@@snowziethepro There’s a lot wrong with this video. But yes it’s suspicious to divide both sides by ln(1). It’s like having an equation 0*x = 0*y, and you say “I’ll divide both sides by 0, giving (0*x)/0 = (0*y)/0. Then you say on the left side that “the zeros cancel each other out” leaving just x, and on the right side “the zeros cancel each other out”, leaving just y. So x=y. Which is clearly not a conclusion we could draw from the original equation of 0*x = 0*y, because x and y could literally be any distinct real numbers and the equation is true.
This video has the same problem as the video from 2 weeks ago that tries to solve the equation 1^x = 5. Namely, the rules for taking the natural log of both sides of the equation are different for the complex plane.
@@8bpiyushn. e^(2 pi i) = 1 and naively taking ln of both sides seems to indicate that 2 pi i = 0. The problem is that exponentiation is not injective for complex numbers, like how squaring is not injective for integers. So inverting it requires us to pick a branch, like how we define sqrt to give back a nonnegative answer. The standard branch is ln(e^(x i)) = y i such that e^(x i) = e^(y i) and 0
correcting you, you cant multiply power of power in complex world, jst like that! sometime it works sometimes it doesn't when you did multiply x in to that wrong step
Dividing by Zero is undefined. Later steps arent Progressing correct way then. Keep the fine line between Math n Fiction. Enough examples proving 1=2 on utube like these.
@@rakeshdwivedi5256 complex numbers aren't real numbers, si they don't exist in our world but they are numbers and answers like these, of you got any farther than High school you should know...
One can run into contradiction w/ carelessly assigning multi-valued arguments to complex numbers. Suppose you wanna test the video's solution. 1^x → e^(2πimx) for m∊ℤ ⇒ 1^x=e^(m/k·ln3)=3^(m/k) which equals 3 only when m=k. So, we have to specify an argument range to keep functions single-valued. In this case, for arguments of complex numbers, θ, (2k-1)π
Applying the ln function to complex numbers as if they were real numbers is a glaring error. the ln function is defined on the set of strictly positive real numbers, if you want to use it elsewhere you must be able to extend its definition, something you have not done.
Exactly, using something that doesn't exist to prove something doesn't exist just doesn't make good sense. Giving these conditions, The sky outside is green, you just have to apply a mythical yellow filter over everything to show it. :) In effect, it's just click bait as to be truthful the equation would have been written as 1^i = 3. In short, this "solution" could have been done in almost any way you could think of, as long as you apply same thing to both sides after he introduced i into the solution as i(any operation)(any value)=i
This procedure found only a subset of the solutions, namely those on the imaginary axis. The other solutions get lost when the author takes the logarithm of both sides, as the complex logarithm needs to be taken, not the real one. Hence, denoting with Log(.) the complex logarithm and with log(.) the real logarithm, remembering that Log(z) = log(|z|) + i (phase(z) + 2m*pi) Log(e^(i 2k*pi*x))= Log(3) -> i 2k*pi * x = log(3) + i (0 + 2m*pi) -> x = m/k - i log(3)/(2k*pi) which, for all pairs of relative numbers (k,m) \in Z^2, gives all the solutions of the complex equation 1^x=3.
You are exactly correct. On the Right Hand Side he invalidly eliminates variable m by choosing it to be 0 and then forms the solution x in terms of integers k on the Left Hand side. Still there are an infinite k values he never discusses. He recognizes 1/ln(1) is infinity but also doesn't show why k not integer 0 but k an infinite set of integers is not related to trying to divide a real number by the real number 0. For the complete set of solutions means considering all integer ms on the right side of the equation (that can include his worked out case problem of m = 0 but now all infinite integers of m also) to be more complete in his already weak proof showing an infinite set of solutions for his ks. This all goes back to solving a set of solutions and forming 0 times a variable = 0 solution that says zero anything is zero. This mirrors proving (1)^k = (1)^m = 1 same proof which is a proof we already have done like 0 times k = 0 times m = 0 same proof also. The error results by this teacher finding infinite solutions for x and calls it a solution! 😂🤣
@@lawrencejelsma8118 The interesting part about this problem is that already the initial "equation" 1^z = 3 is actually not a singlle equation, but an infinite family of equations. This because, while the complex exponential with base e, i.e., f(z)=e^z has single-valued for all complex z, the general complex exponential (with base other than e), for example g(z)=1^z is a multi-valued function because it is defined as 1^z = e^(z*Log(1)) and the complex logarithm is a multi-valued function: Log(1) = ln(1) + i (phase(1)+ 2k*pi) = 0 + i (0+2k*pi) = i 2k*pi. Hence, by the very definition of the general complex exponential the original problem 1^z = 3 amounts to an infinite family of equations e^(i 2k*pi*z)=3, for k= +/-1, +/-2,+/-3,... Note that for k=0 there is no solution as e^0=1 and not 3. Now, for any fixed value of k, we have a single equation whose solutions can be computed by taking the complex multi-valued Log on bot sides: Log ( e^(i 2k*pi*z)) = Log(3), which gives i 2k*pi*z = ln(3) + i (0 + 2m*pi) and the solution z = m/k - i ln(3)/(2k*pi). So every single equation for fixed z has infinite solutions for m== 0, +/-1, +/-2,+/-3,...
