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

  Рет қаралды 411,808

blackpenredpen

blackpenredpen

Күн бұрын

Here are 5 levels of “no answers” in math: Undefined, no solution, no real value, doesn’t exist, and indeterminate. When should we use what?
0:00 Teddy Says Hello
0:30 No Real Value
2:39 No Solution
7:45 Does NOT exist
10:43 Undefined
16:12 Indeterminate
**This video was recorded in May, 2020"
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Пікірлер: 1 200
@richardcao7390
@richardcao7390 3 жыл бұрын
inconsistent
@blackpenredpen
@blackpenredpen 3 жыл бұрын
Oh man, you are right! I missed that one. That one is for systems of equations!!
@richardcao7390
@richardcao7390 3 жыл бұрын
@@blackpenredpen yep! Doing matrix for preCalc right now and instantly drew the connection to this video haha
@justinchen9071
@justinchen9071 3 жыл бұрын
what's that
@ultrio325
@ultrio325 3 жыл бұрын
how do you get inconsistent
@nanamacapagal8342
@nanamacapagal8342 3 жыл бұрын
@@ultrio325 for example, x+y=2, 2x+2y=3. The system has no solution for x and y, so our system is inconsistent.
@JaydentheMathGuy
@JaydentheMathGuy 3 жыл бұрын
Alternative title: How to tell your math teacher "no".
@blackpenredpen
@blackpenredpen 3 жыл бұрын
😆 definitely better
@tomatrix7525
@tomatrix7525 3 жыл бұрын
Lol
@m9l0m6nmelkior7
@m9l0m6nmelkior7 3 жыл бұрын
@Tangent of circle. xD
@sammitramanan6404
@sammitramanan6404 3 жыл бұрын
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
@austinlincoln3414 2 жыл бұрын
Lol
@CardThrower-rb6eg
@CardThrower-rb6eg 3 жыл бұрын
You forgot the final level "I don't know how to solve this"
@fisch37
@fisch37 3 жыл бұрын
Well, that applies in any case where you forgot to study before an exam
@shivamchouhan5077
@shivamchouhan5077 3 жыл бұрын
@@fisch37 😂 lol 😂
@RazorM97
@RazorM97 3 жыл бұрын
"left as an exercise for the reader"
@AgaresOaks
@AgaresOaks 3 жыл бұрын
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
@orlandomoreno6168 3 жыл бұрын
@@AgaresOaks independence of a system of axioms doesn't mean that
@DialecticRed
@DialecticRed Жыл бұрын
“No real value” also happens to be the official mathematical classification for NFTs
@blocks4857
@blocks4857 Жыл бұрын
Value is subjective
@cewla3348
@cewla3348 Жыл бұрын
@@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
@rieldebonk1044 Жыл бұрын
@@cewla3348 You dont do THAT MAN!
@Bobspineable
@Bobspineable Жыл бұрын
@@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
@patrickliberatoalves2931 10 ай бұрын
The value of NFT is imaginary
@Adomas_B
@Adomas_B 3 жыл бұрын
Me on my math exam: The answer is left as an exercise to the reader
@akshatvats7992
@akshatvats7992 3 жыл бұрын
xD
@YourLocalCafe
@YourLocalCafe 3 жыл бұрын
@Lakshya Gadhwal learner*
@idrisShiningTimes
@idrisShiningTimes 2 жыл бұрын
Gives me Reimann Zeta Function vibes
@navaneethshettyb9312
@navaneethshettyb9312 Жыл бұрын
😂😂
@stratonikisporcia8630
@stratonikisporcia8630 Жыл бұрын
@@idrisShiningTimes "I know how to solve this, but I'll only tell you if you give me $500,000"
@somerandomsheep
@somerandomsheep 3 жыл бұрын
when your parents ask if you are lying you can just tell them "it's a complex statement"
@jxdinglol
@jxdinglol 2 жыл бұрын
"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
@saperoi 2 жыл бұрын
Also as contradiction where x is 6 but x has to simultaneously be 9
@arvin390
@arvin390 Жыл бұрын
Yeah, stuff like x = x + 1
@manioqqqq
@manioqqqq Жыл бұрын
then, x=∅ and thus y=∅ ect.
@rynabuns
@rynabuns Жыл бұрын
how about non-Euclidian geometry?
