5 Levels Of “No Answer" (when should we use what?)

[@richardcao7390]

inconsistent

[@blackpenredpen]

Oh man, you are right! I missed that one. That one is for systems of equations!!

[@richardcao7390]

@@blackpenredpen yep! Doing matrix for preCalc right now and instantly drew the connection to this video haha

[@justinchen9071]

what's that

[@ultrio325]

how do you get inconsistent

[@nanamacapagal8342]

@@ultrio325 for example, x+y=2, 2x+2y=3. The system has no solution for x and y, so our system is inconsistent.

[@JaydentheMathGuy]

Alternative title: How to tell your math teacher "no".

[@blackpenredpen]

😆 definitely better

[@tomatrix7525]

Lol

[@m9l0m6nmelkior7]

@Tangent of circle. xD

[@sammitramanan6404]

when she tries to teach you differentiation before limits and you pull up a messed up function because you have studied math the right way

[@austinlincoln3414]

Lol

[@CardThrower-rb6eg]

You forgot the final level "I don't know how to solve this"

[@fisch37]

Well, that applies in any case where you forgot to study before an exam

[@shivamchouhan5077]

@@fisch37 😂 lol 😂

[@RazorM97]

"left as an exercise for the reader"

[@AgaresOaks]

Pfft, that's not even close to the final level. Above that there's "open question" (no one knows how to solve this) and "independent of a given set of axioms" (proven that no one can solve this, but it should have a solution).

[@orlandomoreno6168]

@@AgaresOaks independence of a system of axioms doesn't mean that

[@DialecticRed]

“No real value” also happens to be the official mathematical classification for NFTs

[@blocks4857]

Value is subjective

[@cewla3348]

@@blocks4857 cool, so it's indeterminate, meaning that they are completely worthless? Cool! I have INFINITE BITCOIN, since, to me, bitcoin is worth $0.

[@rieldebonk1044]

@@cewla3348 You dont do THAT MAN!

[@Bobspineable]

@@cewla3348 to someone else it could be worth millions so it doesn’t matter if it worthless to you, you take advantage of what people think and profit off it. That’s how you get something valueless to get value. That’s what our money and jewelry are. Pieces of metal and paper that we perceive to have value.

[@patrickliberatoalves2931]

The value of NFT is imaginary

[@jxdinglol]

"No solution" is used frequently in systems of equations. Two parallel lines have no intersecting points and that is the easiest form of all "no solution" problems to understand.

[@saperoi]

Also as contradiction where x is 6 but x has to simultaneously be 9

[@arvin390]

Yeah, stuff like x = x + 1

[@manioqqqq]

then, x=∅ and thus y=∅ ect.

[@rynabuns]

how about non-Euclidian geometry?

[@ThomasTheThermonuclearBomb]

"No solution" being used there is incorrect, it should be "inconsistent system"

[@somerandomsheep]

when your parents ask if you are lying you can just tell them "it's a complex statement"

[@Adix_Null]

Me on my math exam: The answer is left as an exercise to the reader

[@akshatvats7992]

xD

[@YourLocalCafe]

@Lakshya Gadhwal learner*

[@idrisShiningTimes]

Gives me Reimann Zeta Function vibes

[@navaneethshettyb9312]

😂😂

[@stratonikisporcia8630]

@@idrisShiningTimes "I know how to solve this, but I'll only tell you if you give me $500,000"

[@necrolord1920]

6:51 technically with the definition, the output is always NON NEGATIVE. An absolute value could be 0 :)

[@tperm6695]

Yup, when x=0 absolute value of x is still 0 :)

[@angelmendez-rivera351]

I agree.

[@animatoryGuy]

This guy: (flawlessly explains all the ways an equation can have no answer) My calculator: "NaN"

[@saikitonia]

👵

[@Mg_887]

NaN stands for "Not A Number" in js and in ts

[@SlidellRobotics]

About 15 minutes in, for 2⁰, you could argue/explain the definition (not prove) that 2⁰ = 2¹⁻¹ = 2¹ * 2⁻¹ = 2 * ½ = 1.

[@Shreyas_Jaiswal]

Yes this is how it is defined. My teacher has also taught me this process.

[@陈明年]

I always wonder why 2^0 is a definition

[@angelmendez-rivera351]

This is not how 2^0 is defined, though. Yes, it is true that 2^0 = 2^[1 + (-1)] = 2^1·2^(-1) = 2·2^(-1) = 1, but this is how one motivates the definition for 2^(-1), not for 2^0. You cannot define 2^(-1) reasonably without first defining 2^0. The actual definition of 2^0 is the product of the 0-tuple which, if having elements, would only consist of the number 2. However, the 0-tuple has no elements and is unique, and since its product is 1, 2^0 = 1. In fact, x^0 = 1. This is just a consequence of how exponentiation is defined. There is nothing else to demonstrate.

[@vns1956]

I was taught like this: say you have a^n * a^m, the result is going to be a^(n+m). now, lets plug in 0 for one of the exponents: a^n * a^0 = a^(n+0) = a^n see, you multiplied by something and the value didn't change at all, so the "something" must be 1, when "a" is a value other than 0.

[@tbg-brawlstars]

Right!!!!

[@kevinleugan6037]

Honestly thought this was going to be a lesson on how to stand up for yourself and reject requests you don't want to handle.

[@austinlincoln3414]

Lol

[@danielgammo]

too bad its math class

[@jay-5061]

Nani

[@za1231in]

luckily you stumbled onto something much more useful

[@xenosmoke8915]

It is that lesson, just for tutors not students 🤣

[@mathopediamathexplorer3010]

I love that doll its like Mr . Bean's doll ...

[@SpiderMan-in9rl]

Yeah.. so cute Edit - That doll gives me Nostalgia

[@user-wy8ki2ef1m]

Teddy! His name is Teddy!

[@SpiderMan-in9rl]

@@user-wy8ki2ef1m Yup.. now i remember

[@artandcraftwithmrudhula9206]

🤣🤣🤣

[@Uni-Coder]

The sixth level of 'no answer' is when you are trying to answer the question "is there a sixth level of 'no answer'."

[@kazedcat]

There is a sixth level "undecidable" this is when your axioms are not enough to prove if a statement is true or false.

[@3Jeroen3]

Ah, yes example of this is the arithmoquine function in Gödels proof

[@SG2048-meta]

@IonRuby what, Gödel did make proofs

[@countcrocodile1115]

Not only he explains well, but you can also see how happy he is in his face alone, keep it up man. great video

[@manioqqqq]

What about his statement that √25≠-5? √25=±5

[@alexh.4514]

Plenty of math teachers on YT say the same things as Bprp, but his absolute joy is what makes him such an effective teacher.

