Math also Fails! -2

Most of you shall be knowing about Euler's formula: eiθ=cos(θ)+isin(θ)e^{i\theta}=\cos(\theta)+i\sin(\theta) If you put θ=τ\theta=\tau (where τ\tau is a unit of measuring angle if you don't know about it see this and this) eiτ=cos(τ)+isin(τ)=1+i×0=1e^{i\tau}=\cos(\tau)+i\sin(\tau)=1+i\times0=1 eiτ=1\Rightarrow \boxed{e^{i\tau}=1} Taking ln\ln on both sides ln(eiτ)=ln(1)\ln(e^{i\tau})=\ln(1) iτln(e)=ln(1)\Rightarrow i\tau\ln(e)=\ln(1) iτ×1=ln(1)\Rightarrow i\tau\times1=\ln(1) As ln(1)=0\ln(1)=0 iτ=0\therefore i\tau=0 As i0i≠0 τ=0\therefore {\red{\tau=0}} If you want you can further write 2π=02\pi=0 π=0\Rightarrow \red{\pi=0} Now will you agree with the last statement? Doesn't this means Math also fails!

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Note by Zakir Husain
3 months, 2 weeks ago

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The flaw is that in the complex plane, ez1=ez2⇏z1=z2e^{z_1} = e^{z_2} \not \Rightarrow z_1 = z_2. This is because eiθ=cosθ+isinθe^{i\theta} = \cos \theta + i \sin \theta. The sine and cosine functions each have a period of 2π2\pi, so eiθ=ei(θ+2π)e^{i\theta} = e^{i(\theta + 2\pi)}. This leads to the misleading conclusion that 0=2π0 = 2\pi.

Elijah L - 3 months, 2 weeks ago

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Now, that's the correct explanation!

Zakir Husain - 3 months, 2 weeks ago

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Wow!

Mahdi Raza - 3 months, 2 weeks ago

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@Elijah L although an argunent of 2π is not allowed in argand plane ,but we can generalize by considering periodicity.

Kriti Kamal - 3 months ago

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@Zakir Husain The problem starts when you forget about periodicity of trignometric functions and limits of argument of complex number.

we know that, e^{iθ}=cos(θ)+isin(θ) .The argument of this complex number will be θ and the argument of a complex number should be in the range of (-π,π].So,you can't take θ as 2π,-π etc .

Now,if we want to expand its range we should consider periodicity of trignometric function. e^{iθ}=e^{i2nπ+θ}

I know this question is not for me and neither i have mentioned to answer it.But,as a noob i have tried to figure out something.Please comment if you find any error.

Kriti Kamal - 3 months ago

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@Zakir Husain,please reply am i correct or not?

Kriti Kamal - 3 months ago

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It is also an explanation same as that given by @Elijah L, but stated differently.

Zakir Husain - 3 months ago

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Oh,i see but i am still confused that argument could be 2π or not.

Kriti Kamal - 3 months ago

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@Kriti Kamal It can be, but you need to take this rule with you :

If ez1=ez2:z1C,z2Cz1=z2\boxed{\blue{e^{z_1}=e^{z_2}}:z_1\in C, z_2\in C}\cancel{\Rightarrow}\boxed{\red{z_1=z_2}}

Zakir Husain - 3 months ago

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@Zakir Husain I know it .But,is it a rule?? Many books say that argument has limit(-π,π]

Kriti Kamal - 3 months ago

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@Chew-Seong Cheong, @Mahdi Raza , @Kumudesh Ghosh , @Alak Bhattacharya , @Vinayak Srivastava , @Aryan Sanghi , @Jeff Giff , @Marvin Kalngan

Zakir Husain - 3 months, 2 weeks ago

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@Brilliant Mathematics, I just encountered a bug where i did not get a notification for this mention.

Mahdi Raza - 3 months, 2 weeks ago

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@Zakir Husain, there is a small little typo in the second line as well. There should be a closed bracket instead of 0

Mahdi Raza - 3 months, 2 weeks ago

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@Zakir Husain, Is there a part 1 or 3 of "Math Also Fails"

Mahdi Raza - 3 months, 2 weeks ago

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@Mahdi Raza- Math Also Fails! -1, No part 3 is posted yet

Zakir Husain - 3 months, 2 weeks ago

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Ok

Mahdi Raza - 3 months, 2 weeks ago

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I don't understand much of the proof, but in my opinion there has to be some problem in using 1\sqrt{-1} as I saw some video on Numberphile which says that some properties of numbers are not allowed for 1\sqrt{-1}.

Vinayak Srivastava - 3 months, 2 weeks ago

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@Vinayak Srivastava - I onlly used 22 properties:

  • The euler's identity

  • If ai=0ai=0 then a=0a=0 because i0i≠0

These are all applicable for ii.

Zakir Husain - 3 months, 2 weeks ago

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This mistake is very common, e×πe\times\pi is irrational proof

Zakir Husain - 2 months ago

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@Zakir Husain, I don't agree with the last statement at all. Also (not being rude, this is a good note, don't take it personally), I believe that the Math also Fails series is somewhat undermining mathematics in a way inexpressible. For example, if you're saying π=0\pi = 0, you're in effect, undermining international π\pi day and all the people who have worked hard to help improve the value of π\pi. So, I implore you, although this is a joke and I believe this is a good one, don't put a post that undermines people's work by, in effect, proving mathematics wrong (I check all new discussions every day - that's why I'm saying not to post a note that undermines people's work by, in effect, proving mathematics wrong). You can still post it, as long as it steers away from people's work (i.e. don't do ee, number sequences that people found etc.) and my defensive opinion of people's hard work and mathematics and it's reputation...

A Former Brilliant Member - 3 months, 2 weeks ago

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@Yajat Shamji - Mathematics demands proof. I had just used legal mathematical arguments, if you can find a flaw then find it!

Zakir Husain - 3 months, 2 weeks ago

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Flaw on 44th step...

A Former Brilliant Member - 3 months, 2 weeks ago

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@A Former Brilliant Member Please write the expression so I will be able to understand

Zakir Husain - 3 months, 2 weeks ago

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@Zakir Husain ln(eir)=ln(1)irln(e)=ln(1)ln(e^{ir}) = ln(1) \rightarrow irln(e) = ln(1) is flawed.

A Former Brilliant Member - 3 months, 2 weeks ago

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@A Former Brilliant Member It is a legal mathematical property of logarithm see here for more details

Zakir Husain - 3 months, 2 weeks ago

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Yo I think you're taking this too seriously. Of course it's not seriously questioning the statement that π=3.14\pi = 3.14.

Krishna Karthik - 3 months, 2 weeks ago

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The way I see it, it's not a problem at all. We all know that the proof is false, and it's a sort of entertaining mathematical challenge to find out where the error lies.

Elijah L - 3 months, 2 weeks ago

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Yeah exactly, the guy above can't seem to get it.

Krishna Karthik - 3 months, 2 weeks ago

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