Breaking the rules of math π = 0. Where is the mistake?

[@musicCheckers]

No mistake, Circles are zeros

[@aparnarai3708]

So are donuts...

[@jetx_47]

I love this

[@skytech2501]

tow = 0, pi = half 0

[@smylesg]

​@@skytech2501So pi = ∪?

[@sans1331]

@@smylesgy e s

[@amoswittenbergsmusings]

Presh did not explain *why* this careful handling of complex exponentiation is necessary. Complex exponentiation is basically trigonometric: e^iϑ = cos ϑ + i sin ϑ. Therefore we have e^iϑ = e^(iϑ + 2ikπ). This means that the inverse function log z is multivalued. The faulty proof really boils down to: e^2ikπ = e^0 ==> 2ikπ = 0 ==> π = 0.

[@gdmathguy]

Also, it's best to simplify to avoid these annoying mistakes

[@mikefochtman7164]

I always found it fascinating that anytime you have an exponent with i in it, things get 'weird' really fast. Beginners often try to factor things and move 'i' around just like any other variable. But of course you've just shown that isn't always the case. Wonderful explanation of how 'i' is NOT always 'just another variable'. Thanks for this.

[@BboyKeny]

For imaginary numbers, they're hard to imagine.

[@jonjuathanjoathan2426]

I is not a variable. In fact, it isn't really a constant too.

[@jonjuathanjoathan2426]

i* smh auto caps

[@YuriyDavygora]

​@@jonjuathanjoathan2426How is it not a constant?

[@mzg147]

i is an ordinary number, really. The same mistake can be made using ordinary integers, you can say 2² = (-2)² therefore 2 = -2 - that's not valid! The same with exponents of complex numbers.

[@Jacob.Peyser]

Most of the confusion in the algebra of complex numbers comes down to the branches of the logarithm. If one is simply consistent with a branch cut of the natural logarithm (such as the principle branch), everything works out without discrepancy. Here, I'll make one critique of the video. In the video, it was expressed that ln(exp(2(pi)i+1))=1. This, however, is only true given that it was specified that ln(z) is the principle value of the natural logarithm (i.e. Log(z)). Without context, one could've easily and 'correctly' written that ln(exp(2(pi)+1))=2(pi)i+1, which was indirectly done initially. In general, the natural logarithm is not a function and can be expressed as ln(z)=Log(z)+2(pi)ki, where "k" is some arbitrary integer. So if anything, this video only adds to the confusion surrounding this number system, further mystifying it. Admittedly, I doubt this comment will clear anyone of confusion, but then again, I'm not a math educator.

[@lawrencejelsma8118]

I was thinking it was because going from polar back to rectangular we don't have an angle representing what domain we came from in powers of n (this case power of 2). So 0 to 2π range by squaring moved to angles now in 0 to 4π. So 2π and 4π both now represent e^i(2π). His correction makes sense but e^i∆ = cos∆ +isin∆ real and imaginary conversion of i∆ now has to consider e^u = cos(iu) + (or minus I forgot) sin(iu) that now is not an angle but an imaginary angle. I think this is the interpretation of his correction meaning. 🤔 All my courses never came up with a thinking of wrong thinking contradictions in complex numbers analysis.

[@tunafllsh]

Based on this comment I think he should have said in the video: "π may or may not be equal to 0."

[@globglogabgalabyeast6611]

Yeah, I found it pretty disappointing that the video was just like “you can’t do this” without giving any real justification for why you can’t do it

[@delanmorstik7619]

I am bad at math but at some point its written that e^2ip+1 = e.. so 2ip+1 = 1 so 2ip = 0 so i = 0 ??? Why is that not commented ?? Anyone knows where is my mistake?

[@big_numbers]

​@@vsm1456 You can't remove the e because of the fact that it's using a generalized version of exponentiation that works with imaginary numbers and uses trigonometry. (cos(x)+isin(x)). In this form you can easily see that taking logarithms of stuff like this just doesn't work. This is also why e^(x*i*pi) is a circular function.

[@lesnyk255]

The fallacy occurs at the 4:00 mark, when concluding that -4(pi)^2 = 0 implies that pi=0. The correct conclusion is that -4 = 0. You're welcome.

[@therealsachin]

A simpler one. At 1:00, we have e^(2i x pi) = 1. Now taking natural log on both sides gives us 2i x pi = 0, which means pi = 0.

[@jonasvuillemin9412]

It’s an complexe exponentiel u can’t

[@manuelmatias3772]

​@@jonasvuillemin9412well, it's *supposed* to be wrong.

[@therealsachin]

​@@jonasvuillemin9412 We can do it without taking natural log. e^(2i x pi) = 1 e^(2i x pi) = e^0 If they have to be equal, the exponents have to be equal. Which means, 2i x pi = 0, which means pi = 0.

[@therealsachin]

FYI. Those who are still confused. For a function that is cyclic, there will be many points on the domain which has the same value/image. Equating two different points which has same value and then saying those points are equal is the issue. f(x) = f(y) does not necessarily mean x = y. We can understand it better with an example. cos(2 x pi) = cos(0) 2 x pi = 0 (wrong here!) Hence, pi = 0. :)

[@irrelevant_noob]

@@therealsachin well TBF that *_still is_* "taking natural log." just no longer explicitly so. 🙃 And to the OP:: yeah, that's exactly what i expected to happen at that point. Turns out, the trick was more subtle with complex exponentiation than it would've been with just the log.

[@him21016]

You could have just gone from e^{2pi i}=e^0, “so” 2pi i=0 so pi=0. This makes the same fundamental mistake.

[@nuptvalorant1494]

no it makes a completely different fundamental mistake

[@battlesheep2552]

Not exactly, e^2pi i does equal e^0 without 2pi i equalling zero

[@him21016]

@@battlesheep2552 Yes, I know that. The fundamental error is in assuming complex exponentiation is unambiguously defined, which is an equivalent error to thinking complex exponentiation is injective

[@him21016]

The exponent rule fails for the same reason complex exponentiation isn’t injective; arguments are not unique

[@robertveith6383]

​​@@battlesheep2552 You wrote this wrong. You need grouping symbols: e^(2pi*i).

[@mike1024.]