@ThomasTheThermonuclearBomb
@ThomasTheThermonuclearBomb Жыл бұрын
"No solution" being used there is incorrect, it should be "inconsistent system"
@Th3_Animat0r
@Th3_Animat0r Жыл бұрын
This guy: (flawlessly explains all the ways an equation can have no answer) My calculator: "NaN"
@saikitonia
@saikitonia 6 ай бұрын
👵
@Mg_887
@Mg_887 6 ай бұрын
NaN stands for "Not A Number" in js and in ts
@necrolord1920
@necrolord1920 3 жыл бұрын
6:51 technically with the definition, the output is always NON NEGATIVE. An absolute value could be 0 :)
@tperm6695
@tperm6695 3 жыл бұрын
Yup, when x=0 absolute value of x is still 0 :)
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
I agree.
@SlidellRobotics
@SlidellRobotics 3 жыл бұрын
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
@Shreyas_Jaiswal 3 жыл бұрын
Yes this is how it is defined. My teacher has also taught me this process.
@user-ot4rp8yn8r
@user-ot4rp8yn8r 3 жыл бұрын
I always wonder why 2^0 is a definition
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
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
@vns1956 2 жыл бұрын
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
@tbg-brawlstars 2 жыл бұрын
Right!!!!
@kevinleugan6037
@kevinleugan6037 3 жыл бұрын
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
@austinlincoln3414 2 жыл бұрын
Lol
@danielgammo
@danielgammo 2 жыл бұрын
too bad its math class
@jay-5061
@jay-5061 2 жыл бұрын
Nani
@za1231in
@za1231in 2 жыл бұрын
luckily you stumbled onto something much more useful
@xenosmoke8915
@xenosmoke8915 2 жыл бұрын
It is that lesson, just for tutors not students 🤣
@mathopediamathexplorer3010
@mathopediamathexplorer3010 3 жыл бұрын
I love that doll its like Mr . Bean's doll ...
@SpiderMan-in9rl
@SpiderMan-in9rl 3 жыл бұрын
Yeah.. so cute Edit - That doll gives me Nostalgia
@user-wy8ki2ef1m
@user-wy8ki2ef1m 3 жыл бұрын
Teddy! His name is Teddy!
@SpiderMan-in9rl
@SpiderMan-in9rl 3 жыл бұрын
@@user-wy8ki2ef1m Yup.. now i remember
@artandcraftwithmrudhula9206
@artandcraftwithmrudhula9206 2 жыл бұрын
🤣🤣🤣
@Uni-Coder
@Uni-Coder 3 жыл бұрын
The sixth level of 'no answer' is when you are trying to answer the question "is there a sixth level of 'no answer'."
@kazedcat
@kazedcat 2 жыл бұрын
There is a sixth level "undecidable" this is when your axioms are not enough to prove if a statement is true or false.
@3Jeroen3
@3Jeroen3 2 жыл бұрын
Ah, yes example of this is the arithmoquine function in Gödels proof
@SG2048-meta
@SG2048-meta 6 ай бұрын
@@IonRuby what, Gödel did make proofs
@israelkoiku2076
@israelkoiku2076 3 жыл бұрын
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
@captainpolar2343 Жыл бұрын
nerd alert
@coolskeletondude5902
@coolskeletondude5902 Жыл бұрын
@@captainpolar2343 bro you're watching a math video,dont you think people would be talking about math what a baby
@hcmishra6371
@hcmishra6371 Жыл бұрын
@@captainpolar2343 said the fool to the person commenting about math in a math video
@dogmania2892
@dogmania2892 Жыл бұрын
@@captainpolar2343 mf we are watching math vid stfu
@brightblackhole2442
@brightblackhole2442 Жыл бұрын
@@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"
@chessandmathguy
@chessandmathguy 3 жыл бұрын
3:56 how about the most basic of situations with no solution... Solve x+1=x+2 lol
@nanamacapagal8342
@nanamacapagal8342 3 жыл бұрын
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
@mathsman5219 3 жыл бұрын
X=0 / 0
@paolo6219
@paolo6219 3 жыл бұрын
X+5=x
@_judge_me_not
@_judge_me_not 3 жыл бұрын
x=∞
@aasid2446
@aasid2446 3 жыл бұрын
0 = 1
@aloysiuskurnia7643
@aloysiuskurnia7643 2 жыл бұрын
"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.
