Is x^x=0 solvable?

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  • Опубліковано 23 гру 2024

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  • @blackpenredpen
    @blackpenredpen  3 місяці тому +163

    Finally 0^0 approaches 0:
    ua-cam.com/video/X65LEl7GFOw/v-deo.htmlsi=QRgxgdaL5If2FWnP

    • @costelnica3988
      @costelnica3988 3 місяці тому +3

      BPRP, how solve 1^x = 2?

    • @Player_is_I
      @Player_is_I 3 місяці тому

      @@costelnica3988 I think he has done it

    • @yiutungwong315
      @yiutungwong315 3 місяці тому +2

      0^0 = 1 so i^0 = 1

    • @kyokajiro1808
      @kyokajiro1808 3 місяці тому +1

      case 6 works with something akin to the squeeze theorem
      tl;dr x^x as x approaches -infinity is in the form re^itheta where r approaches 0
      if we rewrite x as re^itheta, as x approaches -infinity thats the same as x equals the limit as r approaches infinity of re^ipi
      (re^ipi)^(re^ipi)=(re^ipi)^-r=(r^-r)(e^(i(-rpi)))
      r^-r is a positive real number that approaches 0 as r approaches infinity, meanwhile e^(i(-rpi)) in the form e^itheta meaning its just polar form with a radius of r^-r
      since its always on the circle on the complex plane with a radius of r^-r, and because the circle gets compressed to a point specifically down to 0, this means limit as x approaches -infinity of x^x should be considered 0
      this is not the most rigorous proof but it works for me

    • @sarmisthadeb5289
      @sarmisthadeb5289 3 місяці тому

      @blackpenredpen Hey bprp. Can u do a formula for a tetrated by x plus b exponentiated by x plus c times x plus d equal to 0. Love ur videos. Bcos of u I learnt and mastered calculus at the age of 10.

  • @bartekabuz855
    @bartekabuz855 3 місяці тому +749

    If x is a complex number other than 0 we can write x=r*e^iθ and then we see that x^x=(re^iθ)^x=r^x e^ixθ
    In order for this expression to be 0 we need either r^x=0 or e^ixθ=0. But since r>0 hence r^x != 0. Likewise e^ixθ != for any x. Thus x^x=0 has no solution in the complex numbers

    • @TNTErick
      @TNTErick 3 місяці тому +10

      wait what

    • @bartekabuz855
      @bartekabuz855 3 місяці тому +5

      @@TNTErick ?

    • @derda3209
      @derda3209 3 місяці тому +12

      probably true, but why can a^x not equal 0 for a!=0? With real numbers it's obvious, but with complex numbers I don't know

    • @bartekabuz855
      @bartekabuz855 3 місяці тому +49

      @@derda3209 If a != 0 then a^x=e^(x*log(a)) which is never 0 (exponential function is never zero over real numbers and complex numbers). Note that log(a) is defined for none zero complex numbers

    • @AndyBaiduc-iloveu
      @AndyBaiduc-iloveu 3 місяці тому +14

      I think, the easiest proof that the solution doesn't exist in the complex world is that there aren't any zero divisors in the complex world, so for x^a to be equal to 0 , x=0 , however contradiction.

  • @chanuldandeniya9120
    @chanuldandeniya9120 3 місяці тому +1241

    He doesn't age man

    • @Bangaudaala
      @Bangaudaala 3 місяці тому +68

      Idk, I really miss his looooong beard

    • @PRIYANSH_SUTHAR
      @PRIYANSH_SUTHAR 3 місяці тому +17

      Maybe we age at the same rate as him 😂

    • @dissuasive5256
      @dissuasive5256 3 місяці тому +15

      He looked the same like years ago I watched him for the first time 😂

    • @bigchungusdriplord2301
      @bigchungusdriplord2301 3 місяці тому

      Well of course, he's asian

    • @siheonseong5920
      @siheonseong5920 3 місяці тому +24

      study math and get eternal life

  • @15silverblade
    @15silverblade 3 місяці тому +763

    For me 0^0 = 1/2, take the average value of 0 and 1, im sure everyone will be happy 😊

    • @fabiobordignon3840
      @fabiobordignon3840 3 місяці тому +28

      😂😂😊

    • @yurenchu
      @yurenchu 3 місяці тому +37

      Nope.
      a^b = c means
      multiplying any constant K by the factor a for a number of b times, is equal to multiplying the same constant K by the factor c.
      Hence,
      0^0 = c means
      multiplying any constant K by the factor 0 for a number of 0 times, is equal to multiplying the same constant K by the factor c.
      The former side gives K (because multiplying K by 0 for a number of 0 times, gives K), the latter side gives K*c (because multiplying K by the factor c gives K*c), so
      K = K*c , valid for any value of K ==>
      c = 1
      Therefore,
      0^0 = c = 1
      Q.E.D.

    • @Oxygenationatom
      @Oxygenationatom 3 місяці тому +87

      @@yurenchuwhat the sigma

    • @petersagitarius4356
      @petersagitarius4356 3 місяці тому +11

      how can 0^0 be 1/2 ??? If we say that "it is defined", and these 2 zeros are identical, then the solution must be 1, because it assymptoticaly goes to 1. But more strictly it is not defined. So in none of case it could be 1/2.

    • @SudhanshuKumarSinghSudhanshuKu
      @SudhanshuKumarSinghSudhanshuKu 3 місяці тому +1

      the average value of 0 to 1 how??

  • @khoozu7802
    @khoozu7802 3 місяці тому +126

    If u want to know about the complex solution, u can find bprp's old video with title "the tetration of (1+i) and the formula (a+bi)^(c+di)"

  • @crsmtl76
    @crsmtl76 2 місяці тому +36

    Naturally, let’s be rational for a moment. This is too complex to be real.

  • @OverLordGoldDragon
    @OverLordGoldDragon 3 місяці тому +148

    You say "not my best video", but I really enjoyed it!

  • @peterciccone620
    @peterciccone620 3 місяці тому +34

    Must be the first vid without "of course, otherwise, how could I make this video?" Love the subversion of expectations

  • @kindafool4083
    @kindafool4083 3 місяці тому +112

    Hey there BPRP, been watching for some 6 years now. Good stuff man

    • @blackpenredpen
      @blackpenredpen  3 місяці тому +25

      Thank you!!

