only length satisfying the triangle requirement (for any three sides a, b, and c, a+b>c) is 1, resulting in a non-right angle. the pythagorean theorem therefore gives you a complex value, going: “heh, show me that length not working!”
I feel I must offer the superior version of that meme template. Lemmy’s not a fan of crowder for reasons likely mirroring those of his ex wife.
Fuck the guy in the original meme, but we need a version for people tired of Calvin.
I get that this is just a meme, but for those who are curious about an actual mathematical argument, it is because Pythagoras’s theorem only works in Euclidean geometries (see proof below). In Euclidean geometry, distances must be real numbers of at least 0.
There exists at least one ∆ABC in a 2-D non-Euclidean plane G where (AB)² + (AC)² ≠ (BC)² and m∠A = π/2
Proof: Let G be a plane of constant positive curvature, i.e. analogous to the exterior surface of a sphere. Let A be any point in G and A’ the point of the furthest possible distance from A. A’ exists because the area of G is finite. Construct any line (i.e. form a circle on the surface of the “sphere”) connecting A and A’. Let this line be AA’. Then, construct another line connecting A and A’ perpendicular to the first line at point A. Let this line be (AA’)’ Mark the midpoints between A and A’ on this (AA’)’ as B and B’. Finally, construct a line connecting B and B’ that bisects both AA’ and (AA’)‘. Let this line be BB’. Mark the intersection points between BB’ and AA’ as C and C’. Now consider the triangle formed at ∆ABC. The measure of ∠A in this triangle is a right angle. The length of all legs of this triangle are, by construction, half the distance between A and A’, i.e. half the maximum distance between two points on G. Thus, AB = AC = BC. Let us define the measure of AB to be 1. Thus, 1² + 1² = 2 ≠ 1². Q.E.D.
I like it.
Read the “1” unit side as “move left 1 unit” and the “i” side as “move up i units”, and the hypotrnuse is the net distance travelled.
The imaginary line is perpendicular to the real line, so “up i unit” is equivalent to “right 1 unit”. The two movements cancel out giving a net distance of zero.
Yep. A vertical line segment above A with length 𝑖 is a horizontal line segment to the left that’s 1 unit long. So, the diagram needs a “not to scale” caveat like a map projection, but there’s nothing actually wrong with it, and the triangle’s BC side is 0 units long.
i= √(-1) = imaginary number (1^2) + (√(-1))^2 = 1 - 1 = 0 7
At least, I thought that was the idea in the OP.
Also, for your version, on a number line or Cartesian plane, the distance from -1 to 1 is 2, not 0
Also, for your version, on a number line or Cartesian plane, the distance from -1 to 1 is 2, not 0
Yeah. I cheated. You have to either deliberately misunderstand how to measure vectors or else drop a minus sign for it to work my way.
(Or, from my previous example, you could just frame it as you’re getting the hypotenuse by measuring between |AB| and -|AC|𝑖 instead of the way I framed it – but that makes it more obvious that you’re fishing for a particular answer.)
“um actually” I guess to properly apply the pythagoras theorem here, you’d need to consider the magnitude of the lengths of each of these vectors in complex space, both of which are 1 (for the magnitude of a complex number you ironically can use pythag, with the real and imaginary coefficients of each complex number.
So for 1 you get mag(1+0i)=root(1^2 + 0^2) and for i you get mag(0+1i)=root(0^2 + 1^2)
Then using pythag on the magnitudes, you get hypotenuse = root(1^2 + 1^2) = root 2, as expected
Shit I meant uhh imaginary number go brr it zero