Similar to taking roots, the logarithm of a negative number results in an imaginary number. So it's only undefined if your range is only the real numbers
@giladperiglass5734
3 ай бұрын
And if you're trying to take the log of 0
@error_6o6
3 ай бұрын
It said in the start of the chapter that the process was being done in the real numbers so yeah it’s undefined
@jonathanl8538
3 ай бұрын
@@error_6o6No, it isn't "said in the start", is it? Where do you find that? I can't find any disclaimer anywhere about only talking about real numbers.
@Kualinar
3 ай бұрын
@@giladperiglass5734 log 0 is undefined in the Real, the Complex, the hypercomplex and the quaternion spaces.
@methatis3013
3 ай бұрын
You still need to be careful. Taking a logarithm can result in infinitely many solutions, or rather, you can no longer treat log as a function, but as a relation. It ultimately comes down to what you mean by "log". You need to define your domain and even what the function does
@KrasBadan
3 ай бұрын
Every operation in math is legal if you're brave enough
@ThoughtThrill365
3 ай бұрын
😂😂 true statement, very true statement
@giustobuffo
3 ай бұрын
The Secret Maths police would like a word
@hassanalihusseini1717
3 ай бұрын
Haha, I bet you are physicist, too! 🙂
@THICCTHICCTHICC
3 ай бұрын
Yeah working with infinity and dividing by 0 are things that ultimately are used in very very niche cases
@FishSticker
3 ай бұрын
Except for log 0
@tfg601
3 ай бұрын
Bro went from easy to hard, and then back to easy 💀
@Mediterranean81
3 ай бұрын
Forgetting the +c in an indefinite integral
@Kier_but_who_cares
3 ай бұрын
And putting +c on definite integral
@linuxp00
3 ай бұрын
It's only a sin in college/university, because you can state c = 0, in some cases.
@lakshya4876
3 ай бұрын
@@Kier_but_who_caresdoesn't matter
@xinpingdonohoe3978
3 ай бұрын
@@linuxp00 until you integrate in two different ways, set constants equal to 0, and get a contradiction. Don't single value your multivalued functions too wanton. ∫(x+1)dx=x²/2+x= ∫(x+1)dx=(x+1)²/2=x²/2+x+1/2 0=1/2
@pedrosso0
3 ай бұрын
@@Kier_but_who_cares the +c on a definite Integral is actually true, it's just that it gets cancelled by a -c of the same constant so they together become 0
@alessiodaniotti264
3 ай бұрын
Most of them are not "unallowed" operations, but simply mistakes (heavy ones) Like the 1/(x+y)=1/x +1/y or sqrt(x²+y²)=x+y)
@ersatz_cats
3 ай бұрын
I'm guessing #2 was square root of a negative number.
@mangouschase
2 ай бұрын
You are imaginating things
@arshia5802
3 ай бұрын
Another illegal operation: Using 2 for counting illegal operations
@canteatpi
3 ай бұрын
6:41 mistake: (x+y)^2 = x^2+y^2 + 2xy
@pedropiata648
3 ай бұрын
It looks so stupid in the vídeo lol
@ThoughtThrill365
3 ай бұрын
thanks for telling me, i wrote "=" instead of "+".
That moment when you mix up what modulo and abs are. Abs is the magnitude of a number and never negative. Modulo is the calculation of remainders.
@booboobaloney
3 ай бұрын
mod can mean magnitude in vector and complex contexts
@stopkillingmemes7259
3 ай бұрын
He said modulus, which is sort of like absolute value but for complex numbers, it's still a metric. Modulo is when you have a (usually algebraic) structure and take a quotient with some equivalence relation. Z/nZ is fundamentally different from |z|.
@methatis3013
3 ай бұрын
When we talk about absolute value, we usually talk about real numbers. When considering complex numbers, we call that value a modulus |z|
@pedrosso0
3 ай бұрын
@@methatis3013I'd still call it the absolute value
@apurvakumardani172
3 ай бұрын
Multiplying infinity with zero is definitely legal under limiting circumstances
@darthvader1793
3 ай бұрын
Where is no. 2 ?
