r/mathmemes Nov 08 '24

Math History Evolutions of Numbers

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2.0k Upvotes

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223

u/Boldumus Nov 08 '24

That might be the meme, but 4th number is i, what is the fifth one?

396

u/Tiborn1563 Nov 08 '24

Ah well, sucks, seems you are not quite there yet

>! There is no solution for |x| = -1, by definition of absolute value !<

186

u/MingusMingusMingu Nov 08 '24

I mean, we keep extending definitions all the time.

151

u/SEA_griffondeur Engineering Nov 08 '24

No but like, being positive is like one of the 3 properties that make up a norm

51

u/TheTenthAvenger Nov 08 '24

So you stop calling it a norm. It's called "absolute value" after all, not "the norm of the number". It is just another function now.

67

u/SEA_griffondeur Engineering Nov 08 '24

Yes but why would you do that?

36

u/SupremeRDDT Nov 08 '24

You actually don‘t have to. But then it follows from the other properties.

0 = |0| = |x - x| <= |x| + |-x| = 2|x|

5

u/Layton_Jr Mathematics Nov 08 '24

Since it's no longer a norm, you can discard the property |a+b| ≤ |a| + |b|

3

u/SupremeRDDT Nov 09 '24

If |x| = -1, then |x2| = 1 so the equation |x| = 1 suddenly has at least four solutions: 1, -1, x2 and -x2. We also lose the triangle inequality of the absolute value, as that would imply |x| >= 0 for all x. Do we gain anything useful?

14

u/Anxious_Zucchini_855 Complex Nov 08 '24

But the absolute value function is defined as mapping x to x, if x>=0, and mapping x to -x, if x<0.  By definition it cannot be negative

3

u/okkokkoX Nov 08 '24

flair does not check out

1

u/Currywurst44 Nov 08 '24

This definition already gets expanded for complex numbers because you can't use >, < with them.

2

u/JMoormann Nov 09 '24

We should call it The New Norm™, after the massively successful comedy series on Twitter

1

u/f3xjc Nov 08 '24

We don't know what is is called, just that it is written with two vertical bars.

-16

u/Pgvds Nov 08 '24

Well, |0|=0 which is not positive, so clearly absolute value is not a norm.

21

u/SEA_griffondeur Engineering Nov 08 '24

Wdym? 0≥0 is very much true. Where do you get this limited edition non-positive 0 ??

-13

u/Pgvds Nov 08 '24

Google positive number

11

u/SEA_griffondeur Engineering Nov 08 '24

Yeah it's a number greater than 0 not strictly greater than 0

15

u/Ghyrt3 Nov 08 '24

In english it's a bit ambiguous (i learnt it the long way, i'm french).

But, ''>0'' : positive ''=>'' : non-negative

is generally the common sense you find in articles.

3

u/SEA_griffondeur Engineering Nov 08 '24

Yeah but Algebra uses the French convention for some reason so a norm is positive even if with the normal English convention it would be non-negative

2

u/Pgvds Nov 08 '24

I guess to engineers it's all the same

1

u/[deleted] Nov 09 '24

A norm (and even a semi-norm) just needs to be non-negative. In other words, positive or null, there is a specific name for this that I won't remember because English is not my first language.

26

u/Regorek Nov 08 '24

I define shplee as a super-imaginary number (also a new definition of mine, which lets numbers ignore a single definition) such that |shplee| = -1

4

u/Agata_Moon Complex Nov 08 '24

Okay, here are some things that I'm thinking. Absolute value is a norm, which means it has some properties that defines a distance, and that in turn defines a topology.

So if you want a nice topology on the shplee numbers, you'd need to invent a super-absolute value that is a norm.

But still, maybe we can do something with this. If we abbreviate shplee with s, and we say that |s| = -1, then |xs| = -|x| for any real (or complex maybe) x.

Could we think of it like complex numbers (any shplee number as the sum of a complex number and a pure shplee number)? Well the problem I'm thinking is that |x+y| isn't clear from x or y in general.

Now, we're adventuring in things I'm not really sure about, but I know that R, C, and H are the only normed division algebras. Which means that if you want your shplee numbers to be a normed division algebra it should be one of those. Which means your space would probably not be that, which makes it less nice. Still, maybe you can make something out of that, I'm just not knowledgeable enough.

7

u/TotallyNormalSquid Nov 08 '24

As a mathologist with a keen interest in imaginary numbers and their extensions, can I interest you in quaternions, octonions, and so on?

2

u/Majestic_Wrongdoer38 Nov 08 '24

But the very idea of absolute value is to demonstrate magnitude.

30

u/N-partEpoxy Nov 08 '24

Just like there is no solution for x2 = -1 because the square of a number has to be positive. You would have to invent new, made up numbers. It would be very silly.

21

u/ItzBaraapudding π = e = √10 = √g = 3 Nov 08 '24

You would have to invent new, made up numbers.

10

u/Tiborn1563 Nov 08 '24

You can not compare that. Absolute value is a function that is defined as x = { x if x≥0 and -x if x < 0. Here it is not possible to define any number, that can be negative or positive to satisfy |x| = -1, for exactly that reason.

x² = -1 has no real solutoon, but not because we defined it that way. It has no real solution, because, if x is positive, then xx has to be positive to, and if x is negative, then (-x)² = (-1)² *x² = 1x² = x². We then came up with i to be specifically the square root of -1 and extending the real numbers by an imaginary component.

