Given , , real numbers.
Any square numbers can be decomposed into two square numbers
Length scales proportionally
Euclidean geometry needs to have Pythagorean theorem to be true
If in a geometry, Pythagorean theorem holds true. Then Pythagorean theorem will also applies for the n-th derivative of the metric.
This is an interesting proof because it doesn’t require any image or drawing. Only definition and set relationship.
Given a right triangle, one of its side is a hypothenuse . The other side is and . A triangle is defined by 3 angles , , and its hypothenuse. All by definition.
A hypothenuse can be dissected with equal parts of right triangle, so any right triangle is composed of two right triangles. Call the dissected triangles and . Call original triangle as
All these triangles is guaranteed to be similar because two of the angles are the same for all triangles.
Assume a function called . It doesn’t matter what the formula to calculate the area is. By function mapping, it should be possible to map these parameter to a certain area.
Angle can be calculated from and because the total has to be constant for all kind of triangles. (can be set to 180 degrees, although any value won’t matter)
Angle is definitely determined to be a constant for any right triangles. (can be set to 90 degrees, although any value won’t matter)
For angle it has to be the same for triangle , , and .
So, actually the entire setup is invariant to above parameters, and only dependent solely on the length of .
Key insight is that:
Any area of triangle should be proportional to a square rectangle of one of its sides. This is because making the square bigger should also make the triangle bigger the same way.
All these triangles uses the same square grid, so the proportionality has to be the same. So just call it .
Finish the proof
Suppose that this holds true
Then we want to proof it that it was necessary for this to be a right triangle.
For ease of convention, rename and . Let , , and be arbitrary function that depends on a shared parameter .
Inspect derivative with respect to
Multiply by and substitute again
Taking a look at how the terms organized in the left side and right side. Coefficient of and must each corresponds. So we got two equations:
If we subsitute this back, we can also get
There should be no preferences between and . Each must changes wrt to the same way. So this satisfy scaling property. If this satisfy scaling property, then and are completely normal/independent/perpendicular. If and has to be perpendicular, then it has to be a right triangle. So it completes the proof.
To illustrate this, use contradiction. I can think of several alternatives.
Suppose at all ,, form a Pythagorean triplet and a right triangle. Call the ratio of the sides that is not hypothenuse as , written:
Now, claim that when and changes, even though it satisfies Pythagorean triplet, the triangle after the changes is not a right triangle.
From the differential equation of and we can derive as a function of alone.
is just a constant so it essentially the same as .
Because the ratio is the same even when the sides changes, then it had to be similar triangle with the initial triangle. Since the initial triangle is a right triangle, then after the changes it has to be a right triangle again.
So the claim is contradictory and Pythagorean triplet has to be true for all right triangles, with any initial constant .
For a right triangle to work, and has to be always perpendicular.
Given an independent parameter , it must change and independently.
Now, claim that and is independent. Meaning, the partial derivatives has to be as follow:
The equation above can be understood this way:
Left side has to be non-zero expression (we already derive it before). If we express and using only , it will mean the right hand side, first term has some expression. But if and is independent.
Meanwhile, using the previous result of the total derivative of Pythagorean theorem, we add extra constraint that doesn’t change.
This is because we want to check only the relation between and given parameter . This will cause
The above differential equation is total and can be solved. But it didn’t tell us much. If we integrate it, we are back to Pythagorean equation. So, we showed it to be dependent (contradiction), but not conclusive because the dependency are back to the original equation.
We try a different approach. Manipulate the equation further, but substitute the constancy of later.
From the original derivation, we subsitute and . For sufficient reason, let be a constant non-zero, so we can divide it.
Unrelated with the proofing method, this is actually quite an interesting corollary. If Pythagorean theorem is true, it was direct consequences that the derivative of each variables with respect to independent parameter is also a Pythagorean triplet. Applying this statement to itself, if the n-th derivative exists, it will also form a Pythagorean triplet. That means Pythagorean theorem is invariant to derivative operator in Euclidean geometry. Then by Fundamental Theorem of Calculus, it would mean that it is also invariant to integral.
Now back with the proof. Since is constant, it cause . The consequence is quite surprising.
This is a square function. In the domain of real number, there is no solution. Meaning and can’t be independent, because we already assume Pythagorean triplets were real numbers. This completes the proof by contradition.
However, there is another way to proof it.
Suppose that Pythagorean theorem can be applied in complex numbers. Proofing it will also proof it in the real numbers. If all , , and complex numbers, then we can solve equation above. One of the solution is just rearranging and taking the square roots. Other solution is just a difference in sign due to absolute values.
The difference with the proof is that we directly proof that is dependent on in a very particular way. In complex number, multiplication by means it rotate the number by 90 degrees.
In other words, is always perpendicular to . So it is dependent in such a way that it guarantees the orthogonality of and , hence it is always a right triangle.
The problem with this line of reasoning is because the construction of Complex numbers usually uses Pythagorean theorem itself. If we have alternative construction using only group theory and definition of from a square root, then it can be used to prove Pythagorean theorem.
I found it from Terrence Tao’s article.
I think it is quite fascinating. The first proof is essentially the same with “Einstein’s proof”.
The second part is interesting because it is handwaived from matrix construction without prior assumption!
Let a matrix exists like this (call it ):
The determinant of the matrix is conveniently
Now choose a second matrix like this (call it ):
The determinant of the matrix is
Now claiming (the Pythagorean theorem) is equivalent in saying that there exists a matrix such that if multiplied in any direction, it will cause both sides to be true, because the determinant is the same. Meaning, these will work:
But for any matrices with non-zero determinant, you can always find by inverting, like this:
Given actual numbers , , , then this is just a numerical works.
The proof is then to cover that these operation is always possible.
First, note that is a diagonal matrix. It will always have an inverse as long as is not zero.
Second, it is always possible to do matrix multiplication.
So will exists.
Third, the determinant can be calculated easily. Meaning determinant of is always 1, following the rule below:
Lastly, if we want to define matrix from a more generic function, there are only some restricted class of function with determinant to be always 1, and always invertible. This is called the rotation matrix which has a more generic form like this:
However, note that it doesn’t really necessary to define the trig function. Just knowing that always exists is enough for the proof. The key to the proof is how we make sure that the matrix imply that each basis is orthogonal from each other (so it became a right triangle).