[@hongkongmapping]

just realised the indeterminate family is on your shirt lmao

[@izjemmr]

In my experience, "indeterminate" is applied whenever it refers to a test, as is "inconclusive". Limit tests, like those in the video, and some primality tests are good examples, but I most often see the term used when it has to do with convergence tests for infinite series. For example, the divergence test or nth-term test proves that an infinite series does not converge to any value if the terms in the series do not approach 0, but does not definitively prove the inverse. There are series that do not converge even though the value of their terms approach 0, so in those cases the nth-term test is indeterminate. In any case, it all just means the test cannot prove an answer and more work must be done.

[@NoActuallyGo-KCUF-Yourself]

That's an okay synonym for inconclusive, but I think inconclusive is a better choice of word for that scenario. To me, inconclusive means that this particular process did not yield a conclusion, but perhaps some other process will. Indeterminate is more like, no, it cannot be determined.

[@indigopari]

⁠@@NoActuallyGo-KCUF-Yourself I agree, I don't feel like indeterminate fits so well with tests; I usually end up using indeterminate for like expressions that mean you have to go back and try solving another way, like if you end up with 0/0 or 0*infinity or something like that. i guess it depends on the context though, whether that means that a meaningful answer does not exist or it just means that you have to try evaluating with a different method. i guess for a really simple example, if you're evaluating f(x)=(x^2-25)/(x-5) at f(5), and you get 0/0, then that would be indeterminate, and you need to go back and try cancelling or smth, though again, i guess it depends on what you're doing whether x+5=10 would even be a meaningful answer in that context. but i've ended up using indeterminate mostly in like calculus/continuous contexts where if you end up with 0/0 or anything like that that just means you need to go try l'hôpital's or smth

[@israelkoiku2076]

You forgot to include "No Nontrivial Solution" since every homogeneous system of equations has at least the trivial solution x=0, e.g., in a system of homogeneous linear equations

[@captainpolar2343]

nerd alert

[@coolskeletondude5902]

@@captainpolar2343 bro you're watching a math video,dont you think people would be talking about math what a baby

[@hcmishra6371]

@@captainpolar2343 said the fool to the person commenting about math in a math video

[@dogmania2892]

@@captainpolar2343 mf we are watching math vid stfu

[@brightblackhole2442]

@@captainpolar2343 did you expect the MATH video to be like "no real value! repeat after me, no real value! that means there's no answer because you don't know anything past natural numbers yet"

[@mrstutoring]

3:56 how about the most basic of situations with no solution... Solve x+1=x+2 lol

[@nanamacapagal8342]

Thank goodness you stuck a +1 there at the start. If you said x=x+1 then -it still does work in computer science as an incrementor- it's not a condition anymore like your normal equation Edit: got absolutely thrashed in the replies, sorry

[@mathsman5219]

X=0 / 0

[@paolo6219]

X+5=x

[@_judge_me_not]

x=∞

[@aasid2446]

0 = 1

[@damianbla4469]

02:35 The case of "No real value" happens also when we calculate the roots of quadratic equation with discriminant (so-called "delta") is negative.

[@blackcat5771]

so cute

[@shrankai7285]

Wouldn’t that be no real solution, as we are solving an equation?

[@dojelnotmyrealname4018]

The solution exists, it's just not a real value. @@shrankai7285

[@johnpaulhumphrey2981]

I was going to close this because it was just another tab, but I loved your pacing, and stuck around. I liked your style and level of explanation. Subscribed

[@JayTemple]

The early explanation of complex numbers reminds me of a Top Ten list I did when I was teaching: Top Ten Lies Math Teachers Tell. It began with, "you can't substract a larger number from a smaller one," and, "you can't divide a smaller number by a larger one," and continued with things like, "You can't take the square root of a negative number." Near the top I had, "20 liters of one substance plus 10 liters of another will always yield 30 liters of the mixture," and the #1 lie was ... "You need to know this."

[@pablopereyra7126]

I understand most of these, but I can't seem to spot is the lie in "20 liters of a substance plus 10 of another will always yield 30 liters" Could you explain that?

[@JayTemple]

@@pablopereyra7126 Depending on how the substances interact, they might actually yield 30 liters, they might only yield 25 liters, or they could explode.

[@mal2ksc]

@@JayTemple Also, if you're adding the contents of a 10 liter gas cylinder to a 20 liter gas cylinder, you still have a 20 liter tank, just at increased pressure.

[@NoActuallyGo-KCUF-Yourself]

Many dissolution processes change the volume due to changing intermolecular forces between the particles. Salt + water is the simplest example. 1.000 L of a 2-molar saline solution mixed with 1.000 L of pure water will not yield a 2.000 L mixture.

[@ThomasTheThermonuclearBomb]

@@pablopereyra7126 Basically, chemistry makes things weird

[@eydum3948]

Im only 1min and 8 sec in and those whitebord skills are slick

[@aloysiuskurnia7643]

"Undefined" also works when you are using a function with an argument outside the domain of its function. Say you have "f(x) = x for x > 0". You can say that f(-4) is undefined.

[@pecugihan]

isn't that should be "doesn't exist"?

[@apia46]

"it has no real value" hey look thats me

[@timothyconnally2167]

As a software developer, I’ve had design discussions about the meaning of “null”. In a database, this is when no value is stored. You’ve supplied a useful set of mathematical meanings. Other non mathematical meanings include “not applicable”, “unknown”, “not yet determined”, “invalid”, “declined to enter”, etc. At first this seems too pedantic, but really it can make a database function better to augment a nullable field with a null reason list to express why a value is missing. Unfortunately databases are not designed to do this easily. Null tends to be the design equivalent of a blank stare.

[@chx1618]

FWIW the inventor of null called it his "billion-dollar mistake".

[@justsomeguy5628]

In programming languages like c#, for example, even "null" and "Null" are two different things, and while they are kinda applied datum types and to field types, respectively, but even then, they don't behave the same. One of the most important things about floating point in computation is that it allows NaN to be, ironically enough, a number.

[@colly6022]

i would say null itself just represents an empty set, and the semantics of what that means are more related to the software's behaviour or programmer's intention rather than being a property of the null field itself. the inverse to this would be "maybe" monads, where they do contain data, but the semantics of how they're used implies there shouldn't be (in some capacity). e.x.: haskell's Maybe, rust's Option, C++'s std::optional, etc..

[@ResonanceHub]

2:39 This has a solution with quaternions, so it'd still be case 1. I get the point though, nice video

[@manaarikicarpentier6038]

At 6:00 ish: For sqrt(x) = -5 x = 25.exp[i(2pi +k*4pi)] Would work (with k as a whole number) I think.

[@angelmendez-rivera351]

It would not. If the square root function is defined as a function of complex numbers whose output it also complex-valued, then -5 is not in the image of C under Sqrt. This is to say, Sqrt : C -> {z in C : z = 0 or Re(z) > 0 or Re(z) = 0 and Im(z) > 0} is a surjection. However, -1 multiplied by Sqrt(25) is equal to -5, and it does solve the equation x^2 = 25, even though it is not true that Sqrt(25) = -5.

