The 7 Indeterminate Forms that Changed Math Forever

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00:00 -- 03:25 Relation Between 1, 0 and infinity
03:25 -- 13:19 Indeterminate Forms
13:19 -- 14:34 Brilliant
14:34 -- 15:01 Book and pdf

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[lagomoof]

It might be important to clarify that while 0 and 1 are considered numbers, infinity is not. Also, while "1/0 = infinity" and "1/infinity = 0" can be useful shorthands to those aware of the more correct limit definitions, it's misleading to those who aren't, and can be the cause of errors either way. I felt like they were thrown around a bit too carelessly at the beginning. Those shorthands also wrongly imply that "infinity·0 = 1" which was neatly sidestepped at that point of the video.

[AlbertTheGamer-gksn]

This is because indeterminate forms are true for ALL values of x. Not just integers, not just rationals, not just reals, not just complex numbers, but the ENTIRE universal set.
There are 3 types of functions:
1. One-to-one function: These functions are STRICTLY increasing or decreasing throughout its domain and has no first-order critical points besides the endpoints if they exist. The inverse function is defined throughout its entire domain.
2. Piecewise inverse function: These functions are not one-to-one, but an inverse exists for a PORTION of the function. These functions have at least one first-order critical point.
3. Singular function: The term "singular" in mathematics refers to anything that are not invertible. Because of that, a singular function is a function that doesn't have an inverse. These functions exist as annihilation functions, where all values in its domain get "annihilated" to a single value as its range. Because of that, a function is singular if and only if (iff) its derivative is 0 throughout its domain, similar to how a singular matrix has a determinant of 0. Let's say you have this function f(x)=1. This means that for every value of x inserted, the function returns 1 every time. Therefore, this graph takes the form of a horizontal line at y=1. If we say we want to invert f(x), we need to switch the domains and ranges, and since this function annihilates all values of x and returns 1, meaning the domain of the function is all numbers, but the range of the function is 1, if an inverse exists, it will then map 1 for the domain and, uh-oh, we have a problem here. A function operates like a time-distance relation: You can be in the same position at 2 different Planck Times, but you, at the same Planck Time, cannot exist in 2 different locations. If f(x)=1, f^-1(x) will have f(n!=1) not making sense, as f(x) cannot not equal 1. However, f^-1(1) maps out all real numbers, meaning that it wouldn't be a function, as at the same time the function simultaneously exists everywhere. This means there doesn't exist any continuous piece of this function that makes sense; it only makes sense for that one point, and even for that point, it violates the basic rules for functions. Graphically, to invert a function, you need to reflect it over the line f(x)=x, and reflecting a horizontal line gives you a vertical line, which violates the vertical line test. Finally, using singular functions, we are able to introduce a concept called indeterminate forms. As you can see, for the function f(x)=1, f^-1(1) gives an indeterminate form due to it being a vertical line at x=1. Given a singular function f(x), f^-1(f^(x)) is indeterminate. Even though applying the inverse function after applying the function should result in the identity function f(x), for annihilation functions, it all has lots of paradoxes. An indeterminate form comes from basic algebra, where a declared variable has no value assigned to it, as variables are MEANT to be indeterminate until you assign a certain value of it or a transformation of it. For example, x is indeterminate. With n variables, you need n equations to make the values determinate. Some equations are degenerate and continue to evaluate an indeterminate form, such as x=x, as it is true for all x. Using that, we can prove the 11 indeterminate forms are indeterminate using these formulas: 0x = 0 for all Aleph-Null x, so 0 / 0 is indeterminate. 