Waste less time on Facebook — follow Brilliant.
×

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

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

Main post link -> https://brilliant.org/assessment/techniques-trainer/proof-that-1-1-and-1-3/

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

No vote yet
22 votes

  Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](http://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \­( ... \­) or \­[ ... \­] to ensure proper formatting.
2 \times 3 \( 2 \times 3 \)
2^{34} \( 2^{34} \)
a_{i-1} \( a_{i-1} \)
\frac{2}{3} \( \frac{2}{3} \)
\sqrt{2} \( \sqrt{2} \)
\sum_{i=1}^3 \( \sum_{i=1}^3 \)
\sin \theta \( \sin \theta \)
\boxed{123} \( \boxed{123} \)

See full formatting guide

Sort by:
Jan 24, 2013
Avi Levy
23, USA

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!

Log in to reply

Jan 27, 2013
O B
18, USA

Are you not agreeing with me?

Log in to reply

Jan 24, 2013
Calvin Lin Staff
30, USA

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.

Log in to reply

Jan 26, 2013
Avi Levy
23, USA

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

Jan 27, 2013
Harshit Kapur
17, India

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 \(3\)

Log in to reply

Apr 6, 2014
Prajwal Kavad
18, India

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

Log in to reply

Feb 4, 2013
Harikrishnan M
19, India

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 \[ n-1=0 \].

(last modified Feb 04, 2013 )

Log in to reply

Jan 27, 2013
Andrea Pagliaricci
36, Italy

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

Jan 24, 2013
O B
18, USA

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.

(last modified Jan 24, 2013 )

Log in to reply

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.

Log in to reply

Jan 27, 2013
Calvin Lin Staff
30, USA

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 \( \theta\).

Log in to reply

Jan 23, 2013
Paras Chadha
18, India

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)

(last modified Jan 23, 2013 )

Log in to reply

Jan 24, 2013
Zi Song Yeoh
14, Malaysia

Log in to reply

Jan 24, 2013
Calvin Lin Staff
30, USA

If you read the blog post, I stated the link there.

@Zi Song, I edited your comment to link directly.

Log in to reply

Jan 24, 2013
Rohan Rao
18, India

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

Jan 24, 2013
O B
18, USA

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?

(last modified Jan 24, 2013 )

Log in to reply

Feb 4, 2013
Harikrishnan M
19, India

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.

Log in to reply

Feb 4, 2013
Calvin Lin Staff
30, USA

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 \(0=1\).

Log in to reply

Apr 8, 2014
Gokul G Menon
18 years old

onnu keri poda

Log in to reply

Jul 16, 2013
Debjit Mandal
20, India

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

Jan 27, 2013
Lotfy Eltemsah
19, Egypt

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.

Log in to reply

Aug 23, 2013
Akshath Singhal
17, India

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

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

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

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

\(1=1\)

Log in to reply

Aug 7, 2013
Arijit Banerjee
19, India

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.....

Log in to reply

Jul 20, 2013
Kalyph Dioquino
25, Philippines

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

Log in to reply

May 27, 2013
Robi Fajar Bahari
20, Indonesia

log b, b >0.

Log in to reply

May 19, 2013
Vishwajeet Karna
24, India

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 in to reply

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.

Log in to reply

Jan 24, 2013
Calvin Lin Staff
30, USA

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

Apr 8, 2014
Gokul G Menon
18 years old

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

Log in to reply

Jan 23, 2014
Uzair Asif
21, Pakistan

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

Log in to reply

Dec 9, 2013
Rachna Singh
31, India

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.

Log in to reply

Oct 14, 2013
Priyesh Pandey
18, India

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.

Log in to reply

Aug 25, 2013
Venkatraman Balaji
16, India

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

Log in to reply

Jul 12, 2013
Jeffer Dave Cagubcob
17, Philippines

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.

Log in to reply

Jul 11, 2013
Osama Fouad
19, Egypt

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

Log in to reply

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...

(last modified Jun 05, 2013 )

Log in to reply

May 28, 2013
Kargar Swati
19, India

wy ... 1=1

Log in to reply

May 27, 2013

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

Log in to reply

Jan 26, 2013
Andrea Pagliaricci
36, Italy

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

(last modified Jan 26, 2013 )

Log in to reply

Jan 26, 2013
Calvin Lin Staff
30, USA

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

Apr 8, 2014
Gokul G Menon
18 years old

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

Log in to reply

Jan 26, 2013
Andrea Pagliaricci
36, Italy

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

(last modified Jan 26, 2013 )

Log in to reply

Jan 26, 2013
Calvin Lin Staff
30, USA

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

Jan 27, 2013
Andrea Pagliaricci
36, Italy

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

Jan 27, 2013
Andrea Pagliaricci
36, Italy

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

Feb 4, 2013
Harikrishnan M
19, India

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

Jan 27, 2013
Andrea Pagliaricci
36, Italy

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?

(last modified Jan 27, 2013 )

Log in to reply

Jan 26, 2013
Luciano Riosa
53, Italy

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

Log in to reply

Jan 26, 2013
Calvin Lin Staff
30, USA

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

Jan 24, 2013
Vladimir Staver
23, Spain

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.

Log in to reply

Jan 24, 2013
Calvin Lin Staff
30, USA

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

Log in to reply

Jul 16, 2013
Debjit Mandal
20, India

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)

Log in to reply

Jun 2, 2013

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

Log in to reply

Jun 7, 2013
Daniel Liu
15, USA

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

Jan 25, 2013
Sahil Anand
17, India

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

Log in to reply

Jan 25, 2013
Vishnu Valsalan
17, India

sqrt(x^2)=|x|

Log in to reply

Jan 25, 2013
Calvin Lin Staff
30, USA

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

Log in to reply

Jan 23, 2013
Mustafa Warsi
19, India

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.

(last modified Jan 23, 2013 )

Log in to reply

Jan 24, 2013
Calvin Lin Staff
30, USA

\( \log (-1)\) 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.

Log in to reply

Feb 1, 2013
Jaydutt Kulkarni
21, India

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

May 20, 2013

step 3 is wrong

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...