@THICCTHICCTHICC
3 ай бұрын
I never really got to study mathematics beyond quite basic high school stuff so it always amazes me that everyone in the comments knows so much stuff about more complex maths
@Vaaaaadim
3 ай бұрын
The people who do know the stuff like to chime in
@BadMathGavin
2 ай бұрын
Bro really be like "lemme just spew off the ideas people fed me my entire life"
@eduardomagalhaes3422
3 ай бұрын
In point 11 you wrote that sqrt( (x + y)^2)) = x + y, but that is false. The correct way of writing this is sqrt( (x + y)^2)) = |x + y|, as sqrt(x^2) = |x| for example
@canteatpi
3 ай бұрын
the square root of a number is always defined to be positive, otherwise the function is not well-defined. So the square root of 49 is 7, not "7 or -7"
@ThoughtThrill365
3 ай бұрын
When you solve the equation: x² = a You will get: x = ±√a Ex: x² = 49 x = ±√49 x = 7, x = -7
@cis5694
3 ай бұрын
√(x)^2= |x| i.e Modulus function And as Modulus Function can never be negative hence the square root of x can also be not negative.
@uggupuggu
3 ай бұрын
@@ThoughtThrill365Still, the sqrt function only returns nonnegative values
@canteatpi
3 ай бұрын
@@ThoughtThrill365 yes, but that alone is not a well defined definition of a function. A function can only have one output
@alessiodaniotti264
3 ай бұрын
@@ThoughtThrill365 the original comment.is right. Even index roots as functions/operators on the Real must give a single output, so they are intended only as the positive number that squared gives the value under the root. The fact that both 7 and -7 squared are 49 is showed by the equation x^2-49=0 But squareroot(49)=7 While -7= -squareroot(49). In fact, if I want to write the negative value that squared gives 2, i must write -sqrt(2), with the explicit minus, and sqrt(2) is only the positive counterpart
@linuxp00
3 ай бұрын
Interestingly, every "disallowed" operation spans a whole new branch of mathematics with their own philosophy to contour this limitations and their applications: 1. Division by zero, Log of zero, negative power of zero, Infinities and ops w/ them, -> Proto-Calculus, Hyperreals/Transfinites/Surreals, Dual numbers 2. Square root of negatives -> Complex Analysis, Quaternions 3. Negative Modulus -> Squares of Time-like Vectors in Special Relativity (SR) and General Relativity (GR) 4. Addition of Scalar and Vector/Matrix -> Linear Algebra (LA), Clifford/Geometric Algebra (GA) 5. Deterninant/Eigen values from rect matrix, Inverse of non-invertable matrix -> Singular Value Decomposition (SVD) 6. Square of rect matrix -> Tensor Algebra (TA) 7. The orders of operations are general guideline, but sometimes you might want to resolve an expression in the denominator or in the base/exponent before doing divs/exps 8. Cancellation Rule (another guideline) violation: (x² + x)/(x² - 1) = x/(-1) = -x if x is a dual number (null-potent matrix) 7. "Exponents" of vectors: LA: (x² + y²) = x² + y² - 2xycos(90°) = x² + y² - 2xy⋅0 = x² + y²; GA: (x² + y²) = x⋅x + y⋅y + x∧y + y∧x = x² + y² + xy + yx = x² + y² + xy - xy = x² + y² if x,y are perpendicular vectors. 8. Adding (juxtaposition or actual summation) unlike terms -> LA, TA, GA, Complex numbers, Quaternions, Hyperreals, Polynomials, SR/GR 9. Splitting denominator (dunno, but there mighr be an exceptional case where it is possible...)
@Fire_Axus
2 ай бұрын
real
@unorthodoxpickle7014
Ай бұрын
It's cause we only say they're disallowed because we don't know enough about them
@ganrimmonim
2 ай бұрын
Why lots of these involve zero is because zero isn't really a number she's just dressed up as one to get into the club. Even negative numbers, by definition are more number-like than zero.
