Everything is possible in math
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- čas přidán 7. 09. 2024
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@higher_mathematics
#maths #math
If you check your 'solution' in Python or the TI-Nspire CX calculator (both of which I checked) you will find that 1^x is 1, not 5. The problem with your solution is that you use some 'identities' that are not always true for complex numbers. In particular, it is not always true that ln(a^b) = b * ln(a) in the complex numbers. This is because the inverse "function" of e^x is a "multivalued function" with infinitely many results (usually), and to make it an actual function we choose just one of the results.
The actual, usual definition of a^b in complex numbers is e^(b*ln(a)) where ln(a) is the principle logarithm of a and no other logarithm. The principle logarithm of 1 is just 0 (or 0 + 0i) so we always get
1^x = e^(x*ln(1)) = e^(x*0) = e^0 = 1.
You can never get 1^x = 5 unless you redefine complex exponentiation. The usual redefinition is to make it multivalued, in which case 5 would be one of the results of 1^x where x is your solution value.
So, the Python is not so powerfull as you thought. The solution is abs. correct.
@@ahojg It is possible that both Python and I (with the TI-Nspire CX calculator and Wolfram Alpha) are wrong in this matter. Do you have any evidence that is so? Can you point to any authoritative method of calculating complex exponentials that yields 1^x = 5 for the video's value of x?
@@rorydaulton6858 in wolfram alpha, take the 1/x power of both sides.
In any case, why trust an algorithm when you can actually do the math?
@@spudmcdougal369 What do you mean? If you type either
1^x where x = -i×log(5)/(2 π) or if x = -i*ln(5)/(2π) what is 1^x
into Wolfram Alpha you get the answer 1.
I'm also not sure what you mean by "take the 1/x power of both sides". Another "identity" that is not true in the complex numbers is (a^b)^c = a^(b*c). For example, check [(-1)^2]^(1/2) which is 1 versus (-1)^[2*1/2] which is -1. So it is not always true that (a^x)^(1/x) = a. I think that is what you are trying to do.
To check a calculation you need to actually check the calculation from the beginning. Try 1^x for any complex value of x that you like. Doing anything else invites errors. That's what I teach my math students to: check the very first line, not later in the argument.
Exp(i*2k*pi)=1*(cos(2k*pi) + i*sin(2k*pi)), so ln(exp(i*...))≠(i*...)*ln(e), but ln(exp(i*...))=ln(cos(2k*pi) + i*sin(2k*pi))=ln(cos(2k*pi))=ln(1)=0. Why? Because ln(exp(kx))=x*ln(k) is valid only for real nums. And for complex nums exp(z)=exp(x + iy)=r*(cosΩ + i*sinΩ), where r=sqrt(x**2 + y**2) & Ω=atan(y/x). And "length" of complex num or radius is r, while cosΩ + i*sinΩ is Ω-angle rotation of complex num radius on complex plain & radius of cosΩ + i*sinΩ is always 1
Everything is possible in math, except your demonstration, which is false. When you modify an equation to find solution (what you does while taking ln of both parts), you should check the solution, which should here be rejected. In complex numbers, ln a = ln b does not imply a=b.
If you accept Cantor's theorem of infinities or the sum 1+2+3+...=-1/12, then you should accept this
@@david-melekh-ysroel Of course I accept Cantor's theorem but not the divergent series (which is just a notation error). In this case here there is a hidden division by zero.
@@tontonbeber4555 Cantor is totally wrong when he said N=Z
Z is clearly equal to 2N
My tutor told me that between any 2 infinities there is an infinity amount of infinities. Take it or leave it
@@david-melekh-ysroel
Cantor is not a real problem. Even the equation x=2x on extended-to-infinity numbers admits 2 solutions : zero and infinity,
But it's even worst than you think because N=Q despite the fact that there are an infinity of Q numbers between two consecutive integers. In fact N=the set of any number you can describe in any language, even if you cannot calculate them. But what is beautiful in Cantor is not that all describable numbers are countable, it is that R is not countable and so R>N
The sol. is abs. correct! Each complex number has infinite versions of itself, including real numbers when treating them in C. So for example 1^i has infinite number of expressions, not just one like 1^x=1 in R. This is how to handle with CN contrary to RN.