@markovermeer1394
@markovermeer1394 2 жыл бұрын
In computing science, we also have "I won't tell you" (no permission or not requested), typically a NULL value.
@gdclemo
@gdclemo 2 жыл бұрын
there is also 𝈜 (upside-down T if Unicode doesn't work) which means that this program never halts.
@technoultimategaming2999
@technoultimategaming2999 2 жыл бұрын
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
@markovermeer1394 2 жыл бұрын
@@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
@dojelnotmyrealname4018 6 ай бұрын
Null usually means "does not exist" which is "no solution" tho.
@hongkongmapping
@hongkongmapping 3 жыл бұрын
just realised the indeterminate family is on your shirt lmao
@JayTemple
@JayTemple 2 жыл бұрын
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
@pablopereyra7126 2 жыл бұрын
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
@JayTemple 2 жыл бұрын
@@pablopereyra7126 Depending on how the substances interact, they might actually yield 30 liters, they might only yield 25 liters, or they could explode.
@mal2ksc
@mal2ksc Жыл бұрын
@@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
@NoActuallyGo-KCUF-Yourself Жыл бұрын
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
@ThomasTheThermonuclearBomb Жыл бұрын
@@pablopereyra7126 Basically, chemistry makes things weird
@damianbla4469
@damianbla4469 3 жыл бұрын
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
@blackcat5771 3 жыл бұрын
so cute
@shrankai7285
@shrankai7285 2 жыл бұрын
Wouldn’t that be no real solution, as we are solving an equation?
@dojelnotmyrealname4018
@dojelnotmyrealname4018 6 ай бұрын
The solution exists, it's just not a real value. @@shrankai7285
@countcrocodile1115
@countcrocodile1115 2 жыл бұрын
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
@manioqqqq Жыл бұрын
What about his statement that √25≠-5? √25=±5
@alexh.4514
@alexh.4514 8 ай бұрын
Plenty of math teachers on YT say the same things as Bprp, but his absolute joy is what makes him such an effective teacher.
@user-vn7ce5ig1z
@user-vn7ce5ig1z 3 жыл бұрын
A simple example for No Solution is "x + 1 = x + 2" It almost looks trivially solvable but obviously isn't, regardless of the system.
@radupopescu9977
@radupopescu9977 2 жыл бұрын
No solution in real and complex numbers!!! BUT, in ordinals numbers, there is another issue.
@thelaststraw1467
@thelaststraw1467 10 ай бұрын
if you REALLY wanted to couldnt you sub in infinity? of course its not a number tho...
@bloomingon6141
@bloomingon6141 7 ай бұрын
@@thelaststraw1467infinity has no value
@thelaststraw1467
@thelaststraw1467 7 ай бұрын
@@bloomingon6141so? how does that imply its not a solution coz it def is
@abigailc8829
@abigailc8829 6 ай бұрын
​@@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).
@izjemmr
@izjemmr 2 жыл бұрын
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
@NoActuallyGo-KCUF-Yourself Жыл бұрын
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
@indigopari 6 ай бұрын
⁠@@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
@johnpaulhumphrey2981
@johnpaulhumphrey2981 Жыл бұрын
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
@SlipperyTeeth
@SlipperyTeeth 3 жыл бұрын
I think indeterminate is also used to refer to certain cases of convergence tests for integrals.
@johnpaulhumphrey2981
@johnpaulhumphrey2981 Жыл бұрын
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.
@scathiebaby
@scathiebaby Жыл бұрын
0⁰ : Could be 1 or 0, but exponent is stronger, so the 1 wins, therefore 0⁰ = 1
@chyawanprash
@chyawanprash 3 жыл бұрын
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
@math_the_why_behind 3 жыл бұрын
Right! The sqrt(x) could also be 0 though :)
@dioniziomorais8138
@dioniziomorais8138 3 жыл бұрын
Correct, but 'twas just examples.
@math_the_why_behind
@math_the_why_behind 3 жыл бұрын
@@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
@dioniziomorais8138 3 жыл бұрын
@Math: The Why Behind ok, I don't have a great understanding in math, I'm not even an native english speaker lol
@beatoriche7301
@beatoriche7301 3 жыл бұрын
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.
