proof of -1 = +1

**mnova** · January 22nd 2009, 12:58 PM

[I hope I did the LaTex correctly.]

We know that $\textit{i}^2 = \sqrt{-1}^2$ and the square root and square operations are inverses so they cancel and $\textit{i}^2 = -1$ .

And we know that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ .
example: $2\textit{i} = \sqrt{4}\sqrt{-1} = \sqrt{4(-1)} = \sqrt{-4}$

So $\textit{i}^2 = \textit{ii} = \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{+1} = +1$ .

So -1 = +1
QED

Where's the problem with this?

**ursa** · January 22nd 2009, 01:09 PM

[I hope I did the LaTex correctly.]

We know that

and the square root and square operations are inverses so they cancel and

.

And we know that

.
example:

So

.

So -1 = +1
QED

Where's the problem with this?

hi
you are only considering one part that is:
(under-root of x^2)=x
but actually it is
(under-root of x^2)=+x or -x

**mnova** · January 22nd 2009, 01:20 PM

Ursa,

Yes, sqrt(1) has two solutions, -1 and +1.

The point is that the +1 answer is just as valid as the -1 answer is.

**Jhevon** · January 22nd 2009, 01:25 PM

Originally Posted by mnova

[I hope I did the LaTex correctly.]

We know that $\textit{i}^2 = \sqrt{-1}^2$ and the square root and square operations are inverses so they cancel and $\textit{i}^2 = -1$ .

And we know that $\sqrt{a}\sqrt{b} = \sqrt{ab}$ .
example: $2\textit{i} = \sqrt{4}\sqrt{-1} = \sqrt{4(-1)} = \sqrt{-4}$

So $\textit{i}^2 = \textit{ii} = \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{+1} = +1$ .

So -1 = +1
QED

Where's the problem with this?

as ursa hinted, there are two square roots to any number--and as such, they cannot be distinguished from each other. in the reals, we distinguish between them in terms of the ordering we have, that is, positives are "greater than" negatives. however, the complex numbers have no such ordering. the proof plays on the fact there are 2 square roots and the fact that the complex numbers are not ordered to give rise to what seems like a contradiction. so really what is going on here is that both 1 and -1 are square roots of 1. so the "equality" is playing on that fact that in the complex numbers, we cannot distinguish between them in terms of order. passing through the complex numbers removes the "natural" ordering we are used to with the real numbers and so, we get the obviously silly statement 1 = -1.

by the way, the $\sqrt{\; \;}$ symbol denotes the principal square root, that is, the positive one. but in idea of "square root" extends beyond that

other views are given here. i do not agree with most of them though

**ursa** · January 22nd 2009, 01:27 PM

Ursa,

Yes, sqrt(1) has two solutions, -1 and +1.

The point is that the +1 answer is just as valid as the -1 answer is.

Noooooo
you simply cant ignore the other part
put -1 instead of +1 and see if you are getting the same.
its maths dear, you have to consider all the situation
one example is enough for contradiction
here you are developing your own moves

**mnova** · January 22nd 2009, 03:31 PM

Jhevon

That link you provided is fascinating.

Among the posts are two posters who cleverly demand that i is defined by the equation $\textit{i}^2 = -1$ and that i is definitely NOT equal to $\sqrt{-1}$ .

That seems suspicious to me, as then how can you justify $2]\textit{i} = \sqrt{-4}$ ?

So I consulted several texts on complex analysis and found:
1- Some authors define, and explicitely state, that $]\textit{i} = \sqrt{-1}$
2- Some authors define math]\textit{i}^2 = -1[/tex] and never bring up the subject of what i is.
3- Some authors don't seem to consider complex numbers at all, but teach it using only the geometric interpretation of i = (0, 1) in the 2-D plane. They seem to be only teaching geometry, although I didn't look into the texts thoroughly.
4- One author doesn't seem to consider complex numbers at all, but teaches it using only vectors in the plane. Again, I didn't look into the text thoroughly.

Only one author, Ahlfors, (of type 2 above) goes further, referring to square roots of positive reals and of complex numbers but not mentioning square roots of negative numbers.

It seems that the conumdrum of what i really is is something mathematicians don't want to talk about.

