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TopNewestBoth of these proofs imprecisely try to extend analysis of functions on the positive real line onto the entire complex plane. For instance, the mistake in the first proof is to implicitly assume that \(\sqrt{\cdot}\) is a univalent function. A proper treatment of these concepts is contained in the theory of branch cuts and Reimann surfaces on the complex plane.

Similarly, the mistake in the second proof is to assume that \(\log(\cdot)\) is univalent on the complex plane. In fact, this is an even "worse" mistake than in the first proof attempt. Whereas \(\sqrt{\cdot}\) has \(2\) branches, \(\log(\cdot)\) has

infinitely many! In fact, it turns out that this function is defined only up to multiples of the complex constant \(2\pi i\).Problems like these show how fruitful Complex Analysis can be, even when an equation doesn't mention anything imaginary at the outset! – Avi Levy · 3 years, 11 months ago

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agreeing with me? – O B · 3 years, 11 months ago

Are you notLog in to reply

As a reminder, us \ ( and \ ) (without spaces) to denote code. – Calvin Lin Staff · 3 years, 11 months ago

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– Avi Levy · 3 years, 11 months ago

Oops, didn't mean to double post. Yeah, I got confused and used $ - on an unrelated note, why does brilliant use \ ( instead? It kind of goes against the flow, I guess. Thanks for deleting the accidental post for me.Log in to reply

I've got a better proof

\(e^{i\pi} = -1\)

\(\Rightarrow e^{2i\pi} = 1\) [squaring both sides]

\(\Rightarrow 2i\pi = 0\) [taking natural log]

\(\Rightarrow i = 0\) [dividing by \(2\pi\)]

\(\Rightarrow -1 = 0\) [squaring both sides]

\(\Rightarrow 1 = 0\) [squaring both sides]

An induction proof will yield that all numbers are equal, not just \(1\) and \(3\) – Harshit Kapur · 3 years, 11 months ago

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– Prajwal Kavad · 2 years, 9 months ago

the mistake is natural log can't take imaginary values because its domains is real numbers in (0,infinity)Log in to reply

– Harikrishnan M · 3 years, 11 months ago

Note a general relation.Which is valid for \[ z\in \mathbb{C} \] \[ \ln { z } =\ln { \left| z \right| } +\left(Arg\left( z \right) +2n\pi\right)\cdot i \\ n\in\mathbb{Z} \] Then your equation simply tells there exist an \[ n \] such that \[ n-1=0 \].Log in to reply

– Andrea Pagliaricci · 3 years, 11 months ago

i = sqrt(-1), not at -1. Is correct? But, I think the substance of the problem does not change or, at least, little changes...Log in to reply

Both proof errors are due to the multi-valued nature of our complex root and logarithm functions. In proof 1, step (3) is indeterminate because \(\sqrt{zw}=\sqrt{z}\sqrt{w}\) and similarly \(\sqrt{\frac{z}w}=\frac{\sqrt{z}}{\sqrt{w}}\) generally

do not hold. We have a very similar problem in step (2) of proof 2, i.e. being inconsistent with which branch of our complex functions we're using and presuming properties that are not guaranteed. – O B · 3 years, 12 months agoLog in to reply

I think the mistake in the second proof is assuming the function log to be one-one over the complex numbers. Note that e^ix= cos(x) + i

sin(x). Putting x=pi, we get the formula e^ix= -1. But if we put x=(2n+1)pi, where n is an integer, we get the formula that e^{i(2n+1)(pi)}= -1. Since there are infinitely many possible values of n, we can conclude that the function e^ix is not one-one.I think here the case is similar to solving a trigonometrical equation. For example, sin(30 degrees)= sin(390 degrees), but 30 is not equal to 390. In general when we solve a trigonometric equation like cos(x)= cos(A), we find infinitely many solutions of the form x= 2n(pi) +-A. Similarly a general solution to e^ix= 1 should be written as 2

pi(2n+1), where n is an integer. I think that the log(-1) which appears at both sides cannot be cancelled out because the value of n in the first log(-1) and the value of n in the second log(-1) are unequal. – Sreejato Bhattacharya · 3 years, 11 months agoLog in to reply