@@italnsd ... I only know this easier understanding of what is going on with this peculiar mathematics that really isn't according to Complex Polar Euler's Formula from my Electrical Engineering that although 1 is expressed as a magnitude 1 at phases 0, 2kπ, 4kπ, 6kπ, etc. from oscilloscope measurements. The 3 is also magnitude 3 at its own 0, 2mπ, 4mπ, 6mπ, etc. With an electronic voltage outputs shown on an oscilloscope produced by electronic voltages inputs on an oscilloscope we still see this infinite wave signal voltages inputs producing infinite wave outputs in voltages. Electrical Engineering figured out that it shows an infinite results vs any particular results in analysis. When he said k not equal to zero situation then in Electrical Engineering it is equivalent to not analyzing the circuit output response where the input voltage is zeroed out! He just has to be a mathematics teacher who teaches math from the sciences in Physics to properly understand what he is teaching without erroneous suggestions. 😂🤣
As 1 can have a non-0 argument in the complex plane, so can 3. So, ln3 → ln3+2πin for n∊ℤ ⇒ x=n/k-i·ln3/2πk, k as in video. If we wanna treat the log function as a single-valued one from the same branch, then n=k. One can run into contradiction w/ carelessly assigning multi-valued arguments to complex numbers. Suppose you wanna test the above solution. 1^x → e^(2πimx) for m∊ℤ ⇒ 1^x=e^(2πimn/k)·e^(m/k·ln3)=3^(m/k)·e^(2πimn/k) which equals 3 only when m=k. So, we have to specify an argument range to keep functions single-valued. In this case, for arguments of complex numbers, θ, (2k-1)π
Además las propiedades de la potencia son aplicables a la definición de potencia considerando la rama principal del logaritmo de la base. De otra forma no se podría operar
El problema no es que divide por cero porque aclara que descarta el caso real. El problema está en que la expresión z exp(w) = e exp (w Ln z) se define para la rama principal del logaritmo de z. Con esa definicion valen las propiedades de operatoria de la potencia. El error esta cuando supone una solycion X y pasa de e exp (2kPi i)=1 a [e exp ( 2kPi i)]exp X=1 exp X Nota: con esa "solución" no se pueden aplicar las propiedades de operatoria de las potencias porque se llegaría a contradicciones del tipo: "Si 1 exp X = 3 entonces 3 = 27"
The k in the Euler Identity can be any integer, not just positive ones. k can be negative. k=0 is valid for the Identity, and in consideration of it, you once again find that you divided by zero.
@@johnmartin03355 1^-1million is equal to 1/1^1million which is 1. The limit as n approaches infinity of 1^n is indeed 1, but saying that 1^infinity = 1 is not true
I’m not comfortable with the step of dividing by ln(1). How about resorting to “dual numbers”, the bigger brother of complex numbers. [X + Y(d), where: d^2 = 0, d does not = 0].
1^(x+ i y)= 1 is true for arbitrary real or complex x and y r. Even if M_k==e^( i 2 k pi) =1 is true for k = 0,1,-1,2,-2 writing 1^z is alwayes interpreted as M_0^z. Why. If we use exponent a^z I have to know a and if a =1 => a=M_0 . Contrary for each number y and M=M_k we may write y^x = (y*M)^x but M_1^x =!= M_2^x thus the notation y^x is completly indefinete. Assumption that y^x might be replaced and discussed as (y M)^x, M= M_k is mathematical NONSENS. Of course we may say that we will discuss the solutions of x for equation M^x= 3 and M equal to M_k if k=!=0. For fixed k=!=0 the problem is standed CORRECTLY.