0 * infinity and infinity / infinity are variants of 0 / 0, as 0^-1 = infinity. infinity + x = infinity for all Aleph-Null x, so infinity - infinity is indeterminate. The infinitieth root of x is 1 for all Aleph-Null x, so 1^infinity is indeterminate. 1^x = 1 for all Aleph-Null x, so log_1(1) is indeterminate. 0^x is either 0, 1, or infinity, so log_0(0), log)_0(infinity), log_infinity(0), and log_infinity(infinity) are all indeterminate. Using these equations, we can prove which forms are NOT indeterminate, to see how many indeterminate forms are there. Calculi operate on indeterminate forms, as a derivative is equal to 0 / 0, and integrals are 0*infinity, so we can see how many calculi are there. For example, the product/geometric calculus consists of the product derivative, which is limit as h goes to 1 of log_h(xh/x), which is log_1(1), and the product integral, which is 1^infinity. Infinity is a fixed-point of X = X+1, so when dealing arithmetic with infinity, we use a variable x to denote the fixed-point, which is useful on checking if forms are indeterminate.
Case 1: 0*infinity, x=0*infinity=0+0+0+0+0+...+0+0+0+0=0+(0+0+0+0+...+0+0+0+0), so x = x+0, and x = x. Indeterminate.
Case 2: 1^infinity, x = 1^infinity = 1*1*1*1*1*...1*1*1*1*1=1*(1*1*1*1*...1*1*1*1*1), so x = 1*x, and x = x. Indeterminate.
Case 3: 1^^infinity, x = 1^^infinity=1^1^1^1^1...1^1^1^1^1=1^(1^1^1^1^1...1^1^1^1^1), so x = 1^x. This only is true if x=1, so 1^^infinity is NOT indeterminate. However, if x=infinity, it equals an indeterminate form, so it is considered indeterminate in the complex world.
Case 4: 0^infinity, x = 0^infinity = 0*0*0*0*0*...0*0*0*0*0=0*(0*0*0*0*...0*0*0*0*0), so x = 0*x, and x = 0. Not indeterminate in the real world.
Case 5: 0^^infinity, x = 0^^infinity = 0^0^0^0^0...0^0^0^0^0=0^(0^0^0^0^0...0^0^0^0^0), so x = 0^x, but NO SOLUTIONS EXIST. This is because 0^^x = 1 if x is even, and 0 if x is odd. Same thing goes with 0^^^x, 0^^^^x, and 0{n}x.
In Boolean algebra, if a function has no solutions, it evaluates to false. If it is an indeterminate form, it evaluates to true. If it is only true for a value or a finite set of values, it returns that value or set of values. Therefore,
Using this, we can derive a Boolean algebra from equations:
Union of solutions: ∪
Intersection of solutions: ∩
Non-solutions: ∥
Exclusive solution: x ⊘ y = x ∩ ∥y
"Xor" solution: x ⊗ y = (x ⊘ y) ∪ (y ⊘ x)
Here are some rules derived from this:
Identity laws: x ∩ I = x, x ∪ ∞ = x where I ∈ indeterminate forms and ∞ = n/0 where n ≠ 0
Domination laws: x ∩ ∞ = ∞, x ∪ I = I
Idempotent laws: x ∩ x = x, x ∪ x = x
Commutative laws: x ∩ y = y ∩ x, x ∪ y = y ∪ x
Associative laws: x ∩ (y ∩ z) = (x ∩ y) ∩ z, x ∪ (y ∪ z) = (x ∪ y) ∪ z
Distributive laws: x ∩ (y ∪ z) = (x ∩ y) ∪ (x ∩ z), x ∪ (y ∩ z) = (x ∪ y) ∩ (x ∪ z)
Negation laws: x ∩ ∥x = ∞, x ∪ ∥x = I
Double negation law: ∥∥x = x
DeMorgan's Laws: ∥(x ∩ y) = ∥x ∪ ∥y, ∥(x ∪ y) = ∥x ∩ ∥y
Using this, we can tell:
For any Boolean algebra, we must have a correctness, an incorrectness, an addition, a multiplication, a negator, and an excludor. Therefore:
Logic: truth ⊤, falsity ⊥, and ∧, or ∨, not ~, implies -->, biconditional <-->, xor ⊕
Set theory: Universality U, Nullity ∅, Intersection ⋂, Union ⋃, Compliment C, Exclusive \, xor ⊕
Arithmetic Boolean algebra: Enabled 1, Disabled 0, Times * ∏, Plus + ∑, Prime ', xor ⊕
Equations: Indeterminate I, Infinity ∞, Intersection of solutions ∩, Union of solutions ∪, Non-solution ∥, Exclusive ⊘, xor ⊗