    • @Player_is_I
      @Player_is_I 3 місяці тому +4

      ​ Lots of love to you and to your iconic pens❤@@blackpenredpen

    • @erickherrerapena8981
      @erickherrerapena8981 3 місяці тому +1

      Yo también lo sigo desde hace 6 años, me puse a pensar en ello al leer tu comentario.

  • @christianstieger279
    @christianstieger279 3 місяці тому +12

    For the limit x^x you can do:
    lim x^x = lim exp(ln(x)*x) = exp(lim ln(x)*x) because exp is a continuous function. Then lim ln(x) * x = lim ln(x) / (1/x) = lim (1/x) / (-1/x^2) = lim 1 / (-1 / x) = lim -x = 0 (using de l'Hopital) and therefore lim exp(ln(x)*x) = lim x^x = 1.

    • @fernandojackson7207
      @fernandojackson7207 3 місяці тому +2

      I think you need to assume x>0 ( If you use x Real) , in order to write x^x= e^{xlnx}, as lnx is defined only in (0, oo) for x Real.

  • @alvinoceanohorsky6093
    @alvinoceanohorsky6093 3 місяці тому +4

    What do you mean not your best video? I enjoyed every single second. Good job

  • @meccamiles7816
    @meccamiles7816 3 місяці тому +8

    This is an example of a pedagogically responsible explanation.

  • @petersagitarius4356
    @petersagitarius4356 3 місяці тому +17

    I understand that mathematician say : "It has no solution, because negative Infinity is not a real or complex number"... But, it is obvious that assymptoticly -inf (in integer) converges to solution. It could be really interesting to make some physical experiment, where the nature say if the -inf is the solution of this equation.... For me, from technicial - engineering aspect, it is solid solution (but not ideal); The only question for me is, if the nature supports the infinity .. :)

    • @badabing3391
      @badabing3391 3 місяці тому +2

      im pretty sure even in complex this works because complex exponentiation doesnt change magnitude, so if it converges to 0 on integers it should converge for the remaining values in between that are bounded by surrounding integer values.

    • @ofeklevy3101
      @ofeklevy3101 2 місяці тому

      thats not true when you write integral from b to inf it's short for lim_(a->inf) integral from b to a you don't realy use infinity

  • @gracek630
    @gracek630 3 місяці тому +32

    Why does function y=x^x does not have any values on the left side of the graph? I mean for x=-1, f(-1)=(-1)^(-1), which would be equal to -1, and so on for other arguments?

    • @stefanalecu9532
      @stefanalecu9532 3 місяці тому +9

      Counter-example: what do you do when x = -1/2? You're getting imaginary numbers for anything but x in Z* (I think)

    • @Qreator06
      @Qreator06 3 місяці тому +16

      It does have values,
      For x that is in the form of even/odd, its |x|^x
      For x in the form of odd/odd, its -|x|^x
      For x in the form of odd/even or irrational, it’s imaginary, if you allow imaginary numbers, forget everything above and just do complex calculations using e^(xlnx). Anyway, overall, the function jumps between positive, negative and imaginary/undefined infinitely many times in any finite region, so it couldn’t be graphed using normal methods, I would graph it is x^x for x = 0 or x>0, ±|x|^x for x < 0

    • @DutchMathematician
      @DutchMathematician 3 місяці тому +7

      There are problems trying to define a^x for non-integer values when a is negative.
      E.g. (-1)^(1/3) should equal (-1)^(2/6) since 1/3 = 2/6. The first expression would give the cube root of -1 which is -1 (note that the cube root function *is* defined for negative real values). However, the second expression would give the 6th root of (-1)^2, which equals 1.
      See en.wikipedia.org/wiki/Exponentiation#Real_exponents for more information.

    • @FocusLRHAP
      @FocusLRHAP 3 місяці тому +5

      Actually, the negative part of the graph is just unconnected points for the integers. Because -1/2 for example.

    • @FocusLRHAP
      @FocusLRHAP 3 місяці тому +5

      Try graphing (|x|)^x or (|x|)^(|x|). The second one is a little weird.

  • @hbfrts5519
    @hbfrts5519 3 місяці тому +98

    My high school teacher once declared 0^0 is undefined because powers can be written as fractions, so 0^0 equals 0/0. I never questioned it until now.

    • @yurenchu
      @yurenchu 3 місяці тому +72

      Your teacher's argument doesn't make sense. By the same logic, 0^1 is undefined, because
      0^1 =
      = 0^(2-1)
      = (0^2)/(0^1)
      = 0/0
      = undefined
      but of course we know that 0^1 = 0 is _not_ undefined.

    • @yashrajtripathi4832
      @yashrajtripathi4832 3 місяці тому +1

      What about 2^2 😂

    • @yurenchu
      @yurenchu 3 місяці тому +3

      @@yashrajtripathi4832 What about it?

    • @yashrajtripathi4832
      @yashrajtripathi4832 3 місяці тому

      @@yurenchu definitely not about your comment bro !

    • @yurenchu
      @yurenchu 3 місяці тому +19

      @@yashrajtripathi4832 Oops! My bad.
      I think what the original commenter's high school teacher meant, is that in general,
      a^b = (a^(b+1)) / a
      which, in the case of a=2 and b=2 , would lead to
      2^2 = (2^(2+1)) / 2 = (2^3)/2 = 8/2
      which is correct (both sides equal 4); so nothing wrong there.
      With a=0 and b=0 , this formula would lead to
      0^0 = (0^(0+1))/0 = (0^1)/0 = 0/0
      which on the righthandside is undefined (and therefore, according to his teacher, 0^0 should be undefined).
      However, the rule a^b = (a^(b+1)) / a is not valid for a=0 , because for example with a=0 and b=1, it leads to
      0^1 = (0^(1+1))/0 = (0^2)/0 = 0/0
      which is clearly wrong, because the lefthandside is _not_ undefined; 0^1 equals 0.
      Therefore, the high school teacher's argument as to why 0^0 is undefined, is wrong, as it doesn't hold water; he applied a rule that shouldn't be applied in this case.

  • @geekonomist
    @geekonomist 3 місяці тому +26

    Love how postmodern math is caught in a fantastic sceptical philosophy. He shows using 10 different reality based ways how x^x cannot be zero but proceeds with conclusion that « in a complex world » it « might ». Bake skepticism into the choice of words to really screw with all brains for 200 years and expect good results.

    • @Kleyguerth
      @Kleyguerth 3 місяці тому +3

      Well, it's easy to prove that I cannot find a solution. It's way harder to prove that there is definitely no solution... Or weirder: that even if a solution exists it is impossible for someone to find it.

