Is this a valid proof of the parallel postulate?

Question

Let x₁ be a line with a slope of tan(n) where 0 ≤ n < π/2.

x₁ = tan((π/2)-n)y, 0 < n ≤ π/2

Let y₂ be a line with a slope of tan(m) where -π/2 < m ≤ 0; that has a y-intercept at b.

x₂ = tan((π/2)+m)y + b, -π/2 ≤ m < 0

Can we use systems of equations to prove that there will be a solution when the interior angles(n and m) don't add up to π/2?

When n = π/2, and m = π/2:

The angles add up to π.

x₁ = tan((π/2)-π/2)y

x₁ = 0

x₂ = tan((π/2)+π/2)y + b

x₂ = b

Solving for the solution of the system:

0 ≠ b, no solutions

Desmos graph

When π/2 > n > 0, and m = π/2:

Angles add upto to less than π.

x₁ = tan((π/2)-n)y

x₂ = tan((π/2)+π/2)y + b = b

Solving for the solution of the system:

tan((π/2)-n)y = b

y = b/tan((π/2)-n), one solution

Desmos graph

When π/2 > n > 0, and π/2 > m > 0:

Angles add upto to less than π.

x₁ = tan((π/2)-n)y

x₂ = tan((π/2)+m)y + b

Solving for the solution of the system:

tan((π/2)-n)y = tan((π/2)+m)y + b

tan((π/2)-n)y - tan((π/2)+m)y = b

y(tan((π/2)-n) - tan((π/2)+m)) = b

y = b/(tan((π/2)-n) - tan((π/2)+m)), one solution

Desmos graph

When n = π/2, and π/2 > m > 0:

The angles add up to less than π.

x₁ = tan((π/2)-π/2)y

x₁ = 0

x₂ = tan((π/2)+m)y + b

Solving for the solution of the system:

0 = tan((π/2)+m)y + b

y = -b/tan((π/2)+m), one solution

Desmos graph

Does this prove that there will be a solution when the interior angles don't add up to π?

Interactive graph: https://www.desmos.com/calculator/laijd4spwx

Thank you

Answer 1

Right from the beginning you speak about lines with particular slopes. But this assumes that you have a rectangular coordinate system to measure the slopes with, and in order to construct such a coordinate system, you need the parallel postulate already.

So your argument does not work as a "proof of the parallel postulate" in the usual sense of a proof that start from the other axioms of Euclidean plane geometry.

Something like your calculations (which I haven't checked in detail) will prove that if we start by defining the plane to be $\mathbb R^2$ with certain operations, then that plane happens to satisfy the parallel postulate. This can definitely be a relevant thing to prove when you want to argue that analytic geometry works in the first place, or that Euclidean geometry is consistent relative to our usual foundations (such as ZFC).

It's just not the problem that "proof of the parallel postulate" is usually taken to be about.