5 Levels Of “No Answer" (when should we use what?)
ฝัง
- เผยแพร่เมื่อ 21 พ.ค. 2024
- 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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inconsistent
Oh man, you are right! I missed that one. That one is for systems of equations!!
@@blackpenredpen yep! Doing matrix for preCalc right now and instantly drew the connection to this video haha
what's that
how do you get inconsistent
@@ultrio325 for example, x+y=2, 2x+2y=3. The system has no solution for x and y, so our system is inconsistent.
Alternative title: How to tell your math teacher "no".
😆 definitely better
Lol
@Tangent of circle. xD
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
Lol
You forgot the final level
"I don't know how to solve this"
Well, that applies in any case where you forgot to study before an exam
@@fisch37 😂 lol 😂
"left as an exercise for the reader"
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).
@@AgaresOaks independence of a system of axioms doesn't mean that
“No real value” also happens to be the official mathematical classification for NFTs
Value is subjective
@@blocks4857 cool, so it's indeterminate, meaning that they are completely worthless? Cool! I have INFINITE BITCOIN, since, to me, bitcoin is worth $0.
@@cewla3348 You dont do THAT MAN!
@@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.
The value of NFT is imaginary
Me on my math exam:
The answer is left as an exercise to the reader
xD
@Lakshya Gadhwal learner*
Gives me Reimann Zeta Function vibes
😂😂
@@idrisShiningTimes "I know how to solve this, but I'll only tell you if you give me $500,000"
This guy: (flawlessly explains all the ways an equation can have no answer)
My calculator: "NaN"
👵
NaN stands for "Not A Number" in js and in ts
when your parents ask if you are lying you can just tell them "it's a complex statement"
"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.
Also as contradiction where x is 6 but x has to simultaneously be 9
Yeah, stuff like x = x + 1
then, x=∅ and thus y=∅ ect.
how about non-Euclidian geometry?
"No solution" being used there is incorrect, it should be "inconsistent system"
6:51 technically with the definition, the output is always NON NEGATIVE. An absolute value could be 0 :)
Yup, when x=0 absolute value of x is still 0 :)
I agree.
About 15 minutes in, for 2⁰, you could argue/explain the definition (not prove) that 2⁰ = 2¹⁻¹ = 2¹ * 2⁻¹ = 2 * ½ = 1.
Yes this is how it is defined. My teacher has also taught me this process.
I always wonder why 2^0 is a definition
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.
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.
Right!!!!
I love that doll its like Mr . Bean's doll ...
Yeah.. so cute
Edit - That doll gives me Nostalgia
Teddy! His name is Teddy!
@@user-wy8ki2ef1m Yup.. now i remember
🤣🤣🤣
3:56 how about the most basic of situations with no solution... Solve x+1=x+2 lol
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
X=0 / 0
X+5=x
x=∞
0 = 1
The sixth level of 'no answer' is when you are trying to answer the question "is there a sixth level of 'no answer'."
There is a sixth level "undecidable" this is when your axioms are not enough to prove if a statement is true or false.
Ah, yes example of this is the arithmoquine function in Gödels proof
@@IonRuby what, Gödel did make proofs
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.
Lol
too bad its math class
Nani
luckily you stumbled onto something much more useful
It is that lesson, just for tutors not students 🤣
just realised the indeterminate family is on your shirt lmao
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
nerd alert
@@captainpolar2343 bro you're watching a math video,dont you think people would be talking about math
what a baby
@@captainpolar2343 said the fool to the person commenting about math in a math video
@@captainpolar2343 mf we are watching math vid stfu
@@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"
In computing science, we also have "I won't tell you" (no permission or not requested), typically a NULL value.
there is also 𝈜 (upside-down T if Unicode doesn't work) which means that this program never halts.
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
@@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.
Null usually means "does not exist" which is "no solution" tho.
"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.
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."
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?
@@pablopereyra7126 Depending on how the substances interact, they might actually yield 30 liters, they might only yield 25 liters, or they could explode.