@douglasstrother6584
3 ай бұрын
The video titled "Indeterminates: the hidden power of 0 divided by 0" by "The Mathologer" (Burkard Polster) is good fun.
@KarlDeux
3 ай бұрын
Vectors cannot be added? Wtf!
@Bfdifan2010
Ай бұрын
5:17 this is something a LOT of people need to know when doing those math questions on the internet
@Inspirator_AG112
2 ай бұрын
*@[**06:01**]:* If this was true, then the Pythagorean Theorem would reduce to an arithmetic sum. (Same goes for the next one.)
@nightmareintegral5593
Ай бұрын
Changing order of limits
@Forcommentingpurposes
2 ай бұрын
Disallowed??? My brother in the Culinary Arts you MADE THE SANDWICH
@Calilasseia
2 ай бұрын
In short ... consider what group, ring, field etc., you're working with, and how the operations for that structure are defined. Remember also that, for example, operations that are commutative in one structure need not be commutative in another, even if broadly compatible definitions exist.
@KarlDeux
3 ай бұрын
6:35, often wrong because (x+y) can be negative, in which case √(x+y)² ≠ (x+y)
@eduardomagalhaes3422
3 ай бұрын
In point 9 you said that if the determinant is zero then the matrix is zero, and that is not true
@stefanbergung5514
3 ай бұрын
You May Not multiply equasions by 0, as then you just get 0=0, which is Always true. Therefore whenever you multiple (or devide) an equasions by a variable it should automaticly be defined as ≠0. Then later you have to Look If the new equasions holts true for lim->0. The root of x is = x^0.5. But the root is defined as Always positive while ^0.5 has two solutions. Therefore there existiert some esealy forgotten Rules that prevent you from getting -1=1.
@xinpingdonohoe3978
3 ай бұрын
The real problem comes about when you multiply both sides by 0 and you *don't* get 0=0.
@ninex9631
2 ай бұрын
Not true Sqrt of (-1) is i and -i
@matttrybus925
4 күн бұрын
Plugging infinity directly into the integral
@pyro4362
2 ай бұрын
Just remember that 0, negative numbers, and infinity used to be put in this category
@saviplayer4546
2 ай бұрын
Square root of number (like 36) = +/- 6. It's only 6 because of the square root function is only positive results. I see many people get mixed with that which is understandable tbf.
@pedrosso0
3 ай бұрын
the derivative of a non-continuous non-smooth function is still defined for all inputs where the function is continuous and smooth.
@AlbertTheGamer-gk7sn
3 ай бұрын
1:08 This is because you cannot have a "negative remainder", as a remainder is always positive or 0. Even though you can have 29 ≡ -1 (mod 6), but you cannot say 29 mod 6 = -1. 1:57 This can also include infinity - infinity, 0 / 0, 1^infinity, infinity^0, infinity / infinity, log_1(1), log_0(0), log_infinity(0), log_0(infinity), and log_infinity(infinity). 2:13 You can add scalars and vectors, but they cannot simplify further, due to object-oriented math saying about how you can add objects and primitives, as a scalar is of int, a primitive, whereas a vector is a subclass of int[], an object. 2:45 This can also include taking inverses of singular annihilation functions like f(x) = 1. This can apply to inverses of singular anything. 3:46. ints, int[]'s, int[][]'s, and int[][][]'s are different things. This is crucial in using for loops to traverse through these arrays. 4:00 This can also include taking the derivative of the absolute value function at x = 0, due to the sharp corner. This also includes cusps as well. 6:08 These are all "freshman's dreams", and if they are true, the entire Mathsverse will collapse due to the Pythagorean Theorem crashing as if a^2+b^2=(a+b)^2, then c^2=(a+b)^2, then c = a+b, which violates the Triangle Inequality.