    • @geekonomist
      @geekonomist 3 місяці тому

      @@Kleyguerth the one with the burden of proof is the one making a claim, an assertion. If you say it is possible, what proof do you present? Until you do, all of the refutations of this statement stand and you must shut up.

    • @Kleyguerth
      @Kleyguerth 3 місяці тому +2

      @@geekonomist Yes, I'm agreeing with you, I just repeated it in simple english lol

    • @Fire_Axus
      @Fire_Axus 3 місяці тому

      YFAI

    • @Tentabrobpy
      @Tentabrobpy 2 місяці тому

      Am I misunderstanding your comment or are you categorizing complex numbers as "postmodern math?" Because they have some quite interesting applications in "reality"

  • @rick4135
    @rick4135 3 місяці тому +3

    I think 💭 the first video I watched from was pi^e vs e^pi.
    Back then I think was entering masters in math/stats and today I’m a year away from PhD in statistics.
    Always inspiring!
    Btw did you ever did a video on Riemann-Stieltjes integration????
    I recall it but can’t find it

  • @i_am_anxious02
    @i_am_anxious02 3 місяці тому +23

    I believe as long as x is real, case 6 with the limit works. lim x->-inf (x^x) = lim x->inf ((-x)^(-x)).
    (-x)^(-x)=(-1)^(-x) * x^(-x)
    The latter factor always approaches zero as x approaches infinity. The former factor is cyclical and the absolute value is always equal to one, so (-x)^(-x) always approaches 0 as x approaches infinity and thus x^x approaches zero as x approaches negative infinity.
    I think there’s an extension of this argument if you allow x to approach negative infinity from paths in the complex plane, but I haven’t fully formulated it yet.

    • @Finkelfunk
      @Finkelfunk 3 місяці тому +1

      This is deceptively simple but it makes sense to me.

    • @DutchMathematician
      @DutchMathematician 3 місяці тому +1

      There are problems trying to define a^x for non-integer values when a is negative.
      E.g. (-1)^(1/3) should equal (-1)^(2/6) since 1/3 = 2/6. The first expression would give the cube root of -1 which is -1 (note that the cube root function *is* defined for negative real values). However, the second expression would give the 6th root of (-1)^2, which equals 1.
      See en.wikipedia.org/wiki/Exponentiation#Real_exponents for more information.

    • @zachansen8293
      @zachansen8293 3 місяці тому +1

      @@DutchMathematician you have to simplify exponent fractions first. Values that you can get when it's not simplified don't matter. He showed this in one of his "1=2 proof" videos.

    • @DutchMathematician
      @DutchMathematician 3 місяці тому +1

      @@zachansen8293
      There is no need to simplify fractions first. If f(x) = a^x is a well-defined function, then f(1/3) = f(2/6) *should* hold, no matter how we supply the same argument to the function f. If we give a function two equivalent expression then the outcome should be the same.

    • @zachansen8293
      @zachansen8293 3 місяці тому +1

      @@DutchMathematician Go watch the video, you absolutely do. That's how he proves that 1=2 or 1=-1 or whatever it was. You have to simplify fractions in an exponent (or you can do them in a specific order if it isn't). If you don't you are likely to get incorrect results.
      edit: Go search for the video named "Hate to be that guy but I need the extra credit! " - it's the last chapter starting at 3:48. He shows that i=1 because i = i^(4/4) = 1
      edit2: I believe if it's not simplified you can do the denominator first, but you cannot do the numerator first. But the easiest thing to remember is to just always simplify fractions that are exponents.
      last edit: and of course you should run it through wolfram alpha to see that it doesn't get your "solution"

  • @phantoniex
    @phantoniex 3 місяці тому +1

    Hello sir, I just wanted to thank you for making these videos because they've helped me out a lot.

  • @JJ_TheGreat
    @JJ_TheGreat 3 місяці тому +3

    2:35 But what about the lim x-> 0 (-) [from the left] of (x^x)?

    • @zachansen8293
      @zachansen8293 3 місяці тому +3

      that's 1. You don't need the 0-, it's 1 from either side.

    • @mitchratka3661
      @mitchratka3661 3 місяці тому

      ​​@@zachansen8293good to know, but wouldn't it be better to just write the limit in general to imply both? I thought specifying one meant the other may or may not be the same

    • @KCrucis
      @KCrucis 3 місяці тому +1

      ​@@mitchratka3661if it's not the same the limit doesn't exist, it's specified during calculations to find exactly that

    • @mitchratka3661
      @mitchratka3661 3 місяці тому

      @@KCrucis I understand that, the limit IS the same from both sides, which is why I think it's disregarding information that could be included for some arbitrary reason

    • @bauchis
      @bauchis 2 місяці тому

      It's because the function is not even defined for negative numbers, that's why you can take the limit from the left side​@@mitchratka3661

  • @adamrussell658
    @adamrussell658 3 місяці тому +3

    I think the lim as x-> negative infinity is valid. You can see that if you try x= -10, -100, -1000 then x^x surely is approaching zero. It could be argued that the function is discontinuous in the negatives, but for going to -infinity at every point the function exists x^x approaches zero as x gets more negative.

  • @EMAngel2718
    @EMAngel2718 3 місяці тому +17

    I'd argue -inf works without djscreetization because the negative ultimately only changes the complex sign and the magnitude still goes to 0 and 0 with any complex sign is still 0

    • @jsonr
      @jsonr 2 місяці тому +2

      yeah it'll be equal to 1/(x^abs(x)) * i which approaches 0 as x -> infinity

  • @Snakeyes244
    @Snakeyes244 3 місяці тому

    This is where new math is to be found. Idk why no one looks here or ever talks about numbers like these. Juat as we thouught i couldnt exist, it does. So this probably does too. Same with |x|=-1

  • @MarkoMikulicic
    @MarkoMikulicic 2 місяці тому +3

    I love the superpower of erasing the board with a knock

  • @rebokfleetfoot
    @rebokfleetfoot 3 місяці тому +5

    would i be correct in thinking that x to the power or x is not a continuous function?

    • @Dreamprism
      @Dreamprism 3 місяці тому +4

      It is continuous on the real interval (0,infinity).
      But if you take two rational approximations of -sqrt(2) arbitrarily close to each other where both have odd denominators but one has an even numerator and the other has an odd numerator then you will get a positive and negative answer more than some specific distance apart, and therefore we know that x^x is not continuous on its maximal domain that keeps the domain and range as subsets of the real numbers.

