Quickly becoming mMathing

@Glaeos

5 ай бұрын

Yh this lol. Not complaining

@NonTwinBrothers

5 ай бұрын

Months ago I had to do a second take on the channel banner including the word "math" The 0.999... vid now has a companion (two actually!)

@vlobdam

4 ай бұрын

cMathing*

Waiting for "43 Nooby Math Habits You Need to Ditch"

0 = -(0x)+0x = -(0x)+(0+0)x = -(0x)+(0x+0x) = ((-0x)+0x)+0x = 0+0x = 0x. This is almost the same as in the video, except it doesn't use the trickery that says "add something to both sides and hopefully doesn't change anything" which is technically a theorem, thus also does need a proof so writing it this way is cleaner in my view.

@user-bbohiv

5 ай бұрын

How do you prove it? I thought it was a property included in the definition of equality: a=b -> f(a)=f(b)

@HoSza1

5 ай бұрын

@@user-bbohiv It appears that you are right, according to this en.m.wikipedia.org/wiki/First-order_logic#Equality_and_its_axioms. In retrospect probably we can say at least that we don't need to rely on that property of equality. Maybe it can be useful when one works with logic that does not include the equality symbol (on the logic level).

@IronLotus15

5 ай бұрын

@@HoSza1 Technically, your proof still relies on that property of equality. For example, although 0 = 0 + 0 directly from 0's definition, we need that property of equality to claim that 0 * x = (0 + 0) * x, and then that -(0 * x) + (0 * x) = -(0 * x) + ((0 + 0) * x), for your second equality. Both mCoding's proof and your proof also relies on the transitivity of equality.

@mskiptr

4 ай бұрын

You're right to generally beware adding something "to both sides". While this does work for addition, a=b -> f(a)=f(b) is one thing, but most of the time you actually want the converse: f(a) = f(b) -> a=b - injectivity! And that absolutely has to be proven for any operation. (here it is not a problem for two reasons actually: you can reverse the addition of a number (it is injective) and a=b -> f(a)=f(b) does go in the direction the proof needs it to.

@mskiptr

4 ай бұрын

@@HoSza1 Also: > logic that does not include the equality symbol (on the logic level). You might be interested in type-theory-based proof assistants like Lean, Agda, Idris and so on. There, equality is typically defined within the standard library and the way that definition interacts with type unification is what gives it all the properties of equality. Here's how you can define your own equality in Idris: data Equals : (ty : Type) -> (a : ty) -> (b : ty) -> Type where EqualsItself : (t : Type) -> (x : t) -> Equals t x x Here, `Equals` is a new data type we're defining (it will represent our proposition) and `EqualsItself` is its only "constructor". So if we now write `Equals Nat 2 3`, that means 2 = 3. But since this is a false statement, we won't be able to write a proof for it. On the other hand, `EqualsItself Nat 1` has a type `Equals Nat 1 1` and so its existence proves that 1 = 1. Btw, the standard definition of equality in Idris uses some language features (like implicit arguments) that make it much nicer to use than this one is. But this will, I hope, be a bit easier to explain. Whenever a value `EqualsItself` is in scope, the unification engine can see that its type has both "sides" of equality filled with the exact same thing (`x`), and so it will consider `a` and `b` the same thing too. As an example, here's a proof for that x=y -> f(x)=f(y): noLongerAnAxiom : {u, v : Type} -> {x, y : u} -> {f : u -> v} -> Equals u x y -> Equals v (f x) (f y) noLongerAnAxiom (EqualsItself u x) = EqualsItself v (f x) The proof for an implication is a function which given one proof, constructs another. It also needs to take a bunch of other arguments, to work for all sorts of values for x, y and f. And here's how it can be used: example : Equals Nat (2 + 2 * 2) (2 + 4) example = noLongerAnAxiom {f = (2 +)} twoTimesTwoIsFour where twoTimesTwoIsFour : Equals Nat (2 * 2) 4 twoTimesTwoIsFour = EqualsItself Nat 4 -- When there are no unknowns, functions like (*) can be fully evaluated -- and so there's no need to manually prove they're equal to their results.

can you please use a dark mode background for these kinds of videos? with someone with significant eye strain (and a strong blue light filter, lower brightness screen and significant background lighting, none of which reduce to pain enough to make it comfortably watchable), I would highly appreciate it

@yash1152

3 ай бұрын

no "pop dark mode" please. use a neutral gray.