The identity property of multiplication should be not used for assumption that 1^x = 5 because this mean that (1*a) it is not necessary equal (a). Therefore, number 1 shall be kept explicitly in multiplications. The full representation of imaginary is r*e^iθ. We have r=1. Therefore, the complex number from minute 8:00 of the presentation is not e^i2πk but 1*e^i2πk. Taking power of x, we have (1^x)(e^i2πkx), not just e^i2πkx. After taking Log we have ln(1^x) +i2πkx, not just i2πkx. Since 1^x = 5, the ln(1^x)=ln(5) and as not equal to zero ln(1^x) cannot be eliminated from the sum. ln(1^x)=0 makes further evaluations not valid.
After assumption that 1^x=5, there is no place for (otherwise proper) assumption that 1^x=1.
next video: x/0 = 5
The problem is ill-posed since the start. I’m not a mathematician but I immediatly noticed that something is missing : in which space do you want to solve for x? R, C, I, N, Z ? Even more fun D, Q, H, O, S ?….
It is pretty clear that there is no sol. in R, but inf# of sols in C. And it is also mentioned in the video, so the comment is weird..
@@ahojgyou are absolutely incorrect. There is 0 solutions in complex numbers
@@methatis3013 u dont know the complex math, sorry
@@methatis3013 Have you seen the video?
@@shxvuls764 yes, I have. And the "solution" in the video is simply incorrect. I posted a relatively lengthy comment on why the solution doesn't work and it ultimately comes down to subtly switching the order of "there exists" and "for all" in the statement of the problem. I have no doubt you will find my comment with a bit of scrolling
Do you know what you are doing? if that's an exam, you will get a zero because you did not solve anything. The simple answer is such an equation has no solution period. No value of x will ever produce 5. 1 to any exponent will always equal 1.
Legit enough like log base 1 (x) is not defined
@@HUNTER-qn5ff Not legit, since no value of x can verify the equation. 1^x is always 1 regardless of x
He incorrectly writes out at the video's 6:50 mark (e^(i2kπ)^x on the left side of the equation as two parts magnitude 1 phase 2k^π and just leaves it as magnitude 5 on the right side of the equation without any 2nπ phase if m and n aren't going to be the same intwgers. With the magnitude of 5 in Euler's Formula for complex numbers it also has to have an Euler phase or angle direction as the left side of the equation was written in.
@@lawrencejelsma8118 The problem is that his solution does verify the equation. 1 to any power is simply 1. Therefore, there is no solution
In his solution he has converted in a complex number but rhs is real actually he should also covert into complex number,ie,5=5+i.0,
log(5+i.0)=1/2 log(25+0)+i.tan^-1(0/25)
algebra just called, says you're wrong
Result is false. Proof is incorrect when you use the incorrect formula for ln of complex number.
Since you're gonna take the ln of both sides of the original equation, you need to specify the branch of the logarithm. Writing 1=e^(2πik) for an integer k, suggests that argument of a complex number, θ, is chosen to lie in the range (2k-1)π
Please check my comment. I pointed out the error in conversions made in the video presentation.
This is wrong. You cannot use the properties of real Ln for a complex numbers. The equation has no solutions in the complex field.
@@oscard2877 u r wrong
Please check my comment. I pointed out the error in conversions made in the video presentation.
Until now, I thought that, unlike mathematics, everything was possible only in politics. He may be a lousy mathematician, but he would make an excellent politician or lawyer. This person is apparently able to provide conclusive proof that he is his own grandfather.
In the question, there is no conditions, but u put your own terms such as k shouldn’t be zero and must be intger
First I changed the notation and changed "x" to "z" to denote a complex number, and assume the general solution is z = x + iy. I also used a slight generalization of Euler's relation: exp(2nπi) = 1 for n = 0, ±1, ±2, ....