@@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.
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.
@@pablopereyra7126 Basically, chemistry makes things weird
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.
so cute
Wouldn’t that be no real solution, as we are solving an equation?
The solution exists, it's just not a real value. @@shrankai7285
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.
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.
@@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
Not only he explains well, but you can also see how happy he is in his face alone, keep it up man. great video
What about his statement that √25≠-5?
√25=±5
Plenty of math teachers on YT say the same things as Bprp, but his absolute joy is what makes him such an effective teacher.
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
A simple example for No Solution is "x + 1 = x + 2" It almost looks trivially solvable but obviously isn't, regardless of the system.
No solution in real and complex numbers!!! BUT, in ordinals numbers, there is another issue.
if you REALLY wanted to couldnt you sub in infinity? of course its not a number tho...
@@thelaststraw1467infinity has no value
@@bloomingon6141so?
how does that imply its not a solution coz it def is
@@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).
6:25 √(x²) = |x|
This would result in |x| = 25
So x = ±5
And ±5 has -5, so there you have your solution
Sqrt ( x) >=0 by def
@@peteradler6005 I never said otherwise
@@peteradler6005 okay but why though. (-5)^2=25 so why not (25)^1/2 = -5
@@peteradler6005 Its okay to veer off "but its defined" and use math to solve problems instead of jerk off about made up rigor
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
I think indeterminate is also used to refer to certain cases of convergence tests for integrals.
0⁰ : Could be 1 or 0, but exponent is stronger, so the 1 wins, therefore 0⁰ = 1
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!
Right! The sqrt(x) could also be 0 though :)
Correct, but 'twas just examples.
@@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 :)
@Math: The Why Behind ok, I don't have a great understanding in math, I'm not even an native english speaker lol
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.
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.
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.
FWIW the inventor of null called it his "billion-dollar mistake".
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.
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..
I am fan of your videos... And your smile when you trying to explain something interesting.. thanks.. for sharing knowledge..
Very nice explanation! As always :D
Excited to watch this video!
Thank you so much for the awesome explanation
Thanks
@@blackpenredpen welcom
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
6:55 you should say non-negative for abs x ccan equal 0
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 ;)
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).
@@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.
Wow this guy is still loving the comments. Salute to you 🙋♂️
I love your videos. Both educational and charmingly presented!
Btw.: When you said "hello" at the very beginning you sounded 100% German.
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.
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.
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.
Exactly
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.
Moivre’s theorem
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.
@@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.
that was decently interesting to read
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.
*[**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.
Misconception here. If you assume 0^0=1 then b^0=empty product. So your argument is circular.
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).
Where would an equation like "x + 1 = x" fit?
No solution.
In programming, increments. In reality, see above
Thinking abstractly: x could be infinity (that obviously isn't a soln) but if you think about it infinity + 1 = infinity
@@wavingbuddy5704 huuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuh
@@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
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.
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 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.
@@radupopescu9977 No, you are wrong. That is not how roots work.
@@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 So all my math professors were idiots... Nice...
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.
how to make mathematicians mald in 4 sentences
Lambert, is that you?
Awesome video. Very helpful ❤👍
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.
Not in Calculus, and that is what he teaches
@@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
However, 0^0=1 implies 0÷0=1
Undefined is the only answer that "works"
@@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
@@vincenzodanello4085 0! = 1 because there's only 1 way to arrange 0 objects.
Everybody gangsta until complex number can't do anything anymore
Nah just throw in + i
Hey, it's better than when they result in periodic solutions.
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.
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.
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
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.
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.
-1 is also its own multiplicative inverse
@@joaquingallardo1728 oh right whoops. whatever there's gonna be a reason to exclude it
@Alejo Sanchez The answer to your comment is given by Iwer Sonsch's reply, right above yours.
@@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.
No real value: *exists*
Complex numbers: let me introduce myself
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
so you're saying 0^2 is undefined?
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.