@MidnightAmethyst
2 ай бұрын
Modulus of a number not being negative is not an operation, its a result
@alexandrevachon541
3 ай бұрын
Some integrals do not have a closed form in the form of elementary functions. This is what is known as Liouville's theorem.
@Wallcraft_Official
2 ай бұрын
Division by zero is possible, but the problem is that our current math system doesn't naturally, coherently, or consistently address the properties of the 4th dimension. We define objects in sets but have no definition for the container which holds those sets. However, you can simulate division by zero in a simple thought experiment: Imagine a jar containing a single object. Performing division on this object in a fractional amount between 0 and 1 yields a greater sized object than the original; the problem stems from when attempting to divide by zero, the object would surpass the boundaries which you have defined (in this case, the undefined space in which this jar exists). 4th dimensional mathematics deals with the understanding and translation of the portions of the object that would fall outside this "outer boundary" of the jar holding the object being divided. But unfortunately, most mathematics past D'alambert's time have little grasp of the true nature of the seven cardinal dimensions that comprise reality, and although they are useful, we simply have not developed significant uses for them, outside of perhaps Quaternions. To understand the 4th dimension in the most simple way; it is an infinite set of data containing all the data of which an object or set is not. It can be thought of as an infinite sea of data, in which the only way to study and understand is to study a small glass of it. Only through defining a frame and viewing the data through this frame, discarding all the rest so that it is no longer an infinitely large data set, can you understand or practically use 4th dimensions. Quaternions accomplish this very thing. Division by zero can effectively be defined as the infinite expansion of the outer boundaries which hold the object in question, and can only be significantly measured when studied against a higher dimension (such as the 5th dimension, time - and the rate of this expansion over time). Mathematically speaking, the creation of the universe was indeed some form of division by zero - causing an infinite expansion of space itself to occur over time. If we consider all objects within the universe in regards to this expansion of the space containing them, we can accurately understand them as infinitely shrinking as well.
@FaizThe-kc6dk
2 ай бұрын
🤔
@cdkw2
3 ай бұрын
Not adding +c after indefinate integration
@finaltheory778
2 ай бұрын
I tried 0 divide 0 on calculator, I'm now in a black hole.
@dbmalesani
2 ай бұрын
This is a nice video. I am perplex about the explanation offered regarding point 13. It is of course true that a non-square matrix has no eigenvalues/eigenvectors. The question is even ill-posed because "𝜆x" and "Ax" have different dimensions so it cannot be that Ax = 𝜆x. But the explanation provided is confusing to me. Coming to square matrices, a non-invertible matrix can certainly have eigenvalues (a trivial example is the zero matrix! But there are many more) - it's just that they need to have 0 as eigenvalue. Also, a matrix can have eigenvalues also when it cannot be diagonalized. In order to be diagonalizable, it's necessary that the algebraic and geometric multiplicities of the eigenvalues coincide. For example, every complex matrix has at least an eigenvalue, but not all of them can be diagonalized.
@johnnysvensson
Ай бұрын
Am i the only one who is fine by division by zero? i mean, if i have 10 beers and divide them up among my buddies(zero) i'll still have 10 beers
@MySerpentine
2 ай бұрын
I feel like anything divided by zero should just be zero, since it's in no pieces.
@FaizThe-kc6dk
2 ай бұрын
≖_≖
@warwithworld11
2 ай бұрын
Such cool would been if equations and formulas had some colors, for visual understanding)
@kales901
2 ай бұрын
3:20 you can if you use imaginary numbers, then log-1(1)=i (or something near that)
@the_linguist_ll
8 күн бұрын
log(-1) = iπ
@johannesvanderhorst9778
3 ай бұрын
Some more illegal operations I can think of: Taking the arc sin or arc cos of a value outside [-1,1]. Or in general, taking the inverse function of a value outside the range of said function. Taking the element "that is closest to the limit" of a set that doesn't contain that limit itself. For example, "the biggest negative number." Rearranging the order of the terms in a series when adding them up, if the series is not absolute convergent. For example, otherwise you can get any answer for 1 - 1/2 + 1/3 - 1/4 + ... Raising 0 to the 0-th power. Taking the integral of a function over an interval that is not a subset of the domain of said function.