This whole thing boils down to the fact that some complex functions (like the power function) are not actually functions unless you include branch cuts. This creates a lot of need for careful watching as you go. For example, when you took the natural log of both sides at the end of your demonstration, you really should have included the coterminal answers also. It's still already incorrect, but that's besides the point.

[@DrEbrahimian]

Nice one! I would be careful in defining z^w as e^(w ln z), which is not quite accurate, as ln z is not well-defined for non real values of z. In other words, ln z is not a function from C to C. It is a multi-valued function. For ln z to be well-defined you need to use an extended notion of the complex numbers called a Riemann surface. Defining ln z as a function would cause all sorts of unintended consequences. For example it cannot be continuous. To see that note that ln 1=0. If you approach 1 from the fourth quadrant of the complex plane the natural log approaches 2 i pi, which is different from zero.

[@Jacob.Peyser]

Agreed. I've found that on KZbin, there is a lot of confusion regarding complex exponentiation and that's because many math content creators fail to properly communicate that not all functions are invertible. Ideas such as branch cuts and principal values need more attention. Hell! Most people don't even fully understand that quadratic equations only have two solutions because quadratic functions cannot be inverted by a function. Forget about periodic functions! This problem is far more deep-rooted. I'd go as far as to say that the school system is primarily to blame.

[@thomasdalton1508]

Yes, it's unfortunate Presh didn't mention the fundamental point here. The reason you get into difficulties is because logarithms of complex numbers are multi-valued. It isn't that log(e^(2πi+1)) doesn't equal 2πi+1. It does. It just doesn't *only* equal that. It also equals 1. And it turns out 1 is the value we need in this scenario. It is like square roots - there are two square roots and if you pick the wrong one you can easily get to contradictions.

[@oelarnes]

To clarify further, the “correct” derivation at 7:10 makes the same mistake as the original derivation, and you could “prove” that pi = 0 by repeating the same steps, since the multiplied exponent is one possible value for the iterated exponent. The real problem is using “ln e = 1” as a step in proving equality, since that is only one possible value of the log at e. We can go all the way back to 1:25 to see that the real error lies in obfuscating the taking of the log on both sides.

[@angela3.14]

Stepping in to ask question since you guys explained it better than the video. If e^(2iPi) = 1 = e^0, as seen at 0:51, why can’t we set the exponents of the same base equal to each other to get 2iPi = 0? I realize this can’t be true since it leads to contradiction. But I haven’t taken any higher math classes on complex numbers to know what the issue with this logic is. Is this an illegal step? Or is the problem one of the steps before it, like squaring both sides, which I know is notorious for creating dubious roots, for instance when x = -3, squaring both sides allows x = 3 and x = -3. But this Pi thing feels different. Help?!

[@DrEbrahimian]

@@angela3.14 you got it right in the last couple sentences. The reason is the same as the reason for why we can’t go from 3^2=(-3)^2 to 3=-3. In math language, exponential function e^x on C is not one-to-one.

[@f5673-t1h]

From the first few seconds of the video, I know where this is going and the answer is: the exponential function is not injective over C, so you can't cancel it. exp(x) = exp(y) implies x-y is an integer multiple of pi, so not necessarily equal.

[@Bibibosh]

I don't care what you have to say. I still believe that Pi equals 0...

[@ShaunakDesaiPiano]

An integer multiple of iπ, but I get your point.

[@RGP_Maths]

@@ShaunakDesaiPiano An even integer multiple of i pi, but I get your point.

[@ShaunakDesaiPiano]

@@RGP_Maths oops yes I stand corrected! 😂

[@Gretchaninov]

@@ShaunakDesaiPiano Isn't it an integer multiple of 2*pi*i?

[@ZacharyVogt]

I've always loved this channel, for years. I always watch the vids, but I also pause the vid early, and solve the problem for myself, then restart the video and get confirmation of my solution. This is my new favorite video, because I completely fell into the trap. I was plain WRONG and needed the video to correct me; and I really like that.

[@trnfncb11]

I think the origin of all trouble lies in the simple fact that the polar representation of a number is not unique. Two numbers in different sheets may be the same but you cannot equate their nat logs (like you would instead equate two real exponents of the same real base) because they in fact differ by a multiple of 2pi.

[@lawrencejelsma8118]

No! He misleads us in his instruction teachings. He forgot in the real number plane the same error happens. We can't use the unity bases of {0, 1} in the real number plane also for envoking y^m = y^n base powers equality implies m = n. He danced around the obvious also in real numbers it fails in base 1 (or 0 to any power let alone itself).

[@masterbeater832]

exactly what i was thinking! this applies in all quadrants too, see that ln 1 = 0 is not absolutely true because if you approach 1 from quadrant 4 you would once again be approaching that factor of 2 i pi

[@mathmachine4266]

The mistake was in the assumption that (a^b)^c always equals a^(bc). This statement holds true for real numbers, but in the case of complex numbers, the actual formula is a little messier: (a^b)^c = a^(bc) * exp(τci*round(-Im(bln(a))/τ)) Where Im denotes the imaginary part, and round denotes the function which rounds numbers to the nearest integer (rounding up if it's a tie). I'm also using τ to denote 2π, not out of preference but to avoid using a confusing number of parentheses. When c is an integer, this is just the same as a^(bc). That's also the case when a and b are real (unless a is negative or zero).

[@nicholassmith9868]

The mistake is @1:40 e^((2ipi+1)*(2ipi+1)) does not equal e. it equals e^((1 + 2 i π)^2). the rules of exponents are slightly different when dealing with e.

[@ShawnPitman]

4:08 your error is that you assumed one constant was equal to zero, and not the other… Everyone knows the value of pi therefore the value of -4 must have been zero

[@cufflink44]

Just learned for the first time that the power rule for complex numbers isn't the same as that for real numbers. Thanks!!!

[@kevinmartin7760]

Around 7:00 it is also important to point out that ln(e^z)≠z in the general case for complex z. The logarithm of a complex number in the general sense has an infinite number of values (just as the inverse tangent function has for real arguments), but like arctan, we define it to produce its principal value whose imaginary part is in the range (-π π] (or perhaps [0 2π), I can't recall) by removing multiples of 2π. Either way ln(e^(2iπ+1)) is 1, not 2iπ+1.

[@johnbyrne1022]

I was wondering about this too, but he does actually point it out at the end of the video.

