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/sci/ - Science & Math


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10827989 No.10827989 [Reply] [Original]

Is this square imaginary or real?

>> No.10827994
File: 126 KB, 1131x622, 1560578891066.jpg [View same] [iqdb] [saucenao] [google]
10827994

>>10827989

>> No.10828043

>>10827994
saved

>> No.10828053

>>10827989
let me know if this is just a bait thread or you actually don't know why that is incorrect and you are curious as to what the answer is

>> No.10828060

>>10827994
based
was about to post something like this

>> No.10828063

>>10827989
Mathematics doesn't model reality, it just shuffles half-thought ideas. Mathematicians are even proud that their ideas are half-assed.

>> No.10828067

>>10828043
newfag

>> No.10828071

>>10827994
Rotate the triangle in a 3rd orthogonal dimension. The length i still exists and the hypotenuse is 0 on the original 2d plane as the triangle is now a line.

>> No.10828076

>>10828053
not OP but geniunely curious anon, please school me

>> No.10828081

>>10827994
http://mathworld.wolfram.com/Lightlike.html

>> No.10828084

>>10828076
he's about to seethingly tell you how his precious math is infallible and that you should in fact use special rules like using the magnitude of [math]i[/math] just so the entire subject doesn't fall apart

>> No.10828086

>>10828076
Use the modulus

>> No.10828133

>>10827994
This is better then the Humor thread.

>> No.10828143

>>10827989
Can you point to a square in real life with an area of -1?

>> No.10828146

>>10827994
top KEK

>> No.10828152

>>10828143
there are no squares or areas in real life nor straight lines for that matter

>> No.10828155
File: 29 KB, 404x242, 1543869737439_2.jpg [View same] [iqdb] [saucenao] [google]
10828155

all math is imaginary

>> No.10828164

>>10828143
>point
that's where you went wrong

>> No.10828172
File: 951 B, 238x127, ddd.png [View same] [iqdb] [saucenao] [google]
10828172

>>10827994
I got it. Such triangle would be looking like that

>> No.10828175
File: 165 KB, 960x720, .jpg [View same] [iqdb] [saucenao] [google]
10828175

>>10828152

>> No.10828184

>>10828172
it has to have a right angle

>> No.10828192

>>10828184
It is right angle. The space is non-euclidean.

>> No.10828440

>>10828071
someone actually gets it thank fucking god

>> No.10828491

>>10828076
The basic geometry you learn in elementary school is all done without a reference system. This means that length can only be a positive real number, not because of some made-up crazy rule but just because it represents a physical observable that is described by a positive real number. If you could observe an imaginary length, than we would have to add that to the rules of the game and you could in fact have an object that occupies "negative area" or volume. This could actually be a useful representation for some physical phenomenon, which is why things like resistivity or the dielectric function can have imaginary parts. Not because there is actually something invisible or imaginary going on, just because the imaginary part describes some dynamic behavior that needs to be set apart from the ordinary real value. But as far as geometry goes, nobody normally needs to describe negative area, so we stick to positive real values.
Now in a reference system this changes a little. In a reference system you have a number of lines (3 for 3D space) that intersect at an origin. In one direction of each line are positive numbers, in the other negative numbers. So if you look at this you might be tempted to, for example, choose the point -3 on the x axis, the point 2 on the y axis and think that you have now a rectangle of area (-3)*2= -6. And yet if you draw this on a squared sheet of paper you can count the number of squares in this rectangle and find to obviously be 6, not -6. This means that, while the point you chose was -3, the length of that side of the rectangle is still 3, not minus -3 as the point would seem to indicate. This is a naive approach to the concept of modulus. Imagine an arrow that, starting from the origin of the RS, points right to the point -3 on the x axis. This is called a vector. The vector is completely described by the point -3, but its length, or modulus, is actually 3.
Now picture a RS with two axis. You should be used to see x and y.