The example of trigonometric functions was going to be my next post for such proofs. Looks like I'd need to create another example to talk about instead. It's a very common mistake made, where students say that \( \sin \theta = \frac {1}{2} \) thus \( \theta = 30^\circ\), without further justifying the domain of \( \theta\). – Calvin Lin Staff · 3 years, 11 months ago

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In the proof of 1 = -1 , step 3 is wrong. We cannot write root (a/b) as root(a)/root(b), when a or b or both are negative. In fact in such case we have root (a/b) as - root(a)/root(b), which gives us the correct result (In this case also) – Paras Chadha · 3 years, 12 months ago

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[Help]Proof that 1 = -1 – Zi Song Yeoh · 3 years, 12 months ago

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@Zi Song, I edited your comment to link directly. – Calvin Lin Staff · 3 years, 12 months ago

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– Rohan Rao · 3 years, 11 months ago

When proving -1 = 1, at one point, you take the square root on both sides. When you do so, shouldn't you get a + - ambiguity? And in the proof of 1 = 3, logarithm of negative numbers is not defined.Log in to reply

howis this better? – O B · 3 years, 12 months agoLog in to reply

The people who are saying that the error is in taking the square root of a negative number are wrong; there is certainly a branch of math for manipulating imaginary and complex numbers. Yes the square root of -1 is imaginary. That does not mean it will cause problems. The real reason lies somewhere between the third and fourth steps. By the fourth step you are already false. sqrt(1) = 1, sqrt(-1) = i, so you have 1/i = i/1 which is not true. 1/i = -i. The reason the math "failed" here is that all complex numbers except zero have two square roots, one negative and one positive, and that when you do this sort of operation there is no clear reason to select one over the other. Square roots commonly give sign errors; square roots with complex numbers are even worse. – Lotfy Eltemsah · 3 years, 11 months ago

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This problem arises since these functions(I am afraid to call them so) are multivalued. And you can call these as functions only under proper domain The following is a simpler example. The below integral is evaluated using integration by parts \[ \int { x\cdot \frac { 1 }{ x^{ 2 } } } dx=x\cdot \frac { -1 }{ x } +\int { 1\cdot \frac { 1 }{ x } } dx \] If \[ x\neq 0 \] \[\int { x\cdot \frac { 1 }{ x^{ 2 } } } dx=\int { 1\cdot \frac { 1 }{ x } } dx \] And hence get \[ 0=-1\] I guess many people would identify the flaw in the above proof . Indefinite integral of a function is not single valued \[ \int { f\left( x \right) dx } =\int { f\left( x \right) dx } +C\]is valid and hence the above proof would only give \[C=-1\] which can be correct. I hope this makes things more clear. – Harikrishnan M · 3 years, 11 months ago

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There goes another way for me to prove that \(0=1\). – Calvin Lin Staff · 3 years, 11 months ago

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– Gokul G Menon · 2 years, 9 months ago

onnu keri podaLog in to reply

– Debjit Mandal · 3 years, 6 months ago

In step 3, x(x-1)=0. Here, the solutions of x are 0 and 1. But we know, that all the solutions of an equation need not satisfy the given conditions. Here, the given condition is x=1. And the solution, x=0 does not satisfies the given condition. So, this value should be eliminated. And, the other value is satisfying the given condition, x=1. So, this is the only solution. Here, the equation does not has two solutions. There is only one solution; x=1[x=0 has already eliminated].Log in to reply

the problem in the first part is in taking square roots, one has to use modulus function while taking sqrt. hence \(\sqrt{1} / \sqrt{-1}=\sqrt{-1} / \sqrt{1}\)

\(|1|/|i|=|i|/|1|\)

\(|1|*|1|=|i|*|i|\)

\(1*1=|i*i|\)