[@Zollaho]

I agree.. Mind masturbation with i, not maths

[@vivada2667]

Thank you! Even though he did point this out at the end of the video, I agree that he should have addressed it earlier.

[@lawrencejelsma8118]

@@vivada2667 ... He still pointed out a wrong hypothesis. It is also a problem with real number planes also that he doesn't mention. Anytime the polar complex angle of the unity exponent is manipulated is just as the case in real numbers. I watched this video again and again getting so confused why e^(-4π^2) was not balancing out. It is because in unity calculations exponent equality fails. In fact, in real numbers, any base not equal to {0, 1} has the powers relationship y^m = y^n implies m = n with the powers so that we can solve those mathematics problems. For example finding x in 3^x = 3^(x^2 - 1) and other powers equality. We can't do this equality for unity 1 nor 0 in the real numbers also.

[@AuroraNora3]

You're exactly right. Math educators on KZbin should really begin to stress this particular nuance of principal values of complex multivariable functions, because I see people make incorrect statements all the time. It should be fairly easy to teach. Start with the familiar fact that the square root is multivariable, but has a defined *principal value* denoted by the symbol "√", then extend that definition for the nth root by introducing the principal branch (-pi, pi], and finally properly define the principal natural log, among other things. Only then should you start to teach how to manipulate and solve complex equations like: e^x = -1 => e^x = e^(Log(-1) + i 2pi n) => x = Log(-1) + i 2pi n where "Log(-1) = i pi" is the principal natural log of -1, and the last term encapsulates all the other natural logs of -1. Now we can use the above to get all the solutions of: x⁴ = -1 => x⁴ = e^(Log(-1) + i 2pi n) => x⁴ = -1 * e^(i 2pi n) => x = ⁴√(-1) * e^(i 2pi n/4) where "⁴√(-1) = e^(i pi/4)" is the principal fourth root of -1, and the other factor encapsulates all the other fourth roots of -1.

[@amritsagarkar7899]

1:07 in the last equation at this time stamp Exp raised to some power yields same exp means the power must be 1 So 2ipi+1=1 So either pi = 0 or i = 0

[@raj.sharma86]

0:52 One can get it directly by this: e^(iπ) = -1 Square both sides ⇒ e^(2iπ) = 1 ⇒ e^(2iπ) = e^0 ⇒ 2iπ = 0 Square both sides ⇒ -4 π^2 = 0 ⇒ π^2 = 0 ⇒ π = 0

[@TomFarrell-p9z]

Reminds me of a favorite Tom Swifty: "Taking the natural log of -1 is as easy as pi," Euler imagined.

[@caicai491]

You have got e=e^(2ipi+1), why don't you let 1=2ipi+1 directly? Then you can derive pi=0 immediately, although the result is ridiculous. I have no idea what you are doing in the next steps and make it so complicated...

[@two697]

He probably thought doing it in a convoluted way is supposed to make it harder to spot the mistake

[@dav3141]

e^x = e^y does not mean x=y for complex numbers

[@caicai491]

@@dav3141 of course not so that the result is ridiculous. Since he just wants to prove pi = 0 in a unreasonable way, this can be much faster.

[@zephyrred3366]

Also think the mistake is different from the one that was shown. When we write "e^z" we usually mean some exponential function defined as either: 1) exp(z) = e^x * (cos(y) + i*sin(y)) 2) exp(z) = 1 + z + z^2 / 2 + z^3/6 + ... Of course, we can show that those definitions are equal. You can notice that if we actually use proper notation, we cannot make the same mistake: exp(2i*pi + 1) = e -- easy to see from 1st definition that it's true However how we cannot make a statement similar to [e ^ (2i * pi)] ^ (2i * pi) = e. On contrast to definition of the exponentioal function, If we want to raise a to the power of b in Complex numbers we shall use definition a^b = exp(b * Ln(a)) , where Ln(a) = ln|a| + i*arg(a) + 2i*pi*k, k any from Z. By that definition e^(2i*pi + 1) = exp((2i*pi + 1) * (1 + 2i*pi*k)) = exp(2i*pi - 4*pi*pi*k + 1 + 2i*pi*k) = / by def (1) / = exp(2i*pi - 4*pi*pi*k + 1). Therfore, e^(2i*pi + 1) has actually infinitelly many values one of which is indeed equal to exp(2i*pi + 1) = e.

[@АртурНасакин]

In fact, at 1:33 the equation is already wrong, wolfram alpha also likes to make mistakes, but if you write a query (e+e-e)^(2*pi*i+1) it will find other values ​​besides e, same for ((e+e-e)^(2*pi*i+1))^(2*pi*i+1). The bottom line is that there is a difference between raising the number e to some power and using the exponential function. The first equation is true if the exponential function is on the left, and the number e is on the right. Thus, replacing one with another in the third line makes no sense and, of course, leads to the mistake.

[@rudy_us]

When e^(2iπ+1) = e Then 2iπ+1 = 1 And from there 2iπ = 0 And π = 0 So it can be done like this and not be wrong

[@TheMofRider2]

I know my "proof" is also false, but I would show the following: At 1:10 we have e^(2i+1) = e. | ln 2i+1 = 1 (bcs e = e^1) | -1 2i = 0 And as neither 2 nor i equals 0 it must be pi = 0.

[@mememastermind3817]

It was so easy to find the mistake. It was there all along. All I had to do was look in the mirror.

[@-ZH]

I managed to figure this one out by seeing 2i(pi) as 0, and then when you started multiplying the complex numbers, I knew something was up.

[@stevenlee8765]

The mistake is in 0:40 As you can see, squaring both side which have different sign (which e^iπ is positive, and -1 is negative) will give a contradiction proof. In other word, imagine when there is 1 = -1 and then you square them up, you will got 1 = 1. Now this is fishy that it is no way that 1 = -1 can be true. You may square both side if : - 1 from 2 equations has square root or (1/2n), n for all integer. - both of them have square root or (1/2n), n for all integer. - 1 from 2 equations, or both of them have absolute equation. (Example for absolute equation : |n + 1| = n and |n + 1| = |n|)

[@DrHans-RudigerZimmermannTheore]

In an Equation one has to do the same operation on both sides. In minute 1.31 in the video effectively only the left side was raised to the power of (2i phi +1), the right side is e and has to be raised with the same power but stayes the same. This is the mistake. Presh inserted in a recursive way e only on the left side! This is not allowed in manipulating equations.