>> No.10828515

>>10828491
Both x and y usually have real values. But this is not a rule. What happens if we put pure imaginary numbers on one of the axis? We get the complex plane, in which every point is a complex number, with a real part that you can read on the axis with real numbers and an imaginary apart that you can read on the imaginary axis. Don't be confused by the term "imaginary": all it means is that these numbers are separate from the real ones you are used to, and so they get an axis of their own. Remember how earlier the point -3 also described a vector, or arrow, that went from the origin to the number -3? It's the exact same now. You can have an arrow that points to any point in the complex plane. But does it mean the arrow has a complex or imaginary length? If it were so, you couldn't measure it with a normal ruler, and yet if you draw the two axis that make up the complex plane and draw an arrow on it you can do exactly that. For example, you can measure the arrow that goes from the origin to the point 2i, and you will find that it has a length of 2 a real positive number.
What happens if instead we put imaginary numbers on both axis instead of just one? Absolutely nothing. The rules remain the same. An arrow pointing to the number -3i on any axis will have a length of three.
So back to the original picture. What you see is nothing but a square that, as a side has the same arrow that in our plane would point to the point i. But we know that that arrow has a length of 1, and that is what we use to calculate the area, a thing that we can observe and measure.

So pretty much this.
>>10828086

>> No.10828845

>>10828192
based

>> No.10828855

the length of a line is always a positive number.

>> No.10828856

>>10827989
true

>> No.10828859

>>10827994
I wish more math memes were like this and not lame

>> No.10828878

>>10827989
I saw someone on here say "all numbers are imaginary" and I liked it

>> No.10828887
File: 8 KB, 486x356, ddd.png [View same] [iqdb] [saucenao] [google]
10828887

>>10828172
Maybe more like this actually

>> No.10829018

It's all in your head bro

>> No.10829032
File: 27 KB, 800x600, ddd.png [View same] [iqdb] [saucenao] [google]
10829032

>>10828887
Or like this. Yeah.. It all make sense now and everything adds up.

>> No.10829037
File: 120 KB, 1024x1024, 1533118641469.png [View same] [iqdb] [saucenao] [google]
10829037

>>10829032
uhhhh

>> No.10829040

>>10829037
No, this is the lengths of sides, not coordinates. The length is positive if you count it clockwise and negative if count it counterclockwise.

>> No.10829042

>>10829037
needs a hat

>> No.10829046

>>10829042
piss off Tooker

>> No.10829055
File: 52 KB, 800x600, ddd.png [View same] [iqdb] [saucenao] [google]
10829055

>>10829040
>>10829037
Let me clarify. Now pythagorean theorem always works.

>> No.10829058 [DELETED] 
File: 629 KB, 3024x4032, 1547767810981.jpg [View same] [iqdb] [saucenao] [google]
10829058

>>10829046
tooker got banned by the cia after this little incident got posted to /sci/

>> No.10829065

>>10829037
>Riemann sphere
Wheels are a type of algebra where division is always defined. In particular, division by zero is meaningful. The real numbers can be extended to a wheel, as can any commutative ring.
The Riemann sphere can also be extended to a wheel by adjoining an element [math]{\displaystyle \bot }[/math], where [math]{\displaystyle 0/0=\bot }[/math]. The Riemann sphere is an extension of the complex plane by an element [math]{\displaystyle \infty }[/math], where [math]{\displaystyle z/0=\infty }[/math] for any complex [math]{\displaystyle z\neq 0}[/math]. However, [math]{\displaystyle 0/0}[/math] is still undefined on the Riemann sphere, but is defined in its extension to a wheel.
The term wheel is inspired by the topological picture [math]{\displaystyle \odot }[/math] of the projective line together with an extra point [math]{\displaystyle \bot =0/0}[/math].

>> No.10829081

>>10829055
You should notice that in this space every line on euclidean 3D correspond to a sphere.

>> No.10829092

>>10829081
Mathlet here, why is this a problem? Don't imaginaries require phasors to make sense? In complex space, why should a line simply be a line?

>> No.10829109

>>10829092
It is suppose to be the answer to >>10827994
it is how you should construct a triangle such that one side of it is of length 1 second of length i and third 1^2 + i^2 = 0. And because we already use imaginary numbers why not also allow sides to be negatives. By doing so you create new non-euclidian space in which there is infinite amount of ways to measure length between two points. But in essence all these lengths lay on a sphere that correspond to a line in euclidean space.

>> No.10829124

>>10827989
The short answer is that negative numbers are just as "imaginary" in the layman sense as the roots of negative numbers. Euclidean geometry simply doesn't have any use for them.