[@angelmendez-rivera351]

@@radupopescu9977 No, you are wrong. That is not how roots work.

[@angelmendez-rivera351]

@@radupopescu9977 sqrt(25) is not defined as the individual solutions to x^2 = 25. That is just not what the symbol sqrt means. You are wrong.

[@angelmendez-rivera351]

@@radupopescu9977 I never said your professors were idiots. They probably tried to communicate the fact that polynomials have multiple roots, but that the corresponding root functions used to find the roots are.... well, functions, but their communication just did not get through to you, and you became confused. Take a look at the quadratic formula. The roots of the polynomial A·X^2 + B·X + C are given by [-B +/- sqrt(B^2 - 4·A·C)]/(2·A). Imagine A = 1, B = -1, C = -6. Hence the roots are given by [1 +/- sqrt(25)]/2. You claim that sqrt(25) = +/-5. This gives you the wrong answer, because after the substitution, the expression is [1 +/- (+/- 5)]/2, and +/- (+/- 5) = 5, hence [1 +/- (+/- 5)]/2 = (1 + 5)/2 = 6/2 = 3. This is only root of the polynomial, and the root given by (1 - 5)/2 = -4/2 = -2. Why is this root missing? Because you incorrectly insisted that sqrt(25) = +/- 5. The correct answer is sqrt(25) = 5, and this gives you both roots, because after the substitution, you get [1 +/- (5)]/2, which is (1 + 5)/2 or (1 - 5)/2, respectively. So yes, the polynomial X^2 - 25 has 2 roots, but only one of the roots is represented by the symbol sqrt(25). The roots are actually -sqrt(25) and sqrt(25), with -sqrt(25) = -5, and sqrt(25) = 5, not sqrt(25) = +/- 5. The same is true if you substitute 25 for any other quantity A^2. The roots of X^2 - A^2 are given by sqrt(A^2) and -sqrt(A^2), not by the symbol sqrt by itself like you think. If A is a real number, then sqrt(A^2) = |A|, while -sqrt(A^2) = -|A|, and {-|A|, |A|} = {-A, A}. The same general principle works for any nth root. In fact, the same principle works for every inverse function. If f : C -> C is surjective, but not injective, then C can be partitioned into disjoint sets C(α) such that for every index α, f|C(α), which denotes the restriction of f to C(α), is injective, but the restriction f|A, with A being a superset of C(α), is not. Then for every index α, f|C(α) has an inverse C -> C(α). For example, this is how the inverse trigonometric funcions are defined. Yes, sin(x) = y has infinitely many solutions, but there is only one value for arcsin(y). Why? Because arcsin is defined as the inverse of sin restricted to [-π/2, π/2]. arcsin(y) is not a symbol that denotes every solution to the equation sin(x) = y. Similarly, sqrt(y) is not a symbol that denotes every solution to the equation x^2 = y. It only denotes one solution, and the other solutions are built/expresses in terms of sqrt(y). In the case of sin(x) = y, you get every solution in the form 2·floor(m/2)·π + (-1)^m·arcsin(y), where m is an integer. In the case of x^2 = y, you get every solution in the form of (-1)^m·sqrt(y), where m = 0, 1. In the case of x^n, you get every solution in the form of exp(2·π·i/n)^m·y^(1/n), where m = 0, 1, 2, ..., n - 1.

[@orangechicken1005]

@@angelmendez-rivera351 sqrt(x) is not a function. |sqrt(x)| is and is usually the one used with sqrt(x)

[@flopsnail4750]

6:08 The square root operation outputs both positive and negative values. Therefore it has not one answer, but two. 5 and -5, making 25 indeed the correct answer.

[@carultch]

That's a common misconception. There are both positive and negative solutions to x^2 = 25, but only one of them is uniquely qualified for the job of *the* square root. By convention, sqrt(x) refers only to the positive square root, or principal square root.

[@wildfire_]

0^0 is undefined because for the x^0 rule, the logic is as follows: when multiplying powers of the same value, you add the power values together, ie. x^a * x^b = x^(a+b) Thus x^0 can be written as x^1 * x^-1. x^1 is just x and x^-1 equals 1/x. x * 1/x = x/x = 1 That means that in the term 0^0, your trying to solve 0/0, which is conflicting because x/0 is undefined, but x/x = 1

[@bruhifysbackup]

so you're saying 0^2 is undefined?

[@M1551NGN0]

"No real value" My life currently:

[@johnpaulhumphrey2981]

19:30 I think 0/0 is inderterminate bc per the long division you used earlier: what, when multiplied by zero is equal to zero? Basically everything. So it is not like 1/0 where we cannot supply a value, it is kinda the opposite, we have too many values.

[@identity5481]

I must be the square root of a negative because my parents say I have no real value

[@markovermeer1394]

In computing science, we also have "I won't tell you" (no permission or not requested), typically a NULL value.

[@gdclemo]

there is also 𝈜 (upside-down T if Unicode doesn't work) which means that this program never halts.

[@technoultimategaming2999]

But null can be an answer. For example set A can be empty, and if someone asks you how many elements are in set A and you say it's empty. That is still a solution

[@markovermeer1394]

@@technoultimategaming2999 That would not be represented as NULL: an empty set would typically be returned as empty array. Your situation is option 2 of the alternatives in this presentation.

[@dojelnotmyrealname4018]

Null usually means "does not exist" which is "no solution" tho.

[@SlipperyTeeth]

I think indeterminate is also used to refer to certain cases of convergence tests for integrals.

[@damianbla4469]

05:40 First of all, the general reason why this equation has no solution is this: The left hand side of the equation is positive and the right hand side of the equation is negative. So easy ;)

[@damianbla4469]

I additionally tell that this rule does not always work. For example for the equation: x^2 = -5 the LHS is positive and the RHS is negative but there are the solutions (two solutions - both are complex: x=5i, x=-5i).

[@Galactic-MathWizard]

Uhm... Shouldn't the complex roots be x= √5i and x= -√5i, since (5i)²= -25≠ -5?

[@teslainvestah5003]

I define w as a number whose square root is -5. I define w as a number whose absolute value is -1. I define w as the limit of sin(x) as x approaches infinity. w is my new favorite number, and it's better than anyone else's favorite number.

[@ferln4]

how to make mathematicians mald in 4 sentences

[@carultch]

Lambert, is that you?

[@chyawanprash]

I disagreed a lot with the √(x) = - 5 but then I came to understand this really well actually. When ever we want both roots, we actually mention ± which means that √x can only be other the positive or negative value. And as far as mathematics is concerned, √x is *_defined_* to give the positive value. Wow, this makes so much sense now!