[@DanGRV]

On the complex plane, every number x+iy (except zero) can be written as r*exp(i*t) where r>0 and t is a real number. It also happens that for every integer n we have x+iy = r*exp(i*t + 2*pi*n) In some sense, it's as if the exponential has infinitely many inverses at each point, one for every integer n. There isn't a unique complex logarithm that can be defined everywhere. However, the "main", "standard" or "usual" complex logarithm is defined as: log(x+iy) = log(r*exp(i*t)) = log(r) + i*t where r>0, log(r) is the real logarithm of r and t is in the interval (-pi, pi) Now, after choosing a logarithm, you can define the complex exponentiation z^w by the rule: z^w = exp(w*log(z*exp(2*pi*n))) where n is an integer such that z*exp(2*pi*n) falls inside the domain of your chosen logarithm.

[@lawrencejelsma8118]

And then after rewatching his erroneous talk about this being a complex only plane problem. It is total nonsense. Anyone equating exponent results in the real number plane also don't find equality in the exponents implied with {0, 1} because the y^m = y^n means the exponents m = n equation is true only if the base is not 1 (nor 0 let alone the undefined 0 with itself).

[@MarcDonis]

But why didn't you just go straight from: e^(2i pi + 1) = e 2i pi + 1 = 1

[@gdmathguy]

In complex analysis, to avoid mistakes like these, always SIMPLIFY first. Also, dividing 0 by things in equasions always leads to the wrong result, so don't do that. e^(2iπ+1) =e*e^(2πi) =e*1 =e Now it makes sense again. Another way to explain why this is a mistake is to not see 2π as a simple number, but an angle instead. 2π therefore is always 0 when deriving it from (radius)e^(i(angle))

[@DrHans-RudigerZimmermannTheore]

Only for real numbers one can equate the exponents. Here we have complex numbers !n an Equation one has to do the same operation on both sides. In minute 1.31 in the video effectively only the left side was raised to the power of (2i phi +1), the right side is e and has to be raised with the same power but stayes the same. This is the mistake. Presh inserted in a recursive way e only on the left side! This is not allowed in manipulating equations.

[@Gretchaninov]

I found the wrong step, but I was quite confused at first. The steps seem legit. But yeah, (e^x)^y isn't always e^xy, basically because ln(z) is not a continuous function for all z. In particular, there's a big discontinuity when z is a positive real number. If you interpret ln(z) as a Riemann surface, or something like that, or perhaps an infinite pile of values (with multiples of 2*pi*i), you might be able to salvage the continuity and keep some of the rules.

[@nicolastorres147]

Euler’s Lemma: Euler’s identity Euler’s Theorem: π = 0 Euler’s Corollary: Circles don’t exist

[@rohangeorge712]

in real life probably not. circles are too perfect to exist in reality

[@nicolastorres147]

@@rohangeorge712 Circles don’t exist in real life almost surely. The probability of drawing a perfect circle is 0.

[@orangenostril]

3:16 less confusing way of doing this step is saying e^(4iπ) = e^(iπ) * e^(iπ) * e^(iπ) * e^(iπ) = (-1) * (-1) * (-1) * (-1) = 1

[@Skandalos]

Step 3: Squaring an equation can change the solution set of the equation.

[@RegisteredLate123]

W

[@sethkingman2118]

Finally, thank you! Yours should be pinned. If anyone doesn't get it: think of pi as a variable in the initial equation, and we're solving for its value. Once we square, we've introduced the potential for extraneous solutions. (It's not clear to me that we lose the correct solution at this step, but it's moot because having taken this step, we have to check our solutions against the original equation regardless.)

[@irrelevant_noob]

Except it's not an "equation" and therefore there's no "solution set" for it...

[@irrelevant_noob]

@@holomurphy22 then why does English have the "equality" word if they're not going to use it for actual equalities? Also, even if i grant you that you _can_ use the "equation" term for mere expressions equaling each other, that still doesn't give it a solution set.

[@yanggang4352]

1:27 i already get that result without the next step. Since you have a common base on both sides, simply equate the 2 exponents: 2i* pi +1 = 1 2i*pi = 0 i*pi=0 pi = 0

[@minebuddiesyt]

Same

[@rohangeorge712]

no that proves either i or pi is = to 0. what if u divided both equations by pi instead? u would get i = 0. so there u could be "dividing by 0" if u will. its kind of goofy.

[@kobalt4083]

@@rohangeorge712 but pi isn't 0 so it isnt division of 0. pi=0 and i=0 are extraneous results, they're not true.

[@rohangeorge712]

@@kobalt4083 yea ik im just saying why that wouldnt work. the proofs are wrong anyways so it doesnt really matter. cause yea ur right pi and i arent 0

[@tejasvishrivastava5888]

You can't equate the complex exponents with same base like e^0 = e^iπ doesn't impliy 0 = π that argument is wrong.

[@dontthrow6064]

2pi = 0 in polar coordinates. But then division is not so simple. You can't simply split tau, otherwise -(2pi)^2 = -0*0 is correct. To divide by 2 correctly, we should think on the whole set of points that represent the same angle [0, 2pi)+2kpi, k integer. 0 is the entire set {2kpi}, divide by 2 => {kpi}. As many others have stated, in the complex plane, we are working with a multivalued function.

[@Alpha_sports890]

You can't square both side , otherswise we can write -1=1 Squaring both sides 1=1

[@EntaroCeraphenine]

Can we not come up with the contradiction much earlier in the process? Like: e^(2*i*pi)=1=e^0. Therefore, 2*i*pi=0. Since 2i is not 0, therefore pi=0. The underlying lesson still holds, tho.

[@129140163]

8:41 ln(e^(2ipi+1))=1 ln(e)=1 Ergo e^(2ipi+1)=e e^1=e Ergo 2ipi+1=1 Ergo 2ipi=0 Ergo i*pi=0 Ergo either i or pi must equal 0, and since pi≠0, therefore i must equal 0.

[@deerh2o]

I learned something about complex exponents -- nice way to motivate that.