    • @GrGalan6464
      @GrGalan6464 3 місяці тому +3

      It is continuous along the positive real numbers. It is true that it is not continuous over the whole domain of real numbers.

    • @rebokfleetfoot
      @rebokfleetfoot 3 місяці тому

      @@Dreamprism sometimes i think that functions such as this require some kind of analytical continuation, i mean the undefined is not necessarily undefined , it's just that we define it that way :)

    • @rebokfleetfoot
      @rebokfleetfoot 3 місяці тому +1

      @@Dreamprism wo.... i never thought of a discontinuance that way, but yes, i guess it is that way :)

    • @rebokfleetfoot
      @rebokfleetfoot 3 місяці тому

      @@Dreamprism wo.... i never thought of a discontinuance that way, but yes, i guess it is that way :)

  • @RealAXork
    @RealAXork 3 місяці тому +1

    I expressed the complex x in the base using euler's form aka re^iphi and the one in the power in the a + bi form, then I got the equation e^(a*ln(r)-bphi)*e^i(aphi + b*lnr), for that to equal zero the term in the first power (that is a*ln(r)-bphi) has to go to negative infinity, so for exact values it is impossible, but if we take the limit of 0 + bi as b goes to infinity, it does equal zero because phi is bounded and the real part of 0 + bi is exactly zero.

  • @SuperDeadparrot
    @SuperDeadparrot 3 місяці тому +2

    I know as x->0, x^x goes to 1, it takes a dip to x=1 then shoots up past that. I know it only exists at negative integers for real solutions. The only way it could possibly get to 0 is at x=-inf. Right? How does it behave in the complex plane?

    • @Tzizenorec
      @Tzizenorec 3 місяці тому

      x^x = e^(x*(ln(-x)+iπ)) = e^(x*ln(-x) + xiπ)
      So as you go into increasingly negative numbers along the real line, you get an ever-shrinking spiral around the x-axis, which... yes, I'm pretty sure converges to 0 as you approach x=-∞.

  • @nocturnalvisionmusic
    @nocturnalvisionmusic 3 місяці тому +3

    Thumbnail said this video isn't your best, but it's still one of your best to me :)

    • @AndyBaiduc-iloveu
      @AndyBaiduc-iloveu 3 місяці тому +1

      The reasoning works for when a is the Gaussian integers or Q[i] but when it is all C I have to deal with x^i =0 so yeah..
      Thx for pointing out. 😊

    • @nocturnalvisionmusic
      @nocturnalvisionmusic 3 місяці тому

      @@AndyBaiduc-iloveu thanks 🤗

  • @Ekhyub
    @Ekhyub 2 місяці тому +1

    0:50
    From one side, 0^0 is 0 because 0^a is 0, from the other, 0^0 is 1, because a^0 = 1. So if we multiply the rules it will be (a^0)(0^a)/(0^a)(a^0)=0/0=1. For a big part of people, 0^0 is equal to 1, for a smaller fraction of them it's 0 or/and undefined, for the smallest one it's superposition between 1 and 0 or/and 0,5 (Corresponds with the mathematical "joke" that says 1+2+3+4...=-1/12).
    I count that 0^0 is undefined, because taking the power of a number is just taking it to the power of 1/(nth root power), so 2nd power is 1/0.5=2. If it's to the power of 0, then it's 1/0, which is undefined, and any operation with undefined gives just undefined in the output.

    • @Qbe_Root
      @Qbe_Root 15 днів тому

      You're making the same thing as what he showed with ln(x² - 1) = ln(0), you're introducing illegal operations and claiming it means the original thing is undefined. In fact by your logic, x^0 is undefined for any value of x, not just 0

  • @tsheringdorjigamming36
    @tsheringdorjigamming36 3 місяці тому

    Here comments is more difficult than videos. Please get this man a medal 🏅🏅. Salute professors.

  • @ttxxxxxxxxxxxxxxt
    @ttxxxxxxxxxxxxxxt 3 місяці тому +8

    (a+bi)^(a+bi) can be rewritten as (a+bi)^a*(a+bi)^bi now we know that either (a+bi)^a=0 or (a+bi)^bi=0 then (a+bi)^bi can be expressed as e^(bi*Log(a+bi)) where Log is the complex log (if f(z)=e^z then f^-1(z)=Log(z)) let's bring the formula for the complex Log: Log(z)=ln(|z|)+arg(z)i now now let's simplify e^(bi*ln(|a+bi|)+arg(a+bi)i) simplify a little bit more (e^bi)^(ln(|a+bi|)+arg(a+bi)i) now simplify
    e^(bi*ln(|a+bi|))*e^(bi*arg(a+bi)i) then just solve the equation e^(bi*ln(|a+bi|))*e^(bi*arg(a+bi)i) = 0 we get (cos(b)+sin(b)i)^ln(|a+bi|)*e(-b*arg(a+bi)) now (cos(b)+sin(b)i)^ln(|a+bi|) can't equal 0 because |cos(b)+sin(b)i|=1 for any b (= R and and for this to equal 0 |cos(b)+sin(b)i| must equal 0 now the second part e^(-b*arg(a+bi)) |e| =e so this is not equal to 0 for the same resons now let's come back to our original problem: either (a+bi)^a=0 or (a+bi)^bi=0 but we proven that (a+bi)^bi/=0 so (a+bi)^a=0 then |a+bi| = 0 then a+bi=0 then a=0 and b=0 then we get 0^0 which is either 1, undefined or has infinitly many values but for all cases 0^0/=0 so (a+bi)^(a+bi) has no solution

    • @ttxxxxxxxxxxxxxxt
      @ttxxxxxxxxxxxxxxt 3 місяці тому +4

      Extra: let's see if equation z^z has any solutions in set denoted as C(epsilon, phi) (I will type j for epsilon and k for phi) where j^2=0 where j /=0 j/j=1 , k=1/j and k^2 is undefined firstly let's define exponents: e^z=sum from: n=0 to: infinity z^n/n! then e^jz=sum from: n=0 to: infinity (jx)^n/n! = 1+jx+0+0+0+... =1+jx now the ln: ln(a+bj)=c+dj then e^(c+dj)=a+bj then simplify (e^c+(e^c)dj) = a+bj then form two equations e^c=a and e^c(dj) = bj solve the first one c = ln(a) substitute a(dj) = bj now ad=b now d = b/a so ln(a+bj)=ln(a)+(b/a)j

    • @ttxxxxxxxxxxxxxxt
      @ttxxxxxxxxxxxxxxt 3 місяці тому +4

      Now I don't want to to waste another 30 minutes of my life solving this equation so if someone sees that please solve this problem

  • @shadismith4549
    @shadismith4549 Місяць тому +2

    1:55 both arguments basically portray when an unstoppable force meets an immovable object

    • @Qbe_Root
      @Qbe_Root 15 днів тому

      Except x^0 = 1 holds whether x is positive, negative, complex, and you even get similar results (as in, the corresponding multiplicative identity) when x is something else like a matrix or a function
      While 0^x = 0 breaks as soon as Re(x) < 0

  • @dogbiscuituk
    @dogbiscuituk 3 місяці тому +2

    Instead of fields like Q, R or C, what about using a ring containing zero divisors? Does x^x=0 have solutions there?