@xinpingdonohoe3978
3 ай бұрын
Depends on your idea of range, from your idea of domain. arcsin and arccos can be applied to everything when you're in C.
@MinhAIPet
7 күн бұрын
Why don't just made up new number system?
@carealoo744
2 ай бұрын
It went from point #1 to point #3...
@prometheus3911
3 ай бұрын
2:47 - actually, you can find so called "pseudoinverted" matrix. Of course it is not the inversion of a matrix as it works with squre matricies, but they satisfying the identity matricies. If you multiply A by A~ (where A~ is pseudoinverted matrix A) you'll get an I.
@methatis3013
3 ай бұрын
Yes, but this is (probably) talking about singular square matrices
@hanihani9029
3 ай бұрын
0:35 This is not accurate, even incorrect at that. Square roots never give a negative value result, ever. Even if you input negative numbers. √(x²) = |x| ≠ ±x. A fundamental rule in mathematics.
@hanihani9029
3 ай бұрын
Great video nonetheless
@ThoughtThrill365
3 ай бұрын
When you solve the equation: x² = a You will get: x = ±√a Ex: x² = 49 x = ±√49 x = 7, x = -7
@hanihani9029
3 ай бұрын
@ThoughtThrill365 EXACTLY dude, ±√a, but √a itself is always positive haha, we just put ± in front of it to combine the 2 solutions. See what you did there? Basically: √a always > 0 -√a always < 0
@hanihani9029
3 ай бұрын
@ThoughtThrill365 if the square root does have 2 values, it will lead to contradictions and uncertainties in simple arithmetic operations. For example: 1+1=2, simple. Whereas √1 + √1 has 4 answers according to your logic, which was never the case in mathematics, you can check the answer of "√1 + √1" on any calculator or software. . Another way you can think of or apply the square root, is to just raise to the power of ½, √a=a^½, and here's the thing, the function a^x is always positive for whatever x, including x=½, proving further that a^½=√a is always, ALWAYS > 0. . I hope I cleared everything up. Cheers, and thanks for your efforts 🌹
@dehsetcimen
3 ай бұрын
Not quite fundamental. It's just that radical symbols are mostly agreed upon to be a respresentative of the principal branch of the square root "function", or just called the principal square root.
@pedrosso0
3 ай бұрын
0:53. False. It's not if and only if, it's just if. Example: x=0. This is an example where |x| = -x and not x
@IsaacDickinson-tf8sf
3 ай бұрын
tan(pi/2)
@Inspirator_AG112
2 ай бұрын
*@[**6:17**]:* Um, you meant a '+' instead of a second '='.
@Azman-bt8yw
2 ай бұрын
i went on google calc and did 1^infinity and it actaully responded with 1😭💀 i think it was a glitch or smth, it doesnt work anymore
@pedrosso0
3 ай бұрын
0^(-x) can be defined if x≤0
@Kamilistgud
3 ай бұрын
Kinda crazy but everything that was in this video is not "Illegal" operations because sooner or later even the most crazy/Random operations that don't make sense could make sense in the future.
@Fire_Axus
2 ай бұрын
some operations are not even undefined, just incorrect you can actually differentiate discontinuous functions, you just need to make the discontinuous points undefined logarithm of a negative number is not undefined the modulus is not an undefined operation
@pedrosso0
3 ай бұрын
2x+3 = 5x is possible for some number x, just not all numbers x.
@DanDart
2 ай бұрын
3:18 "in the reals"
@RSLT
3 ай бұрын
Square root of -x is ix
@angelagonzalez8250
3 ай бұрын
How come you can't divide matrix
@Omer-dv2ef
2 ай бұрын
5:58 if x not equal to 1 or -1
@СеняМорсин
3 ай бұрын
Square root cant be negative. No solution. Not even imaginary
@azaa8128
3 ай бұрын
At 4:18, if a matrix is invertible (non singular) this directly implies that its rows and columns are linearly INdependent??