[@cloudy7937]

Every time I see someone squaring both sides in a proof I immediately get concerned

[@sethkingman2118]

For good reason

[@Nikioko]

0:40: Squaring is no equivalence transformation. Which means that you are generating new pseudo-solutions at this point.

[@marcvanleeuwen5986]

For me this is indicative of a defect of our teaching of rules for exponentiation and the various definitions of the exponentiation operation itself. We teach the multiplicative rule (a^b)^c=a^(bc) (as well as other rules for exponents) in a context where b and c are (usually positive) integers, and then later we expand the definition of exponentiation to allow complex exponents. Never do we stop and consider whether the usual rules for exponents remain valid when we do so; it turns out that (sticking to positive real bases as the expanded definition requires, although this too is often not said explicitly) the additive rule a^b*a^c=a^(b+c) remains valid in general, but the mentioned multiplicative rule is simply no longer valid. To save it one needs to add the condition that the inner exponent b is _real_; however I have never seen a text that states this restriction explicitly. We are simply supposed to understand how complex exponentiation works, and refrain from using the multiplicative rule where it is no longer valid because it is obviously wrong (as the video shows). This oversight is witnessed by the numerous proud presentations of a calculation of i^i, despite the facts that (1) it falls outside the domain of the expanded definition of exponentiation (not a real base) and (2) the computation invariably invokes the multiplicative rule in a case where it is not valid.

[@wiggles7976]

When we proved (a^b)^c = a^(bc), we probably used natural numbers. If the rule was proven for natural numbers, that doesn't mean it works for the integers, or reals, or complex numbers. It may or may not.

[@Ninja20704]

But it does always work for integers. For real rational numbers, as long as the fraction we get from doing bc doesn’t simplify we are good

[@benjaminojeda8094]

That formula works for all complex numbers

[@wiggles7976]

@@benjaminojeda8094 The point of the video was that this rule does not work for all complex numbers.

[@3141minecraft]

Even for complex numbers, I am pretty sure this formula works if c is an integer

[@logosking2848]

I'm at 2:09 and it looks like the problem is that pi is cylclical and you're about to make the mistaken assumption that 1 is the only value for which a^x = a whereas this isn't actually true with imaginary numbers because they function almost like degrees, repeating every two i pi

[@tarmotaipale5704]

7:20 doesn't the natural logarithm become an ambiguous multiple valued function in complex plane? So this might cause some problems, I think Edit: maybe it works out if you define ln(z) for complex z as the principal value of complex natural logarithm?

[@irrelevant_noob]

In this context, it doesn't quite matter, since it's in the exponent and the +2kπi turns out to not have an effect on the result, doesn't it?

[@tarmotaipale5704]

@@irrelevant_noob It does matter especially if the original exponent has a non-zero imaginary part, such as here, I think. And non-integer real exponent might truly change the result if multiplied by 2kpi

[@ThePeterDislikeShow]

it's an extraneous solution, since you squared both sides. It's like solving sqrt(x) = -2.

[@stuff2328]

6:50 but because of the natural logarithm rule, couldn’t you place 2ipi+1 to the left of the ln?

[@КириллЙошкин]

If we use logarithm defined for complex values ln(e) = 1 + 2πni where n is integer. In definition z^w = exp(w*ln(z)) the logarithm is a multivalue function. You can write as well z^w = exp(w *(ln(z) + 2πni))

[@KSJR1000]

e^3ipi = -1 = e^ipi take the ln(), and ignore that ln(-1) is undefined 3ipi = ipi; next divide by ipi 3 = 1; next subtract 1 2 = 0; next multiply by pi/2 Pi = 0 QED

[@yashshukla7985]

restaurant bill: 53 $ meanwhile presh : let me show you how 53$ equals to zero

[@ceo1OO]

🌐 back when I first found out some of the cherished exponential rules didn't work for complex numbers...it shook my mathematical world ...lol 🌐 false proofs like these are good for explaining why they don't work...

[@gordonweir5474]

This video reminds me of another fallacy (the explanation of which I will leave as an exercise for the reader!) June 28 was Tau Day (6.28), analogous to Pi Day (3.14), but celebrated by those who think that the more natural "circle constant" would be the ratio of a circle's diameter to its RADIUS (rather than its diameter). They call this new circle constant TAU and it is mathematically equal to 2π. There are arguments pro and con regarding tau , but the fun begins when you apply Euler's formula to both of these circle constants. It its more familiar π form we have e^(π*i) = -1. If we take natural logs of both sides we have π*i = ln(-1). This is actually a very useful relationship and can actually serve somewhat as a DEFINITION of logs of negative numbers. For example we could say that since -2 is equivalent to (-1)(2), the log of -2 could be considered as being log 2 + πi. Multiplying using logs (which is what we had to do back when I was in high school, as calculators had not yet been invented!) would allow us to substitute addition for multiplication. A problem like X = 2*3 would become log X = log 2 + log 3. If we were to use the above definition of logs of negative numbers, X = (-2)*(-3) would be solved as follows: log X = log (-2) + log (-3) = log 2 + πi + log 3 + πi = (log 2 + log 3 ) + 2πi = (log 6) + 2πi. Taking "antilogs", this would become 6 +0i (since the e^ix function is cyclical) and give us the correct answer of 6. But what about the equivalent tau formula? It can be written as e^(tau*i) = +1. Taking natural logs of both sides we have tau*i = ln 1 = 0. But if a product of numbers = 0, then at least one of the numbers must also = 0, which implies that either tau = 0, or i = 0!

[@ilplolthereturn7525]

i'm very confused but ok...

[@commoncoolchannel8588]

huh didn't know tau = 2

[@gordonweir5474]

tau = 2π @@commoncoolchannel8588

[@singhal.keshav]

Well this was really far fetched. You can straight up write e^(2iπ)= e^(4iπ) [=1]. Then both side to power i and get e^-2π = e^-4π | e^2π= 1 | π=0

[@EvilAxelord19]

Good video, I thought it had made a big mistake by having an additional fallacy of using two different branches of the natural log at 3:38, but I just realized that was just the real natural log function and that the only fallacy was from the power rule misuse, very clever setup.

[@TalysAlankil]

this is really neat! however, there's one thing i think you glossed over a little too quickly: when taking the natural log of a complex number, what's especially important is to pay attention to the imaginary part and make sure it's within the right range (i think it's -π, π? i can never remember LOL). otherwise at around 6:55 you show ln (e^[2iπ+1]) which, without this important explanation, might mislead someone into just saying "well, ln is the reverse of the exponential function, so they cancel out" and get back to the initial result.