Then I rewrite the problem as [exp(2nπi)]^z = 5 ⇒ Therefore we have two equations to deal with: Re{ [exp(2nπi)]^z} = 5 and Im{ [exp(2nπi)]^z} = 0
I found solution of the second equation first ⇒ x = 0, ±1/2n, ±1/n, ±3/2n, ±2/n, ....
The second equation leads to ±1/exp(2nπy) = 5. The negative signed cases are rejected since 5 can only be positive. Taking ln of both sides gives your solution for the imaginary part of z, y = -ln(5)/2nπ. However you seem to have missed the possible real parts of z: x = x = 0, ±1/2n, ±1/n, ±3/2n, ±2/n, ....
Please check my comment. I pointed out the error in conversions made in the video presentation.
Lol what did you do 11 minutes? 1) 1 = exp(2ni*pi) (n is integer); 2) exp(2ni*pi*x)=5; 3) x = ln(5)/(2ni*pi) = -i*ln(5)/(2n*pi) (n is integer but not 0)
He click-baited people with a non-sense equation. I checked the solution with WA and it does not work!
Please check my comment. I pointed out the error in conversions made in the video presentation.
maths made by human and Human made by natural
ln(1) = 0 , so you are dividing by zero in the third step
Yeah,noticed that too
I did that, and it works! The result is my little Pony. Or whatever one wants.
A winning politician would campaign on DIVIDING EVERYTING BY ZERO!!!
Clickbait. 1^z=1 for all real or complex z.
Came here to say the same.
And of course dividing by ln 1 is to divide by zero. Epic fail.
No it is correct. Learn how to calc. with CN.
Well, try this hint:
if z=1/2, there are two well-known sols = 1, -1. Not just one sol. = 1, as you has expressed via 1^z=1.
To be more precise and general, there is always inf# of results when doing a power in CN (including real powers of real nums, but in CN space). In the case of 1/2, due to typical cyclicity, the results seem to be just/only two. But, in general, there is inf# of res. 1, -1, 1, -1, … And for general z in 1^z there is inf# of generally different results.
@@ahojg Give it up man. You must know that you are posting nonsense - fishing for likes...
Nice approach to consider complex power by solving .
It should also be written that x is the element of complex numbers...
With the same mistake of non realizing that the natural logarithm is not bijective in the complex number, some really intelligent people were thinking they have solved the Riemann Hypotheses. We should not be so hard. It is a confusing thing. But before we discuss here anything, we should probably go back to the basics and study the definitions of the real and complex numbers with all the things like the unit element and operations and inverse elements. Then it should become obvious, that this is false….
RN: ln 1 = 0
CN: ln 1 = 0 + i 2pi k
That’s the difference most people here dont realise.
For z=1/2 in the expr. 1^z, there are two well-known sols = 1, -1. Not just one sol. = 1, as you probably think.
To be more precise and general, there is always inf# of results when doing a power in CN (including real powers of real nums, but in CN space). In the case of 1/2, due to typical cyclicity, the results seem to be just/only two. But, in general, there is inf# of res. 1, -1, 1, -1, … And for general z in 1^z there is inf# of generally different results.
Eg. for 1^i the results are = e^-2pi k. Now it is obvious that the res. in video is correct (just use the expr. for 1^i above).
Of course, one should be aware what 1^z really means:
This flow is wrong:
1^z = 1^(x+iy) = 1^x 1^iy = 1^y^i = 1^i.
This flow is correct:
1^z = e^[i 2pi k (x+iy)] = … ->
phase = 2pi k x,
ampl = e^-2pi k y.
Now you immediately see a solution even for complex right side of
1^z = C = |C|e^ic
1^x=1. Otherwise it is a complex non determinabke solution not real one
Please check my comment. I pointed out the error in conversions made in the video presentation.
I don't understand the negative comments: the result is undoubtedly correct. Furthermore, this type of equations reveals a bug in Wolfram Alpha: if you input 1^x=5 (or any other number), it gives no solutions, but if you input 1=5^(1/x) it calculates the exact result.
