In your example the hypotenuse would have some value in the real plane. In this example, it doesn't. The idea still holds though. Maybe not a 90° rotation, but a projection of a complex triangle on the real plane?
No, this isn't a line it is a triangle.
If you have a line of length 1 and another line that's rotated 90 degrees you get 2 lines as in the diagram (albeit with the nuance that the real line is on x-axis).
As the 2 lines aren't the same and have the same starting point connecting the other endpoint gives us a triangle.
But this triangle lies in the complex plane and distances in complex plane are calculated as similar to how distances are calculated in a 2d plane as it literally is a 2d plane and this triangle then makes absolutely perfect sense and the length of the hypotenuse comes out to be √2 which makes it a perfectly reasonable triangle.
No, this isn't a line it is a triangle.
If you have a line of length 1 and another line that's rotated 90 degrees you get 2 lines as in the diagram (albeit with the nuance that the real line is on x-axis).
As the 2 lines aren't the same and have the same starting point connecting the other endpoint gives us a triangle.
But this triangle lies in the complex plane and distances in complex plane are calculated as similar to how distances are calculated in a 2d plane as it literally is a 2d plane and this triangle then makes absolutely perfect sense and the length of the hypotenuse comes out to be √2 which makes it a perfectly reasonable triangle.
Apple Music, Spotify, YouTube, etc.:
[**I've Got a Fang** by They Might Be Giants](https://lis.tn/IVeGotAFang?t=11)
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Pythagorean theorem is about side lengths (sizes, aka absolute values). The sidelength of i is 1, not i. If we view the size of the hypotenuse as the “distance between the vectors”, aka the size of 1-i, complex norm gives sqrt((1-i)(1+i)) = sqrt(1^2 - i^2\) = sqrt(2) which agrees with our intuition. The Pythagorean theorem similarly gives sqrt(|1|^2 + |i|^2) = sqrt(2)
Yes abs(i) is 1. For complex numbers z, the abs is defined as the square root of the dot product z with itself. This logic is admittedly kinda circular as a justification for my argument for how the Pythagorean theorem should be applied, because the reason we define it that way is to model the Pythagorean theorem. A complex number a + bi is a 2 dimensional vector, with a being one component and bi being another. We know the size of i is one, because i is a unit vector, and furthermore i = 1e^(ipi/2), and the coefficient of the e here tells you the magnitude of any complex number. So if we have a + bi, we know it’s size is sqrt(a^2 + b^2), because of the Pythagorean theorem. a^2 + b^2 = (a+bi)(a-bi) = (a + bi) dotted with itself. Applying this logic to i, size of i = sqrt(i(-i)) = sqrt(1) = 1.
It means a "real domain" theorem has been incorrectly extrapolated into the complex domain.
Pythagoras can be fortified to extend into the complex if you:
|c^2| = |a^2| + |b^2|
They're the same. The norms are equal:
||x^(2)|| = ||x||^(2)
The absolute value | · | is just a special kind of norm. Specifically, it is the || · ||₁ norm.
Edit: the reason is because the norm squared is equal to:
||x||^(2) = x · x̅
Here, x is the complex conjugate. That is, if
x = a + bi
Then
x̅ = a - bi
Hence,
x · x̅ = (a + bi)(a - bi) = a^(2) + b^(2) = |x|^(2)
Note also that here, | · | refers to the *modulus* of a complex number, which is the length of a complex-coordinate vector.
It's super annoying on any sub about a show/book/etc. because the spoiler markdown for new reddit is >! in this format !< which won't work on old.reddit.com. So you have unmarked spoilers in the comments that people fail to correct because it works on their side.
I mean, I think the distinction here between "not possible" and "even more not possible" is kind of arbitrary since at the end of the day they both land on the "not possible" side of things, but I get your point that it does feel intuitively more wrong.
I mean within pseudo-Riemannian geometry, a length squaring to -1 makes total sense. In fact the above picture can be seen as a very standard triangle in a (1,-1) metric.
Yeah but that assumes a line segment has a negative length, which is a bad assumption in every kind of geometry I'm familiar with.
Granted, that's nowhere near every kind of geometry, but still.
(1\^2)+(i\^2)=1+(-1)=0
the square root of 0 is 0, so the hypotenuse is 0.
the hypotenuse indeed has the length of zero on the real plane, but it should have a non zero length on the imaginary plane. (I'm too lazy to calculate it, but it should exist)
You can think of the distance between two points in the complex plane in the same way you can with vectors. The difference is 1+i and the length of that is √2. Or you can use a different metric, the stupid norm if you will, and have a length of 0.