@beatoriche7301
@beatoriche7301 3 жыл бұрын
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
@edgar4887 3 жыл бұрын
Moivre’s theorem
@user-hb6ro7ic2t
@user-hb6ro7ic2t 3 жыл бұрын
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
@beatoriche7301 3 жыл бұрын
@@user-hb6ro7ic2t 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.
@Strategic. 3 жыл бұрын
that was decently interesting to read
@popularmisconception1
@popularmisconception1 2 жыл бұрын
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.
@timothyconnally2167
@timothyconnally2167 2 жыл бұрын
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
@chx1618 Жыл бұрын
FWIW the inventor of null called it his "billion-dollar mistake".
@justsomeguy5628
@justsomeguy5628 6 ай бұрын
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
@colly6022 6 ай бұрын
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..
@pancito3108
@pancito3108 Жыл бұрын
6:25 √(x²) = |x| This would result in |x| = 25 So x = ±5 And ±5 has -5, so there you have your solution
@peteradler6005
@peteradler6005 Жыл бұрын
Sqrt ( x) >=0 by def
@pancito3108
@pancito3108 Жыл бұрын
@@peteradler6005 I never said otherwise
@ostepolsegudensprofet
@ostepolsegudensprofet 6 ай бұрын
​@@peteradler6005 okay but why though. (-5)^2=25 so why not (25)^1/2 = -5
@WingMyWay
@WingMyWay 5 ай бұрын
​@@peteradler6005 Its okay to veer off "but its defined" and use math to solve problems instead of jerk off about made up rigor
@xinpingdonohoe3978
@xinpingdonohoe3978 4 ай бұрын
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
@Nebula_ya
@Nebula_ya 3 жыл бұрын
Where would an equation like "x + 1 = x" fit?
@blackpenredpen
@blackpenredpen 3 жыл бұрын
No solution.
@thebloxxer22
@thebloxxer22 3 жыл бұрын
In programming, increments. In reality, see above
@wavingbuddy5704
@wavingbuddy5704 3 жыл бұрын
Thinking abstractly: x could be infinity (that obviously isn't a soln) but if you think about it infinity + 1 = infinity
@rhaq426
@rhaq426 3 жыл бұрын
@@wavingbuddy5704 huuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuh
@wavingbuddy5704
@wavingbuddy5704 3 жыл бұрын
@@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
@b4ttlemast0r
@b4ttlemast0r 2 жыл бұрын
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
@Pacvalham
@Pacvalham 3 жыл бұрын
I think 0^0 should be 1 because the exponent (0) says that you do not multiply the base (0), so you are not multiplying by 0.
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
Exactly
@reciprocal4286
@reciprocal4286 2 жыл бұрын
I am fan of your videos... And your smile when you trying to explain something interesting.. thanks.. for sharing knowledge..
@rockinroggenrola7277
@rockinroggenrola7277 3 жыл бұрын
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
@MrRogordo 3 жыл бұрын
Not in Calculus, and that is what he teaches
@brunolevilevi5054
@brunolevilevi5054 3 жыл бұрын
@@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
@fgvcosmic6752 3 жыл бұрын
However, 0^0=1 implies 0÷0=1 Undefined is the only answer that "works"
@vincenzodanello4085
@vincenzodanello4085 3 жыл бұрын
@@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
@MementoMoriGrizzly 3 жыл бұрын
@@vincenzodanello4085 0! = 1 because there's only 1 way to arrange 0 objects.
@shashwat4920
@shashwat4920 3 жыл бұрын
Wow this guy is still loving the comments. Salute to you 🙋‍♂️
@math_the_why_behind
@math_the_why_behind 3 жыл бұрын
Excited to watch this video!
@TheMrFrukt
@TheMrFrukt 3 жыл бұрын
Very nice explanation! As always :D
@Inspirator_AG112
@Inspirator_AG112 Жыл бұрын
*[**14:22**]:* Misconception here. The equation b⁰ collapses to an empty product when considering repeated multiplication, which is 1 by definition. This is also applicable to 0⁰ = 1.
@peteradler6005
@peteradler6005 Жыл бұрын
Misconception here. If you assume 0^0=1 then b^0=empty product. So your argument is circular.