We know that 1^n = 1 (always!). Nevertheless try to solve the equation 1^x = 5 using logarithms:
1^x = 5
x = log(5)/log(1)
but log(1) = 0 => log(5)/log(1) is impossible (division by 0)
let's try with a value VERY close to one: 0.999999
x = log(5)/log(0.999999)
x = -1609437.10
but 1^-1609437.10 = 1 and not 5
■ conclusion: THERE IS NO SOLUTION FOR 1^x = 5
🙂
The error is relatively subtle, but you are transforming the equation into
1^x=5
There exists x such that 1^x=5
There exists x such that
for all k in Z:
e^i2kxπ=5
There exists x such that
for all k in Z:
i2kπx=ln5
There exists x such that
for all k in Z:
x=ln5/(i2kπ)
This is not possible, though, because from this, we have following:
There exists x such that
x=ln5/i4π and x=ln5/i2π
From this, we have 2=4 which is a contradiction.
Always check your soultions
@@methatis3013 lol😹 people are so creative but also dummy. They know nothing about complex math, but are trying to show how they understand it and how they are clever 😹
Firstly learn and then solve.
Btw. if a quadrtic eq. has two sols x=a and x=b, it means a=b or what???!!!
Always check yourself, then others.
@@ahojg alright then, point out where I made a wrong step
@@methatis3013 You might be confusing some things:
The task is to search for all x ∈ ℂ, such that 1^x = 5, so x is defined before you start transforming the given equation.
Usually well known definitions are assumed to have been defined before we start the first transformation in an order to avoid referencing sth undefined; so real natural logarithm (ln), complex natural logarithm (Ln) and complex exponentiation are already defined prior to the task.
Usually the definitions are given as follows (or similar):
Let be z := r*e^(iα), r ∈ ℝ_0+, α ∈ ℝ, then we define complex Argument, complex natural logarithm and complex exponentiation as
Arg(z) := (α + 2 π n) i,
Ln(z) := ln(|z|) + i Arg(z) = ln(r) + (α + 2 π n) i,
b^z := e^(z*Ln(b)), b, z ∈ ℂ.
(I don't like using the same label for the complex and real natural logarithm.)
It's not technically wrong per se, to repeat variable definitions (such as "There exists x such that for all k in Z:"), but I'm not sure that you understood that everytime you do that, you create a new variable that only shares the label(=name in this case) with the older ones, but are completely independent and different variables; that also comes with an implied scope, which is limited by implied parentheses.
So when you write something like
"∃ x: ∀ k ∈ ℤ: e^i2kxπ=5
∃ x: ∀ k ∈ ℤ: i2kπx=ln5
...",
then that technically means:
"∃ x: ∀ k ∈ ℤ: e^i2kxπ=5 (∃ x: ∀ k ∈ ℤ: i2kπx=ln5 ... )";
note that within that parentheses you only can access the inner defined variables labeled with x and k, while the outer variables are different, despite the same name. And outside that parentheses, you cannot access the inner x and k, while the outer ones are used instead.
But in this case it is unneccessary, to redefine the needed variables everytime we transform the equation.
Because we silently assume that in the result x depends on k, the usual way is to define k first (to avoid variable x beeing equal to some undefined mathematcial object, though that's not really neccessary here):
∀ k ∈ ℤ, x ∈ ℂ : ( 1^x = 5
e^(x*Ln(1)) = 5
x*Ln(1) = ln(5)
x = ln(5)/Ln(1)
x = ln(5)/(ln(1) + (0 + 2 π k) i)
x = - ln(5)/(2 π k) i
)
Note that this is also a massively shortened representation of the transformation; for example I didn't incorporate the definitions given above (which isn't really neccessary, while at the same time much nicer to read).