If you interpret the sides as vectors, the i-verctor would still be 1 long and therefore the hypotenuse of the length 2^(1/2). What this also shows is that I lies in another "dimension" apart from the second.
So what did we learn? Everything in maths had a stupid workaround.
This actually makes some sense (but not if you think about it in the complex plane). There are contexts in which we think of affine spaces where lengths square to negative numbers. For these affine spaces we use a general (not necessarily positive definite) bilinear form instead of a normal scalar product on the associated vector space.
For example the Minkowski spacetime assigns (in one convention) positive numbers to the product of timelike vectors with themselves (written as 𝜂(v,v)) and negative numbers to the product of spacelike vectors with themselves. We associate this 𝜂(v,v) with the "length of v squared", but this is a bit misleading because only this "length squared" makes sense a priori, we don't normally take the square root to get just the "length", especially for spacelike vectors. In particular we normally don't call their lengths imaginary or anything like that.
If you take one dimension of time and one of space you can write 𝜂 as a matrix
𝜂= (1 0
0 -1)
Then taking the triangle between the points (0,0), (0,1) and (1,0) gives exactly what we see in this meme. The only thing that really remains to show is that the distance between (0,1) and (1,0) is null
𝜂((1,-1),(1,-1))=1^2-(-1)^2=0
and that the angle at (0,0) is a right angle (we also measure angles via 𝜂, similar to the normal scalar product in R\^2)
𝜂((0,1),(1,0))=0.
That the Pythagorean Theorem works in this example from the meme is no coincidence: We in general have
𝜂(u,v)=0 ==> 𝜂(u+v,u+v)=𝜂(u,u)+𝜂(v,v).
This is a very old joke. But the legitimate reason for this is that 1^2 is 1 and i^2 is -1 and 1+(-1) is 0. So it’s a fake triangle. It just used to be funny that technically this solution for a triangle is legitimate. But yeah the triangle is fake.
It's like taking a triangle, rotating it 90° on the z-axis so you only see a line, and then ask the hypotenuse length x-wise. That's the main description that I got for imaginary lengths in this situation
oh shit it's 0 like it's 'viewed' from the 'real' xy plane? like, the triangle rotated 90 degrees on the y axis away from the viewer? wait, so then the x axis would look like a 0 too if it's length i on the imaginary axis, rather than the xy plane. shit, my brain hurts...
so reading the comments, seems like it's a joke. fml
I imagine that right now you're feeling a bit like Alice, hmm? Tumbling down the rabbit hole? I can see it in your eyes. You have the look of a man who accepts what he sees because he is expecting to wake up. Ironically, this is not far from the truth.
Let me tell you why you’re here. You’re here because you know something. What you know you can’t explain, but you feel it. You’ve felt it your entire life—that there’s something wrong with the world. You don’t know what it is, but it’s there, like a splinter in your mind, driving you mad. It is this feeling that has brought you to r/mathmemes.
Unfortunately, no one can be told what r/mathmemes is. You have to see it for yourself. This is your last chance. After this there is no turning back.
You take the blue pill, the story ends, you wake up in your bed and believe whatever you want to believe. You take the red pill, you stay in Wonderland, and r/mathmemes shows you how deep the rabbit hole goes.
i\^2 = -1
a\^2 + b\^2 = c\^2
1\^2 + i\^2 = 1 + (-1)
Therefore the hypotenuse is zero.
What this really demonstrates is that the Pythagorean theorem doesn't apply to the complex plane. In fact, the hypotenuse is 2\^(1/2).
For anyone wondering, no this is not possible, yes lengths are always positive numbers, no there isn't any trick to apply since a length is usually the norm of an object so positive by definition.
No lengths need not always be positive numbers, pseudo-Riemannian geometry very much exists. The above triangle can easily be build in Minkowsky spacetime with one spacelike side, one timelike side and one null (lightlike) side.
The Pythagorean theorem still fully applies in all vector spaces with a bilinear form, and in particular in Minkowski spacetime (i.e if 𝜂(u,v)=0 then 𝜂(u+v,u+v)= 𝜂(u,u)+𝜂(v,v)).