Hi! R=Thanks for your demonstration! Very interesting. I would only correct your designation for a set of solutions. You state that k ∈ Z and k ≠ 0 but the set Z of whole numbers includes 0. So a better way is to say k is positive and negative whole numbers or Z \ {0}.
Do you mean that to start studying math first I have to study law or something else? I think that you mean that certain problems are not for beginners.
Is (a^b)^c=a^(bc) only true when b and c are real numbers? x=ln3/ln1=ln3/0 This is not "no real solution." It is "no solution." No matter how you transform it by introducing imaginary numbers, the denominator of x is always 0.
As a suggestion, when determining no real solutions in this instance, stop the proof at xLn1 = Ln3 which is a falsehood, therefore, no real solutions. Dividing by Ln1 is not advisable practice as is cancelling Ln1/Ln1 as it can lead to illogical results that is shown in the renown proof of 1=2!! Good derivation of complex solutions. Nice work!
You forgot a whole degree of freedom. You should have used ln(3*e^(2πmi)) for all integers m instead of simply ln(3). So x = (2πm - ln(3)i)/2πk for all integers m and k, I believe
This is false, and it is a circular reasoning. He claims to demonstrate that 1^x can be equal to 3 (which is false). Then he says that since e^2kPi = 1 then (e^2kPi)^x = 3. He uses his false hypothesis as validation of his calculation. For me this error is voluntary and intended to provoke comments and thus increase the ranking of this video, so it is highlighted by TH-cam therefore more clicks therefore more money.
in the third line you devided the both sides dy ln(01) we knew that ln(1)=0 and it's not allowed to devide by zero However, the result will always be undefined because ln(1) = 0. Dividing any number by 0 is mathematically undefined please waiting ur replying to this issue
@@waelalsaleh75 bros threatening a teacher trying to explain a question just for fun if ur not entertained ur more than welcome to unsubscribe its not like anythings gonna change if u unsubscribe lmfaao
He doesnt want that, It isnt and answer, it's like posing a question and as solution saying "the solution of the question Is the answer of the question"
This is geometrically and algebraically senseless. Division by zero is not allowed in both reals and complex numbers. Raising 1 to any number whether real or complex will only be one in magnitude whether real or complex. The only difference between reals and complex numbers is in the orientation but not in the magnitude in this case. Further, k doesn’t even have to be a whole number. Still nothing would be different if we assumed otherwise.
I loved the video and was annoyed by the amount of people that wasn't able to keep listening from "divide both sides by ln(1)" on. I wonder if it would have any impact to state that ln() is a *multibranch function* in the complex from the very beginning (the emphasis is not for you, but rather for all the "but ln(1) is zero!" people) and that all other *non-zero* possible values of ln(1) could actually divide both sides, so all this frisson about dividing both sides by 0 could not dominate the comment section. Math constrained to the real number is so deeply ingrained in people's mind that they could not even keep listening from that point on. However, on the bright side, all that fuss generated engagement to the video, so kudos anyway 😆
Either this is a bad joke, or there is time to learn to respect 1 as the multiplication identity and learn how to use logarithms of complex numbers!!!! For complex number in form r*e^(iθ) the logarithm is: Ln(re^(iθ)) = ln(r) + i(θ + 2πk), not just (iθ). In this case r = 1, and we can choose θ = 0 and k = 0, and we are back with ln(1). The ‘magic’ of complex number disappears and there is no way to continue with this nonsense. There is no such a thing like a complex number with selected values of k so elimination of k=0 is not possible.
This is for a reason called "higher mathematics". Complex numbers cannot be visualised using ordinary mind. This can be compared to non-Euclid ^n dimensional geometry. If you can imagine 4th, 5th,... Nth dimension, then you can imagine space of complex numbers. So yes, 1^x=3 can be solved, but not in "our world", hope that make sense..
It's a big mistake from the begining. (ln 1 / ln 1 = 0/0). You can't demonstrate nothing using that. The best thing you should do is erasing this video.
