Every Forbidden Operation in Math

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  • Опубликовано: 26 июл 2024
  • These are some of the operations which are mostly not allowed.
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    Timestamps
    0:00 Division by 0
    0:48 Logarithm of zero
    1:08 Modulus of a Number will be positive
    1:34 Raising zero to a negative number
    1:57 Multiplying infinity with zero
    2:13 Addition of scalar and vector
    2:28 Determinant of a rectangular matrix
    2:45 Inverse of a singular Matrix
    3:21 Log of a negative number
    3:46 Scalar number with a matrix
    3:59 Derivative of a non-differential function
    4:15 Eigenvalues of non-square matrix
    4:44 Square of a non-square matrix
    5:25 Order of operations
    6:08 Cancellation
    6:24 Solving exponents
    6:45 No parenthesis
    7:03 Adding unlike terms
    7:42 Splitting the denominator
    ------
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    - DISCLAIMER -
    This video is intended for entertainment and educational purposes only. It should not be your sole source of information. Some details may be oversimplified or inaccurate. My goal is to spark your curiosity and encourage you to conduct your own research on these topics.

Комментарии • 138

  • @Dravignor
    @Dravignor 23 дня назад +125

    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
      @giladperiglass5734 23 дня назад +4

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

    • @error_6o6
      @error_6o6 23 дня назад +5

      It said in the start of the chapter that the process was being done in the real numbers so yeah it’s undefined

    • @jonathanl8538
      @jonathanl8538 22 дня назад +1

      ​@@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
      @Kualinar 22 дня назад

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

    • @methatis3013
      @methatis3013 22 дня назад +3

      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
    @KrasBadan 23 дня назад +176

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

    • @ThoughtThrill365
      @ThoughtThrill365  23 дня назад +24

      😂😂 true statement, very true statement

    • @giustobuffo
      @giustobuffo 23 дня назад +8

      The Secret Maths police would like a word

    • @hassanalihusseini1717
      @hassanalihusseini1717 22 дня назад +4

      Haha, I bet you are physicist, too! 🙂

    • @THICCTHICCTHICC
      @THICCTHICCTHICC 22 дня назад +2

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

    • @FishSticker
      @FishSticker 22 дня назад

      Except for log 0

  • @tfg601
    @tfg601 22 дня назад +35

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

  • @Mediterranean81
    @Mediterranean81 23 дня назад +54

    Forgetting the +c in an indefinite integral

    • @Kier_but_who_cares
      @Kier_but_who_cares 22 дня назад +5

      And putting +c on definite integral

    • @linuxp00
      @linuxp00 22 дня назад +1

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

    • @lakshya4876
      @lakshya4876 21 день назад

      ​@@Kier_but_who_caresdoesn't matter

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 21 день назад

      ​@@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
      @pedrosso0 20 дней назад

      ​@@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
    @alessiodaniotti264 23 дня назад +37

    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)

  • @canteatpi
    @canteatpi 23 дня назад +33

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

    • @pedropiata648
      @pedropiata648 23 дня назад +2

      It looks so stupid in the vídeo lol

    • @ThoughtThrill365
      @ThoughtThrill365  23 дня назад +14

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

    • @someone9927
      @someone9927 22 дня назад

      ​@@ThoughtThrill365(a+b)² = a²+b²

    • @pedrosso0
      @pedrosso0 20 дней назад

      ​@@someone9927only for ab=0

    • @davidbrown7142
      @davidbrown7142 8 дней назад

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

  • @ersatz_cats
    @ersatz_cats 22 дня назад +10

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

  • @arshia5802
    @arshia5802 21 день назад +6

    Another illegal operation:
    Using 2 for counting illegal operations

  • @kliqx
    @kliqx 23 дня назад +14

    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
      @booboobaloney 23 дня назад

      mod can mean magnitude in vector and complex contexts

    • @stopkillingmemes7259
      @stopkillingmemes7259 22 дня назад

      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
      @methatis3013 22 дня назад +2

      When we talk about absolute value, we usually talk about real numbers. When considering complex numbers, we call that value a modulus |z|

    • @pedrosso0
      @pedrosso0 20 дней назад

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

  • @darthvader1793
    @darthvader1793 22 дня назад +3

    Where is no. 2 ?

  • @eduardomagalhaes3422
    @eduardomagalhaes3422 23 дня назад +4

    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

  • @apurvakumardani172
    @apurvakumardani172 21 день назад +3

    Multiplying infinity with zero is definitely legal under limiting circumstances

  • @linuxp00
    @linuxp00 22 дня назад +2

    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...)

  • @douglasstrother6584
    @douglasstrother6584 22 дня назад +3

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

  • @THICCTHICCTHICC
    @THICCTHICCTHICC 22 дня назад +2

    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
      @Vaaaaadim 21 день назад

      The people who do know the stuff like to chime in

  • @eduardomagalhaes3422
    @eduardomagalhaes3422 23 дня назад +2

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

  • @badmath9099
    @badmath9099 2 дня назад

    Bro really be like "lemme just spew off the ideas people fed me my entire life"

  • @canteatpi
    @canteatpi 23 дня назад +29

    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
      @ThoughtThrill365  23 дня назад +6

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

    • @cis5694
      @cis5694 23 дня назад +16

      √(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
      @uggupuggu 23 дня назад +17

      @@ThoughtThrill365Still, the sqrt function only returns nonnegative values

    • @canteatpi
      @canteatpi 23 дня назад +13

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

    • @alessiodaniotti264
      @alessiodaniotti264 23 дня назад +10

      ​@@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

  • @Inspirator_AG112
    @Inspirator_AG112 16 дней назад +1

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

  • @MidnightAmethyst
    @MidnightAmethyst 5 дней назад

    Modulus of a number not being negative is not an operation, its a result

  • @pyro4362
    @pyro4362 10 дней назад

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

  • @stefanbergung5514
    @stefanbergung5514 23 дня назад +3

    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
      @xinpingdonohoe3978 21 день назад

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

    • @ninex9631
      @ninex9631 7 дней назад

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

  • @MySerpentine
    @MySerpentine День назад

    I feel like anything divided by zero should just be zero, since it's in no pieces.

