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# [Blog post] Proof that -1=1 and 1=3

#### Shared by Calvin Lin 30, USA · 2 years, 3 months ago

Can you identify the subtleties which caused these proofs to be incorrect?

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Jan 24, 2013

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!

Jan 27, 2013

Are you not agreeing with me?

Jan 24, 2013

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.

Jan 26, 2013

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.

Jan 27, 2013

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 $$3$$

Apr 6, 2014

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

Feb 4, 2013

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 $n-1=0$.

Jan 27, 2013

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

Jan 24, 2013

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.

Jan 27, 2013

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.

Jan 27, 2013

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 $$\theta$$.

Jan 23, 2013

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)

Jan 24, 2013

Jan 24, 2013

Jan 24, 2013

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.

Jan 24, 2013

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?

Feb 4, 2013

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.

Feb 4, 2013

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 $$0=1$$.

Apr 8, 2014

onnu keri poda

Jul 16, 2013

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].

Jan 27, 2013

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.

Aug 23, 2013

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

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

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

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

$$1=1$$

Aug 7, 2013

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.....

Jul 20, 2013

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

May 27, 2013

log b, b >0.

May 19, 2013

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.

Jan 24, 2013

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.

Jan 24, 2013

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

Apr 8, 2014

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

Jan 23, 2014

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

Dec 9, 2013

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.

Oct 14, 2013

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.

Aug 25, 2013

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

Jul 12, 2013

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.

Jul 11, 2013

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

Jun 5, 2013

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...

May 28, 2013

wy ... 1=1

May 27, 2013

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

Jan 26, 2013

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

Jan 26, 2013

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

Apr 8, 2014

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

Jan 26, 2013

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

Jan 26, 2013

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.

Jan 27, 2013

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.

Jan 27, 2013

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?

Feb 4, 2013

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.?

Jan 27, 2013

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?

Jan 26, 2013

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

Jan 26, 2013

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.

Jan 24, 2013

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.

Jan 24, 2013

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

Jul 16, 2013

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)

Jun 2, 2013

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

Jun 7, 2013

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

Jan 25, 2013

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

Jan 25, 2013

sqrt(x^2)=|x|

Jan 25, 2013

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

Jan 23, 2013

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.

Jan 24, 2013

$$\log (-1)$$ 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.

Feb 1, 2013

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!!!!

May 20, 2013

step 3 is wrong