@@onkotonkobluMaths is human creation to explain reality. With a real number of something there is no point to equation. Essentially the question is what is the answer to infinity multiplied by zero? Anything multiplied by zero is zero. But anything multiplied by infinity is infinity. All this video equation shows is that infinity multiplied by zero can be any number… including 3. Think of the universe. The size of space tends to infinity. Most of everything is nothing… yet here we are.
Horrible title. Math has to be "for beginners" to attract new enthusiasts. Are you intentionally promoting math as an elitist undertaking, only for those with pre-existing expertise? May I suggest "This Problem is not for Beginners"?
isnt it already wrong to divide by ln1????
No not wrong
We are using to find if x exist or not,by dividing we find x has a complex root.
@@snowziethepro There’s a lot wrong with this video. But yes it’s suspicious to divide both sides by ln(1). It’s like having an equation 0*x = 0*y, and you say “I’ll divide both sides by 0, giving (0*x)/0 = (0*y)/0. Then you say on the left side that “the zeros cancel each other out” leaving just x, and on the right side “the zeros cancel each other out”, leaving just y. So x=y. Which is clearly not a conclusion we could draw from the original equation of 0*x = 0*y, because x and y could literally be any distinct real numbers and the equation is true.
It is wrong ln1 is zero and we can not divide by zero
Yeah, but the divide by 0 wasn't really a part of the procedure to the supposed answer they found, so this might be looking at the wrong issue.
@@Aristotle000001 True but he did divide by zero to “show” that there are no real solutions. So yeah, a broken clock is right twice a day I guess.
This video has the same problem as the video from 2 weeks ago that tries to solve the equation 1^x = 5. Namely, the rules for taking the natural log of both sides of the equation are different for the complex plane.
Please ELABORATE
Yes u are right I've watched that
@@8bpiyushn. e^(2 pi i) = 1 and naively taking ln of both sides seems to indicate that 2 pi i = 0. The problem is that exponentiation is not injective for complex numbers, like how squaring is not injective for integers. So inverting it requires us to pick a branch, like how we define sqrt to give back a nonnegative answer. The standard branch is ln(e^(x i)) = y i such that e^(x i) = e^(y i) and 0
@@8bpiyushn. It’s really too complicated to fully explain in a TH-cam comment, but @deityblah gives the summary.
I agree. This is false math, like the 1^x = 5 video. There is no complex solution to 1^x = 3.
I recall watching a mathematician prove 1=2. He afterwards pointed to a particular step where he made the sleight of hand.
correcting you, you cant multiply power of power in complex world, jst like that! sometime it works sometimes it doesn't when you did multiply x in to that wrong step
What doesnt work is a^(b^c) = a^c^b. It may or may not work for complex numbers. I haven’t seen the video, too simple for me
Yeah
Diving both sides by zero is futile.
Yes, he said it
@onkotonkoblu why
Dividing by Zero is undefined.
Later steps arent Progressing correct way then.
Keep the fine line between Math n Fiction.
Enough examples proving 1=2 on utube like these.
@@rakeshdwivedi5256 complex numbers aren't real numbers, si they don't exist in our world but they are numbers and answers like these, of you got any farther than High school you should know...
1^X = 1
X should be all Real and Complex Number
How do you divide both sides by zero at 0:57
One can run into contradiction w/ carelessly assigning multi-valued arguments to complex numbers. Suppose you wanna test the video's solution.
1^x → e^(2πimx) for m∊ℤ ⇒ 1^x=e^(m/k·ln3)=3^(m/k)
which equals 3 only when m=k. So, we have to specify an argument range to keep functions single-valued. In this case, for arguments of complex numbers, θ,
(2k-1)π
can you please check your found solution in the main equation (1^x=3) ?
You can never divide by "ln 1" or "0" without corrupting the set of solutions!
Man, I need that pen
in wolframalpha, the result of 1^x=3 is "False", 1 power to anything except infinity is 1.
At infinity also it is 1
Applying the ln function to complex numbers as if they were real numbers is a glaring error. the ln function is defined on the set of strictly positive real numbers, if you want to use it elsewhere you must be able to extend its definition, something you have not done.
actually you can go through "illegal" steps if you check the solution afterwards, so it makes some sense to proceed like this.