[@dlp778]

It was a very important point that you mentioned it well The truth is, I had not paid attention to this point until now. Thanks dear Prof.❤

[@Geenimetsuri]

(e^(2·i·pi+1))^(2·i·pi+1) to e^(-4·pi^2+4·i·pi+1) changes the log-branch. The correct general form would be e^(-4·pi^2·n+2·i·pi·n+2·i·pi+1) which for n=0 reduces to e.

[@EllipticGeometry]

(x^y)^n=x^(ny) is actually valid when n is an integer, including in our case of (e^(iπ))^2=e^(2iπ). This is because x^n is not sensitive to the branch cut. Indeed, you can just multiply repeatedly without ever taking a logarithm. To the extent that a branch cut still applies to (x^y)^n=x^(ny), it’s determined solely by x which is equal on both sides. This result is much more general than one little coincidence.

[@isaacricker9108]

Is there a mathematical formula to find a number with the most factors? In the case of a math problem there would be constraints, like "what number has the most factors between 1 and 100?" Is there a way to solve that question with something like a formula without taking every number and counting the factors? (The answer in that question is 60, with 12 factors, but I had to check every number to make sure, there has to be a more streamlined way.) The reason I have this question is because my family has an obsession with prime numbers, so I want to be able to find the "least prime" number.

[@goldensnitch5492]

Your story is interesting

[@isaacricker9108]

@@goldensnitch5492 they full on microwave in only prime numbers, set alarms in prime numbers, etc. its a contest for them, seeing who can do it for the longest

[@sophiastern2719]

I don't know of a formula, but this wikipedia page should be up your alley: en.wikipedia.org/wiki/Highly_composite_number

[@sriramn1809]

At 7:00 The usual log rule says u can take out the exponent and write it outside the log. Ln x^y = y Ln x U saying "we know that is e" just means ur undoing the thing that is supposed to prove it wrong, basically saying we know its e so its e. That said. I cant really find where the contradiction starts. Except if taking out the exponent isnt something u can do in these situations. Given the calculator said this is where the value changes.

[@The-EJ-Factor]

5:15 I felt a wave of power flush over me. As he said where I thought something went wrong, actually was the part when it went wrong.

[@mikemian]

The fundamental problem is exp(2ipi) =exp(0) so when you raise 0^0 you have an undefined result.

[@Qermaq]

The fundamental problem is we as a species long for both equality and wealth, which are mutually exclusive ideals. But I agree with the rest.

[@timmydirtyrat6015]

​@Qermaq Wealth is not incompatible with equality, limitless overconsumption and abhorrent greed is.

[@Qermaq]

@@timmydirtyrat6015 Those two are a pickle, I agree. But the desire for wealth is by definition the desire for more than others can have, which is the antithesis of equality.

[@kenderpl]

@@Qermaq Desire for wealth =/= desire for more than others can have. That would be more true if you specified desire for relative wealth. And even then it's not contradictory for everyone to be equally wealthy. I don't think your statement is defensible.

[@Airwave2k2]

@@Qermaq what a low level materialistic thinking

[@lmr4403]

You could have shown this fallacy without adding the extra exponent (2ipi+1), but just instead straight away pretend to take the log of both sides: exp(2i*pi+1)=exp(1) ->log(exp(2i*pi+1))=log(exp(1)) -/>2i*pi+1=1 ->2i*pi=0 ->pi=0.

[@lmr4403]

@@DrHans-RudigerZimmermannTheore this relation comes from the assumption (=given theorem) exp(2i*pi+1)=exp(1), so you put both sides to the power (2i*pi+1), then the right side becomes exp(2i*pi+1) which is simply exp(1) again.So this is not the mistake.

[@DrHans-RudigerZimmermannTheore]

@@lmr4403 Yes you are right. I did not express myself clearly. I meant that only for real numbers the normal expotential laws are valid, not for complex numbers. this is the error. I think one must treat both sides of an equation equally and presh is only operating on the left side. In a recursive way ! He shoud have written it out, so one can calculate what you have. In theory of equations one can show that when manipulating endlessly only one side Presh will have in the exponent infinety times the "wrong" multiplication of the complex numbers (2pi+1). Do the calculation and heartly wishes .

[@DrHans-RudigerZimmermannTheore]

@@lmr4403 Yes, you are right. I failed to paste my full commentary. : Only with real numbers one can apply all the known potencial laws, not with complex numbers. Preh manipulates only the left side and could repeat this left-side recursion until infinity giving always different answers. Heartly wishes

[@lmr4403]

@@DrHans-RudigerZimmermannTheore e^(2i*pi+1)^(2i*pi+1)^......^(2i*pi+1)=exp(1) still holds.

[@DrHans-RudigerZimmermannTheore]

@@lmr4403 yes, but trivial. But if you out multiply the exponentes (not allowed) you get always larger complex numbers. (2pi+1)(2pi+1)(2pi +1) ...... Okay, thanks and let stop here

[@biswajitmukherjee9888]

Please indicate where the Rule of mathematics is broken. e^(i *Pi) = -1 Squaring both side, we will get e^(2*Pi*i) = +1 Again, it's true that e^(0) = +1 So, we can take that e^(2*i*Pi) = e^(0) => 2*i*Pi = 0 As, 2*I is not zero, then the value of Pi is zero. But, this is an absurdity. Like that , if anyone wants to find the point of inter section between the line, and a circle, whose equations are respectively Y = b, and X ^2 + Y ^2 = r^2, and if the b > r, the point of intersection will be complex. But, this type of intersection is simply impossible. That means the result is an absurdity. Thanks a lot.

[@ahmed.d3289]

I think there is an earlier mistake at 1:27 We can't substitute an equation into itself we will end up with both sides of the equation to be equals Like if we have the following equation y=x+2 equ1 Then x=y-2 equ2 Then we substitute equ2 in equ1 y=y-2+2 ---> y=y

[@bentaylor25]

1:12 - e^(2 i pi + 1) = e or e^(2 i pi + 1) = e^(1) Does this not suggest that: (2 i pi + 1) = 1 [because a^b = a^c -> b = c] Therefore 2 i pi = 0 i pi = 0 Then either i = 0 or pi = 0, so there's already been a mistake by this point. (My background is Computer Science, not Maths; please let me know if I've done something wrong). (Perhaps the a^b = a^c -> b = c rule doesn't apply to complex numbers?)