  • @pedrosso0
    @pedrosso0 20 дней назад

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

  • @KarlDeux
    @KarlDeux 22 дня назад +1

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

  • @finaltheory778
    @finaltheory778 4 дня назад

    I tried 0 divide 0 on calculator, I'm now in a black hole.

  • @cdkw2
    @cdkw2 23 дня назад +3

    Not adding +c after indefinate integration

  • @KarlDeux
    @KarlDeux 22 дня назад +1

    Vectors cannot be added?
    Wtf!

  • @saviplayer4546
    @saviplayer4546 13 дней назад

    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.

  • @alexandrevachon541
    @alexandrevachon541 22 дня назад

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

  • @ganrimmonim
    @ganrimmonim 14 дней назад

    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.

  • @carealoo744
    @carealoo744 7 дней назад

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

  • @azaa8128
    @azaa8128 21 день назад

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

  • @AlbertTheGamer-gk7sn
    @AlbertTheGamer-gk7sn 22 дня назад

    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.

  • @IsaacDickinson-tf8sf
    @IsaacDickinson-tf8sf 22 дня назад +1

    tan(pi/2)

  • @DanDart
    @DanDart 15 дней назад

    3:18 "in the reals"

  • @Inspirator_AG112
    @Inspirator_AG112 16 дней назад +1

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

  • @pedrosso0
    @pedrosso0 20 дней назад

    0^(-x) can be defined if x≤0

  • @johannesvanderhorst9778
    @johannesvanderhorst9778 22 дня назад

    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
      @xinpingdonohoe3978 21 день назад

      Depends on your idea of range, from your idea of domain. arcsin and arccos can be applied to everything when you're in C.

  • @angelagonzalez8250
    @angelagonzalez8250 19 дней назад

    How come you can't divide matrix

  • @warwithworld11
    @warwithworld11 11 дней назад

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

  • @pedrosso0
    @pedrosso0 20 дней назад

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

  • @Fire_Axus
    @Fire_Axus 3 дня назад

    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

  • @XanderAnimations
    @XanderAnimations 22 дня назад

    isn't the logarithm of zero negative infinity?

  • @Azman-bt8yw
    @Azman-bt8yw 16 дней назад

    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

  • @roygalaasen
    @roygalaasen 22 дня назад

    Did you save number 2 for last? 😂

  • @muhammadniohastungkoro1763
    @muhammadniohastungkoro1763 22 дня назад

    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
      @tali64squared 22 дня назад

      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
      @azaa8128 21 день назад

      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

  • @pedrosso0
    @pedrosso0 20 дней назад

    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

  • @Omer-dv2ef
    @Omer-dv2ef 10 дней назад

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

  • @prometheus3911
    @prometheus3911 22 дня назад

    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
      @methatis3013 22 дня назад

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

  • @RSLT
    @RSLT 19 дней назад

    Square root of -x is ix

  • @PlanesAndGames732
    @PlanesAndGames732 21 день назад

    What about #2?

    • @pedrosso0
      @pedrosso0 20 дней назад

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

  • @SparkySywer
    @SparkySywer 22 дня назад

    You do not know what modulus is

  • @ICR_Producer
    @ICR_Producer 22 дня назад

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

    • @MichaelRothwell1
      @MichaelRothwell1 22 дня назад

      Or the base is negative...

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 21 день назад

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

  • @Kamilistgud
    @Kamilistgud 22 дня назад

    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.

  • @badmath9099
    @badmath9099 2 дня назад

    Just get better at math, and you, too, can divide by zero. 😉

  • @user-rg3vh9hb5h
    @user-rg3vh9hb5h 23 дня назад

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

  • @ScienceOnly-y2n
    @ScienceOnly-y2n 23 дня назад +2

    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

  • @hanihani9029
    @hanihani9029 23 дня назад +5

    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
      @hanihani9029 23 дня назад +2

      Great video nonetheless

    • @ThoughtThrill365
      @ThoughtThrill365  23 дня назад +1

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

    • @hanihani9029
      @hanihani9029 22 дня назад +1

      @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
      @hanihani9029 22 дня назад +1

      @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
      @dehsetcimen 22 дня назад

      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.

  • @khoitan4758
    @khoitan4758 23 дня назад +1

    I'd love you teaching calculus.

  • @ScienceOnly-y2n
    @ScienceOnly-y2n 23 дня назад +2

    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 😅

  • @RISHABHKUMAR-zk1fu
    @RISHABHKUMAR-zk1fu 23 дня назад +3

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

    • @ScienceOnly-y2n
      @ScienceOnly-y2n 23 дня назад

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

    • @RayMyName
      @RayMyName 22 дня назад

      are u just dumb, deaf or stubborn

    • @hanihani9029
      @hanihani9029 22 дня назад +1

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

    • @tali64squared
      @tali64squared 22 дня назад +1

      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
      @MichaelRothwell1 22 дня назад

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

  • @annxu8219
    @annxu8219 6 дней назад

    round(x)(does not ∈) Z

  • @laprankster3264
    @laprankster3264 14 дней назад

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

  • @SysFan808
    @SysFan808 22 дня назад

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