Sidenote:
The task is given somewhat imprecise, because it doesn't specify whether or not that 5 there is supposed to be a real number (which is the underlying assumption in the video) or a (representant of the) complex multivalue constant (representing that complex multivalue constant), in which case you might solve the task as follows instead:
1^x = 5
e^(x*Ln(1)) = e^Ln(5)
e^(x*(ln(1)+i*2*k*π)) = e^(ln(5)+i*2*l*π)
x*(i*2*k*π) = ln(5)+i*2*l*π
x = l/k - ln(5)/(2*k*π) i
But if you raise 1 to the 'solution' you found, you will still get 1, so you don't have a solution. Also, the very expression 1^x=5, when you take natural logarithms from both sides, is equivalent to ln(1)*x=ln(5), or 0=ln(5), which is a false equality. And any equation which is equivalent to an invalid equality or an illegal expression, is an invalid expression by itself, hence has no (real or complex) solutions.
Lol...the way I imagine that being possible is to rewrite 1 as a power of e^i(theta), etc, using De Moivre or something, and then find how to make it equal to 5...
I'm sure this has been said already, but this false proof is a good example of why one needs to be careful with the rules of logarithms when using complex exponents. In particular, the rule [log x^a = a log x] doesn't hold. In very particular, [log e^(2iπ) ≠ 2iπ].
The proof is correct.
log a^b = b log a stays valid in CN.
The only problem is u dont know what log of CN is.
There are posted proofs that 1^x=3 and 1^x=5. Autor shall go little further: using the same method it can be proven that 1^x= y (where y >0). Then it will be possible to proof that any real number >0 is equal to any other real number >0. I wonder if this proof can be further extended to all real number (w/o or w/ ‘0’) and even extended to equality off all complex numbers. Than statement than "Everything is possible in math" is no longer valid--math no longer exists.
Other option: please read my previous comment.
I LIKE THIS CONTAIN, I THINK U R THE BEST I EVER MET
the trick is erase the 1 of the equation with the identity e^(i*2*k*pi)
Your mistake is making the leap from: e^(i.2k.pi) = 1
to
( e^(i.2k.pi) )^x = 5
1 to the power anything is still 1. Setting it = 5 doesn't make it valid. You could have set it = 1,000,000 and rearranged the equation to get x, but it would still be wrong.
Plug your answer into the original question and the answer is still 1 not 5, as expected.
U r not right. U use wrong assumptions. Eg. 1^1/2 is not always 1 as u say. In CN space it has two results +-1.
@@ahojg no, it doesn't. Just because C allows for complex solutions (results of negative roots) doesn't extend the domain of a general function. This basically means that the square root of 1 is always 1, even in C. Because in C, the square root of -4 is always 2i, not -2i. This is obvious since the function of a variable x or z always has one value per x/z value.
@@danielebeltrame3184 Correct.
@ahojg is correct in that 1^½=1 or -1, but he is wrong about 1^x=5 being possible, abs(1^x) always equals 1, but 1^x has multiple solutions if x has a fractional part
So unless you divide by 0, you won't be able to get 5 from that
Just by the looks of it... 1x1x1x1x1..... is 1 so there is no solution for x as 1 is not equal to 5. No need for those exotic equations.. Actually you can also stop from the x times ln1(=0) because any real number multiplied by 0 is zero.
Too many missteps. First, as others have pointed out, taking the log of a complex isn't done correctly here.
And restricting k to be an integer makes that equality incorrect.
Indeed, here's a proof that 2=1.
1^2 = 1.
2^2 = 2 + 2
3^2 = 3+3 + 3
4^2 = 4 +4 + 4+ 4 .. . . . . .
n^2= (n+n+n+......n) [n times]. upon differentiating,
2n = (1 + 1 + 1 + 1+ . . . .) [ntimes[ or 2n = 1n = >
2 = 1.
Although incorrect, mathematical sophistry is still interesting.
Show what You say
Please, learn how to calc. with CN.
@@ahojg he is right. 1^x=1 for all complex numbers x
1^x=5 because 1=5 based on the equation I saw in the video, so this video is an invalid source, division by 0 isn't valid, thus, 1^x=5 has no (valid) solutions
It's not really natural log, though...
It is the change of base formula...
x=ln(5)/ln(1).
P. S. I think so!
1 raised to this solution could NEVER = 5
1 raised ro anything is STILL 1!!
