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?

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?

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 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

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

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

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) + 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.

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

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

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

Jan 24, 2013

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

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 \(0=1\).

just now

onnu keri poda

Jul 16, 2013

Jan 27, 2013

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

Jul 20, 2013

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

May 27, 2013

log b, b >0.

May 19, 2013

Jan 24, 2013

Jan 24, 2013

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

Oct 14, 2013

Aug 25, 2013

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

Jul 12, 2013

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

Apr 8, 2014

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

Jan 27, 2013

Jan 27, 2013

Feb 4, 2013

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

Jan 24, 2013

Jan 24, 2013

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

Jul 16, 2013

Jun 2, 2013

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

Jun 7, 2013

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

May 20, 2013

step 3 is wrong

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