this post was submitted on 20 Apr 2024
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Math Memes

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Memes related to mathematics.

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1: Memes must be related to mathematics in some way.
2: No bigotry of any kind.

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[–] [email protected] 59 points 2 months ago (1 children)

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.

[–] [email protected] 17 points 2 months ago (2 children)

Fuck the guy in the original meme, but we need a version for people tired of Calvin.

[–] SpaceNoodle 12 points 2 months ago

No, let's filter out those people too.

[–] NateNate60 41 points 2 months ago* (last edited 2 months ago) (1 children)

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.

[–] [email protected] 1 points 2 months ago

The pythagorean theorem works in any complex hilbert space

[–] [email protected] 15 points 2 months ago (3 children)

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.

[–] [email protected] 1 points 3 days ago

If so moving down the imaginary line should be equivalent to miving left but then the answer must be 2 units long

But (-i)² is also -1 and it still results in 0

[–] [email protected] 13 points 2 months ago (1 children)

there is... A lot wrong with that

[–] [email protected] 5 points 2 months ago

Why do you hate fun

[–] [email protected] 7 points 2 months ago* (last edited 2 months ago) (1 children)

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.

[–] [email protected] 1 points 2 months ago* (last edited 2 months ago) (2 children)

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

[–] [email protected] 1 points 2 months ago* (last edited 2 months ago)

Yeah. We were making a joke about the complex plane -- you could say that measuring the hypotenuse of a triangle is equivalent to measuring the distance between points |AB| and |AC|𝑖 on the complex plane. That definition actually makes quite a bit of sense, and I think by sheer coincidence it's possible to misunderstand how to do it and wind up with a way of looking at it where the hypotenuse of a right triangle with sides 1 and 𝑖 would work out to exactly 0. Which brings it back into concordance with OP's (also wrong) Pythagorean presentation of it.

It obviously doesn't really work that way, but it's hard to see necessarily anything wrong with it, which makes it a fun math thing.

[–] [email protected] 1 points 2 months ago (1 children)

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.)

[–] [email protected] 1 points 2 months ago

I liked your other reply better, but either way I still have more to learn. ie, I had no idea what the complex plane is.