>> No.10829126
File: 55 KB, 800x600, ddd.png [View same] [iqdb] [saucenao] [google]
10829126

>>10829109
If you will stretch this sphere to an ellipse connecting A and B then in a limit you will approach a line and get your normal euclidean space.

>> No.10829137
File: 47 KB, 644x486, TIMESAND___404-762-2019.jpg [View same] [iqdb] [saucenao] [google]
10829137

>>10828175
Isn't the real point of the allegory of the cave that someone from the outside tells the prisoners what is really going on, and then they kill him for being an asshole?

>> No.10829149
File: 70 KB, 907x755, ddd.png [View same] [iqdb] [saucenao] [google]
10829149

>>10829126
Now you can even answer to OP's original question and construct a square with area of -1 (shaded by red).

>> No.10829150 [DELETED] 

>>10829058
I got banned when they saw that I am above them in the grand scheme of things and when they saw that I would use my position above them in the grand scheme things to mercilessly destroy them for their wickedness. This other math stuff is pseudo-incidental in that it is truly the cause of all of it but they had already banned me before they saw it.

>> No.10829156

>>10829150
that really makes you think

>> No.10829172

>>10829149
Did anyone got what I mean?

>> No.10829274

>>10829172
no

>> No.10829333

>>10827994
So we get an area of [math]\frac{i}{2}[/math] with the [math]A = \frac{hb}{2}[/math] , [math]h[/math] being [math]i[/math] in right triangle's case: [math]A = \frac{1 * i}{2} = \frac{i}{2}[/math]
Interestingly enough, we get the exact same result applying Heron's formula for area: [math]\sqrt{s * (s-a)*(s-b)*(s-c)}[/math]:
>[math]s = \frac{a+b+c}{2} = \frac{i+1+0}{2} = \frac{i + 1}{2}[/math]
>[math]s - a = \frac{i + 1}{2} - i = \frac{1-i}{2}[/math]
>[math]s - b = \frac{i + 1}{2} - 1 = \frac{i-1}{2}[/math]
>[math]s - c = \frac{i + 1}{2} - 0 = \frac{i+1}{2}[/math]
>[math]\frac{i + 1}{2} * \frac{1-i}{2} * \frac{i-1}{2} * \frac{i+1}{2} = -\frac{1}{4}[/math] https://www.wolframalpha.com/input/?i=(i%2B1)%2F2+*+(1-i)%2F2+*+(i-1)%2F2+*+(i%2B1)%2F2
>[math]\sqrt{-\frac{1}{4}} = i\sqrt{\frac{1}{4}} = \frac{i}{2}[/math]
https://www.wolframalpha.com/input/?i=solve+sqrt(s((s-a)*(s-b)*(s-c)))+where+s%3D(a%2Bb%2Bc)%2F2,+a%3Di,+b%3D1,+c%3D0

It's quite interesting that it remains consistent throughout both of these, even if one of the formula goes through such a long length to compute it. Also using the cosine formula we can derive the angle between the legs: https://www.wolframalpha.com/input/?i=solve+(c%5E2+%3D+a%5E2+%2B+b%5E2+-+2*a*b*cos(x))+where+a+%3D+i,+b+%3D+1,+c+%3D+0 , which means that [math]x = 90^\circ[/math]
However, the other two are indeterminate.
Quite interesting shitpost actually.

>> No.10829418

>>10829333
Furthermore, the radius of the inscribed circle is https://www.wolframalpha.com/input/?i=solve+sqrt(((s-a)(s-b)(s-c))%2Fs)+where+s%3D(a%2Bb%2Bc)%2F2,+a%3Di,+b%3D1,+c%3D0 , i.e [math]\frac{1+i}{2}[/math]
Using the alternative formula for a right triangle, [math]\frac{a+b-c}{2}[/math], we get [math]\frac{1+i}{2}[/math] again
According to https://en.wikipedia.org/wiki/Law_of_cotangents , [math]\frac{\cot(45^\circ)}{\frac{1+i}{2}} = \frac{1}{r}[/math], which is also correct, but also meaning that we can extend this to the other two:
>[math]\frac{\cot(\frac{\alpha}{2})}{\frac{1-i}{2}} = \frac{1}{\frac{1+i}{2}}[/math]
i.e the only way that this can be true is if [math]\cot(\frac{\alpha}{2})[/math] equals to [math]-i[/math]: https://www.wolframalpha.com/input/?i=solve+x%2F((1-i)%2F2)%3D1%2F((1%2Bi)%2F2) which means that [math]\alpha[/math] is some magical angle that we're not yet aware of, possibly the indeterminate form of [math]\cot(0)[/math]
But here's the catch, [math]\cot(\alpha) = \frac{b}{a}[/math], or [math]\frac{1}{i}[/math], and it just so happens that [math]\frac{1}{i} = -i[/math]
Following this,
>[math]\frac{\cot(\frac{\beta}{2})}{\frac{i-1}{2}} = \frac{1}{\frac{1+i}{2}}[/math]
with a solution of [math]\cot(\beta) = i[/math] and [math]\frac{i}{1} = i[/math]