    • @rainerzufall42
      @rainerzufall42 3 місяці тому

      Nice one! What about p-adic numbers?

    • @dogbiscuituk
      @dogbiscuituk 3 місяці тому +1

      @rainerzufall42 Ooh you've taken that too far 😃

    • @dogbiscuituk
      @dogbiscuituk 3 місяці тому +1

      I guess so. Integers modulo 4 for example. Let x=2. Then...
      x^x=2^2=4=0 modulo 4.

    • @rainerzufall42
      @rainerzufall42 3 місяці тому +1

      @@dogbiscuituk Yeah, this example was pretty obvious...

  • @dectorey7233
    @dectorey7233 3 місяці тому +12

    2 AM blackpenredpen? Dont mind if i do!

    • @craftcrewtv8094
      @craftcrewtv8094 3 місяці тому +3

      Woah my time when it got released was 9 am. I guess you're somewhere close to United States while I am in Europe

  • @magma90
    @magma90 3 місяці тому +1

    The magnitude of x^y for real numbers y, is |x|^y, and if we have the magnitude of a limit approaches 0, then we can safely say that the limit is zero. the limit as x-> -\infty of |x|^x, is also equal to the limit as x->\infty x^{-x} = 0, meaning that the magnitude is 0, therefore the limit as x-> -\infty x^x = 0

  • @FocusLRHAP
    @FocusLRHAP 3 місяці тому +1

    What about x^(x-a)=0 when a is a positive integer?

  • @pradipkumardas3060
    @pradipkumardas3060 3 місяці тому +2

    Please make a video on how to manually calculate complex answers of Lambert w function

  • @mathepunk
    @mathepunk 3 місяці тому +1

    With x = a + ib we have z = x^x = (a + ib)^(a+ib) = exp([½ a ln(a²+b²) - b theta] + [½b ln(a²+b²) + a theta] i), where theta = arg(a,b). We want │z│= exp(½ a ln(a²+b²) - b theta) = 0. To me it seems this is the case if a goes to minus infinity independent of b. (Or possibly if b goes to plus infinity ???). So my tentative answer would be x = - infinity + i b with b arbitrary. I checked (-50)^(-50) is very small, i.e. close to zero. (-50)^(-50) = 1.126 · 10^(-85). So it seems plausible.

  • @Angi_Mathochist
    @Angi_Mathochist Місяць тому

    ​@yurenchu it's the Banach limit. I'm not actually sure you can apply it to 0^0. It applies to sequences. The Banach limit of {0,1,0,1,...} is 1/2. This sequence does not converge under the usual sense of limits, but Banach limits extend the usual limits to non-convergent sequences. Not all sequences have uniquely defined Banach limits.

    • @yurenchu
      @yurenchu Місяць тому

      Thanks for your reply. I didn't receive a reply notification because your reply wasn't posted within the same thread, but luckily I happened to revisit this comment section and noticed it by accident.
      Interesting to read about the Banach limit, I'm not familiar with it, so thanks for that. It leaves me puzzled though how it can be applied to a function as described in your previous post (i.e. a real function f(x) that's 1 for rational x , and 0 for irrational x, and then show that the Banach limit of f(x) equals 1/2 for x approaching infinity).

  • @BruininksBart
    @BruininksBart 3 місяці тому

    Exploring Solutions in Modular Arithmetic: If we think about our number system as being modular (clock) arithmetic, it’s easy to show that there is a solution to `x^x=0`. Simply use the `modulus x`, because when `x` is raised to the power of `x`, it will always be congruent to `0 modulo x`. In other words, if we work in `modulo x`, then `x^x≡0`.
    No one specified what kind of numbers we were talking about, right? 😄

  • @Mosux2007
    @Mosux2007 3 місяці тому

    I wonder if there are any hypercomplex solutions? Or matrix solutions?

  • @RandyKing314
    @RandyKing314 3 місяці тому +3

    nice exploration of techniques!

  • @laitinlok1
    @laitinlok1 Місяць тому

    Can this be solved with dual numbers (e^2=0 and e≠0)?

  • @fernandojackson7207
    @fernandojackson7207 3 місяці тому +2

    Well, if you consider x>0, then we can use x^x= e^{x*lnx}, and e^y is never 0. Same goes for the Complex Exponential; 1 is the only value e^z doesn't take. I think this does it for X^x=0.

  • @kmsbean
    @kmsbean 3 місяці тому

    what if, we instead use (2n+1)? lim (2n+1) {[2n+1]^(2n+1} also approaches 0 (from the negative side), just as 2n approaches from the positive side

  • @DK-fn6xr
    @DK-fn6xr Місяць тому

    z^z = exp(z*Logg(z)) in complex numbers, where Log(z) is the principal branch of the complex logarithm. However, the modulus
    |z^z| = exp(Re(z*Log(z)))
    = exp((x/2)*ln(x^2 + y^2) - y*ArcTan(y/x))
    The expression in the exponent seems to have a local minimum for x = -1/e and y = 0, so it seems
    |z^z| >= 1/e, for all complex z = x + 1j*y

  • @klausolekristiansen2960
    @klausolekristiansen2960 3 місяці тому +7

    I have seen people on Quora get really angry when told that 0^0 is not one.

    • @ronaldking1054
      @ronaldking1054 3 місяці тому +3

      They should as the definition of the power is 0 multiplications from the original definition of 1. This is a formulation for the inductive proof of every single number. The fact that every other positive power of 0 is 0 is irrelevant to what the zeroth power of zero is.