[@math_the_why_behind]

Right! The sqrt(x) could also be 0 though :)

[@dioniziomorais8138]

Correct, but 'twas just examples.

[@math_the_why_behind]

@@dioniziomorais8138 Right, they were just examples. I just mentioned it because he said it was defined to be positive, but it could also be zero, and zero isn't positive :) But yes for that specific example the answer is defined to be the positive one :)

[@dioniziomorais8138]

@Math: The Why Behind ok, I don't have a great understanding in math, I'm not even an native english speaker lol

[@beatoriche7301]

Well, the real answer is that, for positive real values other than 0, the equation x^2 = a actually has two solutions; we want sqrt(x) to be a function, which means it has to yield a single output value, and so the square root function is defined to be the positive solution to that equation. It's similar in the complex numbers - for every nonzero complex number a, the equation z^2 = a has two distinct solutions. However, in this case, there is no such obvious criterion to latch onto; the square root function is inherently a multi-valued function, which has all sorts of implications for things like power series expansions. There are ways to restrict the output range of the square root multifunction so as to make it a proper function; for example, one common convention is to define the square root of a number to always have positive real part and to be located on the positive imaginary axis for negative numbers. A similar thing occurs when you measure the angle (often called the argument, or arg for short) a nonzero complex number makes with the real axis; obviously, adding any number of full turns will still yield a valid angle to describe that complex number. Here, the usual convention is to restrict the angle to lie in the interval (-π, π]. These types of situations are quite common in complex analysis, and these functions with their naturally but still, in essence, arbitrarily restricted output ranges are known as the principal branches of those functions. However, restricting multifunctions to their principal branches comes with a whole bunch of problems - for instance, general theorems such as arg(z_1*z_2) = arg(z_1) + arg(z_2), the famous multiplication rule for complex numbers, do not hold anymore when the argument is replaced with its principal value. The principal branch of the argument is also not continuous, making it not terribly useful for more advanced analytical purposes. The bottom line, these situations require great care, and conventions are tricky; 5 is the value of the real square root function at 25, but the complex square root - a multifunction - evaluated at 25 has two values, namely 5 and -5 (and so -5 is indeed a square root of 25). By contrast, 5 is the principal square root of 25, which means that, in a sense, the equation sqrt(z) = -5 is indeed not solvable if the square root symbol is referring to the principal root.

[@beyondtherice8277]

Alternative thumbnail caption: "5 levels of dehydration as seen in piss"

[@xenosmoke8915]

Imagine walking into this class without knowing he’s holding a mic.

[@alexh.4514]

I seriously thought it was just a cute prop XD

[@randomname9291]

I like how I’m the video you say that 0 to the power of zero is undefined while your shirt shows it as indeterminate

[@user-vn7ce5ig1z]

A simple example for No Solution is "x + 1 = x + 2" It almost looks trivially solvable but obviously isn't, regardless of the system.

[@thelaststraw1467]

if you REALLY wanted to couldnt you sub in infinity? of course its not a number tho...

[@bloomingon6141]

@@thelaststraw1467infinity has no value

[@thelaststraw1467]

@@bloomingon6141so? how does that imply its not a solution coz it def is

[@justyouraveragecorgi]

​@@thelaststraw1467 Because if it's not a number of some sort, it can't be a solution. I can't go and say the answer is "triangle" or "purple" or "ham sandwich" because that isn't how math works. An earlier commenter mentioned ordinal numbers, which is essentially what you're getting at, but infinity isn't an ordinal number - ordinal numbers are essentially used to represent infinity, from my (very limited) understanding of them. You could put in the ordinal solution ω (omega), but that isn't the same thing as infinity because ω + 1 sort of equals ω (except also not really? Ordinal numbers are...weird).

[@thelaststraw1467]

@@justyouraveragecorgi first of we both know what im getting at so why would you call me out on not using formal terminologies if you wanna debate in good faith. and second could you define ur "number" for me since for the current knowledge of maths i possess "1" is as much a concept as "infinity" is

[@CannedWatermelon-fd3my]

"They have been causing trouble to all the calculus students" killed me

[@b4ttlemast0r]

you can prove 2^0=1 and the undefinedness of 0^0 if you define exponentiation as x^1=x; x^(n+1) = x^n * x; x^(n-1) = x^n / x; this means that 2^0= 2^1 / 2= 2 / 2 = 1, and for any x it means that x^0 = x / x, which is equal to 1 for nonzero x, but undefined for 0^0 because division by 0 is undefined edit: by this definition, inf^0 is also equivalent to inf / inf

[@pancito3108]

6:25 √(x²) = |x| This would result in |x| = 25 So x = ±5 And ±5 has -5, so there you have your solution

[@peteradler6005]

Sqrt ( x) >=0 by def

[@pancito3108]

@@peteradler6005 I never said otherwise

[@ostepolsegudensprofet]

​@@peteradler6005 okay but why though. (-5)^2=25 so why not (25)^1/2 = -5

[@WingMyWay]

​@@peteradler6005 Its okay to veer off "but its defined" and use math to solve problems instead of jerk off about made up rigor

[@xinpingdonohoe3978]

Although I guess he purposefully restricted the domain to take only a single branch of the multivalued function √, and made sure to choose the bit where √(x>0)>0

[@MarcoMa210]

You forgot "No agreed upon answer" (eg. 0^0)

[@baacloud]

after 20 years thank you SO much for being the person to teach me precisely why anything divided by zero is called "undefined." it's sweet justice for my young self who kept hearing the same "well what if you had six slices of pizza and had to distribute it to zero people???" with no elaboration

[@bienvenidos9360]

There will be no transaction. Nothing is taking place. No slices can be distributed if there isn't a value to distribute them to. The pizza rots because no one's there to take it??

[@shivaudaiyar2556]

Thanks for such a great content with love from India

[@HN-vu8pp]

Didn’t you learn all this at age of two?

[@RishaadKhan]

@@HN-vu8pp this is so unfunny its funny

[@yashkrishnatery9082]

@@HN-vu8pp well, we did but revision is necessary. 😂😂

[@thephoenix8728]

@@HN-vu8pp its complicated bro the teaching pattern here is kinda terrible like we learn differentiation a year before limits so....

[@greengreninja5725]

The indeterminate family is just a code name for things that make my calculator scream.

[@asbil3642]

I study maths with arabic and french but i don t know why that man make maths easy with that innocent smile .😘

[@Strategic.]