@sunmichoi6888
@sunmichoi6888 3 жыл бұрын
Thank you so much for the awesome explanation
@blackpenredpen
@blackpenredpen 3 жыл бұрын
Thanks
@tago3860
@tago3860 2 жыл бұрын
@@blackpenredpen welcom
@damianbla4469
@damianbla4469 3 жыл бұрын
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
@damianbla4469 3 жыл бұрын
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).
@radupopescu9977
@radupopescu9977 2 жыл бұрын
@@damianbla4469 You are right, so it has 2 real solution, and 2 complex one's. I didn't think of the complex one's. Sqrt(25) = -5 and Sqrt(25) = 5. In fact any square root of positive something has 2 values, except for 0.
@BryndanMeyerholtTheRealDeal
@BryndanMeyerholtTheRealDeal Жыл бұрын
For the no real value, I usually use a+bi
@wildfire_
@wildfire_ 6 ай бұрын
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
@bruhifysbackup 26 күн бұрын
so you're saying 0^2 is undefined?
@winter_c
@winter_c 3 жыл бұрын
6:55 you should say non-negative for abs x ccan equal 0
@ff15damage86
@ff15damage86 3 жыл бұрын
Everybody gangsta until complex number can't do anything anymore
@alexh.4514
@alexh.4514 8 ай бұрын
Nah just throw in + i
@justsomeguy5628
@justsomeguy5628 6 ай бұрын
Hey, it's better than when they result in periodic solutions.
@MattMorgasmo
@MattMorgasmo Жыл бұрын
I love your videos. Both educational and charmingly presented! Btw.: When you said "hello" at the very beginning you sounded 100% German.
@apia46
@apia46 Жыл бұрын
"it has no real value" hey look thats me
@LeeAnne750
@LeeAnne750 3 жыл бұрын
So, for computations 0^0 is undefined. For limits, 0^0 is indeterminate .
@huhneat1076
@huhneat1076 2 жыл бұрын
The main difference between "undefined" and "does not exist" is that anything that "does not exist" still has a definition. The lim(x→∞) sinx is defined, it's [insert definition of limit] for sinx when x approaches infinity, but when you attempt to compute it, it happens that no value can be the answer.
@omp199
@omp199 2 жыл бұрын
I don't think that is the distinction. 6 / 0 is "defined" in the same sense that the above limit is "defined". It is defined as the unique number x such that 0x = 6. It just so happens that there is no such number x.
@AwesomepianoTURTLES
@AwesomepianoTURTLES 2 жыл бұрын
@@omp199 But the expression 6/0 has no definition. Sin(x) has a definition and if you evaluate the limit quantitatively you will get numbers back as your x increases since sin is defined across the reals. There’s just no answer to the limit itself because it never converges to one number, therefore it doesn’t exist. The question itself is defined very well, while 6/0 doesn’t even mean anything. Asking how many times does 0 go into 6 is nonsensical, but asking if the y value on a unit circle converges to a single number as your angle increases indefinitely makes a lot of sense but has no answer
@omp199
@omp199 2 жыл бұрын
@@AwesomepianoTURTLES I can define 6 / 0 as the unique number a such that 0a = 6. I can define the limit of sin(x) as x tends to infinity as the unique number b such that for any ε greater than 0, there exists a number k such that for all x > k, the absolute value of sin(x) - b is less than ε. There. I have given definitions for both. It just so happens that there is no number a that satisfies the first definition, and no number b that satisfies the second definition. So what's the difference?
@NirateGoel
@NirateGoel 2 жыл бұрын
@@omp199 0a=6 would define 6 as equaling 0 though. That's not a definition.
@omp199
@omp199 2 жыл бұрын
@@NirateGoel No, it wouldn't. I didn't give a definition of the number 6. I defined the _expression_ "6 / 0" as the unique number a such that 0a = 6. That is not a definition of the number 6. It is a definition of the _expression_ "6 / 0". As it happens, there is no number a that satisfies that definition, just as there is no number b that satisfies the definition of the limit of sin(x) as x tends to infinity that I gave in my comment above.
@gammasennin3304
@gammasennin3304 Жыл бұрын
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
@joschistep3442 Жыл бұрын
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.
@flopsnail4750
@flopsnail4750 Жыл бұрын
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
@carultch Жыл бұрын
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.
@Mothuzad
@Mothuzad 3 жыл бұрын
"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
@angelmendez-rivera351 3 жыл бұрын
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
@Mothuzad 3 жыл бұрын
@@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.