5:50 sqrt(25)= + or - 5, meaning that 5 and -5 are solutions
sqrt is always positive, x²=25 is what you are looking for
I am surprised lots of students don't know this lol...
@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.
0⁰ is 1; it's literally defined.
18:57 everybody happy made me lought
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.
1/2 is an even number, and therefore defined
I study maths with arabic and french but i don t know why that man make maths easy with that innocent smile .😘
French sucks
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.
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)
1^infinity, can be any value, or no value, and it is case by case. (-1)^infinity is also interesting....
@@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
"it has no real value"
hey look thats me
12:07 Absolutely hilarious. Good video anyway
24:18
Thanks for such a great content with love from India
Didn’t you learn all this at age of two?
@@HN-vu8pp this is so unfunny its funny
@@HN-vu8pp well, we did but revision is necessary. 😂😂
@@HN-vu8pp its complicated bro the teaching pattern here is kinda terrible like we learn differentiation a year before limits so....
The first example given for "no solution" (#2) is exactly the same kind of example as for "no real solution" (#1).
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.
So, for computations 0^0 is undefined.
For limits, 0^0 is indeterminate .
"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.
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.
@@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.
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.
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
Im only 1min and 8 sec in and those whitebord skills are slick
Teddy is adorable 💖
I would love to have the indeterminate family t-shirt jajajaja
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.
6:25 I would like to raise the possibility of x = 25i^4, since Sqrt(i^4) = i^[4*(1/2)] = i^2 = -1
Weird and probably faulty reasoning but I thought it to be at least a little fun to think about.
Imagine walking into this class without knowing he’s holding a mic.
I seriously thought it was just a cute prop XD
Another kind of “no solution” equation could be 2x+1=2x+5
You've basically written down x=x+1 in a camouflaged way
Actually, you've written 1 = 5, and thats a wrong statement.
0=±4, of course
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
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??
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.
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
The contradiction, I think, is that it would follow that sqrt(25) = -5, which is not true.
@@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.
Can you solve this integral
:
Integral of t^n/(1+t)^n dt,
t from 0 to infinity .
Does not converge. You could try t/(1+t)^n which is 1/(n-1)(n-2) for n greater than 2
Since lim [t^n/(1+t)^n] as t-->infinite = 1 is not zero, then the integral diverges
Awesome like usual
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
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.
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.
@@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
@@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?
@@omp199 0a=6 would define 6 as equaling 0 though. That's not a definition.
@@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.
The ultimate level that finish your homework instantly: "I don't know"
2:39 This has a solution with quaternions, so it'd still be case 1. I get the point though, nice video
You forget about bicomplex numbers... I like them because they permit zero divisors.
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.
√-x = i√x, where x is bigger than 0 and i is the square root of -1.
11:42 :the answer is in wheel theory
YAY teddy!!!
So you would say, no complex value or something like that 🤣
@Lakshya Gadhwal read about en.wikipedia.org/wiki/Wheel_theory
@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
@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.
For the no real value, I usually use a+bi
NRV was always the shorthand my teachers understood was me saying "no real value" when in school...
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).
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.
Using L'Hospital on lnx^x I got infinity , but directly got 1
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.
Well, in means principal value. In real numbers the principal value is the positive root.
Is 3i not a positive complex number?
@@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.
@@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)
@@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.
The reason for 'undefined' is because the multiplicative inverse of zero has no definition. Therefore division by zero has no answer.
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.
@@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.
In the complex space, the root function returns a set of values rather than a single value, and there are complex roots to -5 (one example would be e^(ln5+iπ)), so square root of -5 should actually fall into the first category (no real solution).
6/0 = 17 with remainder 6 🌚
LOL
Can somebody tell me a function in which the limit 1^(inf) will differ from 1?
Edit:
(1+1/n)^n -> e
(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.
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.
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
Complex solution for sqrt(x)=-5:
Sqrt(25[(i^2)^2]) ==> 5(i^2) ==> -5
やあ老師
とてもわかりやすかったです😍😍😍