Love your videos, you definitely put a math spin on every programming thing you do, and it seems that works the other way around too!

In Conway and Knuth's "Surreal Numbers", infinity (countable ordinal infinity - called Omega) times 0 is defined. And, IIRC, it is 0.

What, are you math explainer now? 😄 ThreeBlueOneBrown beware

@prawnydagrate

5 ай бұрын

He does math, C++, and Python. And btw a real 3b1b fan knows it's 3Blue1Brown, not ThreeBlueOneBrown

@Joel-zi6pt

5 ай бұрын

🤓​@@prawnydagrate

love this! keep bringing the math vids - love em all

I was not told there would be math on this channel

In the ring axioms you don't have to take commutativity for addition. If you look at different ways to open the bracetes in (x+y)(1+1) you can see that it is derived from the other axioms

@mCoding

4 ай бұрын

Great observation! It is usually included in the definition of a ring, but as you have noted, it is not necessary!

One line explanation: because by definition 0 is the additive unit, and all of those examples are Rings, you can apply the ring axioms to show that 0x = 0 for all x. The deeper question is why are Rings so ubiquitous

@Impatient_Ape

4 ай бұрын

"one line 'explanation' " = go to wikipedia and learn what a "ring" is.

Tldr, because reals (R) is a field (it being a ring would've sufficed too, so it works for Z as well). So, 0a = (b - b)a = ba - ba = 0.

This is the question I didn't know I needed an answer to

Michael Penn uploaded a video right after yours, about the exact thing you mentioned at the end 😂

can you explain what 0^0 (zero to the power of zero) is? It could be 0 or 1 depending how you approach it, if either the exponent is variable and the base is 0 or if the base is variable and the exponent is 0.

@stapler942

5 ай бұрын

My understanding is that if one must define it, then it is generally agreed that it must be 1. Often it is left undefined, but occasionally defining 0^0 = 1 can be useful for theorems. Defining 0^0 = 0 does not seem to be mainstream or recommended at all.

@RepChris

5 ай бұрын

If it isnt one, then you start running into issues with a surprising amount of theorems. Even in rngs (rings without an identity (1)), it is in effect 1, at least if you want basic stuff like the binomial theorem to work. It comes down to how exactly you define exponentiation, but pretty much all of the sensible ways lead to it being 1

@brauggithebold7956

4 ай бұрын

0^0 = 1 reason: 0^0 means that you multiply 0 with itself 0 times. Performing any operation 0 times means that you get te neutral element of that operation, which in the case of multiplication is 1. Forget about arguing with limits, they prove nothing. They only describe what a function does in the neighbourhood of a point, not what it does at a point. Functions are allowed to be discontinuous and have jumps, e.g. the heaviside function has a major discontinuity in 0.

@RepChris

4 ай бұрын

@@brauggithebold7956 I mean it even works similarly in a rng, where there isnt a 1; you can prove the binomial theorem for those(cant remember if they have to be commutative), and 0^0 = 0 would throw a wrench into that, as we have to assume that x^0y^n = y^n, i.e. the x^0 "falls away", which would be a pretty severe contradiction if it were 0. So even if there is no such thing as 1 (multiplicative identity), we still wind up with a "virtual" 1 for the lack of a better term You could argue that x^0 is ill defined if there isnt a multiplicative identity, but it kinda just works

0:59 what is the variable s in this case?

@ABDTalk1

5 ай бұрын

it stands for sex

@thomassynths

5 ай бұрын

It's what is being integrated over due to the `ds`, where `s` varies from `-inf` to `inf`. This is the definition of a convolution between two functions.

@noxus7462

5 ай бұрын

It’s a “dummy variable”, it’s just notation to show what the integration is with respect to, with endpoints -infinity and infinity.

As a side note, 0 x infinity is only undefined if "0" is a function whose limit is 0, and not 0 itself. If you had f(x) = x and take its limit to infinity and multiply it by 0, then the answer IS defined, its 0. The undefined comes from when you multiply it with a function that tends to 0, not 0 itself.

The proof James gives here for 0 · x = 0 is what math students see in the 1st or 2nd week of a "Real Analysis" university course.

1:35 how do we know or assume that that number has a negative?

@mr.vladislav5746

5 ай бұрын

We assume that, as mentioned at 2:04.

@RepChris

5 ай бұрын

Without that assumption math doesnt exactly work

@mr.vladislav5746

4 ай бұрын

@@RepChris What's that supposed to mean? Math can work just fine even in domains where there are no negatives (additive inverses), such as the natural numbers. But maybe you meant "the math" referring to specifically the computations in this video and not math in general?