[simonwillover]

1:40 and that's why call x/0 undefined for real xs, complex xs, and even matrix xs. Becuase, (x/0)'s value is not real, it's not complex, and it's not a matrix. It would belong in a different number system if you defined it, and in that number system, there would exist some y = x/0, where y*0 != 0, but instead y*0 = x. This makes no sense in any common number system, and the few mathematicians who do play around with the number systems that allow this have not found a use for it yet.

It's not a fact that you can't divide by 0. It is a fact that defining the value you get from dividing by 0 is useless. Hopefully, in the future, someone will find a reason to divide by 0, but no one has found it yet. For now we will be stuck with undefined and inifinity being the only answers (fun fact: modern computers define 1/0 as infinity).

[hirotempest]

Idk what I had one day, but I made some weird shit that gave me as a result that: "infinity equals 1" I need to find that paper, but, yeah, I was probably drunk or some shi

[ianfowler]

You have to very careful when you say that the limit "= +inf". This is simply an agreed upon shorthand notation used as a stand in for " increases without bound" and that should be made clear before we start using an = sign in this context. When we write lim[f(x)] = +inf, the equal sign, in this special case, does not have the same meaning as lim[f(x)] = L. In such a case the limit does not exists, but the symbol +inf indicates the behaviour of the function as x -->a. Expressions like: "it's unclear what the limit of this infinity is" and "the x becomes infinity" and "it's unclear how fast each infinity is reached" are not correct. Infinity is not a destination. It cannot be reached. A real variable cannot become infinity. A limit cannot equal infinity - it can tend toward infinity. Again, "=" +/ inf is just a short hand stand-in notation.

I get what you are trying to explain in the video and I applaud your efforts. We just all have to be careful with our exact wording and be clear about what symbols we use actually mean.

[ValidatingUsername]

You should do a video on base 0, base 1, and 0*ln(0).

2:15 The additive identity kicks in too.

[markwrede]

Nature works upon slopes, best characterized by certain primes. These are quickly exhausted under multiplication, however, slopes more flat than multiplication can produce may be characterized by slopes of immense Pythagorean triangles where the hypotenuses are but one or two integers larger than the longest leg, i.e. b+1 or b+2. While b+3 is congruent, no integers come to satisfy it.

[ONRIPRESENCE]

This video further motivates my interest in linguistical mathematics and descriptive linguistics.

[Alice_Sweicrowe]

Since Graham's number is so large that it can't be physically written we could potentially use that in place of Infinity. Or, to be cheeky, we could use 10^88, which is the approximate number of atoms in the universe. Some would argue that that would be limitless as well. We'll have to wait until somebody can get there to find out.

[TimTuinman-ed]

pretty sure 8 divides 8 evaluates to true

[Alice_Sweicrowe]

You know, a sign for infinitesimals would solve a couple issues here.

[JessicaMelissaValderramaDolore]

Can you do L' hospat rule on those forms? Also, What is L' hospat rule? I've heard of it but not knowing what it is.

[maverickvalderrama]

Are 1/0 and 1/infinity indeterminate forms?

[victoramezcua]

For God's sake, DON'T write 1 + 1 = 2 + 1 = 3 + 1 = 4 + 1 = ... as stupidly is taught in elementary school

[ethanbartiromo]

There’s so much stuff in this video that seems like you guys are trying to explain something that you don’t TRULY understand, because you’re just giving examples and using quite bad notation. I believe you guys believe these things are true because of these examples, but don’t truly KNOW that these things are true. I know this is for non math people, but would you rather have people tell someone else something they heard, but have no clue why, or would you rather have people be able to back it up? If you heard someone say something that a politician said, but had nothing to back it up, then would you believe them just because they said that they know it’s true because so and so said it?

[TheHebrewMathematician]

The video was great!
But in the last section when you talked about 0⁰ what you said isn't true. By definition we know that 0⁰=1, you may argue that there are limits which go towards an expression of the form 0⁰ and they're value isn't 1, but it doesn't mean that 0⁰ is undefined.

[crigsbe]

Please, just learn to reason like a mathematician. You only present a plausibel explanation for n x 0 = 0. This is not proof at all !!!
Assume n to be any chosen number. Now n x 0 = n x (0 + 0) = n x 0 + n x 0 therefore nx0 = nx0 + nx0 Subtract of both side nx0 and you get 0 = nx0. Reasoning: The first = mark is valid, because 0 is the neutral element of addition. That is: 0 + y = 0 for any y. Here we have y = 0 and it follows: 0 + 0 = 0. The second = mark is valid, because the multiplication is distributive over the additon, that is: n x (0 + 0) = nx0 + nx0. The following = mark is valid, because each number has an inverse element corresponding to additon, that is: The number plus it's inverse add up to the neutral element of additon. Therefore nx0 + (--nx0 ) = 0. It follows: The left side is 0 and the right side is nx0 of the = mark. Summary: n x 0 = 0 for any chosen number n. QED

[ravenecho]

Inf - inf = (-inf, 0, inf) + c
It just does

1/0 = +/- inf
It just does

0 / 0 = c
It just does

C/inf = 0+

+/-0*inf = +/-1
sign(c) = sign(0)*inf
it just does

[Shadramelk]

I have a bone to nitpick with your video setup — the depth of field is too small, and the earrings keep swinging in and out of focus as you speak. It feels incredibly annoying in an otherwise interesting video

[perpetgholl]

the infinite is not a value.
that's why everything szid in this video, is BS.