French sucks

[@mrstaemin7958]

WAIT IS THE MICROPHONE IN THE BEAR

[@shashwat4920]

Wow this guy is still loving the comments. Salute to you 🙋‍♂️

[@runnow2655]

here's why I think 0^0 should = 1 x^0 (x/=0) = 1 0^x (x/=0) = 0 so if any number to the 0 gives 1, we need to know where the 1 came from if we have 2^0 we're multiplying by 2 0 times, therfore nothing is happening, which means there is at base a 1 preexisting before the exponentiation happens, this works for other exponents to, for example 3^2 = 3 * 3 by definition, but assuming there's an implicit 1 already there doesn't change that value, 1 * 3 * 3 = 3 * 3 because 1 is the multiplicitive identity and therefore doesn't change anything, yet if this 1 wasn't there the first rule wouldn't make sense, because if you have no starting argument x^0 would always be undefined, it couldn't be 1 because there would be no ones put into the equation, and it couldn't be 0 because that would mean there's a 0 there, but that's not possible because 0 * 3 * 3 /= 3 * 3. Therefore, there has to be an implicit 1 not written in the equation. 0^x (x/=0) is always 0 because anything other than zero multiplies that 1 by a zero, making the equation always equal to zero; however, 0^0 would equal 1 because the preexisting 1 isn't being multiplied by any amount of zeros because there are 0 0s

[@SuperLemons-ve69o]

Let's work out some : We have 2² . It means that we are repeatedly multiplying 2 two times.. If we take 2³ , we are again multiplying 2 to 2² .. Now let's do it backwards shall we... Let's take 3⁵ which we know is 3 multiplied by itself 5 times... Taking 3⁴ is just dividing 3⁵ by 3 one time ... If we thus take a number to its zeroth power , we can say that that the number has been divided by its unit power once that is n¹ is divided by n once .. Let's take 0 per say : We know that 0 raised to any power except 0 is 0 (right..?) . So if we take 0¹ , we get 0 .. Now let's divide it by itself ... Ahh ! We get 0/0 which is undefined... Hence we simply can not define the value for 0⁰.... Hope it might have helped you out...

[@runnow2655]

@SuperLemons-ve69o but by that logic you can't find 0¹ either because dividing 0² by 0 is also undefined, but that logic doesn't work because the division pattern isn't how it's defined, it judt happens to emerge. For example, the pattern 1, 2, 4, 8, 16, you've probably seen this before but if we continue the pattern we get 32, but since this series is actually the amount of sections in a circle for each line you add you get 31 instead. You can't use patterns that happen to emerge as disproof for when that pattern doesn't work.

[@sunmichoi6888]

Thank you so much for the awesome explanation

[@blackpenredpen]

Thanks

[@tago3860]

@@blackpenredpen welcom

[@user-xb5xn6dg1x]

6:10 square root of 25 can give you 5 or -5 both work, why is it wrong?

[@matthewhunter2443]

5:50 sqrt(25)= + or - 5, meaning that 5 and -5 are solutions

[@semicolumnn]

sqrt is always positive, x²=25 is what you are looking for

[@Leo-gb1mo]

I am surprised lots of students don't know this lol...

[@semicolumnn]

@turbo Yes. The solutions to x²-25=0 is both 5 and -5. A function can only have one value. sqrt() only returns the positive root.

[@colly6022]

great video! if i'm being nitpicky, i would say the definitions could be a bit better than "for ...". for example, 1. no real value = the solution can't be expressed as a fraction; not in the set of real numbers. 2. no solution = there is no input that would satisfy the equation. 3. doesn't exist = if the solution could be a number, it is outside the given domain. 4. undefined = there is no solution achievable given the type of problem. 5. indeterminate = the solution relies on a "no answer" problem having an answer; if it is the solution to a problem, the indeterminate can only be represented by itself.

[@math_the_why_behind]

Excited to watch this video!

[@antoa.2942]

18:57 everybody happy made me lought

[@inertiasquared6667]

for 6:00 if you let x=25i^4 (25 * 1) then sqrt(x) = 5i^2 = -5, wouldn't this count as a complex solution? I know its kind of playing a technicality but I can't find any way to contradict it

[@guanglaikangyi6054]

The contradiction, I think, is that it would follow that sqrt(25) = -5, which is not true.

[@inertiasquared6667]

@@guanglaikangyi6054 Yes if we're working with real numbers. But in complex space you can avoid the contradiction by letting x = 25 * 1, sub 1 for i^4, then when you square root, you get 5 * i^2 which is 5 * -1 i.e. -5. The assertion does in fact have a complex solution.

[@jellomochas]

0^x is undefined for negative x (equivalent to 1/(0^-x) = 1/0). 0^0 is indeterminate, and when the exponent zero is a discrete value and not a limit, it is convenient to define all x^0 := 1, including 0^0 (this is used in expressions of polynomials as summations, for example).

[@neilgerace355]

1:12 If the symbol means "positive square root", then no, there is no positive square root of -9, even in complex numbers. 3i is not a positive number, as positive numbers are real numbers.

[@adb012]

Well, in means principal value. In real numbers the principal value is the positive root.

[@fgvcosmic6752]

Is 3i not a positive complex number?

[@neilgerace355]

@@fgvcosmic6752 There's no ordering of complex numbers, so we don't know which ones are greater than zero, unless the number is purely real.

[@adb012]

@@fgvcosmic6752 Nope. It is a complex number with a positive imaginary part (which, by the way, in the number 3i, or -2+3i for the sake of it, the imaginary PART is 3, the real number that goes wit the i, not 3i)

[@angelmendez-rivera351]

@@neilgerace355 There is no total ordering on the complex numbers for which complex addition and complex multiplication are isotonic binary functions, but this is irrelevant. The symbol, by definition, refers to the positive-real-part-or-positive-imaginary-part-or-zero-square root. In other words, define C+ := {z is an element of C: 0 = z or 0 < Re(z) or 0 = Re(z) and 0 < Im(z)}. Consider the function sq : C+ -> C, z |-> z·z = z^2. sq is a bijection, and therefore, there exists an inverse function sq^(-1) = sqrt. This is the function which mathematicians, by consensus, call the square root function in complex algebra, and it has codomain C+. This codomain serves as an extension of the idea of "nonnegative real numbers" to complex numbers, albeit with no total ordering. In fact, this idea is useful even outside the topic of nth roots in complex analysis.

[@AaronWGaming]

NRV was always the shorthand my teachers understood was me saying "no real value" when in school...

[@winter_c]

6:55 you should say non-negative for abs x ccan equal 0

[@senshtatulo]

The first example given for "no solution" (#2) is exactly the same kind of example as for "no real solution" (#1).

[@ff15damage86]

Everybody gangsta until complex number can't do anything anymore

[@alexh.4514]

Nah just throw in + i

[@justsomeguy5628]

Hey, it's better than when they result in periodic solutions.

[@reciprocal4286]

I am fan of your videos... And your smile when you trying to explain something interesting.. thanks.. for sharing knowledge..