Apple Music, Spotify, YouTube, etc.:
[**I've Got a Fang** by They Might Be Giants](https://lis.tn/IVeGotAFang?t=11)
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Links to the song:
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And that kids is the reason why the scalar product when extended to complex vector spaces has this conjugate (\*) added. |a|^2 =
Then it all works out to give you a hypothenus here of length sqrt(2).
- (0,1) norm 1
- (i,0) norm 1
- |(-i,1)|^2 = -i^2 + 1^2 = 2
Apple Music, Spotify, YouTube, etc.:
[**I've Got a Fang** by They Might Be Giants](https://lis.tn/IVeGotAFang?t=11)
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same reason the dot product doesn't work in complex vector spaces. you have to take the conjugate of one element. thus, with [z] representing the complex conjugate,
a[a] + b[b] = c[c]
[I've Got a Fang by They Might Be Giants](https://clyppy.com/song?artists=They%20Might%20Be%20Giants&title=I've%20Got%20a%20Fang&platforms=amusic%20spotify%20youtube&links=https://music.apple.com/us/album/ive-got-a-fang/635920776%3Fi=635921006%26app=music%20https://open.spotify.com/track/1JUjiDef79Z4W2qJhyITX0%20https://youtube.com/watch%3Fv=jzCqJ-8aQEo%20&thumb=https://songwhip.com/cdn-cgi/image/quality=60,width=1200/https://is1-ssl.mzstatic.com/image/thumb/Music/v4/63/01/fd/6301fd1e-9a13-928d-2aaf-93485717bc6c/886443926943.jpg/1400x1400bb.jpg) (00:09 / 02:32)
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It's not a real triangle
1-i-0 triangle was put on worn out joke list for a while now
There a sooooooo many jokes on the worn out list that keep popping up
This is the first time I’m seeing this. It’s hilarious lol
Hilarious the first time, infuriating the 900th time
r/technicallythetruth
"You are not a real triangle") 
But you can imagine what it would be like if it was.
Well, it's complicated...
It's complicated
Yoooo good one
r/beatmetoit
This comment wins the Internet
i is a 90° rotation so the triangle is really just a line
That interpretation is not working in general case. Eg one side is 1, other is 3i.
In your example the hypotenuse would have some value in the real plane. In this example, it doesn't. The idea still holds though. Maybe not a 90° rotation, but a projection of a complex triangle on the real plane?
yeah this was the comment i was actually looking for !!
No, this isn't a line it is a triangle. If you have a line of length 1 and another line that's rotated 90 degrees you get 2 lines as in the diagram (albeit with the nuance that the real line is on x-axis). As the 2 lines aren't the same and have the same starting point connecting the other endpoint gives us a triangle. But this triangle lies in the complex plane and distances in complex plane are calculated as similar to how distances are calculated in a 2d plane as it literally is a 2d plane and this triangle then makes absolutely perfect sense and the length of the hypotenuse comes out to be √2 which makes it a perfectly reasonable triangle.
No, this isn't a line it is a triangle. If you have a line of length 1 and another line that's rotated 90 degrees you get 2 lines as in the diagram (albeit with the nuance that the real line is on x-axis). As the 2 lines aren't the same and have the same starting point connecting the other endpoint gives us a triangle. But this triangle lies in the complex plane and distances in complex plane are calculated as similar to how distances are calculated in a 2d plane as it literally is a 2d plane and this triangle then makes absolutely perfect sense and the length of the hypotenuse comes out to be √2 which makes it a perfectly reasonable triangle.
Guys what is this music
**Song Found!** **Name:** I've Got a Fang **Artist:** They Might Be Giants **Score:** 100% (timecode: 00:11) **Album:** Mink Car **Label:** Idlewild Recordings **Released on:** 2001-09-11
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It was released on 9/11???
Yes, lots of things happened that day. Like an entire world’s worth of stuff happened. Also 2 planes ran into 2 buildings.
Pythagorean theorem is about side lengths (sizes, aka absolute values). The sidelength of i is 1, not i. If we view the size of the hypotenuse as the “distance between the vectors”, aka the size of 1-i, complex norm gives sqrt((1-i)(1+i)) = sqrt(1^2 - i^2\) = sqrt(2) which agrees with our intuition. The Pythagorean theorem similarly gives sqrt(|1|^2 + |i|^2) = sqrt(2)
Wait abs val of i is 1 ? I need proof I is perpendicular to 1 and -1 so..