At this point we can pretty much conclude it: this triangle probably does actually exist as it has violated nothing so far, even going into trigonometric functions. The question is, why has no one addressed this and why is no one researching it?

>> No.10829566

>>10828859
>le memes are funny images
can't wait for summer to be over

>> No.10829583

>>10829566
Not that anon but are you retarded?

>> No.10829592

>>10827994
We would have to divide by zero when finding the sin and cos for this triangle. Am i wrong?

>> No.10829758
File: 12 KB, 500x487, 111.png [View same] [iqdb] [saucenao] [google]
10829758

>>10827989

>> No.10829772

>>10829109
It's wrong to assume the pythagorean theorem holds tho

>> No.10829897

>>10828063
>Mathematics doesn't model reality

By the way, have we even confirmed, if the Reals are real?

>> No.10829967

>>10829772
It was my starting point. So it have to hold from definition, and as >>10829333 >>10829418 showed it do holds for it anyway.

>> No.10829970

>>10829897
No. We can't even confirm that Natural numbers real.

>> No.10829981

>>10829566
The imaginary triangle is a meme, not a "funny image"

>> No.10829984

>>10827989
Imaginary, since both imaginaries and negatives are imaginary

>> No.10829995

>>10829149
I got it.

>> No.10830003

>>10829566
How many high school children do you believe are browsing /sci/ - Science and Mathematics and deciding to discuss whether or not a square with an area of -1 is real?

>> No.10830025

>>10828859
i.e. not actually requiring an understanding of math to get them

>> No.10830222

>>10827989
that's not the square of a complex/imaginary number tho.

>> No.10830299

>>10829592
You're not. We run into all kinds of division by zero and such a triangle must have a solution to [math]\sin(x)^2+cos(x)^2 = 0[/math]. Talking about sines, the area of the triangle must be equal to [math]\frac{1}{2}a(b\sin(\frac{\pi}{2}))[/math], which is [math]\frac{i}{2}[/math] which proves those calculations further >>10829333
However it also implies something according to this https://en.wikipedia.org/wiki/Law_of_sines#Proof : that it is equal to [math]\frac{1}{2}c(a\sin(\beta))[/math], which means that [math]0*(i\sin(\beta))[/math] must equal to [math]\frac{i}{2}[/math]. But on the other hand, [math]\frac{1}{2}b(c\sin(\alpha))[/math] must also equal [math]\frac{i}{2}[/math], which means that [math]0*\sin(\alpha) = i[/math]. Perhaps some kind of a new meme number that we can call anti-zero? Conversely, [math]\sin(\alpha) = \frac{i}{0}[/math] which legitimizes division by zero without having to pull infinities or other meme answers like 1 or 0.

If only this triangle was proven to be consistent with math, it'd light up an entire new room of undiscovered identities that we've stubbornly ignored for quite a while

>> No.10830413

>>10830025
Most math memes are lame and don't even require you to know what an imaginary number is
This is relatively advanced since it assumes you've passed high school

>> No.10830428

>>10830299
By the way, taking the identity [math]\frac{a}{\sin(\alpha)} = \frac{b}{\sin(\beta)} = \frac{c}{\sin(\gamma)}[/math], where [math]\frac{c}{\sin(\gamma)} = 0[/math], it should therefor follow that [math]\frac{i}{\sin(\alpha)} = 0[/math] as derived again in the post above through different means, further confirming that [math]\frac{i}{0} = \sin(\alpha)[/math]