[@beatoriche7301]

Technically, the square root of a complex number is a multi-valued function; whilst the real square root of 25 is 5 by definition (as the square root of a real number, if it exists, is defined to be the positive number whose square yields that number), 25 has two complex square roots, namely 5 and -5. In fact, any nonzero complex number has two distinct square roots. Also, 1/0 can obviously be defined to be whatever you like - be it 17, -3, or even a newly invented number such as ∞. In that case, 17*0 would be 1 by definition; the trouble, however, is that this is not consistent with the algebraic structure of a field, as the distributive law would yield 1 = 17(0+0) = 17*0 + 17*0 = 1 + 1 = 2. It would also mean that the multiplicative inverse of a number is not uniquely determined and all sorts of other stuff - if you're willing to make that trade-off, though, you are free to do so, as mathematicians can literally do whatever they want. Similarly, 0^0 can be defined to be 0, 1, or whatever you want, and in fact, there are contexts where 0^0 is defined to be 1 by convention; this is often done, for instance, to avoid cumbersome situations in general formulae such as the binomial theorem. The trouble is that 0^0 cannot be defined in the real or complex numbers while remaining consistent with familiar properties of limits, such as multiplicativity. This is a crucial point; as long as you're not touching limits, you're fine doing whatever you want with 0^0. In fact, you're even fine with limits as long as you formulate all your theorems about limits while excluding all 0^0 type situations. Of course, that's a lot of work, which makes it an unusual convention. It is critical to realize that definitions can mean whatever we want them to mean; the point of definitions is to capture the essence of a certain object, aid learners in understanding a given subject, and make theorems and proofs as brief as possible. You may, for instance, define 1 to be a prime number, but if you're doing number theory afterwards, it would make your theorems longer because, as 1 has very different structural properties from what we generally consider to be the primes, you would have to keep considering it as a special case and potentially exclude it. Of course, this is all just a bunch of sounds coming out of our mouths that we decide means something, and in general, you should always be able to say "wale" instead of "prime number," "eyeshadow" instead of "cardinality," and "lightbulb" instead of "angle." All of it is arbitrary, after all. This thought is brought to its logical conclusion in predicate logic, which, simply put, is a purely syntactical type of language that starts out with only very few basic symbols. One nice way to picture translating your statement into predicate logic is that you feed it to a computer, whom you have previously given a few abbreviations (e. g. A ∧ B is a shorthand for ¬(¬A ∨¬B), A → B is a shorthand for ¬(A ∧¬B), etc., where, if you haven't seen these symbols before, ∧ means "and," ∨ means "or," ¬ means "not," and → means "implies" - normally, you'd use a different type of arrow for that last one, but the typographical limitations of my device don't allow for that), and that the computer basically "unravels" all of it by substituting in what you wrote these things should stand for and makes a rather long mess out of it. Of course, no mathematician actually thinks in those terms; however, the good thing is that all of it is unambiguous and can be deciphered and even checked based on axioms and inference rules that you are to first declare as valid or invalid.

[@edgar4887]

Moivre’s theorem

[@ΘεΘεερ]

That's incredibly interesting, I never thought of mathematics as so... constructed. For me, this raises a broader question of truth within mathematics if definitions can bend around exceptions, which they essentially have to if they are to include all situations (ie 0^0). Is there any direction you could point me for more education in this area? It would be much appreciated.

[@beatoriche7301]

@@ΘεΘεερ The whole area of math philosophy deals exactly with these types of questions; a whole range of mathematicians and philosophers has given all sorts of different answers as to whether or not mathematical statements are objective and/or correspond to the real world in some way, when (if in any case at all) we can reasonably call a statement "true," and so on and so forth. Personally, I align very strongly with the ideas expressed by the English mathematician G. H. Hardy (who summarized his thoughts on the role of mathematics in society in his work _A Mathematician's Apology_) and, more recently, in Paul Lockhart's similarly named essay _A Mathematician's Lament._ If anything, I would personally call myself a mathematical hedonist (that's not like an accepted term or anything, though); I believe mathematics is a purely artistic endeavor limited in scope only by our collective imaginations and that mathematics is valuable insofar as it provides pleasure and entertainment. Basically, it's all a fiction going on in our heads, and we should do it as long as it's fun.

[@Strategic.]

that was decently interesting to read

[@popularmisconception1]

compare: sqrt(25) = x 25 = x^2 there's a slight difference between asking how much sqrt(25) is and asking what numbers multiply themselves to 25. that's why powering your equation to two is not an equivalent transformation. yes it is a matter of definition, but there is a practical reason why thing are defined the way they are. otherwise you could say a length of a hypotenuse is a negative number. so no, square root is not a multivalued function, because functions are not multivalued. but equations can have multiple solution. any time you need a multivalued function, perhaps you should rephrase your problem as an equation.

[@kfibcudwiefjw7428]

13:24 The other distinction between DNE and undefined is that undefined values are literally that: undefined. We have not defined what x/0 is. Mathematicians haven’t settled on it. DNE is defined however, namely that it simply does not exist. Sin(x) does not approach anything and therefore we define it as DNE. We don’t say does not exist for x/0, because there is no mutual agreement on that it does not exist.

[@mahdial-harafsheh2170]

Can you solve this integral : Integral of t^n/(1+t)^n dt, t from 0 to infinity .

[@linggamusroji227]

Since lim [t^n/(1+t)^n] as t-->infinite = 1 is not zero, then the integral diverges

[@yodaqwq]

You can show that a^0 = 1, a = a^(1+0) = a*a^0 then divide by a on both side and you get a^0 = 1.

[@angelmendez-rivera351]

This proof only works if a = 0 is false: you cannot divide by 0. However, it is still correct that 0^0 = 1, just not for the reasons you describe.

[@yodaqwq]

@@angelmendez-rivera351 I know this doesn't work for a=0, the point was to show that a^0=1for a≠0, also 0^0≠1 because 0^0=0^(1-1)=0^1×0^(-1)=0/0 which is undefined.

[@angelmendez-rivera351]

@@yodaqwq No, your argument does not prove 0^0 is undefined. Using 0^(1 - 1) = 0^1·0^(-1) is invalid. By that logic, 0^2 is undefined, because 0^2 = 0^[3 + (-1)] = 0^3·0^(-1), and the right hand side is undefined as well.

[@rockinroggenrola7277]

In many contexts, 0^0 is actually defined as 1 since it obeys more algebraic rules than if we were to define it as 0.

[@MrRogordo]

Not in Calculus, and that is what he teaches

[@brunolevilevi5054]

@@MrRogordo isnt it especially in calculus that it's defined to be 1? Like if you just take the taylor series of e^x = x^n/n! , if you want to know whats f(0) dont you have to assume that 0^0 is 1? Maybe assume isnt the right word, but 0^0 being equal to 1 makes more sense than like 1^(infinity) being equal to 1 or being equal to infinity

[@fgvcosmic6752]

However, 0^0=1 implies 0÷0=1 Undefined is the only answer that "works"

[@vincenzodanello4085]

@@fgvcosmic6752 why would it imply that? 0^0 means that you multiply 0, 0 times, so basically you don't do any operation. And "doing nothing" in a multiplication = 1. It's the same as 0!. When you do 0!, you don't do any operation since you don't multiply anything at all. Hence why 0! = 1. Same reasoning for the 0^0

[@MementoMoriGrizzly]

@@vincenzodanello4085 0! = 1 because there's only 1 way to arrange 0 objects.