@asbil3642
@asbil3642 3 жыл бұрын
I study maths with arabic and french but i don t know why that man make maths easy with that innocent smile .😘
@Strategic.
@Strategic. 3 жыл бұрын
French sucks
@TheWifeHaver436
@TheWifeHaver436 6 ай бұрын
0^0 = 1 Because You can always put a "1 •" before something and it doesnt change For example with 2 1 • 2^2 = 1 • 2 • 2 =4 1 • 2^1 = 1 • 2 = 0 1 • 2^1 = 1 As you can see its always a "• 2" less So with 0 its this 1 • 0^2 = 1 • 0 • 0 = 0 1 • 0^1 = 1 • 0 = 0 1 • 0^0 = 1
@eydum3948
@eydum3948 6 ай бұрын
Im only 1min and 8 sec in and those whitebord skills are slick
@HedinnBjornsson
@HedinnBjornsson 2 жыл бұрын
Another kind of “no solution” equation could be 2x+1=2x+5
@P4R5
@P4R5 Жыл бұрын
You've basically written down x=x+1 in a camouflaged way
@joschistep3442
@joschistep3442 Жыл бұрын
Actually, you've written 1 = 5, and thats a wrong statement.
@corruptconverter2616
@corruptconverter2616 Жыл бұрын
0=±4, of course
@piadas804
@piadas804 6 ай бұрын
0⁰ is 1; it's literally defined.
@QuirkyView
@QuirkyView Жыл бұрын
That's the first time I've heard long division explained like that. I always thought of it as "How many times does the outside number fit into the inside number?" So when dividing by 0 the answer would be infinity, which is not a number and so calculators could and should say NaN. The way you described division changes that. Though I'd honestly say "no solution" rather than undefined still.
@avyakthaachar2.718
@avyakthaachar2.718 Жыл бұрын
Awesome video. Very helpful ❤👍
@xenosmoke8915
@xenosmoke8915 2 жыл бұрын
Imagine walking into this class without knowing he’s holding a mic.
@alexh.4514
@alexh.4514 8 ай бұрын
I seriously thought it was just a cute prop XD
@inertiasquared6667
@inertiasquared6667 2 жыл бұрын
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
@guanglaikangyi6054 Жыл бұрын
The contradiction, I think, is that it would follow that sqrt(25) = -5, which is not true.
@inertiasquared6667
@inertiasquared6667 Жыл бұрын
@@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
@jellomochas Жыл бұрын
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).
@matthewhunter2443
@matthewhunter2443 2 жыл бұрын
5:50 sqrt(25)= + or - 5, meaning that 5 and -5 are solutions
@semicolumnn
@semicolumnn 2 жыл бұрын
sqrt is always positive, x²=25 is what you are looking for
@Leo-gb1mo
@Leo-gb1mo 2 жыл бұрын
I am surprised lots of students don't know this lol...
@semicolumnn
@semicolumnn 2 жыл бұрын
@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.
@tzonic8655
@tzonic8655 4 жыл бұрын
No real value: *exists* Complex numbers: let me introduce myself
@shivaudaiyar2556
@shivaudaiyar2556 3 жыл бұрын
Thanks for such a great content with love from India
@HN-vu8pp
@HN-vu8pp 3 жыл бұрын
Didn’t you learn all this at age of two?
@RishaadKhan
@RishaadKhan 3 жыл бұрын
@@HN-vu8pp this is so unfunny its funny
@yashkrishnatery9082
@yashkrishnatery9082 3 жыл бұрын
@@HN-vu8pp well, we did but revision is necessary. 😂😂
@thephoenix8728
@thephoenix8728 3 жыл бұрын
@@HN-vu8pp its complicated bro the teaching pattern here is kinda terrible like we learn differentiation a year before limits so....
@senshtatulo
@senshtatulo 6 ай бұрын
The first example given for "no solution" (#2) is exactly the same kind of example as for "no real solution" (#1).
@abdulahshahzad1732
@abdulahshahzad1732 Жыл бұрын
When you were talking about the non existent you gave the example of an equation including -5. Now square root of 25 is 5 but it is also -5, solution of square root of 25 is +-5. So square root of 25 gives the true result -5 and an extra result +5 so solution exists
@manaarikicarpentier6038
@manaarikicarpentier6038 3 жыл бұрын
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
@angelmendez-rivera351 3 жыл бұрын
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.