[@littlecousin5630]

Another problem is that e^{x}=1 doesn’t imply x=0 over the complex numbers. It implies that x=2ipi n as n ranges over the integers. The logarithm “function” over the complex numbers isn’t a function (it’s a so called multivalued function). It is defined as Log(x)=ln|x|+i(arg(x)+2pi n) where ln here is just the normal logsrithm over the reals, and Log is just ro highlight the fact we are taking Log over the complex numbers. The way of looking at it is that the function sending a complex number x to e^x isn’t bijective.

[@sohomsen2922]

i^i = e^(-π/2). Here we convert i=e^(iπ/2). Then we have to multiply the two powers

[@HenrikMyrhaug]

Whenever you have e (or any other real number) raised to some complex exponent z of the form z = a+ib, where a and b are real, a will dictate the modulus (size) if the complex number, while b will dictate the argument (rotation in the complex plane). Because any angle of θ and θ+2π, are equivalent, this can give results where the exponentioation of e to a complex number z=a+ib is equal to the exponentiation of e to a different complex number x=a+i(b+2πn), and so it looks like the complex numbers are equal when they are not. This in my opinion is the core issue of the proof, or at least the root thereof.

[@PASHKULI]

1:00 min. from here we can see that if the bases are equal, so must be the exponents. Thus the exponent 2iπ + 1= 1, therefore 2iπ = 0. No need for the other transform after that 1:00 min. point. Means the mistake happened before that 1:00 min. point in the video.

[@TheMathManProfundities]

The problematic step was actually the one before the one you highlighted. The issue is that a=b doesn't mean that aˣ=bˣ. By replacing e with e^(2πi+1) you have introduced a branch switch which produces the failure. This should not be considered an inconvenience, it is a necessary component of maths to extract all solutions. For example, if we wish to calculate all the cube root of 1 we could use z={e^(2kπi)}^⅓, k∈ℤ. With the method you have outlined this would only result in one solution. You didn't say which log should always be used but assuming the principal log then the only solution produced would be z = 1. Another issue with this approach is that basic mathematical laws fail. For example, ln{(-1)²} ≠ 2ln(-1): ln{(-1)²}=ln1 = 0 but 2ln(-1)=2(ln1+πi) = 2πi. The correct approach is to establish which branch we are using (by explicitly including arguments) and sticking to it, still using aᵇ=e^{bln(a)}. If we do this then ln(eˣ)=x and (aᵇ)ᶜ=aᵇᶜ (a≠0). For example, with our ln{(-1)²}, 2ln(-1) case we would have ln{(-1)²}=ln[{e^(iπ)}²]=ln{e^(2πi)} = 2πi and 2ln(-1)=2ln{e^(πi)} = 2πi. In your example 'correct' solution the branch switch introduced by the substitution was changed back by another branch switch in the log function to give the correct answer. Sometimes two wrongs can make a right. Note that there will be infinitely many branches, even though roots may repeat, their corresponding branches still cannot be interchanged.

[@WindStreak_]

This was a great video and introduction to the power rule of complex numbers for me, but I got confused with the formula. Does the e in (e^x)^y = e^(y ln(e^x)) refer to the base, like z, or always Euler's number? In the future, it might be better to use a less ambiguous variable, or to specify which one e is

[@omintionpg3d652]

3:06 the equation is using pi as 3.14…and not 0 right

[@cbhorxo]

1:11 From here, we could also do this: e^2ipi=e^1 So 2ipi=1 i×pi=1/2 i×pi=0.5

[@TheReaverOfDarkness]

I came up with a simple method of resolving these X = any Y type proof: seek the point at which the entire operation is multiplied by 0. This is a reset which gives control over everything and makes the outcome determined by the design of the equation after that point, with any prior inputs being lost. Now that I have written this, I will watch the video and attempt to use my method to spot the error.

[@TheReaverOfDarkness]

I'm not following the math very well, but already at 1:03 if e^x = e then x = 1. If we accept that i is a known value and pi is unknown, then the only way to satisfy i*pi+1=1 is i*pi=0 or pi = 0. In this particular instance, pi is essentially being used as a proxy for 0, in order to zero out the entire equation and make X = to any Y.

[@sethkingman2118]

This is good thinking but it's not the only way to work it: sometimes you add solutions by a certain step. So π=0 is an extraneous non-solution, but if you're careful, π=π is also a solution.

[@redtoxic8701]

Actually in general, ln(exp(2iπ+1)) can equal 2iπ+1. It can equal 2kiπ+1 for any integer k. What this "paradox" showed is that in this particular problem, k=0 was the only valid branch

[@ianrobinson8518]

I think the following question leads to similar issues: What is i^(4i) ? Firstly, we develop this general formula and result: x^i = e^(i.lnx) = cos(lnx) + i. sin(lnx). Thus 1^i = 1. So is it = (i^i)^4 = [e^(-π/2)]^4 = 0.2078… ^4 = 0.016… ? OR (i^4)^i = 1^i = 1 ??? OR is it both?

[@vishalmishra3046]

*Key Takeaway from this video* cos and sin are periodic functions with period = 2 π. Therefore, e ^ (i 2 π n) = 1 for not just n = 0 but also for all integers n. Therefore, log is also a periodic function, such that ln 1 = i 2 π n (which is 0 only when n = 0 but not for all other non-zero integers).

[@saritanjay]

Simple rule whenever you find something not possible that means some of the real number rule not applied you can easily find them by analysing your process of doing.

[@nicolassigwalt2209]

What's really intersting here is that the mistakes comes from a property of the log we're supposed to be able to use thanks to an other property of it, which is that on the set of real numbers, the log • exp (x) = x, and that's possible for exp and log are both bijections from one set to an other. However, this doesn't work on the complex field for exp(i2n×pi) = 1 for any given integer n, which shows exp is no longer a bijection and as such cannot verify the first property.

[@chessrapper888]

in 2:54 you can't cancel the e's.