I call 'reductio ad absurdum' on this one!! 🤣🤣🤣
You are wrong, because x is not 0. Stop promoting wrong maths thought and method on CZcams, please. You should think about (Mv − Rvτ + ∆)^2 = (Mv)^2
It is correct. U R wrong.
@@ahojg You are wrong, it's not mathematics, it's a thought of Nazi-Communist "lie repeatedly", and you are promoting worship nothingness and abolishing the principles of geometry. But the right thought and correct theory of math is to explain the process of physics and chemistry.
*Infinite logs of unity (1 real and all the others complex/imaginary)*
Trigonometric functions are periodic and exponential in complex numbers, so just like there are n nth roots of unity, similarly there are infinite logs of unity - 1 is real and all other log
log(1) = 2 pi k i = 0 for k = 0 (the only real log of unity)
There is only 1 real log of unity which is zero. All other logs of unity are imaginary numbers (i 2 pi K for K not equal to zero). So x = log(5) / log(1) = infinite complex (imaginary) numbers.
1 exp x= 1 dla x należących do zbioru Z. To już udowodnił Euler. Komu wyjdzie inny wynik, ten popełnił błąd! Wyprowadził antynomię. Reszta to tylko manipulacją dla ludzi nie znających matamatyki i pozwalających wprowadzać się w błąd, albo tych, którzy szykają w matematyce sensacji
No solution exist, neither real nor complex
I tried a couple of values of k in Octave and it gave me 1^x = 1, not 5
It is not a solution because it doesn't work. This is "there's a hole in my bucket, dear Lisa" mathematics. Because the solution is still the problem!
Ну, если мужчине боксёру можно на Олимпиаде биться с женщиной, то и так можно.
Инклюзивность, она такая))
Каким мне хочется, таким и будет ответ.
Hello from Russia)
x= i ln5 / ( 2k * pi )
Everything is possible in math, except 1 x 1 = 2
@@attiylanen it is 2 according to Terrriology
Gracias a esta resolución, se demuestra que el movimiento perpetuo existe, también la piedra filosofal y el elixir de la eterna juventud
Incluso que se puede superar la velocidad de la luz
Y que el cero absoluto tiene valores negativos
I really want to know how much did this dude pay ahojg to write the comments he is writing(they make no sense(maybe ha was drunk(for an entire week)))
I would do a video on solving 1^x=n, that saves you a lot of time...
And 0x = n as well.
Hello to every hater and obviously no mathematicians here. Please learn first how to calc. with CN - yes, it is a bit different compared to RN.
For instance, 1^x=1 in R, but 1^i has an infinite number of solutions / expressions in C, not just one! Try to figure out them first, and then write some comments..
1^i does not have solutions, it's not an equation
still no solutions for x=x+1 😢
Дуракам закон не писан.
Если писан, то не читан.
Если читан, то не понят.
Если понят, то то не так...
X is complex vector - 90 degrees
sérieusement une seconde pour trouver,..bravo pour la démonstration de la complexité de la simplicité
And where is a reverse check solution?
I showed it in a reply to a comment. The solution is checked.
Do it by ur own - it is very easy.
e^(i×pi)=-1
This is totally wrong what's wrong with mathematicians these days
@@XHunter-lm6qp he is not a mathematician
You are not mathematicians :)
@@ahojg this is wrong and I am a mathematician
@@ahojgAre you even aware of what you're saying, 1 raised to any real number or complex number is 1.
@@XHunter-lm6qpSorry, but wrong. In C there is infinite number of expressions of 1^i.
If you dont believe to math, use eg. Wolfram Alpha
Are you sure?
Forcing a chicken to get you eggs. .ln(1)=3. So it will work 😂😂😂😂
The solution is absolutely true in the complex
With imaginary number everything is possible
Excellent!!! - Merci,
Higher Mathematics !
Puro cuento. Es facilmente demostrable que 1 elevado a cualquier número incluido el cero da uno.
Just as we have fantasies in music, there are fantasies in math like this one for example.
x=ln5/ln1=ln5/0= infinity not 0
As k approaches infinity, x approaches 0
But stays imaginary what changes the game.