@radupopescu9977
@radupopescu9977 2 жыл бұрын
@@angelmendez-rivera351 Sqrt(25) = -5. is true! Because Sqrt(any number except 0)=+/-(other number). Sqrt(25) = -5 and Sqrt(25) = 5. In practice we neglect negative value. There are also complex values. Remember nth root of a number, except 0, has n values.
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
@@radupopescu9977 No, you are wrong. That is not how roots work.
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
@@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.
@radupopescu9977
@radupopescu9977 2 жыл бұрын
@@angelmendez-rivera351 So all my math professors were idiots... Nice...
@ArchimedesBC
@ArchimedesBC 3 жыл бұрын
In algebra and combinatorics, it's generally agreed that the value of x to the zero power = 1. It is in calculus and mathematical analysis that the expression is left undefined. The polynomial "x to the zero power" is the identity element of a polynomial ring. Defining x to the zero power = 1 is necessary for many polynomial identities like the binomial theorem.
@antoa.2942
@antoa.2942 2 жыл бұрын
18:57 everybody happy made me lought
@GayAnnabeth
@GayAnnabeth 3 жыл бұрын
one of my favourite examples with the 1^infinity I.F. is when you have the limit of 1 to some function f(x), where f(x) is approaching inifinity, because to deal with it you change from lim 1^f(x) to 1^lim f(x)
@radupopescu9977
@radupopescu9977 2 жыл бұрын
1^infinity, can be any value, or no value, and it is case by case. (-1)^infinity is also interesting....
@GayAnnabeth
@GayAnnabeth 2 жыл бұрын
@@radupopescu9977 well in the case of just a normal 1, the result should be one it's when we approach 1 that things get wonky
@neilgerace355
@neilgerace355 3 жыл бұрын
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
@adb012 3 жыл бұрын
Well, in means principal value. In real numbers the principal value is the positive root.
@fgvcosmic6752
@fgvcosmic6752 3 жыл бұрын
Is 3i not a positive complex number?
@neilgerace355
@neilgerace355 3 жыл бұрын
@@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
@adb012 3 жыл бұрын
@@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
@angelmendez-rivera351 3 жыл бұрын
@@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.
@iwersonsch5131
@iwersonsch5131 3 жыл бұрын
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
@iwersonsch5131 3 жыл бұрын
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
@joaquingallardo1728 3 жыл бұрын
-1 is also its own multiplicative inverse
@iwersonsch5131
@iwersonsch5131 3 жыл бұрын
@@joaquingallardo1728 oh right whoops. whatever there's gonna be a reason to exclude it
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
@Alejo Sanchez The answer to your comment is given by Iwer Sonsch's reply, right above yours.
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
@@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.
@randomname9291
@randomname9291 Жыл бұрын
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
@AaronWGaming
@AaronWGaming Жыл бұрын
NRV was always the shorthand my teachers understood was me saying "no real value" when in school...
@mahdial-harafsheh2170
@mahdial-harafsheh2170 3 жыл бұрын
Can you solve this integral : Integral of t^n/(1+t)^n dt, t from 0 to infinity .
@guitar_jero
@guitar_jero 3 жыл бұрын
Does not converge. You could try t/(1+t)^n which is 1/(n-1)(n-2) for n greater than 2
@linggamusroji227
@linggamusroji227 3 жыл бұрын
Since lim [t^n/(1+t)^n] as t-->infinite = 1 is not zero, then the integral diverges
@brunolevilevi5054
@brunolevilevi5054 3 жыл бұрын
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
@pkmnfrk 3 жыл бұрын
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
@brunolevilevi5054 3 жыл бұрын
@@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
@angelmendez-rivera351 3 жыл бұрын
"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.
@bugsyms
@bugsyms 5 ай бұрын
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
@albertlau867
@albertlau867 3 жыл бұрын
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
@Nameless-qe9hu Жыл бұрын
1/2 is an even number, and therefore defined
@guidomazzeo3767
@guidomazzeo3767 3 жыл бұрын
I would love to have the indeterminate family t-shirt jajajaja
@sailingteam1minecraft124
@sailingteam1minecraft124 3 жыл бұрын
12:07 Absolutely hilarious. Good video anyway 24:18
@baacloud
@baacloud Жыл бұрын
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
@bienvenidos9360 Жыл бұрын
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??