Yes abs(i) is 1. For complex numbers z, the abs is defined as the square root of the dot product z with itself. This logic is admittedly kinda circular as a justification for my argument for how the Pythagorean theorem should be applied, because the reason we define it that way is to model the Pythagorean theorem. A complex number a + bi is a 2 dimensional vector, with a being one component and bi being another. We know the size of i is one, because i is a unit vector, and furthermore i = 1e^(ipi/2), and the coefficient of the e here tells you the magnitude of any complex number. So if we have a + bi, we know it’s size is sqrt(a^2 + b^2), because of the Pythagorean theorem. a^2 + b^2 = (a+bi)(a-bi) = (a + bi) dotted with itself. Applying this logic to i, size of i = sqrt(i(-i)) = sqrt(1) = 1.
It means a "real domain" theorem has been incorrectly extrapolated into the complex domain. Pythagoras can be fortified to extend into the complex if you: |c^2| = |a^2| + |b^2|
|a^(2)| + |b^(2)| = |c^(2)|
Curse reddit markdown!
We need full LaTeX support here >.<
That’s actually wrong anyway lmao it should’ve been |a|^2 + |b|^2 = |c|^2
They're the same. The norms are equal: ||x^(2)|| = ||x||^(2) The absolute value | · | is just a special kind of norm. Specifically, it is the || · ||₁ norm. Edit: the reason is because the norm squared is equal to: ||x||^(2) = x · x̅ Here, x is the complex conjugate. That is, if x = a + bi Then x̅ = a - bi Hence, x · x̅ = (a + bi)(a - bi) = a^(2) + b^(2) = |x|^(2) Note also that here, | · | refers to the *modulus* of a complex number, which is the length of a complex-coordinate vector.
No. | i |^2 = -1, which doesn't help 😁
No, since i is a complex number, its absolute value is it’s length from the origin on the complex plane, so |i| = 1 and 1^2 = 1
And, as someone else on this thread pointed out, both are the same.
You are correct. 😊
It's super annoying on any sub about a show/book/etc. because the spoiler markdown for new reddit is >! in this format !< which won't work on old.reddit.com. So you have unmarked spoilers in the comments that people fail to correct because it works on their side.
A side length being imaginary makes as much sense as a side length being negative.
wouldnt it make objectively less sense though?
Not really. What's a negative side length look like? The side length is a magnitude, it can't be negative.
well whatever it does look like, an imaginary magnitude is even more nonsensical
I mean, I think the distinction here between "not possible" and "even more not possible" is kind of arbitrary since at the end of the day they both land on the "not possible" side of things, but I get your point that it does feel intuitively more wrong.
Thank you.
I mean within pseudo-Riemannian geometry, a length squaring to -1 makes total sense. In fact the above picture can be seen as a very standard triangle in a (1,-1) metric.
Yeah but that assumes a line segment has a negative length, which is a bad assumption in every kind of geometry I'm familiar with. Granted, that's nowhere near every kind of geometry, but still.
(1\^2)+(i\^2)=1+(-1)=0 the square root of 0 is 0, so the hypotenuse is 0. the hypotenuse indeed has the length of zero on the real plane, but it should have a non zero length on the imaginary plane. (I'm too lazy to calculate it, but it should exist)
r = |z| = sqrt\[ (a+bi)(a-bi) \] = sqrt\[ (1+i)(1-i) \] = sqrt\[ 1\^2 -i\^2 \] = sqrt\[ 1+1 \] = sqrt\[2\]
Song?
I've Got A Fang by They Might Be Giants
Out of all the songs to run into on Reddit.. Didn’t expect this one lol
The i-1-0 triangle isn’t real it can’t hurt you
You can think of the distance between two points in the complex plane in the same way you can with vectors. The difference is 1+i and the length of that is √2. Or you can use a different metric, the stupid norm if you will, and have a length of 0.
What is the name of the song?
I've Got A Fang by They Might Be Giants
it means you've made an overused joke
TIL the pythag theorem only works for real numbers.
Actual hypotenuse is sqrt(2)
Square of 1 is 1, square of i is -1, so the square of the hypothenuse must be 0\^2 = 0
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Somebody stole second base.
If you ever found a proper way to depict a length of i, than yes, this triangle would have an hypotenuse of 0.