[@s4623]

6:13 square root of 25 is both positive and negative 5 (the WHOLE point of putting ± or ∓ there). Whether either of them work depends on the context of whether it is required to be positive or negative for the rest of the problem if there is any.

[@moustachescarz]

the symbol for sqrt implies the principal root, where we take the positive value (otherwise it could not be considered a function) therefore, to reverse the process of squaring a value while maintaining logical equivalency, we use +-. in the case he shws, there is no +-, hence it is accurate to say it has no solution

[@brunolevilevi5054]

hey blackpenredpen, I'm still kinda confused, isn't 0/0 by itself indeterminate? Since if you have 0/0 = x then 0x = 0, therefore x can be any number, but if you have 1/0 just saying it is equal to a number doesnt make sense, so its undefined. Or is 0/0 only indeterminate in the context of limits?

[@pkmnfrk]

Indeterminate means that the formula, as written, does not give a sensible answer. However, as you have noted, for 0x=0, x can be all numbers. That's not a useful result, and none of the infinite number of answers can be said to be _the_ answer. Thus, undefined. (Contrast to, eg f(x) = sqrt(x) for x = 4, which also has multiple answers, +2 and -2, but they are finite and definite)

[@brunolevilevi5054]

@@pkmnfrk sqrt x only gives postive values, its a function, so f(4) is 2 and not 2 and -2. That would be the case if you were talking about y^2 = x

[@angelmendez-rivera351]

"Indeterminate" is a mathematical description that applies only to expressions containing limits. 0/0 is not indeterminate. lim f(x)/g(x) (x -> c), with lim f(x) (x -> c) = lim g(x) (x -> c) = 0, is indeterminate. 0/0 is not indeterminate. 0/0 is an abbreviation for 0·0^(-1), where 0^(-1) is the symbol representing the multiplicative inverse of 0. Since the multiplicative inverse of 0 does not exist in any of the standard mathematical structures we work with, the symbol 0/0 is just said to be "undefined," although it is well-defined if you work in a wheel.

[@ПавелХороших-в8г]

6:35. Please explain this to me. As i thought if you sqrt 25 you get +-5, isn't it?

[@Felixr2]

An easy way to look at 0^0 is by just looking at the general pattern with exponents. An exponent is in the form a^b. Every time we increase b by 1, we multiply by a, and every time we decrease b by 1, we divide by a. We also say that a^1 = a. Using this, we can determine that a^0 = a/a, so 0^0 = 0/0, which is undefined. Note that for every a =/= 0, a/a = 1, which is consistent with the definition of a^0 (and arguably is where the definition comes from).

[@zacknattack]

0^0 is an empty product, just like any other number to the 0th power. The result of an empty product is 1. The real reason 0^0 is an issue is because a^b is discontinuous at 0^0, so l'hopital's rule must be used if that is the result.

[@timelsen2236]

Using L'Hospital on lnx^x I got infinity , but directly got 1

[@gammasennin3304]

I would like to start this by saying that I absolutely love your channel and videos, you have inspired me to learn and enjoy math for years and so thank you! I do have a bit of confusion with the “no solution” part though, specifically the “sqrt(x) = -5” part as if you rework the equation as “sqrt(x) = i² • x” then square both sides you end up with “x = i⁴ •5²” if you take the fourth root of this you end up with the expression “quartic root(x) = Z = 0 + i•sqrt(5)” which resembles a complex number. I am absolutely no expert on this matter by any means so there is a very high probability that I made a few mistakes along the way, this may not even be a valid solution but I thought about it as soon as I saw the equation so if you could be so kind as to clarify this, it would mean the world to me as learning a new thing, especially from someone as talented and kind as yourself, is a graceful opportunity for me.

[@joschistep3442]

Nice idea, but you end up with the same "fake-solution". After you squared them to x = i⁴ *5², you don't have to bother taking the root, just calculate it. You'll end up with i² *i² * 25 = -1 * -1 *25= 25. As said in the video, sqrt(25) ≠ -5, therefore it's a fake-solution.

[@karryy01]

The ultimate level that finish your homework instantly: "I don't know"

[@reelbeenz]

'has no real value' hit too close to home

[@sailingteam1minecraft124]

12:07 Absolutely hilarious. Good video anyway 24:18

[@tom-on]

In my opinion, 0⁰=1. I don't know why, but i just felt like it

[@SudhasagarGollapalli]

OK...let's 0⁰=x apply log on b.s will give 0log0 base e equals logx then solve it that is equal to 1

[@iwersonsch5131]

x^-a = 1/x^a by definition (for positive a, negative a, and even a=0) => x^0 = 1/x^0 => 0^0 = 1/0^0 => 0^0 is its own multiplicative inverse => 0^0 = 1, as there is no other real number that is its own multiplicative inverse.

[@iwersonsch5131]

For nonnegative integer exponents, there is also another rule for powers: x^a = product_1:a(x). The empty product is 1 by definition, regardless of whether the factor it doesn't contain is zero.

[@joaquingallardo1728]

-1 is also its own multiplicative inverse

[@iwersonsch5131]

@@joaquingallardo1728 oh right whoops. whatever there's gonna be a reason to exclude it

[@angelmendez-rivera351]

@Alejo Sanchez The answer to your comment is given by Iwer Sonsch's reply, right above yours.

[@angelmendez-rivera351]

@@iwersonsch5131 Actually, fixing your argument is quite easy. 0^0 satisfies the equation x = 1/x AND it satisfies the equation x^2 = x. The solutions to the first equation are x = -1 or x = 1. The solutions to the second equation are x = 0 or x = 1. Only x = 1 satisfies both. Therefore, 0^0 = 1.

[@albertlau867]

undefined is often for function, when the input is not in the domain. define f(x) = 3x+1 if x is odd; x÷2 if x is even. we can see that the domain of f is integer. f(27)=82 f(82)=41 f(0.5) is undefined.

[@Nameless-qe9hu]

1/2 is an even number, and therefore defined

[@Nebula_ya]

Where would an equation like "x + 1 = x" fit?

[@blackpenredpen]

No solution.