@J3ff_K1ng
@J3ff_K1ng Жыл бұрын
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?
@octavylon9008
@octavylon9008 3 жыл бұрын
Why sqrt(x) = -5 doesnt work ? sqrt(25) = ± 5 , not just +5
@semicolumnn
@semicolumnn 2 жыл бұрын
i think the square root always gives positive results, and if you want plus or minus 5 you should have the equation x²=25
@carolinamarcmar
@carolinamarcmar 3 жыл бұрын
Teddy is adorable 💖
@teslainvestah5003
@teslainvestah5003 2 жыл бұрын
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
@ferln4 2 жыл бұрын
how to make mathematicians mald in 4 sentences
@carultch
@carultch Жыл бұрын
Lambert, is that you?
@thebloxxer22
@thebloxxer22 3 жыл бұрын
Now thinking about it, DIV0 is a way to force a remainder, as from long division, if the last subtraction results in a non-zero positive integer, that integer becomes the remainder. Possible exploitation from NULL Remainder Numerator?
@liab-qc5sk
@liab-qc5sk 3 жыл бұрын
11:42 :the answer is in wheel theory YAY teddy!!!
@kenrickchung8176
@kenrickchung8176 3 жыл бұрын
So you would say, no complex value or something like that 🤣
@liab-qc5sk
@liab-qc5sk 3 жыл бұрын
@Lakshya Gadhwal read about en.wikipedia.org/wiki/Wheel_theory
@liab-qc5sk
@liab-qc5sk 3 жыл бұрын
@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
@angelmendez-rivera351 3 жыл бұрын
@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.
@Felixr2
@Felixr2 3 жыл бұрын
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
@zacknattack Жыл бұрын
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
@timelsen2236 9 ай бұрын
Using L'Hospital on lnx^x I got infinity , but directly got 1
@johnnolen8338
@johnnolen8338 3 жыл бұрын
The reason for 'undefined' is because the multiplicative inverse of zero has no definition. Therefore division by zero has no answer.
@paulchapman8023
@paulchapman8023 3 жыл бұрын
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
@angelmendez-rivera351 3 жыл бұрын
@@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.
@robloxplayer0003
@robloxplayer0003 2 жыл бұрын
When you thought it was going to be a meme video converting real life to math equations but it's actually a helpful video of math terms.
@karryy01
@karryy01 3 жыл бұрын
The ultimate level that finish your homework instantly: "I don't know"
@WindowsXP_YT
@WindowsXP_YT 3 жыл бұрын
√-x = i√x, where x is bigger than 0 and i is the square root of -1.
@chaincat33
@chaincat33 6 ай бұрын
There's a bit more rigor for why dividing by 0 is undefined instead of DNE. Consider taking the limit of dividing a number by X as X approaches 0. On the positive side, you are approaching infinity. On the negative side you are approaching negative infinity. Aside from the fact of how multiplying by 0 works, the limit is contradicting, positive and negative infinity. It's because of the limit that we say undefined instead of DNE.
@eolapade
@eolapade Жыл бұрын
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.
@michaelroditis1952
@michaelroditis1952 3 жыл бұрын
Can somebody tell me a function in which the limit 1^(inf) will differ from 1? Edit: (1+1/n)^n -> e
@paulchapman8023
@paulchapman8023 3 жыл бұрын
(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
@Ferraco05 3 жыл бұрын
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
@popularmisconception1 2 жыл бұрын
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
@spockfan2000
@spockfan2000 3 жыл бұрын
Programmers are also used to "NaN" = Not A Number.
@Pacvalham
@Pacvalham 3 жыл бұрын
0/0 is (results in) NaN but 0/0 is not equal to NaN.
@Theraot
@Theraot 3 жыл бұрын
@@Pacvalham NaN is not equal to NaN.
@Pacvalham
@Pacvalham 3 жыл бұрын
x!=x What is x? Math: 1 Computer science: NaN
@kfibcudwiefjw7428
@kfibcudwiefjw7428 Жыл бұрын
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.
@reelbeenz
@reelbeenz Жыл бұрын
'has no real value' hit too close to home
2 legit proofs & 1 false proof
11:25
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