[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?

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- January 22nd 2009, 12:58 PMmnovaproof of -1 = +1
[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? - January 22nd 2009, 01:09 PMursaQuote:

[I hope I did the LaTex correctly.]

We know that http://www.mathhelpforum.com/math-he...e3c90701-1.gif and the square root and square operations are inverses so they cancel and http://www.mathhelpforum.com/math-he...4d3f1e01-1.gif .

And we know that http://www.mathhelpforum.com/math-he...b54f0fae-1.gif .

example: http://www.mathhelpforum.com/math-he...1a190073-1.gif

So http://www.mathhelpforum.com/math-he...00032a69-1.gif .

So -1 = +1

QED

Where's the problem with this?

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 - January 22nd 2009, 01:20 PMmnova
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. - January 22nd 2009, 01:25 PMJhevon
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 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 - January 22nd 2009, 01:27 PMursaQuote:

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.

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 - January 22nd 2009, 03:31 PMmnova
Jhevon

That link you provided is fascinating.

Among the posts are two posters who cleverly demand that i is defined by the equation and that i is definitely NOT equal to .

That seems suspicious to me, as then how can you justify ?

So I consulted several texts on complex analysis and found:

1- Some authors define, and explicitely state, that

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. - January 22nd 2009, 03:36 PMJhevon
that is the part i do not agree with

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

this is one of the reasons i do not like that view. saying is not 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, 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 can be defined in a meaningful way

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

Quote:

So I consulted several texts on complex analysis and found:

1- Some authors define, and explicitely state, that

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.

Quote:

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

- January 22nd 2009, 08:39 PMThePerfectHacker
The mistake is !

- January 22nd 2009, 09:04 PMJhevon
- January 22nd 2009, 10:09 PMChop Suey
The property is defined for . And hence, as TPH mentioned, your step here:

is erroneous. - January 23rd 2009, 07:23 AMThePerfectHacker
- January 23rd 2009, 01:54 PMJhevon
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 false for ? etc

- January 24th 2009, 07:18 AMlebanonhi
there is an error...........and -1 is undifined

- January 24th 2009, 07:20 AMjanvdl
- January 24th 2009, 08:09 AMConstatine11