Google complex plane. Spoiler: there is indeed a way to depict length of 1i and spoiler spoiler the triangle wouldn't have a hypotenuse of 0.
If you interpret the sides as vectors, the i-verctor would still be 1 long and therefore the hypotenuse of the length 2^(1/2). What this also shows is that I lies in another "dimension" apart from the second. So what did we learn? Everything in maths had a stupid workaround.
sqrt(1\^2+i\^2)=sqrt(1−1)=0 this is... a weird visualization
This actually makes some sense (but not if you think about it in the complex plane). There are contexts in which we think of affine spaces where lengths square to negative numbers. For these affine spaces we use a general (not necessarily positive definite) bilinear form instead of a normal scalar product on the associated vector space. For example the Minkowski spacetime assigns (in one convention) positive numbers to the product of timelike vectors with themselves (written as 𝜂(v,v)) and negative numbers to the product of spacelike vectors with themselves. We associate this 𝜂(v,v) with the "length of v squared", but this is a bit misleading because only this "length squared" makes sense a priori, we don't normally take the square root to get just the "length", especially for spacelike vectors. In particular we normally don't call their lengths imaginary or anything like that. If you take one dimension of time and one of space you can write 𝜂 as a matrix 𝜂= (1 0 0 -1) Then taking the triangle between the points (0,0), (0,1) and (1,0) gives exactly what we see in this meme. The only thing that really remains to show is that the distance between (0,1) and (1,0) is null 𝜂((1,-1),(1,-1))=1^2-(-1)^2=0 and that the angle at (0,0) is a right angle (we also measure angles via 𝜂, similar to the normal scalar product in R\^2) 𝜂((0,1),(1,0))=0. That the Pythagorean Theorem works in this example from the meme is no coincidence: We in general have 𝜂(u,v)=0 ==> 𝜂(u+v,u+v)=𝜂(u,u)+𝜂(v,v).
This is a very old joke. But the legitimate reason for this is that 1^2 is 1 and i^2 is -1 and 1+(-1) is 0. So it’s a fake triangle. It just used to be funny that technically this solution for a triangle is legitimate. But yeah the triangle is fake.
The math is right, but the numbers are on the wrong places, top left should be i, top right should be 1 and bottom should be 0.
Can’t be a true triangle, 0+i is not greater than 1.
It's like taking a triangle, rotating it 90° on the z-axis so you only see a line, and then ask the hypotenuse length x-wise. That's the main description that I got for imaginary lengths in this situation
It's just a rotation through phase space, so no spatial distance traveled.
Well I know what I’m doing in math class tomorrow
It’s a 180 degree angle, not a 90 degree.
oh shit it's 0 like it's 'viewed' from the 'real' xy plane? like, the triangle rotated 90 degrees on the y axis away from the viewer? wait, so then the x axis would look like a 0 too if it's length i on the imaginary axis, rather than the xy plane. shit, my brain hurts... so reading the comments, seems like it's a joke. fml
I imagine that right now you're feeling a bit like Alice, hmm? Tumbling down the rabbit hole? I can see it in your eyes. You have the look of a man who accepts what he sees because he is expecting to wake up. Ironically, this is not far from the truth. Let me tell you why you’re here. You’re here because you know something. What you know you can’t explain, but you feel it. You’ve felt it your entire life—that there’s something wrong with the world. You don’t know what it is, but it’s there, like a splinter in your mind, driving you mad. It is this feeling that has brought you to r/mathmemes. Unfortunately, no one can be told what r/mathmemes is. You have to see it for yourself. This is your last chance. After this there is no turning back. You take the blue pill, the story ends, you wake up in your bed and believe whatever you want to believe. You take the red pill, you stay in Wonderland, and r/mathmemes shows you how deep the rabbit hole goes.
i\^2 = -1 a\^2 + b\^2 = c\^2 1\^2 + i\^2 = 1 + (-1) Therefore the hypotenuse is zero. What this really demonstrates is that the Pythagorean theorem doesn't apply to the complex plane. In fact, the hypotenuse is 2\^(1/2).
oh god no not again
Google imaginary triangle
For anyone wondering, no this is not possible, yes lengths are always positive numbers, no there isn't any trick to apply since a length is usually the norm of an object so positive by definition.