As others have pointed out, more or less, when you take the log on both sides the LHS is (x times) zero while the RHS is log 3.
The answer is undefined. Stop playing nonsense. 1 power to anything is just 1.
Exactly, using something that doesn't exist to prove something doesn't exist just doesn't make good sense. Giving these conditions, The sky outside is green, you just have to apply a mythical yellow filter over everything to show it. :) In effect, it's just click bait as to be truthful the equation would have been written as 1^i = 3. In short, this "solution" could have been done in almost any way you could think of, as long as you apply same thing to both sides after he introduced i into the solution as i(any operation)(any value)=i
This procedure found only a subset of the solutions, namely those on the imaginary axis. The other solutions get lost when the author takes the logarithm of both sides, as the complex logarithm needs to be taken, not the real one. Hence, denoting with Log(.) the complex logarithm and with log(.) the real logarithm, remembering that Log(z) = log(|z|) + i (phase(z) + 2m*pi)
Log(e^(i 2k*pi*x))= Log(3) -> i 2k*pi * x = log(3) + i (0 + 2m*pi) -> x = m/k - i log(3)/(2k*pi) which, for all pairs of relative numbers (k,m) \in Z^2, gives all the solutions of the complex equation 1^x=3.
You are exactly correct. On the Right Hand Side he invalidly eliminates variable m by choosing it to be 0 and then forms the solution x in terms of integers k on the Left Hand side. Still there are an infinite k values he never discusses. He recognizes 1/ln(1) is infinity but also doesn't show why k not integer 0 but k an infinite set of integers is not related to trying to divide a real number by the real number 0.
For the complete set of solutions means considering all integer ms on the right side of the equation (that can include his worked out case problem of m = 0 but now all infinite integers of m also) to be more complete in his already weak proof showing an infinite set of solutions for his ks. This all goes back to solving a set of solutions and forming 0 times a variable = 0 solution that says zero anything is zero. This mirrors proving (1)^k = (1)^m = 1 same proof which is a proof we already have done like 0 times k = 0 times m = 0 same proof also.
The error results by this teacher finding infinite solutions for x and calls it a solution! 😂🤣
@@lawrencejelsma8118 The interesting part about this problem is that already the initial "equation" 1^z = 3 is actually not a singlle equation, but an infinite family of equations. This because, while the complex exponential with base e, i.e., f(z)=e^z has single-valued for all complex z, the general complex exponential (with base other than e), for example g(z)=1^z is a multi-valued function because it is defined as 1^z = e^(z*Log(1)) and the complex logarithm is a multi-valued function: Log(1) = ln(1) + i (phase(1)+ 2k*pi) = 0 + i (0+2k*pi) = i 2k*pi. Hence, by the very definition of the general complex exponential the original problem 1^z = 3 amounts to an infinite family of equations e^(i 2k*pi*z)=3, for k= +/-1, +/-2,+/-3,... Note that for k=0 there is no solution as e^0=1 and not 3. Now, for any fixed value of k, we have a single equation whose solutions can be computed by taking the complex multi-valued Log on bot sides:
Log ( e^(i 2k*pi*z)) = Log(3), which gives i 2k*pi*z = ln(3) + i (0 + 2m*pi) and the solution z = m/k - i ln(3)/(2k*pi).
So every single equation for fixed z has infinite solutions for m== 0, +/-1, +/-2,+/-3,...
@@italnsd ... I only know this easier understanding of what is going on with this peculiar mathematics that really isn't according to Complex Polar Euler's Formula from my Electrical Engineering that although 1 is expressed as a magnitude 1 at phases 0, 2kπ, 4kπ, 6kπ, etc. from oscilloscope measurements. The 3 is also magnitude 3 at its own 0, 2mπ, 4mπ, 6mπ, etc. With an electronic voltage outputs shown on an oscilloscope produced by electronic voltages inputs on an oscilloscope we still see this infinite wave signal voltages inputs producing infinite wave outputs in voltages. Electrical Engineering figured out that it shows an infinite results vs any particular results in analysis. When he said k not equal to zero situation then in Electrical Engineering it is equivalent to not analyzing the circuit output response where the input voltage is zeroed out! He just has to be a mathematics teacher who teaches math from the sciences in Physics to properly understand what he is teaching without erroneous suggestions. 😂🤣
@@italnsd x= [ ln(3)+2kπi ]/2nπi , ,n ∈ Z ∖{0}, k ∈ Z by chatgpt, although I have to point out to chatgpt that n can not be 0
As 1 can have a non-0 argument in the complex plane, so can 3. So,
ln3 → ln3+2πin for n∊ℤ ⇒ x=n/k-i·ln3/2πk, k as in video.