@RepChris

4 ай бұрын

@@mr.vladislav5746 poor phrasing on my part, i was more talking about "math" as how you usually know it. Just being reserved to the natural numbers (excluding the expansion you can do on them to get to places like the integers or rationals) stops you from doing a lot of things that are vital to a large chunk of what people classically think of as math. Of course you can do math on the natural numbers, just not all of it, including the one in the video. Having addition be a group is rather useful

@mr.vladislav5746

4 ай бұрын

@@RepChris Good points and clarifications.

I believe this one is quicker, if I did not make any mistake: Choose any number y, then it has a negative -y, that cancels it. 0 x = (y-y) x = y x - y x = 0

@davidpement

5 ай бұрын

But then there wouldn't be a video

@IronLotus15

5 ай бұрын

I think you still need to prove that (-y * x) is the additive inverse of (y * x). Edit: I could not find a non-circular proof that this is the case. I have a feeling this approach is a dead end. On the other hand, assuming that 0 * x = 0 already, we can prove this property by reversing the steps in your comment: (-y * x) + (y * x) = (-y + y) * x = 0 * x = 0. (By distributivity, definition of inverse, knowing that 0 * x = 0 for any x.) And in particular, we see as a corollary that -1 * x is the additive inverse of x: (-1 * x) + x = (-1 * x) + (1 * x) = -(1 * x) + (1 * x) = 0. (By definition of multiplicative identity, knowing that (-y * x) = -(y * x) for any x and y, and definition of inverse.)

@shadamethyst1258

5 ай бұрын

I don't think you can prove that `(-y) * x = -(y * x)` from only the additive inverse and multiplication distribution rules alone. Usually the distributivity of negation with multiplication is stated as a rule, but if you assume that multiplication is associative, then `-y = -1 * y` also lets you prove it, since `(-y) * x = (-1 * y) * x = -1 * (y * x) = -(y * x)`

@sbares

5 ай бұрын

As @IronLotus15 mentioned, there is a subtlety here namely that the distributive property for subtraction (i.e. "(x-y)*z = x*z - y*z") is not usually one of the ring axioms (unlike the distributive property for addition), so one would have to prove it first. However, I'm pretty sure one would first need to show that 0*x = 0 to prove this property, thus making the argument circular.

@japedr

5 ай бұрын

Thanks for the comments! Then it looks like this approach had an important oversight.

I didn’t know I was subscribed to 3m1coding

mAth

@xshadow-0

5 ай бұрын

I didn't see no physics here yet.

And if you want to know, how zero was invented: kzread.info/dash/bejne/h2isvI-BiJjfmbg.html

discord light mode be like

สวัสดีครับ ขอบคุณครับ

Why doesn't this break down for the natural numbers then? Where there doesn't exist a negative. And yet the natural numbers are closed under multiplication and if you include zero its still zero times anything is zero. I imagine we just temporarily extend to the integers and contract back down similar to solving quadratics with complex numbers even if we only care about real answers?

@MuffinsAPlenty

4 ай бұрын

Great question! The reasoning does _not_ hold in the natural numbers, even though 0*n = n*0 = 0 is true for all natural numbers n. I think you have provided a reasonable explanation, but let me give you another one, which I personally find to be better. In the context of the natural numbers, multiplication is repeated addition (this is not true in other contexts, but in the natural numbers, it _is_ true). If one takes n*m = m+m+...+m (n addends), then we get two scenarios: n*0 = 0+0+...+0 (n addends) = 0 by the additive identity property of 0. 0*n = ____ (0 addends). Here, we have the "empty sum", a sum of no addends. We define the empty sum to equal 0 since 0 is the additive identity, and the identity is the only value an empty operation could have which makes it consistent with the associative property of of that operation.

So you're becoming a math channel too now. Maybe you could start a 2nd channel called mathCoding :D

If every egg carton contains 0 eggs and you buy x egg cartons, how many eggs would you get?

@thechessmaster9291

3 ай бұрын

No answers, because common sense is lost on nerds ....., bunch of dumfks , PROOF me wrong ........-;) ..,

so, it's a consequence of the axioms 0+x=0 addition distributes over multiplication every number has a negative associativity of addition But AFAICT these axioms are actually consequences of the axioms of ZFC? would be cool if you went into that (if set theory is your thing)

@shadamethyst1258

5 ай бұрын

These are usually not axioms, but rather rules you require structures to have so that you can work with them with those tools. The set-theoretic natural number construction does follow some of these rules (it's a monoid with zero), but you can easily construct a structure with multiple zeroes that would break the rule analyzed in the video (0*x=0) while satisfying the other rules of a ring.

