# [Blog post] Proof that -1=1 and 1=3 Can you identify the subtleties which caused these proofs to be incorrect? Note by Calvin Lin
8 years, 6 months ago

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Both 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!

- 8 years, 6 months ago

I see you got the formatting right in this post, and I'd delete the other one :)

As a reminder, us \ ( and \ ) (without spaces) to denote code.

Staff - 8 years, 6 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.

- 8 years, 6 months ago

Are you not agreeing with me?

- 8 years, 6 months ago

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 $3$

- 8 years, 6 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 $n-1=0$.

- 8 years, 6 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...

- 8 years, 6 months ago

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

- 7 years, 4 months ago

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.

- 8 years, 6 months ago

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) + isin(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 2pi(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.

- 8 years, 6 months ago

That's great.

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 $\theta$.

Staff - 8 years, 6 months ago

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)

- 8 years, 6 months ago

- 8 years, 6 months ago

Staff - 8 years, 6 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.

- 8 years, 6 months ago

Are you sure? According to you, $\sqrt{\frac1{-1}}=-\frac{\sqrt1}{\sqrt{-1}}=-\frac1i$ and $\sqrt{\frac{-1}1}=\frac{\sqrt{-1}}{\sqrt1}=-\frac{i}1$... now we have $-\frac1i=-\frac{i}1$ -- tell me, how is this better?

- 8 years, 6 months ago

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.

- 8 years, 6 months ago

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.

- 8 years, 6 months ago

Indeed. Many students forget the constant $C$ when doing integration (and hence lose 1 mark per problem).

There goes another way for me to prove that $0=1$.

Staff - 8 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].

- 8 years ago

onnu keri poda

- 7 years, 3 months ago

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.

- 8 years, 6 months ago

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

Staff - 8 years, 6 months ago

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.

- 8 years, 2 months ago

log b, b >0.

- 8 years, 2 months ago

haha nice try...but log(-1) is kinda erhm.....math error

- 8 years ago

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.....

- 7 years, 12 months ago

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

$|1|/|i|=|i|/|1|$

$|1|*|1|=|i|*|i|$

$1*1=|i*i|$

$1=1$

- 7 years, 11 months ago

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.

- 8 years, 6 months ago

What about $e^{i \pi} = -1$?

Staff - 8 years, 6 months ago

The $\sqrt[n]{A}$ has n roots and in both case you chose the "wrong" ones...

- 8 years, 6 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.

Staff - 8 years, 6 months ago

For problem 1: the wrong is at step 3 -> 4: sqrt(-1)^2 is not i but i^2!

- 8 years, 6 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 $\sqrt{-1}$.

Staff - 8 years, 6 months ago

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

- 8 years, 6 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.

Staff - 8 years, 6 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.

- 8 years, 6 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?

- 8 years, 6 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?

- 8 years, 6 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.?

- 8 years, 6 months ago

ninnod njan orayiram pravashyam chodichille choru veno choru veno ennu...,aksharam onnum maari poyittillallo..??

- 7 years, 3 months ago

in simple words :: I guess for the second problem we can't take the log for a negative number

- 8 years, 2 months ago

wy ... 1=1

- 8 years, 2 months ago

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...

- 8 years, 2 months ago

you can't take log(-1) it is a math error :)

- 8 years ago

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.

- 8 years ago

from step 3 i.i=/1./1 and not 1

- 7 years, 11 months ago

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.

- 7 years, 9 months ago

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.

- 7 years, 7 months ago

In second problem there is a mistake in 2nd step. We can't take log of negative numbers

- 7 years, 6 months ago

both proofs are erroneous; in proof 1,error is in 3rd step; in proof 2,error is in 2nd step;

- 7 years, 3 months ago

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.

- 8 years, 6 months ago

$\log (-1)$ can be defined.

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.

Staff - 8 years, 6 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!!!!

- 8 years, 6 months ago

step 3 is wrong

- 8 years, 2 months ago

SINCE 1=1,(1)^2=(1)^2 (-1)^2=(1)^2 -1=1 Hence proved

- 8 years, 6 months ago

sqrt(x^2)=|x|

- 8 years, 6 months ago

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

Staff - 8 years, 6 months ago

we can't apply squareroot for negative numbers,so we can't say 1st proof is correct

- 8 years, 2 months ago

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

- 8 years, 2 months ago

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)

- 8 years ago