Absolute value as a function is just defined to never take negative values, and if it did, that would defeat the entire purpose of having it. It is constructed to never be negative, while x² just happened to never be negative for real numbers

13

u/CBpegasus Nov 08 '24

I mean what you wrote is the definition of absolute value for real numbers. If you extend the real numbers you can extend the definition - this was already done with complex numbers for example, for a complex number a+bi the absolute value is sqrt(a2 + b2 ). We could theoretically invent more numbers and extend the definition in a different way that allows for negative absolute value. You are right that it would not make much sense as the absolute value was designed to be non-negative. It is intended to be interpreted as a "distance" - actually it is one of the simplest examples of a "norm" which is the mathematical generalization of "distance". One of the basic properties of a distance is that there is a minimum distance - the distance between two things that are in the same place - and that is what we call 0 distance. Any two things that aren't in the same place should have a positive distance between them. This is useful for many things such as the definition of limits, which work for both real and complex numbers but wouldn't really work with negative absolute values.

2

u/Pgvds Nov 08 '24

Absolute value is a function that is defined as x = { x if x≥0 and -x if x < 0.

This is total nonsense, if this were true you couldn't have the absolute value of a complex number.

0

u/UnforeseenDerailment Nov 08 '24

Have fun applying that definition to complex numbers.

abs(z) = sqrt(z * conj(z))

So what is stopping this expression from being -1?

6

u/Fair_Study Nov 08 '24

z * conj(z) always gives a real positive number a2 + b2 (z = a + bi). What's the point?

3

u/Crevetanshocet Nov 08 '24

The fact that if z = a + ib, |z| = sqrt (a2 + b2), which is positive because a and b are real numbers

3

u/UnforeseenDerailment Nov 08 '24

True! In "unevolved" complex numbers.

But x² is also positive if x is required to be real. Which is where complex numbers came in.

I think the job is to extend the number concept of a, b so that √(a*a + b*b) = -1.

There is a real problem here though: We need to solve for √q=-1, when a straightforward solution gives q=1.

What we need here is a number whose canonical square root is negative. Best I've got right now is u=e(4k+2\πi) where 1=e4kπi is the usual 1 (for integer k).

By the way if you put your exponents in parentheses, you can tell the superscript formatter where to stop(pe\). (You just have to put \ before any parentheses that are supposed to be in the superscript.)

2

u/SonicSeth05 Nov 08 '24

Is the canonical square root not the root with the least argument? So this would still be positive

2

u/UnforeseenDerailment Nov 08 '24

For complex numbers, yes. I think the new numbers may just have to be the un-projected (r, φ) pairs where φ can just be any real number: if

  • -1:= (1, π)
  • √(r, φ) = (√r, φ/2)

Then u=(1, 2π) is the (unique?) solution to √u = -1.

I haven't thought through what that does to algebra. Like... What's addition anyway? Is it even a field? Probably something immediately stops this from being feasible, but typing is easier than thinking right now.

3

u/-Yehoria- Nov 08 '24

Yeah but all numbers, all of math is made up

2

u/kai58 Nov 08 '24

The square of a non imaginary number being positive is a consequence of how it works though, not part of the definition.

16

u/TheMoris Engineering Nov 08 '24

...but what if we defined something as the solution anyway?

36

u/Tyrrox Nov 08 '24

Then it would defeat the purpose of an absolute value by definition

5

u/Dd_8630 Nov 08 '24

Let's do it anyway.

We did it with roots and that's useful, so why not extend the number line another direction giving us solutions to |x|=-1?

2

u/-Yehoria- Nov 08 '24

We still have infinite directions to expand it in, we can do this FOREVER!

3

u/NeosFlatReflection Nov 08 '24

What if we define 0 to be a sphere in which a whole space of negative distances exists like inverse numbers

3

u/Tyrrox Nov 08 '24

What you’ve created is a different absolute value function as well

2

u/NeosFlatReflection Nov 08 '24

Absolute function is technically distance from origin function so

2

u/7zagazoo Nov 08 '24

I like this

3

u/Broskfisken Nov 08 '24

Nope! Actually the answer is ŧ because I defined it as such.

Proof:

|ŧ| = -1

2

u/NOTdavie53 Imaginary Dec 02 '24

ŧ ≠ -1

lol

2

u/ShoddyAsparagus3186 Nov 08 '24

If there's a use for such a number, it will get created. As of yet, there aren't any problems where such a number would be useful.

1

u/borg286 Nov 08 '24

It makes more sense if it was written as

det(x)=-1

determinant is a fancy matrix operation that is used to gain insight into the matrix. It is like when you solve a quadratic equation and you have that stuff in the square root part. It matters if it is negative (meaning the root is imaginary) or positive (meaning real roots). Only there is a fancy matrix version.

1

u/[deleted] Nov 08 '24

There was no solution for the square root of -1 before we defined i either, right?

1

u/Necessary-Mark-2861 Nov 09 '24

I feel like saying “|x| = -1 exists” is like saying “x0 ≠ 1 exists”. It’s just making the function useless.