\(1=1\)– Akshath Singhal · 3 years, 4 months agoLog in to reply

how can anyone write log(-1) it doesn't even exist .... its undefined..... its just like multiplying both sides of a given proof by 0..... – Arijit Banerjee · 3 years, 5 months ago

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haha nice try...but log(-1) is kinda erhm.....math error – Kalyph Dioquino · 3 years, 6 months ago

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log b, b >0. – Robi Fajar Bahari · 3 years, 7 months ago

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Imaginary Numbers are special because when squared, they give a negative result. Normally this doesn't happen, because when you square a positive number you get a positive result, and when you square a negative number you also get a positive result (because a negative times a negative gives a positive). But just imagine there is such a number, because we are going to need it! So, our conception of real and imaginary numbers contradict here, so one needs to give a new theory or case and conditions in existing concept to prove the same way. – Vishwajeet Karna · 3 years, 8 months ago

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You can't raise ten to the anything-th and get neg 1, so you can't take the log of a negative # in the second one. And for the first one, b/c you introduced the sq root, isn't the root of 1 plus or minus one? So the assumption that the square root of one is the positive root is incorrect. – Mary Jane Porzenheim · 3 years, 12 months ago

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– Calvin Lin Staff · 3 years, 11 months ago

Recall euler's formula that \( e^{i \pi} = -1\). So we can raise 10 to some (complex) power and obtain \(-1\).Log in to reply

both proofs are erroneous; in proof 1,error is in 3rd step; in proof 2,error is in 2nd step; – Gokul G Menon · 2 years, 9 months ago

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In second problem there is a mistake in 2nd step. We can't take log of negative numbers – Uzair Asif · 2 years, 12 months ago

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the problem with second proof is that the log of negative number does not exist, i.e. we cannot apply log function to a negative number. – Rachna Singh · 3 years, 1 month ago

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This is for your proof that 1=0; X=1; X^2=1; This implies, X=1/X; giving the result x can take value =1 and not 0 since 1/0 is not defined. – Priyesh Pandey · 3 years, 3 months ago

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from step 3 i.i=/1./1 and not 1 – Venkatraman Balaji · 3 years, 4 months ago

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Step 5 is wrong. We cannot cross-multiply it since it is not an equality. Number 4 only shows that 1/i is equal to i/1. – Jeffer Dave Cagubcob · 3 years, 6 months ago

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you can't take log(-1) it is a math error :) – Osama Fouad · 3 years, 6 months ago

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sir i don't know if i'm wrong but plz check that log(-ve nos.) is not defined as domain condition is not satisfied... – Bhanu Pratap Sharma · 3 years, 7 months ago

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wy ... 1=1 – Kargar Swati · 3 years, 7 months ago

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in simple words :: I guess for the second problem we can't take the log for a negative number – Masochistic Psychopath · 3 years, 7 months ago

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For problem 1: the wrong is at step 3 -> 4: sqrt(-1)^2 is not i but i^2! – Andrea Pagliaricci · 3 years, 11 months ago

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– Calvin Lin Staff · 3 years, 11 months ago

I do not understand you. At no point do we take \( \sqrt{(-1)^2}\) or \( (\sqrt{-1})^2\). We only have \( \sqrt{1}\) and \( \sqrt{-1} \).Log in to reply

– Gokul G Menon · 2 years, 9 months ago

ninnod njan orayiram pravashyam chodichille choru veno choru veno ennu...,aksharam onnum maari poyittillallo..??Log in to reply

– Andrea Pagliaricci · 3 years, 11 months ago

(sqrt(-1))^2 = i^2... ok? step 3 -> step 4 is do: sqrt(-1) -> i... is wrong! or not?Log in to reply

– Calvin Lin Staff · 3 years, 11 months ago

Yes, so the key is to explain why \( \sqrt{-1} = i \) is wrong. In particular, why would you say that it's wrong in this 'proof', whereas most people would normally say that is correct.Log in to reply

– Andrea Pagliaricci · 3 years, 11 months ago

Obviously, despite not having identified which step was the error, the sarcasm and the lack of courtesy in responding are not functional and useful to the discussion.Log in to reply