**Jhevon** · January 22nd 2009, 03:36 PM

Originally Posted by mnova

Among the posts are two posters who cleverly demand that i is defined by the equation $\textit{i}^2 = -1$ and that i is definitely NOT equal to $\sqrt{-1}$ .

that is the part i do not agree with

i think both definitions are equally valid. the latter is just saying that $i$ is one of the roots to the equation $x^2 + 1 = 0$ . that is, it is one of the square roots of -1 (which can be found using basic methods in complex analysis)

Originally Posted by mnova

That seems suspicious to me, as then how can you justify $2]\textit{i} = \sqrt{-4}$ ?

this is one of the reasons i do not like that view. saying $i$ is not $\sqrt{-1}$ is so limiting. it really makes it a pain (for no reason) for doing a lot of problems involving algebra in the complex numbers. as far as i can see, and several professors i have spoken to about it, $i = \sqrt{-1}$ makes perfect "sense". it gets the job done, and is consistent with the way things work in the complex numbers. writing complex numbers in polar form, for instance, makes it clear that something like $\sqrt{-1}$ can be defined in a meaningful way

(and 2i is only one of the square roots of -4)

So I consulted several texts on complex analysis and found:
1- Some authors define, and explicitely state, that $]\textit{i} = \sqrt{-1}$
2- Some authors define math]\textit{i}^2 = -1[/tex] and never bring up the subject of what i is.
3- Some authors don't seem to consider complex numbers at all, but teach it using only the geometric interpretation of i = (0, 1) in the 2-D plane. They seem to be only teaching geometry, although I didn't look into the texts thoroughly.
4- One author doesn't seem to consider complex numbers at all, but teaches it using only vectors in the plane. Again, I didn't look into the text thoroughly.

i do not know what texts you are referring to, or what goals those texts had in mind when they were dealing with complex analysis, so i cannot comment intelligibly on that. but i will say that complex numbers haven't always been popular. the notion of $i$ was as controversial as Cantors notion of "infinite sets can be of different sizes, one can be bigger than another and the line has as many points as the plane etc etc etc" once upon a time. depending on when these texts were written, it may have some lingering touches of that "timidness" to venture into the world of unknown math or math that many were not comfortable with at the time.

It seems that the conumdrum of what i really is is something mathematicians don't want to talk about.

it is not that they don't want to talk about it, it is just that math has to make sense and be consistent and follow certain rules. unless mathematicians are convince beyond the shadow of a doubt that a certain definition complies with this, they will hesitate to move forward with it, and talk about it in ways as "official" as writing a text. today, however, is not so bad. a lot of research has been done and is being done in complex analysis, and i would be as bold as to say $\sqrt{-1}$ is A-OK with modern mathematics as far as most mathematicians today are concerned

**ThePerfectHacker** · January 22nd 2009, 08:39 PM

The mistake is $\sqrt{ab} = \sqrt{a}\sqrt{b}$ !

**Jhevon** · January 22nd 2009, 09:04 PM

Originally Posted by ThePerfectHacker

The mistake is $\sqrt{ab} = \sqrt{a}\sqrt{b}$ !

i don't recall doing the proof that this does not work in complex analysis. what is it?

**Chop Suey** · January 22nd 2009, 10:09 PM

The property $\sqrt{ab} = \sqrt{a} \sqrt{b}$ is defined for $a, b \geq 0$ . And hence, as TPH mentioned, your step here:
$<br /> \textit{i}^2 = \textit{ii} = \sqrt{-1}\sqrt{-1} = \sqrt{(-1)(-1)} = \sqrt{+1} = +1<br />$

is erroneous.

**ThePerfectHacker** · January 23rd 2009, 07:23 AM

Originally Posted by Jhevon

i don't recall doing the proof that this does not work in complex analysis. what is it?

The example in Post #1 as Chop Suey said is an example when this does not work.

**Jhevon** · January 23rd 2009, 01:54 PM

Originally Posted by ThePerfectHacker

The example in Post #1 as Chop Suey said is an example when this does not work.

yes, proof by counter-example is obvious here. i was hoping for something else. because this seems to imply a larger problem, namely, the distribution of powers over a product. if it doesn't work for the 1/2 power, that would mean it probably won't work for all or some other powers, right? a proof i would like to see is when this move is illegal, or is it always illegal? is $(ab)^3 = a^3b^3$ false for $a,b \in \mathbb{C}$ ? etc

**lebanon** · January 24th 2009, 07:18 AM

there is an error...........and -1 is undifined

**janvdl** · January 24th 2009, 07:20 AM

Originally Posted by lebanon

look i knew that if x^2=16
then, x= redical 16=4
and if i^2=-1
then, i= redical -1

$i$ is defined as $\sqrt{-1}$

and

$i^2 = -1$

$i$ is a complex number. You don't do it on your level.

**Constatine11** · January 24th 2009, 08:09 AM

Originally Posted by lebanon

look i knew that if x^2=16
then, x= redical 16=4

If

$x^2=16$

then

$x=\pm\sqrt{16}=\pm4$

.

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