    • @yiutungwong315
      @yiutungwong315 3 місяці тому

      ​@@ronaldking1054
      0^0 = 1 so i^0 = 1

    • @mhm6421
      @mhm6421 3 місяці тому +6

      @@ronaldking1054 0^0 is not a calculateable thing, you just define it, there is no real answer unless everyone agrees on it.

    • @ronaldking1054
      @ronaldking1054 3 місяці тому +3

      @@mhm6421 When the definition of power by simple integer is set up with 1 as the basis, it makes little sense to claim that 0^0 is not 1. You'd be putting in a branch for no reason. In other words, it is simpler.

    • @somenerd8139
      @somenerd8139 3 місяці тому +2

      @@ronaldking1054That definition was not just randomly defined, it comes from a pattern seen with exponents. For example, 2^4 = 2*2*2*2. 2^3 = 2*2*2. Based on this, we could assume that x^n = (x^(n+1))/x. By this logic, 0^0 = 0^1 / 0. Division by zero is undefined, and so this has no answer. We could define one, but not everyone will agree on the definition.

  • @roccocrocko1835
    @roccocrocko1835 2 місяці тому +1

    What actually changes from X to 2N?

  • @thexoxob9448
    @thexoxob9448 3 місяці тому

    I want you to discuss the hailstone sequence and Collatz conjecture and your opinion about it

  • @makermaker11
    @makermaker11 3 місяці тому

    Hey what if we make a new complex number especially for 0⁰? Same for the square root of negative 1 it also inteecept so why not make a new imaginary number for it?

  • @saaah707
    @saaah707 3 місяці тому

    how do you do that tap erase thingy? it's like a jump cut but your marker doesn't even move at all. mind blown

  • @aljawad
    @aljawad 3 місяці тому

    Use hypercomplex (dual) numbers?

  • @harshsaxena347
    @harshsaxena347 3 місяці тому

    when you talk about log(0), shouldn't it be negative infinity and not undefined?

  • @janbormans3913
    @janbormans3913 3 місяці тому

    Nice video, a bit in the style of your videos of a few years ago. Thank you!

  • @joyis9638
    @joyis9638 3 місяці тому

    Intuitively I knew that before your attempts to find a solution. My Calc teacher may have gone over this with me in college and I am just recalling the lesson.

  • @LichKingg23
    @LichKingg23 3 місяці тому

    Yeah, we arrived at the same conclusion. Lim -> - infinity but I didnt think of rational numbers, saying x belongs to Z solves it the best we can do. Now I have a question, cam we do lim in Complex numbers?

  • @chrisjust7445
    @chrisjust7445 2 місяці тому

    So I was thinking maybe you could try: x^x = 1 - 1 or maybe x^((x-1) + 1) = 0 or what if you take the xth root of both sides?

  • @maxrs07
    @maxrs07 3 місяці тому

    wouldnt any complex number z with r

    • @yurenchu
      @yurenchu 3 місяці тому

      No, because
      lim_{r--> 0} z^z = 1

  • @SharkyShocker
    @SharkyShocker 3 місяці тому +1

    Funnily enough if you put in Desmos x = 0 and 0^(1/x), then you end up with 0^(1/x) = 0
    I suppose from a coding perspective, Desmos sets priority that 0^anything overwrites the calculation that 1/0 is undefined. Just an interesting mention!

  • @omtux9235
    @omtux9235 Місяць тому

    subtract 1 from x^x and from 0, square both sides, take x^x as t, solve the quadratic equation and get that x^x = 1 +- i . multiply x^x and 1+- i by x^-1 and get that x = 1 +- i

  • @MitchBurns
    @MitchBurns 3 місяці тому

    I suspect it would be some kind of complex number. I’d have to look up the exact formula since it’s been a few years, but raising something to a complex power breaks down into the cos(x) + sin(x)i or something like that. Seems like the direction you will need to go to find a complex solution.

  • @skillhunter4804
    @skillhunter4804 3 місяці тому +1

    There is no finite complex solution
    If z = r* e^it, r and t real
    z^z = r^z * e^itz
    So the equation becomes r^z=0 or e^itz=0
    Case 1
    r^z=0=>|r^z|=0
    =>r^r =0, which has no solution as r is a positive real number
    Case 2
    e^itz=0
    => |e^itz|=0
    =>e^-ity=0
    Since t is between - pi and pi, y has to be - inf, thus no finite complex solution
    This also requires

  • @bluerendar2194
    @bluerendar2194 3 місяці тому

    For the original #6: it feels like the argument should be makeable based on that the magnitude of said expression feels like it should approach zero. If we restrict to rational values, then it should be trivial to argue - for irrational values, perhaps we can still show any branch approaches zero, and argue that way?

  • @geogeo6071
    @geogeo6071 3 місяці тому +3

    This is probably the best video you have ever done. It’s every way to approach the solution to an equation disguised as a search for the answer to x^x=0. Bravo. Beautiful.

  • @NotBroihon
    @NotBroihon 3 місяці тому

    I was thinking about higher dimensional number systems that have zero divisors and how you can prove that for example the complex numbers can't have them. Could you prove that no "zero exponents" can exist in these higher dimensional systems?

  • @mark.slater
    @mark.slater 3 місяці тому

    What about using dual numbers?

  • @almightysapling
    @almightysapling Місяць тому

    X= -infinity seems like a good solution to me. Even if we have to use complex numbers to define powers of negatives, they still approach 0 in absolute value

  • @TheCoroboCorner
    @TheCoroboCorner 3 місяці тому

    As for more (fruitless) ways to solve it, there were a couple I came up with (assuming you're using the extended complex plane)
    The first is pretty simple: if z^z = 0, then e^(z ln z) = 0, and thus z ln z = -infinity (we're in the extended complex plane so this is fine)
    The only value this works for is z = -infinity, so that's our solution
    The second is just substitution; set z = x+1, so (x+1)^(x+1) = 0
    Then (x+1)^x * (x+1) = 0, then x * (x+1)^x + 1 * (x+1)^x = 0
    So x * (x+1)^x = -(x+1)^x, meaning ln(x) + x ln(x+1) = ln(-1) + x ln(x+1)
    Assuming x =/= -1, we can remove x ln(x+1) from both sides, giving us ln(x) = ln(-1), which lets us conclude x = -1, which is a contradiction of our assumptions and thus the equation has no solutions

    • @methatis3013
      @methatis3013 2 місяці тому

      Why would
      z*ln(z) = -♾
      necessarily imply
      z = -♾?
      If we assume
      ln(-♾) = ln(-1) + ln(♾) = iπ + ♾, then we have
      -♾*ln(-♾) = -♾*ln(-1) + -♾² =
      = -i*♾ - ♾
      We end up with complex infinity here, or rather, unsigned infinity, not -♾

  • @thexoxob9448
    @thexoxob9448 3 місяці тому

    actually for no. 6, the 2n limit, it even works for x = n for n integer, but it approacges 0 from up then down, up the down, and so on (much like sin(x)/x as x approaches infinity

  • @TheReaperZX
    @TheReaperZX 2 місяці тому

    I understand almost nothing about these types of videos yet they interest me lmao

  • @SynaTek240
    @SynaTek240 Місяць тому +1

    why 2n and not just n?