[@thebloxxer22]

In programming, increments. In reality, see above

[@wavingbuddy5704]

Thinking abstractly: x could be infinity (that obviously isn't a soln) but if you think about it infinity + 1 = infinity

[@rhaq426]

@@wavingbuddy5704 huuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuh

[@wavingbuddy5704]

@@rhaq426 infinity doesn't increase in size when you add to it, it's infinity after all. it's not really a mathematically rigorous way of putting it as x+1 = x really doesn't have any solutions I was just being annoying tbh XD

[@J3ff_K1ng]

This video gives me another question, how computer do limits? For example the sin doesn't have a solution and other things are just see what it seems to approximate so how computers see that?

[@WindowsXP_YT]

√-x = i√x, where x is bigger than 0 and i is the square root of -1.

[@johnnolen8338]

The reason for 'undefined' is because the multiplicative inverse of zero has no definition. Therefore division by zero has no answer.

[@paulchapman8023]

In case further explanation is necessary: The multiplicative inverse property says that any number multiplied by its reciprocal (or multiplicative inverse) equals 1. The zero product property says that any number multiplied by 0 equals 0. These two properties would lead to a contradiction if the reciprocal of 0 were defined, since 1 does not equal 0. Therefore, the reciprocal of 0 must be undefined.

[@angelmendez-rivera351]

@@paulchapman8023 That logic is not actually correct. The property that 0·x = 0 for every complex number x is true for, well, only the complex numbers x. Nothing is stopping us from declaring the existence a new type of number ψ that is not a complex number, and defining it implicitly by the equation 0·ψ = 1. This does not cause any contradictions: the claim 0·x = 0 would still be true for every complex number x, since ψ is not a complex number. There is no reason to a priori demand that 0·ψ = 0 also be true, except for unreasonable stubborness. The problem is that doing this creates a structure in which multiplication no longer distributes over addition and it is no longer associative, and in addition, ψ would have no additive inverse in this structure, hence only pushing back the problem we wanted to solve. So it is not a very appealing solution, and so mathematicians have decided to not use this approach. Working with a field is much better, so it is perfectly fine to not actually try to invent the multiplicative inverse of 0.

[@tzonic8655]

No real value: *exists* Complex numbers: let me introduce myself

[@SuryaBudimansyah]

6. "It depends"

[@angelmendez-rivera351]

"It depends" is really the only correct answer here, and it encapsulates all 5 of the points in the video.

[@guidomazzeo3767]

I would love to have the indeterminate family t-shirt jajajaja

[@eolapade]

In those cases 1, 2, and 3, I think it makes sense to say that a solution of the equation f(x) = 0 does not exist in the real numbers or that the limit of a function as x approaches some number does not exist as a real number or that the value of a function evaluated x does not exist in the real numbers (perhaps because the function is not defined at x). For cases 4 and 5, the fact that a function is not defined at x does not mean that the function cannot be defined at x. An example is the reciprocal function a/x for some fixed real number a. There are some applications where you can define a/0 to be 0. While there may be some ambiguity in defining a/x at 0, we should not interpret "undefined" and "indeterminate" as "cannot be defined" and cannot be determined, respectively.

[@Nailgy]

I want a gf that would be looking at me as this guy looks a 0^0 (14:20)

[@DavidRichfield]

Would she also say "it's kinda weird"?

[@marbangens]

good luck!

[@bugsyms]

For the indeterminate section, when you get a limit that evaluates to 0/0, you can just apply L’Hopital’s rule and get the same answer

[@liab-qc5sk]

11:42 :the answer is in wheel theory YAY teddy!!!

[@kenrickchung8176]

So you would say, no complex value or something like that 🤣

[@liab-qc5sk]

@Lakshya Gadhwal read about en.wikipedia.org/wiki/Wheel_theory

[@liab-qc5sk]

@Lakshya Gadhwal sorry best thing that i can to you is reading about algebric structures like groups and rings than maybe you will understand better

[@angelmendez-rivera351]

@Lakshya Gadhwal The value of 8/0 in the complex wheel is equal to /0. There is no simpler way of expressing this number using other complex numbers, because /0 is not a complex number: it is its own number in the wheel... much like how i = sqrt(-1) is its own number in the complex numbers, not more simply expressible using real numbers alone.

[@BryndanMeyerholtTheRealDeal]

For the no real value, I usually use a+bi

[@carolinamarcmar]

Teddy is adorable 💖

[@saumitrachakravarty]

24:49 Why did you cross out zeroes from both numerator and denominator? Shouldn't we use L'Hôpital's rule instead for limits with indeterminate forms?

[@michaelroditis1952]

Can somebody tell me a function in which the limit 1^(inf) will differ from 1? Edit: (1+1/n)^n -> e

[@paulchapman8023]

(2^(1/x))^x -> 2 The x-root of 2 (or any greater-than-zero constant, for that matter) approaches 1 as x approaches infinity, but if you raise it to the x power, it cancels out the root and leaves you with the constant.

[@Ferraco05]

If your function is f(x) = 1^x, then the limit of f(x) as x approaches infinity is 1. But if your function f(x) approaches 1^x, for example, f(x) = (1+1/x)^x, then the limit may very well be different from 1.

[@popularmisconception1]

slightly above 1 and (1+eps)^inf is infinite. slightly below 1 and (1-eps)^inf is zero. infinitesimally close to 1 and (sth approaching 1)^(sth approaching inf) can be anywhere from 0 to infinity, because you can more or less think of a^b as continuous even if b=inf and thus a=1 can be any spot where you can connect the resulting infinity if a>1 to zero if a

[@munko_koni]

10:05 guys my brother didn't said why he wrote open circle on Y line. Because it means x=0 and in sinx/x, x cant be 0.

[@Mothuzad]

"DNE" is a negation of a quantifier in logic, whereas "undefined" refers to any operation which is given an argument outside its domain. This is consistent with what he says in the video, but more general.

[@angelmendez-rivera351]

Eh... yes, but really, no. "Undefined" is not actually a word mathematicians have ever really used in their publications. "Undefined" is a buzzword that was basically invented by mathematics teachers and that only really has meaning inside the classroom, not in mathematics in general. What it means is that the answer to the problem in question cannot be given in the specific setting being worked on, for one reason or another. "Undefined" has no meaning outside of the classroom, and as I said, you will never see a mathematician talk about this in a publication, because it not actually a mathematical idea, it is just a tool for teaching.

[@Mothuzad]

@@angelmendez-rivera351 "Undefined" is a common term in academics within the realm of computer science, especially dealing with language specifications. That's just a fun fact, not directly relevant to your reply. In mathematics, the concept of "undefined" still exists for professional mathematicians, I'm sure, but everyone at that level of expertise already knows not to use operands outside the domains of the functions they're using. It's like a competent adult already knowing to look both ways before crossing the street. It's too juvenile to be worth mentioning. But of course, the classroom is where they teach that lesson in the first place.

[@Pokefan462]

6:00 I thought this was exactly why you had to put the plus or minus when solving a square root in an equation? Positive five is a false answer, but negative five is correct? Am I misremembering?