for such things to emerge more strongly by themselves.