  • @mrmc2345
    @mrmc2345 3 місяці тому

    I enjoyed this video a lot!!

  • @SiddharthNilakantan
    @SiddharthNilakantan 2 місяці тому

    that was some fun analysis!

  • @arthurclifford8290
    @arthurclifford8290 2 місяці тому

    I would think that it depends on whether you consider 0 a digit or an un-digit. If it is an un-digit then in base 0 the only digit is the un-digit 0. So 1 in base zero is the the un-digit 0. If however you think that 0 is a digit then base 0 has no digits and therefore there is no way to represent 1 in base 0 which I suppose is undefined.
    That might suggest that base 0 is not a base with no digits but rather a an unary base like base 1, except its only digit is 0 (nonary?). If we represent the base by prefixing the number (like in coding) with 0n (n for nonary) then 0n0 is nonary 1, 0n00 would be nonary 2, and 0n000 would be nonary 3 but would still evaluate to 0 in decimal because the value of the nonary value would be 0 times the number of digits which will always be 0.
    I think that at the very least one hast to define what a base 0 system is before using it in expressions.

  • @PakShuMan
    @PakShuMan Місяць тому

    My unserious part had a laugh at 7:50 ... I'll go stand in the corner... 😅

  • @excentrisitet7922
    @excentrisitet7922 3 місяці тому

    To me 0^0 is undefined even if we take a seventh grade approach.
    5^0 = 5^(m-m) = 5^m / 5^m.
    If we take 0 instead of 5 we get:
    0^m / 0^m
    m can be any number yet we still have undefined fraction of 0/0 which could be be approached if we have some context, like in sinc function: sin(x)/x when x -> 0. But here we don't have that context, so it remains undefined.
    P.S. I also was amazed that in English this limit doesn't have a special name. For example in Russian this limit is called "Первый замечательный предел" or "first remarkable (or wonderful) limit".
    And the second such limit is (1+1/x)^x when x-> infinity.

  • @KarlDeux
    @KarlDeux 2 місяці тому

    5:41, what does he say?

    • @hush3956
      @hush3956 2 місяці тому

      "can we conclude that it has no solution? Again, ehhh maybe not"

    • @KarlDeux
      @KarlDeux 2 місяці тому

      ​@@hush3956 ah OK, because I got focused on "eehjhhhhh" only.

  • @AntoninoGeraci-j3m
    @AntoninoGeraci-j3m 3 місяці тому +19

    I have a solution for this question, rewrite 0 as ln(1) and rewrite 1 as e^2πin, and the equation become x^x=2πin, and now you can clearly solve it whit W lambert function and take natral log on both side and become xln(x)=ln(2πn)+ln(i), and using lambert W function it become ln(x)=W(ln(2πn)+e^i(π/2+2πm)) and teh solution of the equazion x^x=0 is x=e^W(ln(2πn)+e^i(π/2+2πm)) with n,m integer

    • @Zeblinz
      @Zeblinz 3 місяці тому

      What happens when n and m are 0?

    • @AntoninoGeraci-j3m
      @AntoninoGeraci-j3m 3 місяці тому +4

      n cannot be equal to 0 because ln(0) is undefined

    • @xinpingdonohoe3978
      @xinpingdonohoe3978 3 місяці тому +19

      But that doesn't work because you've changed branch. You need the branch corresponding to n=0 to solve the initial equation.

    • @AntoninoGeraci-j3m
      @AntoninoGeraci-j3m 3 місяці тому

      @@xinpingdonohoe3978 i don't know bro but wolfram alpha said this is allowed

  • @hariommishra6543
    @hariommishra6543 3 місяці тому

    Hey man can you provide solution of imo problem 5

  • @xinpingdonohoe3978
    @xinpingdonohoe3978 3 місяці тому +2

    Take two complex numbers z and w. If z,w≠0, then z^x=w can be solved for x. If z=0 and w=0, then x>0. If z=0 and w≠0, x has no solutions. If z≠0 and w=0, x has no solutions.
    x^x=0 should have no solutions. Sure, if we take the limit as x→-∞, where -∞=inf R travelled along the negative real axis and not the general complex ∞, then we'd get a complex number of magnitude decreasing to 0, but we want numbers, not extended numbers.

  • @kevinlobjois5110
    @kevinlobjois5110 3 місяці тому +3

    I have a question concerning 0⁰ :
    We know that :
    X¹ = X, and X⁻¹ = 1/X.
    We also know that :
    X¹ * X⁻¹ = X⁽¹⁻¹⁾ = X⁰
    And X¹ * X⁻¹ = X¹ / X¹ = X / X
    So, X⁰ = X / X.
    And I've learned that there are convention saying that 0⁰ = 1.
    So, does that mean that 0 / 0 can be solved, using that logic specifically ?
    Or is something wrong here, like the proof that 1 = 2 ?

    • @p0gr
      @p0gr 3 місяці тому +2

      x^(-1) does not exist for x=0, so your argument fails at the very beginning.

    • @kevinlobjois5110
      @kevinlobjois5110 3 місяці тому

      @@p0gr Okay, thanks for the answer.

  • @bossm60
    @bossm60 Місяць тому

    my intuition before I watch the video, not sure if this will get anywhere: I learned early on in calculus that if you have an equation f(x) with no constants, you can rewrite it as e^ln(f(x)) to solve that issue.