@muhammadniohastungkoro1763
3 ай бұрын
I always think of this. If a number is divided by zero, the result is the number itself. Since it is not divided by anything, it will remain the same, just like addition and subtraction. Any thoughts on this?
@tali64squared
3 ай бұрын
There's a problem with assuming that x/0 = x. Say we're trying to simplify x/(1/y). To divide a fraction, we multiply by its reciprocal; specifically, the reciprocal of (1/y) is y, so x/(1/y) = xy. If y is greater than 1, xy will always be greater than x, since we're multiplying x by a value greater than 1. As y grows larger, so does xy; as y grows larger, (1/y) shrinks. In mathematical terms, the limit of (1/y) approaches 0 as y tends to infinity; at this point, one could make an argument that x/0 = ±∞. This is the reason why x/1 = x; x/(1/1) = 1x = x.
@azaa8128
3 ай бұрын
x/0 = x would imply that x=0 if you multiply both sides by 0. So not true for all x. This is quite an horrendous reasoning but yeah your idea does not work
@ScienceOnly-y2n
3 ай бұрын
when you divide a number by a small number, the number what you get is huge, and because of that, I think you can Achieve infinity and also a negative infinity, because when you look at a graph, the infinity goes to the positive, and negative in the same time, so 1/0 can be negative infinity and positive infinity as a answer 😅
@roygalaasen
3 ай бұрын
Did you save number 2 for last? 😂
@XanderAnimations
3 ай бұрын
isn't the logarithm of zero negative infinity?
@ScienceOnly-y2n
3 ай бұрын
Bruh you can teach better then my math teacher in School 😂😂 and Also how the hell i can understand this Complicated math i thought that im a dumbo but now ive realized that im not so dumb after all
@BadMathGavin
2 ай бұрын
Just get better at math, and you, too, can divide by zero. 😉
@PlanesAndGames732
3 ай бұрын
What about #2?
@pedrosso0
3 ай бұрын
There is no REAL number that outputs from √(-1), however the result is the complex number i.
@SparkySywer
3 ай бұрын
You do not know what modulus is
@ICR_Producer
3 ай бұрын
What about log of any number but the log is base 0
@MichaelRothwell1
3 ай бұрын
Or the base is negative...
@xinpingdonohoe3978
3 ай бұрын
log0(x)=ln(x)/ln(0) As a limit, 1/ln(0)=0 This reflects how 0⁰ can take any value.
@khoitan4758
3 ай бұрын
I'd love you teaching calculus.
@ThoughtThrill365
3 ай бұрын
Working on it!
@RISHABHKUMAR-zk1fu
3 ай бұрын
7:13 2x+3 = 5x then x = 1 you are wrong!!!!!
@ScienceOnly-y2n
3 ай бұрын
You are dumb don't you see that the equation it Is not Equal
@RayMyName
3 ай бұрын
are u just dumb, deaf or stubborn
@hanihani9029
3 ай бұрын
Bruh he isn't solving an equation, watch the video again and understand it.
@tali64squared
3 ай бұрын
That section of the video is trying to explain that 2x + 3 does not equal (2 + 3)x most of the time; adding a constant to a coefficient and making the result the new coefficient is invalid in math.
@MichaelRothwell1
3 ай бұрын
The criticism is valid. He should have written 2x+3≢5x (not equivalent to, not identical to) rather than ≠ (not equal to).
@annxu8219
2 ай бұрын
round(x)(does not ∈) Z
@laprankster3264
2 ай бұрын
The factorial of a negative integer cannot be taken, since x! approaches ∓︎∞︎ as x approaches any negative integer.
@SysFan808
3 ай бұрын
PEMDAS? PEMA. division = x * [1/y] addition = x + [-y]
Пікірлер: 153