If we wanna treat the log function as a single-valued one from the same branch, then
n=k.
One can run into contradiction w/ carelessly assigning multi-valued arguments to complex numbers. Suppose you wanna test the above solution.
1^x → e^(2πimx) for m∊ℤ ⇒ 1^x=e^(2πimn/k)·e^(m/k·ln3)=3^(m/k)·e^(2πimn/k)
which equals 3 only when m=k. So, we have to specify an argument range to keep functions single-valued. In this case, for arguments of complex numbers, θ,
(2k-1)π
Además las propiedades de la potencia son aplicables a la definición de potencia considerando la rama principal del logaritmo de la base. De otra forma no se podría operar
El problema no es que divide por cero porque aclara que descarta el caso real. El problema está en que la expresión z exp(w) = e exp (w Ln z) se define para la rama principal del logaritmo de z. Con esa definicion valen las propiedades de operatoria de la potencia.
El error esta cuando supone una solycion X y pasa de
e exp (2kPi i)=1
a [e exp ( 2kPi i)]exp X=1 exp X
Nota: con esa "solución" no se pueden aplicar las propiedades de operatoria de las potencias porque se llegaría a contradicciones del tipo:
"Si 1 exp X = 3 entonces 3 = 27"
The value of x you find is also valid for negative integers also then why you sholud take only k as positive integer only
Shouldn't you check your working with open ai (CGT)? Sometimes it gets answers wrong, but it's a big help in checking an answer.
The k in the Euler Identity can be any integer, not just positive ones. k can be negative. k=0 is valid for the Identity, and in consideration of it, you once again find that you divided by zero.
You should have stopped the moment you had ln 1 in the denominator. Everything after that was wrong.
i believe that 1^infinity is still 1
Prove it
@@ivanmolnar9306 well 1^1million=1x1x1x1..... into 1 million...
okay 1^-1million will give us fractions it's less than one😂😂
@@johnmartin03355 1^-1million is equal to 1/1^1million which is 1.
The limit as n approaches infinity of 1^n is indeed 1, but saying that 1^infinity = 1 is not true
@@ivanmolnar9306 pls go and read ur book,,..
this is common sense🤦🤦
@@ivanmolnar9306 you prove it then
Wouldn't it be useful to write 3 in polar form?
your solution is correct, Just a small osbervation, when you have x*Ln1=Ln 3 can't divide both sides by Ln1 becase Ln1= 0 and that is undefined.
undefined in normal world, but in complex world, its real.
If x is an any complex number, we have 1^x = e^(xln1).
Since xln1= 0 so e^0 =1, therefore 1^x = 1
when I replaced x by its value it gave me 1 not 3.
It must be an error somewhere.
I’m not comfortable with the step of dividing by ln(1). How about resorting to “dual numbers”, the bigger brother of complex numbers. [X + Y(d), where: d^2 = 0, d does not = 0].
Do you know what an integer is?
This article is right within complex analysis.
Absolutely right, very tricky question,
After cancelling i2kpi on LHS where did “en e” gone
1^(x+ i y)= 1 is true for arbitrary real or complex x and y r. Even if M_k==e^( i 2 k pi) =1 is true for k = 0,1,-1,2,-2 writing 1^z is alwayes interpreted as M_0^z. Why. If we use exponent a^z I have to know a and if a =1 => a=M_0 . Contrary for each number y and M=M_k we may write y^x = (y*M)^x but M_1^x =!= M_2^x thus the notation y^x is completly indefinete.
Assumption that y^x might be replaced and discussed as (y M)^x, M= M_k is mathematical NONSENS. Of course we may say that we will discuss the solutions of x for equation M^x= 3 and M equal to M_k if k=!=0. For fixed k=!=0 the problem is standed CORRECTLY.
You should run for Nobel Prize in Mathematics!
