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

17 күн бұрын

And if you're trying to take the log of 0

@error_6o6

17 күн 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

17 күн 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

16 күн бұрын

@@giladperiglass5734 log 0 is undefined in the Real, the Complex, the hypercomplex and the quaternion spaces.

@methatis3013

16 күн 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

Every operation in math is legal if you're brave enough

@ThoughtThrill365

17 күн бұрын

😂😂 true statement, very true statement

@giustobuffo

17 күн бұрын

The Secret Maths police would like a word

@hassanalihusseini1717

17 күн бұрын

Haha, I bet you are physicist, too! 🙂

@THICCTHICCTHICC

17 күн бұрын

Yeah working with infinity and dividing by 0 are things that ultimately are used in very very niche cases

@FishSticker

16 күн бұрын

Except for log 0

Bro went from easy to hard, and then back to easy 💀

Forgetting the +c in an indefinite integral

@Kier_but_who_cares

17 күн бұрын

And putting +c on definite integral

@linuxp00

16 күн бұрын

It's only a sin in college/university, because you can state c = 0, in some cases.

@lakshya4876

15 күн бұрын

​@@Kier_but_who_caresdoesn't matter

@xinpingdonohoe3978

15 күн 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

14 күн 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

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)

6:41 mistake: (x+y)^2 = x^2+y^2 + 2xy

@pedropiata648

17 күн бұрын

It looks so stupid in the vídeo lol

@ThoughtThrill365

17 күн бұрын

thanks for telling me, i wrote "=" instead of "+".

@someone9927

17 күн бұрын

​@@ThoughtThrill365(a+b)² = a²+b²

@pedrosso0

14 күн бұрын

​@@someone9927only for ab=0

@davidbrown7142

3 күн бұрын

​@@someone9927 actually (a+b)^2 = (a+b)(a+b) = a^2 + b^2 + 2ab

I'm guessing #2 was square root of a negative number.

@mangouschase

3 күн бұрын

You are imaginating things

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

17 күн бұрын

mod can mean magnitude in vector and complex contexts

@stopkillingmemes7259

17 күн 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

16 күн 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

14 күн бұрын

​@@methatis3013I'd still call it the absolute value

Another illegal operation: Using 2 for counting illegal operations

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...)

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

Multiplying infinity with zero is definitely legal under limiting circumstances

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

16 күн бұрын

The people who do know the stuff like to chime in

Where is no. 2 ?

The video titled "Indeterminates: the hidden power of 0 divided by 0" by "The Mathologer" (Burkard Polster) is good fun.

*@[**06:01**]:* If this was true, then the Pythagorean Theorem would reduce to an arithmetic sum. (Same goes for the next one.)

In point 9 you said that if the determinant is zero then the matrix is zero, and that is not true

the derivative of a non-continuous non-smooth function is still defined for all inputs where the function is continuous and smooth.

Just remember that 0, negative numbers, and infinity used to be put in this category

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

17 күн бұрын

When you solve the equation: x² = a You will get: x = ±√a Ex: x² = 49 x = ±√49 x = 7, x = -7

@cis5694

17 күн 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

17 күн бұрын

@@ThoughtThrill365Still, the sqrt function only returns nonnegative values

@canteatpi

17 күн бұрын

@@ThoughtThrill365 yes, but that alone is not a well defined definition of a function. A function can only have one output

@alessiodaniotti264

17 күн 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

6:35, often wrong because (x+y) can be negative, in which case √(x+y)² ≠ (x+y)

Not adding +c after indefinate integration

Some integrals do not have a closed form in the form of elementary functions. This is what is known as Liouville's theorem.

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.

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

15 күн бұрын

The real problem comes about when you multiply both sides by 0 and you *don't* get 0=0.

@ninex9631

Күн бұрын

Not true Sqrt of (-1) is i and -i

At 4:18, if a matrix is invertible (non singular) this directly implies that its rows and columns are linearly INdependent??

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.

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.

Vectors cannot be added? Wtf!

It went from point #1 to point #3...

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

15 күн 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.

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

0^(-x) can be defined if x≤0

How come you can't divide matrix

2x+3 = 5x is possible for some number x, just not all numbers x.

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

16 күн бұрын

Yes, but this is (probably) talking about singular square matrices

Such cool would been if equations and formulas had some colors, for visual understanding)

*@[**6:17**]:* Um, you meant a '+' instead of a second '='.

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

3:18 "in the reals"

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

17 күн 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

15 күн 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

tan(pi/2)

5:58 if x not equal to 1 or -1

isn't the logarithm of zero negative infinity?

Square root of -x is ix

What about #2?

@pedrosso0

14 күн бұрын

There is no REAL number that outputs from √(-1), however the result is the complex number i.

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.

What about log of any number but the log is base 0

@MichaelRothwell1

16 күн бұрын

Or the base is negative...

@xinpingdonohoe3978

15 күн бұрын

log0(x)=ln(x)/ln(0) As a limit, 1/ln(0)=0 This reflects how 0⁰ can take any value.

Did you save number 2 for last? 😂

Square root cant be negative. No solution. Not even imaginary

You do not know what modulus is

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

17 күн бұрын

Great video nonetheless

@ThoughtThrill365

17 күн бұрын

When you solve the equation: x² = a You will get: x = ±√a Ex: x² = 49 x = ±√49 x = 7, x = -7

@hanihani9029

17 күн 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

17 күн 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

17 күн 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.

I'd love you teaching calculus.

@ThoughtThrill365

17 күн бұрын

Working on it!

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

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 😅

7:13 2x+3 = 5x then x = 1 you are wrong!!!!!

@ScienceOnly-y2n

17 күн бұрын

You are dumb don't you see that the equation it Is not Equal

@RayMyName

17 күн бұрын

are u just dumb, deaf or stubborn

@hanihani9029

17 күн бұрын

Bruh he isn't solving an equation, watch the video again and understand it.

@tali64squared

17 күн 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

16 күн бұрын

The criticism is valid. He should have written 2x+3≢5x (not equivalent to, not identical to) rather than ≠ (not equal to).

round(x)(does not ∈) Z

The factorial of a negative integer cannot be taken, since x! approaches ∓︎∞︎ as x approaches any negative integer.

PEMDAS? PEMA. division = x * [1/y] addition = x + [-y]