[@WebMafia]

if you look at complex numbers as vectors, you will see that 2π _is_ actually zero, if it's measured in radians. a complex number z can be defined as either z = re^iθ (exponential) or z = r (cos θ + i sin θ) (Cartesian). euler's identity is e^iπ = -1. it represents a vector (-1, 0i) in complex field with Cartesian coordinates, while its polar counterpart is a vector (1, π). this may not be a proof of anything, just some clarification from another viewpoint. at the end, 2π = 0, while π = 180°.

[@InPursuitOfCuriosity]

Videos like these confirm that math is not certainly not one of my strengths.

[@azricon]

beta male fixed mindset

[@irrelevant_noob]

Neither are double negatives...

[@knotwilg3596]

Adding 2pi to the angle of a polar representation of a number doesn't change the number's value, i.e. e^z = e^(z+i*2pi) It is equally well known you can't infer the exponents are equal, because indeed that would mean i*2pi = 0 And that's another way of saying the logarithm doesn't act as the exponential's inverse in the complex field like it does in the real field. The featured "proof" camouflages that very well though by eliminating i before the logarithm is taken. Nice!

[@meyes1098]

But with the thing at 1:20, where you substitute e with e^(2iPi+1), can't you go on forever with this? e^(2iPi+1) = e (e^(2iPi+1))^(2iPi+1) = e ((e^(2iPi+1))^(2iPi+1))^(2iPi+1) = e (((e^(2iPi+1))^(2iPi+1))^(2iPi+1))^(2iPi+1) = e And so on.

[@DrHans-RudigerZimmermannTheore]

Great ! In an Equation one has to do the same operation on both sides. In minute 1.31 in the video effectively only the left side was raised to the power of (2i phi +1), the right side is e and has to be raised with the same power but stayes the same. This is the mistake. Presh inserted in a recursive way e only on the left side! This is not allowed in manipulating equations.

[@meyes1098]

@@DrHans-RudigerZimmermannTheore the way he explains it, he's not raising both sides to (2iPi+1), he's substituting e on the left side for (2iPi+1), as it's an identity.

[@jeremyc6054]

You could get to basically the same result in a two-liner: exp(2*pi*i) = 1 = exp(0) hence 2*pi*i = 0, so pi = 0. (Technically, one or more of pi, or 2, and i, must be equal to zero.) The answer is that the complex exponential (unlike the real exponential) is not one-to-one so exp(x) = exp(y) even if x != y. (Specifically, x-y = 2*n*pi*i, so the "difference" is one complete circle around the complex plane, returning you to the same point.) Of course all the extra steps in the video help to conceal what's actually happening...

[@rauf_aze]

Mistake actually begins from where you started to square both sides.

[@irrelevant_noob]

Nah, everything was still good up to that point, and for quite a while after that too.

[@twoduece]

so he gives a mathematical understanding of the issue, but i wanted to give a more intuitive one: complex exponentiation for powers of e can be understood as rotation around a circle in the complex plane. for e^(a+bi), e^a gives the magnitude of the output vector, or the radius of the circle, and b is how many radians you move around the circle. this is why e^(1+2pi(i)) is e, because e^a (e^1 here) is e, and b radians (2pi) just takes you back to the start of the circle. the top equation at 5:30 is true because its just a repeating process of moving 2pi radians around a circle of radius e in the complex plane, but the bottom equation is false because, if you multiply the exponents in the way described, you effectively are mixing together the a and b coefficients. since these coefficients are what determine the radius and distance around the circle you travel, mixing them together ruins the almost iterative way that higher exponents act like they did in the first equation. what tripped me up for a second as someone thats worked a good bit with complex powers is that i spot checked the first time he used the flawed rule he discusses at 8:00 in my head and it worked out, so it took me a few minutes to go back and consider the second time he used it. i was able to figure out the whole w ln(z) thing a few years ago on my own, but i guess since i didnt have a teacher that showed me i never consciously recognized how the regular exponent rule doesnt quite work until now.

[@kamalsaleh6497]

We could take the logarithm of both sides on 0:52 as another path.

[@vishalmishra3046]

You can shorten the proof to this identity => Since e^(i 2 pi) = 1, Therefore, e^(-4 pi^2) = [e^(i 2 pi) = 1] ^ (i 2 pi) = 1 This also means ln 1 = ln [ e ^ (i 2 pi) ] = i 2 pi

[@admingalaxy6628]

Wait wait wait, go to 1:16 and stop the video, The equation there, can't we say that the base is equal to the base so the power is equal to the power and then say that [ 2 × i × pi + 1 = 1 ] and by subtracting 1 from both sides from 1 we get that [2×i×pi = 0] and then all of the 2 and i and pi are equal to 0? I know it is absurdly wrong but I want an actual explanation

[@sethkingman2118]

Not absurdly wrong at all! You remembered the rule correctly; it's just not valid for complex exponents, which is essentially what this video is getting at.

[@admingalaxy6628]

@sethkingman2118 I see, thanks for clarifying

[@Mejayy]

Other perplexing results that i found in my Complex Analysis course were: Cos and sin are not bounded. ln is not holomorph on the whole plabe e^1/z = a has an infinity of solutions (i think) You can solve real integrals using complex analysis

[@geoinmot]

At the step where you had: e^2iπ+1= e Since you have the same base (e) on ether side of the equation, are the exponents not equal to each other? Therefore: e^2iπ+1 = e^1 2iπ+1 = 1

[@konstantinsotov6251]

It's the same as proving that 1 = -1. Just square the both sides and see the results!

[@mssm9495]

This is a circular argument. You'd also need to explain why ln(e^z)z

[@tordjarv3802]

The problem is more related to that the logarithm is not an analytic function over the entire complex plane but it has a branch cut between zero and infinity. When you evaluate (a^b)^c = e^(c*ln(a^b)) you have to consider what branch you evaluate the logarithm on.

[@laurendoe168]

8:27 I don't understand the difference (I'm not disputing it, just saying I don't understand it). Am I correct in the belief that ln (e^x) = xln(e)? If so, doesn't this equal x, since ln(e) = 1?

[@ManioqV]

Complex shenanigans methinks.

[@timangar9771]

cos(pi) = cos(0) Trivially, it can be shown that pi = 0.

[@dattatreyadas]

It is like saying sin (0) = sin(pi), so pi = 0