Hi! R=Thanks for your demonstration! Very interesting. I would only correct your designation for a set of solutions. You state that k ∈ Z and k ≠ 0 but the set Z of whole numbers includes 0. So a better way is to say k is positive and negative whole numbers or Z \ {0}.
Do you mean that to start studying math first I have to study law or something else?
I think that you mean that certain problems are not for beginners.
No solution
Also you divide by zero
@@cal18338u can't
@@cal18338: If you watch the entire video, you’ll see that he doesn’t divide by zero.
Is (a^b)^c=a^(bc) only true when b and c are real numbers?
x=ln3/ln1=ln3/0 This is not "no real solution." It is "no solution."
No matter how you transform it by introducing imaginary numbers, the denominator of x is always 0.
As a suggestion, when determining no real solutions in this instance, stop the proof at xLn1 = Ln3 which is a falsehood, therefore, no real solutions. Dividing by Ln1 is not advisable practice as is cancelling Ln1/Ln1 as it can lead to illogical results that is shown in the renown proof of 1=2!! Good derivation of complex solutions. Nice work!
You forgot a whole degree of freedom. You should have used ln(3*e^(2πmi)) for all integers m instead of simply ln(3).
So x = (2πm - ln(3)i)/2πk for all integers m and k, I believe
One too many assumptions. Simply assume 1^x = 3.
That will reduce the labour you put.
The numbers like 1,2,3,...,n... are N ( natural) not Z.
THE ARE ALSO Z BRO, N Is contained in Z
@@onkotonkobluand?
this video is of the kind: find the error!
1^X = 1
X should be all Real and Complex Number...
"In math everything is possible." You just showed that to be false. You tried to divide a number by zero and correctly said that's not possible.
You state k is from Z, but then list natural numbers
incomplete solution.
full solution is x= k2/k -i ln(3)/2*k*pi with k2 in Z and k in Z*
Beautiful problem
Are you dividing zero by zero?
Greetings. Thank you very much.
1^x=3 1^(3)=(1)^3=(1)^3=1 (x ➖ 1ix+1i)
How can you divide ln1 as it's value is 0 isn't it wrong
He said that fucking hell
@@onkotonkoblubros said bad word
wolframalpha says that 1^x = 3 is false.
This is false, and it is a circular reasoning.
He claims to demonstrate that 1^x can be equal to 3 (which is false).
Then he says that since e^2kPi = 1 then (e^2kPi)^x = 3. He uses his false hypothesis as validation of his calculation.
For me this error is voluntary and intended to provoke comments and thus increase the ranking of this video, so it is highlighted by TH-cam therefore more clicks therefore more money.
wolframalpha says that 1^x = 3 is false.
1^x=3 is false, but what if u enter 1=3^(1/x)
Fine until making i^2 at end
in the third line you devided the both sides dy ln(01)
we knew that ln(1)=0
and it's not allowed to devide by zero
However, the result will always be undefined because ln(1) = 0. Dividing any number by 0 is mathematically undefined
please waiting ur replying to this issue
explain it to me before I unsubscribe
@@waelalsaleh75 bro who are u unsubscribe lmao
@@waelalsaleh75 bros threatening a teacher trying to explain a question just for fun if ur not entertained ur more than welcome to unsubscribe its not like anythings gonna change if u unsubscribe lmfaao
Tell me you don't know math without telling me, you don't know math
X=ln(3)/[i(2kpi)]
0:15
"In maths, everything is possible."
What 10/0 ?
No solution, by inspection. 🧐
This is an example of.... OXYMORON.... equation.
How do you devide by ln1 😂
sorry, i checked in matlab and not a single k matches...
He’s STILL putting “no real solutions” where the answer should be “undefined.” I give his channel an F minus.
He doesnt want that, It isnt and answer, it's like posing a question and as solution saying "the solution of the question Is the answer of the question"
@@onkotonkobluwat
This is geometrically and algebraically senseless. Division by zero is not allowed in both reals and complex numbers. Raising 1 to any number whether real or complex will only be one in magnitude whether real or complex. The only difference between reals and complex numbers is in the orientation but not in the magnitude in this case. Further, k doesn’t even have to be a whole number. Still nothing would be different if we assumed otherwise.
X = max
Divide by Zero n all is gone.
@biscuit is that you??
Please stop circulating eye catching but incorrect math "proof".