@RepChris

5 ай бұрын

They are axioms in the sense that they are how addition and multiplication is *defined* You could of course call some other binary operations addition and multiplication, but addition being an abellian group, and multiplication being a half group which distributes over addition are needed (rng/ring axioms) are needed if you want them to behave anything like what we usually call addition/multiplication

@shadamethyst1258

5 ай бұрын

@@RepChris By that logic, any function is an axiom, since you have to define it before using it. Natural numbers are usually defined using two constructors: zero and the successor function. Addition is then defined recursively using those condtructors, as is multiplication, and you can then prove properties about natural numbers without introducing any other axiom other than injectivity and induction

@RepChris

4 ай бұрын

@@shadamethyst1258 Fair enough, i was a bit vague. The natural numbers dont contain inverses. If you want addition over the integers, or any other ring/field, you need the inverse element. And a function is, axiomatically, defined how its defined. You wouldnt usually call the definition an axiom, but it still is a given which you cant prove. The inverse element existing is one of the so called group axioms, and addition being an (abellian) group is part of how rings and fields are defined. Again, it is called a group *axiom*.

Dark background please. This is a coding channel not an eye burning speedrun channel

Wondering if all the axioms used can be proved without assuming that 0x = 0.

@jonathan3488

5 ай бұрын

The axioms define the behavior of the field, hence they don't require proof. For instance, the 0 element is defined as a neutral element to addition and therefore by definition a+0=a. You can prove that the axioms apply to groups so they can be considered fields under certain operations, but you don't prove the axioms.

@Quzori

5 ай бұрын

what? axioms aren't something to prove -- they're assumed true as the foundation of mathematics. 0x = 0, on the other hand, is something that can be proven just from axioms.

@Ostap1974

5 ай бұрын

@@jonathan3488 My point was that can we assume that (x + y) * z = xz + yz, if we do not know that 0 * x = 0;

@jonathan3488

5 ай бұрын

@@Ostap1974 It's an axiom so yes. Just like we assume a+0=a, that's the properties of a field. I don't understand what you mean

@Ostap1974

5 ай бұрын

@@jonathan3488 My point was that proving smth very trivial we must be really careful not using some knowledge that is actually based on what we are trying to prove. For example: 0x = (1 - 1)x = x - x = 0

Because you can't have a nil amount of things and you can't have an amount of nil things. You either have something or nothing. That's how numbers work with the concept of 0.

mMaths

By definition?

Because you have nothing to multiply

Works for any rng

Well, obviously, Infinity times Zero equals NaN.

@mCoding

4 ай бұрын

Seasoned take!

pls moar math

zero is equivalent of Lord Shiva in Hindu religion

dislike. white background :////

@Eurotool

5 ай бұрын

Agreed.

Dude, stick to coding. Math isn't your strong point.

Uhm... We can also refer to common sense, or even the philosophy behind the concept of zero. Zero means notning. So zero of anything is also nothing. Where's the big deal? XD

No, regarding infinity, what about ordinal numbers? 0 * ∞ translates to 0 * ῳ, which means "an infinite amount of empty sets all in a row". Which is equal to the empty set {}, which is 0!

You gotta be insane to do all this shit to prove something known. It's like trying to prove 0+0=0. Get a life !😅

@MemeConnoisseur

5 ай бұрын

that's how math work, you just can't assume stuffs. You need to have proofs of things that appears to be common sense.

@juicedelemon

5 ай бұрын

it took a paper to prove 1+1=2 math is life, math is love

@thomassynths

5 ай бұрын

This is literally something every college level 101 pure math class does unironically.

@xSyn08

5 ай бұрын

That's a bizarre response, and clearly you have no interest or knowledge about maths

@HoSza1

5 ай бұрын

You don't have to go in deep effort to prove that, because x+0=x for every possible x (that's ground truth for many types of numbers/matrices/functtions/mathstuff... aka an axiom). Now just apply this axiom for the case when x=0 so x+0=x turns into 0+0=0. Simple, right? 😂 These kind of proofs are useful to illustrate how mathematical proofs work like, when doing them by taking baby steps. In practice, many of those small steps are omitted.