– Andrea Pagliaricci · 3 years, 11 months ago

Proof that 1 = -1 was not your theorem that I was your trying to disassemble dramatically, but it is a small mathematical game which we know already departing that there is an error and, the error, must be found... If one fails, it has not hurt anyone, or not?Log in to reply

– Harikrishnan M · 3 years, 11 months ago

Square root is a multivalued function when the domain becomes negative real numbers Hence you can't conclude \[ \sqrt{-1} =i \] But \[ \sqrt{-1} =\pm i \] In case of positive real numbers it has been a convention to take \[ \sqrt{x} \] as positive with \[ x \] as a positive. But in case of -ve real numbers whose square root is complex where there does not exist anything like positive or negative. I hope that explains the flaw in first problem.?Log in to reply

– Andrea Pagliaricci · 3 years, 11 months ago

Obviously, I meant that (sqrt(-1))^2 is not equal to i but is equal to i^2... But, of course, isn't this the whole point of the problem... The error, I think is in 2 -> 3 step ... Rather, the extent to which it is legitimate to compare or even match a real number to a complex number?Log in to reply

The \(\sqrt[n]{A}\) has n roots and in both case you chose the "wrong" ones... – Luciano Riosa · 3 years, 11 months ago

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– Calvin Lin Staff · 3 years, 11 months ago

Yes, that's something that's often not realized. When taking roots, we have to be really careful about which root we end up taking. Especially when taking multiple square roots, there is no guarantee that the relevant root will be chosen for each of them.Log in to reply

In the first proposal proof (1=-1), there are a mistake in the third step. The mistake is that square(-1)*square(-1) is not equal at 1. Respect at the second proposal proof the mistake is that the number (-1) can't be solution of e^x=-1 or in other form log(-1) don't exist. – Vladimir Staver · 3 years, 11 months ago

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– Calvin Lin Staff · 3 years, 11 months ago

What about \( e^{i \pi} = -1\)?Log in to reply

In the step 4, the square roots are taken, but the negative value is not taken. We know that, when we take the square roots, then two values come out, one is positive and the other one is negative. But it is not necessary that, all of them satisfies. At least one of them must satisfy. In the step 4 the positive value has taken. Here the positive does not satisfy. So we should take the negative value. Here the negative value satisfies. And then, we shall get, 1=1, not 1=(-1) – Debjit Mandal · 3 years, 6 months ago

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we can't apply squareroot for negative numbers,so we can't say 1st proof is correct – Avinash Sakalabakthula · 3 years, 7 months ago

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– Daniel Liu · 3 years, 7 months ago

we can apply square root for negative numbers, just the multiplicative property of square roots does not apply for negative numbers.Log in to reply

SINCE 1=1,(1)^2=(1)^2 (-1)^2=(1)^2 -1=1 Hence proved – Sahil Anand · 3 years, 11 months ago

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– Vishnu Valsalan · 3 years, 11 months ago

sqrt(x^2)=|x|Log in to reply

– Calvin Lin Staff · 3 years, 11 months ago

That's good, that's another way of expressing it. And it fails for the same reason too.Log in to reply

Well, the proof of 1 = 3 is faulty is so, because there exists no log (-1). That is undefined, and so you can’t divide both sides by log (-1). It’s like dividing by 0, I’d hazard. – Mustafa Warsi · 3 years, 12 months ago

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Just because you haven't seen it yet doesn't mean that it doesn't exist. For example, when you started learning about natural numbers and then rational numbers, it might be reasonable to assume that everything must be the ratio of 2 integers, like the ancients did. However, irrational numbers do exist, and you need to find a way to explain their existence. – Calvin Lin Staff · 3 years, 11 months ago

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– Jaydutt Kulkarni · 3 years, 11 months ago

I think the step where the proof of no.2 lost all sense was when you assigned a single value to a one-to-many function as -1.Very cunning!!!!Log in to reply

– Vasanth Balakrishnan · 3 years, 8 months ago

step 3 is wrongLog in to reply