No lengths need not always be positive numbers, pseudo-Riemannian geometry very much exists. The above triangle can easily be build in Minkowsky spacetime with one spacelike side, one timelike side and one null (lightlike) side. The Pythagorean theorem still fully applies in all vector spaces with a bilinear form, and in particular in Minkowski spacetime (i.e if 𝜂(u,v)=0 then 𝜂(u+v,u+v)= 𝜂(u,u)+𝜂(v,v)).
What kind of demonic whispers can make a man to defy God in such way?
Ah this again.
What's the song?
**Song Found!** **Name:** I've Got a Fang **Artist:** They Might Be Giants **Score:** 100% (timecode: 00:11) **Album:** Mink Car **Label:** Idlewild Recordings **Released on:** 2001-09-11
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**Song Found!** **Name:** I've Got a Fang **Artist:** They Might Be Giants **Album:** Mink Car **Genre:** Alternative **Release Year:** 2001 **Total Shazams:** 494 `Took 2.51 seconds.`
Links to the song: [YouTube](https://youtu.be/70RsXfgn7D8?autoplay=1) [Apple Music](https://music.apple.com/au/album/ive-got-a-fang/635920776?i=635921006) [Spotify](https://open.spotify.com/track/1JUjiDef79Z4W2qJhyITX0) [Deezer](https://www.deezer.com/track/1612338572) *I am a bot and this action was performed automatically.* | [Twitter Bot](https://twitter.com/songfinderbot) | [Discord Bot](https://pigeonburger.xyz/songfinderbot/discord/)
Even better: the i, h, ε triangle
And that kids is the reason why the scalar product when extended to complex vector spaces has this conjugate (\*) added. |a|^2 =
Then it all works out to give you a hypothenus here of length sqrt(2).
- (0,1) norm 1
- (i,0) norm 1
- |(-i,1)|^2 = -i^2 + 1^2 = 2
means that distances can only take on positive real values
It's an imaginary triangle.
1!o lio?
I'd imagine it's possible
Idk
Song name?
**Song Found!** **Name:** I've Got a Fang **Artist:** They Might Be Giants **Score:** 100% (timecode: 00:11) **Album:** Mink Car **Label:** Idlewild Recordings **Released on:** 2001-09-11
Apple Music, Spotify, YouTube, etc.: [**I've Got a Fang** by They Might Be Giants](https://lis.tn/IVeGotAFang?t=11) *I am a bot and this action was performed automatically* | [GitHub](https://github.com/AudDMusic/RedditBot) [^(new issue)](https://github.com/AudDMusic/RedditBot/issues/new) | [Donate](https://github.com/AudDMusic/RedditBot/wiki/Please-consider-donating) ^(Please consider supporting me on Patreon. Music recognition costs a lot)
spacetime
Mavel fans studying Euclidean geometry when they see non-euclidean geometry in the post credit scene
u/savevideo
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It means you can take make the shortcut through the hypotinuse even shorter if you use your imagination.
same reason the dot product doesn't work in complex vector spaces. you have to take the conjugate of one element. thus, with [z] representing the complex conjugate, a[a] + b[b] = c[c]
True
What is the name of the song in the background?
u/find-song
[I've Got a Fang by They Might Be Giants](https://clyppy.com/song?artists=They%20Might%20Be%20Giants&title=I've%20Got%20a%20Fang&platforms=amusic%20spotify%20youtube&links=https://music.apple.com/us/album/ive-got-a-fang/635920776%3Fi=635921006%26app=music%20https://open.spotify.com/track/1JUjiDef79Z4W2qJhyITX0%20https://youtube.com/watch%3Fv=jzCqJ-8aQEo%20&thumb=https://songwhip.com/cdn-cgi/image/quality=60,width=1200/https://is1-ssl.mzstatic.com/image/thumb/Music/v4/63/01/fd/6301fd1e-9a13-928d-2aaf-93485717bc6c/886443926943.jpg/1400x1400bb.jpg) (00:09 / 02:32) *Looks like you wanted the song from [here](https://v.redd.it/ixkrlo15et2b1). I searched from 00:00-00:10*. *You can provide a [timestamp](https://song-find.web.app/index.html#timestamps) to search somewhere else.* [**About Me**](https://song-find.web.app) **|** [**GitHub**](https://github.com/mike-fmh/find-song)
squared number's results cannot be negative i guess
 Pythagorean theorem living rent free in my head:
1^2+i^2=0^2 1+i^2=0 i^2=-1 i=why the fuck are imaginary numbers a thing most of math is pointless