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

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!

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 $3$

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.

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.

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)

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.

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.

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.

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.

log b, b >0.

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

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.....

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

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

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

$1*1=|i*i|$

$1=1$

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.

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

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

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

wy ... 1=1

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...

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

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.

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

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.

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.

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

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

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.

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

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

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)