SRL Archive - Discussions & Debates

I interestingly ran into this problem when pondering exponential/logarithmic calculations with complex/imaginary numbers.
e^(Pi * i) = -1
This is the Euler's identity, which can be proven. I will not go to that in this thread though, just believe it.

From the sentence we can see that
ln(-1) = Pi * i
Which can be checked from calculators etc. too.

Therefore
Pi = ln(-1) / i

Because
ln(a^b) = b * ln(a)
And
-1 = (-1)^(-1)
Follows
Pi = ln(-1^(-1)) / i = -ln(-1) / i

And because we already stated that
Pi = ln(-1) / i
we can say that
ln(-1) / i = -ln(-1) / i
Pi = -Pi


And by multiplying both sides of the last equation with any number, we can say that a = -a.

Pi = -Pi || *(3/Pi)
(3/Pi)*Pi = -(3*Pi)/Pi
3 = -3

Pretty neat, ay?

Lol?

Abs(1) = Abs(-1)

From: IceFire908
Lol?

Abs(1) = Abs(-1)

This has nothing to do with abs :p
[Abs(3) = Abs(-3)] <> [3 = -3]

I can't see any of the pictures.. :(

From: Zyt3x
I can't see any of the pictures.. :(

They don't seem to load :(

cant see pictures :(

He took the pictures from Wolfram Alpha. They no longer exist at the address he took them from.

Now it might be a bit harder to read, but I wrote the stuff without wolfram :rolleyes:

You don't need all that, just this one:
ln(a^b) = b * ln(a)

In one particular case (a=-1 b=-1), it turns out that a^b=a (also a=b and b=a^b) meaning:
ln(a)=b*ln(a)
so
x=b*x in that particular case (b=-1) supposedly making -1 a multiplicative identity. The other particular case is a=1 b=1, but we already knew 1 was the multiplicative identity. I'm hoping that in some math book it says
ln(a^b) = b * ln(a) where a<>b
or something like that.

From: Boreas
You don't need all that, just this one:
ln(a^b) = b * ln(a)

In one particular case (a=-1 b=-1), it turns out that a^b=a (also a=b and b=a^b) meaning:
ln(a)=b*ln(a)
so
x=b*x in that particular case (b=-1) supposedly making -1 a multiplicative identity. The other particular case is a=1 b=1, but we already knew 1 was the multiplicative identity. I'm hoping that in some math book it says
ln(a^b) = b * ln(a) where a<>b
or something like that.

But the rule does apply if a and b are both positive real numbers
ln(4^4) = 4*ln(4)

I love this pseudomathematics stuff…but proving that 1 = -1 is alot simpler than you made it out to be :p

And for the record, all indice/log laws only apply in the case of real numbers. Log (-1) is not real.

From: Santa_Clause
I love this pseudomathematics stuff…but proving that 1 = -1 is alot simpler than you made it out to be :p

And for the record, all indice/log laws only apply in the case of real numbers. Log (-1) is not real.

Yea that's how it's supposed to be but I haven't seen anyone say the "when x is R" when talking about logarithms. :p
And also, if there are exponential calculations for complex numbers there must be logarithmic calculations too! :)

And I'd like to see you proof -1=1 without complex numbers o_O

a = b, so a - b = 0.

Also: 3a = 3b, so 3a - 3b = 0.

Therefore: a - b = 3a - 3b, a - b = 3(a - b), 1 = 3.

Subtracting 2 from both sides: -1 = 1.

There you go :p

I think we've been over this. Math is flawed and this is a novel trick. If something doesn't make fucking sense it doesn't make fucking sense.

From: R0b0t1
I think we've been over this. Math is flawed and this is a novel trick. If something doesn't make fucking sense it doesn't make fucking sense.

I appreciate your wisdom in pointing out that 1 and -1 are in fact not equal and what we're saying is a joke. I promise we wouldn't have figured it out without your divine intervention.

From: Santa_Clause
a = b, so a - b = 0.

Also: 3a = 3b, so 3a - 3b = 0.

Therefore: a - b = 3a - 3b, a - b = 3(a - b), 1 = 3.

Subtracting 2 from both sides: -1 = 1.

There you go :p

Clever, BUT that's not correct. (a - b) = 0, you can't divide by that!
Gotcha! ;)

This is only valid within the realm of Imaginary numbers. So no, it is not at all interesting or surprising. :)

From: Wizzup?
This is only valid within the realm of Imaginary numbers. So no, it is not at all interesting or surprising. :)

They are numbers too, just like 1, 2 or Pi :rolleyes:
Well not exactly, because you can't say which is bigger, 2i or (i + 1)
But still!

From: marpis
They are numbers too, just like 1, 2 or Pi :rolleyes:
Well not exactly, because you can't say which is bigger, 2i or (i + 1)
But still!