  • @joshfotsch3314
    @joshfotsch3314 3 місяці тому

    if you put it into desmos, it will show a weird looking graph on the negative infinity side, and if you move it in any way it disappears!!! idk why

  • @prodromoskonstandas155
    @prodromoskonstandas155 3 місяці тому +1

    What I thought of is multiplying both sides by x so then we have x^x•x=0•x =>. x^(x+1)=0 but x^x is also 0 so x^(x+1)=x^x take ln both sides then cancel the lnx because x can't be one and then we have x+1=x witch definitely has no solutions

    • @nakellold
      @nakellold 3 місяці тому

      you cannot multiply both sides by x when one side is 0, if you could, nothing would make sense anymore, take x^2-1=0 and multiply both sides by x and you got x^3-x=0 and that doesnt work

    • @prodromoskonstandas155
      @prodromoskonstandas155 3 місяці тому

      @@nakellold if you don't include zero in the solutions Im pretty sure you can do that

    • @yurenchu
      @yurenchu 3 місяці тому

      You cannot take the ln at both sides when both sides equal 0 .
      For example, solve (x-1)^x = 0 . Clearly x=1 is a solution. But if we raise both sides to the power of 3:
      (x-1)^(3x) = 0 = (x-1)^x
      (x-1)^(3x) = (x-1)^x
      Take ln at both sides:
      ln( (x-1)^(3x) ) = ln( (x-1)^x )
      3x * ln(x-1) = x * ln(x-1)
      2x * ln(x-1) = 0
      2x = 0 OR ln(x-1) = 0
      x = 0 OR x-1 = 1
      x = 0 OR x = 2
      which are clearly not valid solutions to the original equation. Instead, these are extraneous solutions that were introduced by raising both sides to the power of 3 (before taking the ln ); but that's not the problem. The real problem is that the solution x=1 that we wanted to find, is now suddenly _gone_ . And this disappearance is caused by the step of taking the ln (when both sides equal 0).

  • @pleappleappleap
    @pleappleappleap 3 місяці тому

    Can it be proven whether or not y=x^x 8s continuous except at 0?

  • @MykhailoIvancha
    @MykhailoIvancha 2 місяці тому

    Maybe we should try making up numbers to solve this. Previous big updates were caused by addition, multiplication and exponentiation, while the problem in the video is caused by tetration. And the solution doesn’t have to be -♾️, it could be an orthogonal unit just like i

  • @jennymarx9228
    @jennymarx9228 Місяць тому +1

    Sir what if we take x = - ♾️
    Since x^x = - ♾️ ^ - ♾️
    = - 1
    _______
    ♾️^ ♾️
    = -1/ ♾️
    = Approaching 0 from negative side right??

  • @JJ_TheGreat
    @JJ_TheGreat 3 місяці тому +9

    1:16 Yeah, but then you also have 0^x = 0 - so don’t people also make an argument that it could also equal 0?

    • @tsouphs
      @tsouphs 3 місяці тому +1

      0^x=0 for x≠0

    • @JJ_TheGreat
      @JJ_TheGreat 3 місяці тому +4

      @@tsouphs The same thing for x^0...
      x^0=1 for x≠0, which this video was discussing...
      So I was making a point that you can also look at it as 0^x...

    • @andreimiga8101
      @andreimiga8101 2 місяці тому

      @@JJ_TheGreat x^0 is only undefined for x=0, it works with anything else. 0^x is only defined for positive real numbers, I think that's why it's not taken into account that much. 0^(-1) is undefined (it would literally mean 1/0), as well as 0^i (=e^i*ln(0)=e^-i*inf=cos(-inf)+i*sin(-inf) and that's undefined) or whatever else you come up with. In contrast, (-1)^0=1 and i^0=1.

  • @davidsaintjohn4248
    @davidsaintjohn4248 Місяць тому

    Please cover the lambert function

  • @keltan6511
    @keltan6511 3 місяці тому

    Pow(x,x ) is undefined for all Re(x)0 , will be defined ob Re(x)

  • @martyl1313
    @martyl1313 3 місяці тому

    There are 2 complex numbers that satisfy the equation. Before I state what they are, is it worth publishing?

  • @yashtelgar2883
    @yashtelgar2883 3 місяці тому

    if 1*log(1)=0
    but we know a*log (b) = b^a
    then 1*log(1) = 1^1= 1
    how this is possible please explain

    • @yurenchu
      @yurenchu 3 місяці тому

      You wrote " a*log(b) = b^a ", which is not correct. The correct identity is
      a*log(b) = log(b^a)
      which is valid for positive values of b.

  • @brunobuzzacchi1108
    @brunobuzzacchi1108 2 місяці тому

    Shouldn't you use
    lim (2n)^2n = 0
    n→-∞

  • @RatolokoMemo
    @RatolokoMemo 3 місяці тому +2

    no solutions in the complex, because if z is not 0 then by definition z^z = e^(z*ln z), and e^w is nonzero for any w complex

  • @Eichro
    @Eichro 3 місяці тому +1

    Been yelling "MINUS INFINITY" two minutes in. Glad I got close enough, too bad i hate complex numbers

  • @yousefmagableh3854
    @yousefmagableh3854 2 місяці тому

    Multiply both sides by x, and you reach a contradiction where you equate powers to get 1+x=x

  • @connorlaurings6367
    @connorlaurings6367 2 місяці тому

    In my mind 0⁰ is undefined because take 2, divide it by 2 you get 2/2 which is also equal to 2¹‐¹ =2⁰ (sorry for the poor notation I'm on mobile) the same can then be said for 0, take 0 divide it by 0 and get 0/0 which is undefined if I remember correctly and can be rewritten as 0⁰. That's my two cents but I may be wrong

    • @connorlaurings6367
      @connorlaurings6367 2 місяці тому

      You can use induction to prove that any number n divided by n is equal to n⁰ so 🤷‍♂️

  • @leodoms4682
    @leodoms4682 2 місяці тому

    What if we substitute 0^0 = 0^(1-1)
    0^0 = 0^1.0^(-1) = 0/0 . This is undefined, so 0^0 is also undefined ?

  • @nailedthepotato1941
    @nailedthepotato1941 3 місяці тому

    Why can't we use logarithmic definition to conclude there is no solution?
    Loga (b) = c a^c = b
    So
    x^x = 0 logx (0) = x
    And since log of 0 is undefined, then there are no solutions.

  • @AlucardNoir
    @AlucardNoir 2 місяці тому

    Those imaginary numbers made me think and I have a question for you: what id we define the answer? Imaginary numbers come from us defining the square root of -1 as i. What if we define a number m so that m multiplied by itself m times is 0. I mean, just at first glance, if we did that and kept the n root of 0 is zero it would mean raising something to a power and roots would no longer be whatever the math term is for their current relationship.
    Just a thought there. Feel free to ignore me though.