Another identity, [math]\tan(\alpha) = \frac{a\sin(\gamma)}{b-a\cos(\gamma)} = \frac{i\sin(\frac{\pi}{2})}{1} = i[/math]
[math]\tan(\alpha) = i = \frac{\sin(\alpha)}{\cos(\alpha)} = \frac{i}{0\cos(\alpha)}[/math], or [math]0\cos(\alpha) = 1[/math]

So let's plug them into [math]sin(x)^2 + cos(x)^2 = 0[/math] which supposedly has no solutions
[math](\frac{i}{0})^2+(\frac{1}{0})^2 = 0[/math]
[math]-\frac{1}{0}+\frac{1}{0} = 0[/math]

hmm, now let's try to derive [math]0\cos(\alpha) = 1[/math] through other means
[math]a^2 = b^2 + c^2 - 2bc\cos(\alpha)[/math]
If we take the derivation above that [math]c\cos(\alpha) = 1[/math]:
[math]-1 = 1 + 0 - 2*1*1[/math]
[math]-1 = -1[/math]

So in short:
[math]\frac{i}{0} = \sin{\alpha}[/math]
[math]\frac{1}{0} = \cos{\alpha}[/math]
Where [math]\alpha[/math] is some unknown abnormal angle

>> No.10830457

>>10830003
its mostly cs and engineering majors they’re not very creative or intelligent people

>> No.10830477

>>10830428
Also again going back to that identity on the top line, [math]\frac{b}{\sin(\beta)} = \frac{1}{\sin(\beta)} = 0[/math], which means that [math]\frac{1}{\sin(\beta)} = \frac{1}{\cos(\alpha)}[/math], or [math]\sin(\beta) = \cos(\alpha)[/math], which is actually true in normal triangles as well.
So this should automatically imply that [math]\sin(\alpha) = \cos(\beta)[/math], let's try it out:
[math]b^2 = a^2 + c^2 - 2ac\cos(\beta)[/math]
If [math]\sin(\alpha) = \cos(\beta)[/math] then [math]0\cos(\beta) = i[/math]
[math]1 = -1 + 0 - 2i*i[/math]
[math]1 = -1 + 0 + 2[/math]
[math]1 = 1[/math]

>> No.10830524

>>10827994
>a triangle with one side zero in length
Congratulations asshole, you invented a line segment.

>> No.10830832

>>10830477
Another consistent identity that includes division by zero:

[math]1 + \tan(\alpha)^2 = \frac{1}{\cos(\alpha)^2}[/math]
[math]1 - 1 = \frac{1}{\cos(\alpha)^2}[/math]
[math]0 = \frac{1}{(\frac{1}{0})^2}[/math]
[math]0 = \frac{0}{1}[/math]

In addition for tan/cot, going back to the cot derivations here >>10829418 , [math]\cot(\alpha) = -i[/math], and [math]\cot(x)[/math] being [math]\frac{1}{\tan(x)}[/math], following the conclusion here >>10830428 that [math]\tan(\alpha) = i[/math], then [math]\frac{1}{\tan(\alpha)}[/math] should be [math]-i[/math], which is true as [math]\frac{1}{i} = -i[/math]
I think that these are enough overlapping confirmations in any possible form that this triangle does in fact exist and most importantly, that imaginary lengths do exist as well. The other thing it confirms is that division by zero exists as well, and is defined as the sine and cosine of this triangle around the zero hypotenuse, not infinity, zero or 1, and also that there are two versions of that number that are supposedly different: imaginary and real, [math]\frac{i}{0}[/math] and [math]\frac{1}{0}[/math], almost like a whole new complex number

So now what? How can we apply this to test it further? Physics perhaps?

>> No.10830886

>>10827994
Unironically true in Minkowski space/special relativity
https://en.wikipedia.org/wiki/Spacetime#Spacetime_interval
https://en.wikipedia.org/wiki/Pseudometric_space

>> No.10832542

>>10829970
No, I mean if the space(time) we occupy is continuous.
A lot of cumbersome infinities would disappear, if space and time were integer-like rather than "real".

>> No.10832898

>>10830428
>sin(x)2+cos(x)2=0

you serious nigga

>> No.10832929

>>10829065
How are operations defined on ⊥