1^x = 1 ln1=0
Log1(3)
So you said that all k in z set is solution
And it gave the same result for x 😂😂😂😂😂😂😂😂 wtf
Thank You for watching! Have a great day! A great question with complex numbers!
I loved the video and was annoyed by the amount of people that wasn't able to keep listening from "divide both sides by ln(1)" on.
I wonder if it would have any impact to state that ln() is a *multibranch function* in the complex from the very beginning (the emphasis is not for you, but rather for all the "but ln(1) is zero!" people) and that all other *non-zero* possible values of ln(1) could actually divide both sides, so all this frisson about dividing both sides by 0 could not dominate the comment section.
Math constrained to the real number is so deeply ingrained in people's mind that they could not even keep listening from that point on.
However, on the bright side, all that fuss generated engagement to the video, so kudos anyway 😆
you cannot divide by ln1 ( ZERO).
No real solution. 1 raised to any real power = 1. Not 3.
YES THE ANSWER ISNT REAL DINGUS
@@onkotonkoblume TOOOO!!
ln 1 = 0
Either this is a bad joke, or there is time to learn to respect 1 as the multiplication identity and learn how to use logarithms of complex numbers!!!!
For complex number in form r*e^(iθ) the logarithm is: Ln(re^(iθ)) = ln(r) + i(θ + 2πk), not just (iθ).
In this case r = 1, and we can choose θ = 0 and k = 0, and we are back with ln(1). The ‘magic’ of complex number disappears and there is no way to continue with this nonsense.
There is no such a thing like a complex number with selected values of k so elimination of k=0 is not possible.
A proposta dessa igualdade é uma piada !😅
Hi! 1 = exp^(i*2*k*π) and your solution looks like true! But!!! Also, 1 = n^(i*2*k*π/log(n)) and if n is not exp, your solution is wrong!
How can 1 power something be another value. It's impossible. There is no solution for this problem.
If you divide by 0 like this guy, then everything is possible.
@@vadimkern5836which guy is it?
This is for a reason called "higher mathematics". Complex numbers cannot be visualised using ordinary mind. This can be compared to non-Euclid ^n dimensional geometry. If you can imagine 4th, 5th,... Nth dimension, then you can imagine space of complex numbers. So yes, 1^x=3 can be solved, but not in "our world", hope that make sense..
@@stjepannikolic5418n dimensional geometry is Euclidean
Getting a bit of déjà vu...
C is a field: 1^z=1, for z € C
Solution is ∞
No
@@onkotonkobluhow do I know
No, this is not a tricky question at all. And it is essentially the same problem that you did a short while ago. Another 10 minutes wasted.
It's a big mistake from the begining. (ln 1 / ln 1 = 0/0). You can't demonstrate nothing using that. The best thing you should do is erasing this video.
The division by zero is just to conclude that there are no real numbers that satisfy the equation, it is not an actual solution
Impossibile
you can't divide by in1 by reason in1=0
So all is wrong y bro,
HE SAID IT'S WRONG AT THE BEGGINNING OF THE FUCKING VIDEI, WHAT IS UP WITH YOU PEOPLE?
He said the bad word again.
Infinity.
Not a number
@@onkotonkobluExactly.
@@guitartommo2794 it's not AN ANSWER, the answer Is complex
@@onkotonkobluso?
@@onkotonkobluMaths is human creation to explain reality. With a real number of something there is no point to equation. Essentially the question is what is the answer to infinity multiplied by zero? Anything multiplied by zero is zero. But anything multiplied by infinity is infinity. All this video equation shows is that infinity multiplied by zero can be any number… including 3. Think of the universe. The size of space tends to infinity. Most of everything is nothing… yet here we are.
Horrible title. Math has to be "for beginners" to attract new enthusiasts. Are you intentionally promoting math as an elitist undertaking, only for those with pre-existing expertise? May I suggest "This Problem is not for Beginners"?
Vous posez des problèmes bizarres vous savez que log1=0 vous cherchez midi à quatorze heure
Log 1 is 0, explanation is wrong
Great sir
Nonsense! You divided by zero 🤣
I feel this is just wrong, completly nonsence. The solution should be undefined.
no, just stop
You speak to fast
No useful
Great solutions😮😊😅😢
NAtural....and NOT NAYtural !!! Why can't your pronunciation be correct.