@@ahojg infinitesimally imaginary.
X= infinity 🚶🚶
What makes anyone think that Theta is zero here?
To the speaker.....STOP SAYING " NAY tural "...its NATURAL !!!
x=-(iln(5))/2kp is not a solution
I kind of remember another video from this guy that was false. Maybe this satire math
what in division by zero makes you think the domain is real? :) this has no solutions in any domain, I think your algebra has a problem when you apply formula (a^b)^c = a^(bc) to complex numbers when it is not true in general case, only for real numbers
There is no div by 0
That identity is valid in CN as well, of course. The Math is not about what one think, but what is proven.
This note for foolish,SO please You may go awsy from this.
This guy has such a heavy accent and talks so fast that most of what he says is unintelligible.
It's to distract you from the fact that he used division by 0, he didn't want everyone to see that he used an invalid equation
Because an exponent is the number of times a base multiplies by itself, 1 multiplied by 1 any number of times will always gives you one.
Therefore, the final answer is no solution.
LOL i think you are also high when you do higher mathematics. 😬
You did the same with 2 at least thrice and 3 at least thrice, let's do something we don't know! :D
He is a one trick pony, he can't do something else. Also, his solution is wrong.
Isn't 1=5 if we divide both sides with ln(1)?
1/ln(1)=5/ln(1) = anything btw.
ln(1) is zero so you're dividing by zero. It's like saying 1 = 2, divide both sides by ln(1) to get 1/ln(1) = 2/ln(1), and boom, 1 = 2.
@@shxvuls764 I've taken a little look at it, and it is actually allowed in some cases, like this! I don't quite get it, but it has to do with complex numbers being periodical. 0 is like 2pi.
Mathematic analysis LOL.
ln1=0 so ln5/ln1=infinite
x/0≠∞, x/0=y+x remainder if you are doing remainder, else, you get invalid syntax
k should be a natural number, but the restrictions are correct as the complex plane allows the impossible lol
sqrt(x) = -1?
We've now descended into trolling for comments. -> unsub
An shoking excersise!!!
Absolut nonsense!!! 1 is a real number, also is 1^x = 1
i ❤ Mathematics
Бессмысленная запись
Sorry. If we use Imaginary digits, and 1= exp (i2Pi×K), then 5 equal to 5× exp (i2Pi×N), and more full absver would be x=( ln5 + i2Pi×N)/i2Pi×K
Il manque la vérif' !!
Ø for x from R.
Apart from the number five, every other intiger number gives the same result, that is, x does not become a specific value. What is your opinion?
Exactly!
I put my pony into X and it then told me that its name is Theta and that it is not Zero, thank you very much.
Wrong. The res. is correct. U probably dont understand to CN and namely complex log.
@@ahojg Is division by ln(1) really allowed? There's no real solution, but why would there be a complex solution? And the k2pi in the "solution" is suspicuous. And as said above, isn't the same "true" for any figure and not specifically for 5? Isn't that typical after the division by zero error, that the same solution fits any formulation of the problem?
@@ahojg I think you're right! Those complex numbers are, well, more complex than I thought. I actually asked ChatGPT4 and it explained very well for me why one cannot always devide by 0=ln(1) to find complex solutions. ChatGPT3 was rubbish at math, it once told me that sqrt(5)=2 and was frequently confusing units. But ChatGPT4 seems to be quite something different. I'll test its maths abilities more, for sure. And thank you for challenging my nonchalance!
Not every thing is possible in math. This is again proven by your „calculation“.
❤❤❤❤❤
Sigma
Now I'watched thid "elände" to the end. And he never explains that it was a joke.
That's bad!
Even I thought that I learned something about maths watching ideos this YT channel. But obviously not. This is bad bad! I'm unsubscribed.
Beyond my capabilities, cannot understand the meaning of solutions though having no problem to follow the calculation.
When I see One is power to the something is not equal to One I automatically expect an "i" in the solution.
Super Zoum-Zoum !
1-sinflarga urgatganday sóyladinggo, mazgi