Well, only partially. Like I've said in my other post (other thread) it really depends on what kind of numbers you work with. In daily life, Complex/Imaginary numbers aren't really numbers, they are often too abstract to be used in daily use. There's a reason they are only introduced in the last year of high school or university. They're fun (and useful) to play with, but if you do, you also accept certain axioms that are not valid for real numbers. Taking a square root of a negative number for example.

From: marpis
Clever, BUT that's not correct. (a - b) = 0, you can't divide by that!
Gotcha! ;)

REALLY?! I seriously thought that 1 = -1 :p

Oh and the proof of Euler's identity is perhaps one of my favourites :)

From: Wizzup?
There's a reason they are only introduced in the last year of high school or university.

I learned about them pretty early on in school, and probably before high school. :confused:

From: Method
I learned about them pretty early on in school, and probably before high school. :confused:

So before you became 12 years old?

From: Method
I learned about them pretty early on in school, and probably before high school. :confused:

I doubt you did! :p
In Finland only those who chose advanced maths use imaginary numbers.
Or elictrician students.

We learned a bit about Imaginary Numbers a few months ago, at some yr 10 conference…

The only bits I found interesting were Fractals and Recurrsion :D

From: Wizzup?
So before you became 12 years old?

Yes, I'm pretty sure we were introduced to them towards the end of middle school (grades 7 and 8 for me). Either way, it certainly wasn't introduced to us towards the end of our high school careers.

From: marpis
I doubt you did! :p
In Finland only those who chose advanced maths use imaginary numbers.
Or elictrician students.

I don't live in Finland…

From: Method
Yes, I'm pretty sure we were introduced to them towards the end of middle school (grades 7 and 8 for me). Either way, it certainly wasn't introduced to us towards the end of our high school careers.

I find it funny you get Complex numbers even before you get algebra.

From: Method
Yes, I'm pretty sure we were introduced to them towards the end of middle school (grades 7 and 8 for me).

If some single teacher "btw-mentions" a word "imaginary number" it doesn't mean they've been introduced to you… And I don't know what age you are on 7th or 8th grade, so they don't tell much to me.

From: Wizzup?
I find it funny you get Complex numbers even before you get algebra.

Algebra 1 was offered to me in sixth grade, and I probably learned a bit about complex numbers in seventh or eighth.

From: marpis
If some single teacher "btw-mentions" a word "imaginary number" it doesn't mean they've been introduced to you… And I don't know what age you are on 7th or 8th grade, so they don't tell much to me.

I'd like to think I know what the word "introduce" means, thank you very much.

From: Method
Algebra 1 was offered to me in sixth grade, and I probably learned a bit about complex numbers in seventh or eighth.



I'd like to think I know what the word "introduce" means, thank you very much.

Wait, so how old were you when complex numbers were first taught to you? As in complex roots of unity and so on…

From: Santa_Clause
Wait, so how old were you when complex numbers were first taught to you? As in complex roots of unity and so on…

I don't think I've been formally taught the nth roots of unity concept, but I was introduced to complex numbers and i when I was 12 or 13, which was later expanded on in high school.

From: Wizzup?
I find it funny you get Complex numbers even before you get algebra.

The majority of students learn algebra in year 5 (and most likely above) :p

since when does ln(e^x) = x work for complex numbers?

From: Santa_Clause
a = b, so a - b = 0.

Also: 3a = 3b, so 3a - 3b = 0.

Therefore: a - b = 3a - 3b, a - b = 3(a - b), 1 = 3.

Subtracting 2 from both sides: -1 = 1.

There you go :p

a - b = 0, and you division by 0 is undefined in algebra

From: Method
I learned about them pretty early on in school, and probably before high school. :confused:

same; i learned about imaginary and complex numbers in the 6th grade

I only know (-1^2) = (1^2)

[QUOTE=Dan's The Man;664131]The majority of students learn algebra in year 5 (and most likely above) :p[/QUOTE]

lol algebra in year 5, i don't remember learning anything in year 5.. all i remembered was that i played on the computers everyday

From: super_
since when does ln(e^x) = x work for complex numbers?

Since when it hasn't? And why wouldn't it? :rolleyes:
"It messes things up and shows false answers" is not good enough of a reason to deny using logarithms with complex.

After reading the equation, I'm starting to wonder what to do with my life…

From: Woodiewoo7
After reading the equation, I'm starting to wonder what to do with my life…

I've seen you post several posts which add nothing to the thread only to get your 10 posts I presume, I now reported you as I'm tired of it, hopefully your post count gets reset and you have to post something valuable.

Pi = ln(-1^(-1)) / i = -ln(-1) / i
FALSE!
http://www.wolframalpha.com/input/?i=Pi+%3D+ln%28-1^%28-1%29%29+%2F+i+%3D+-ln%28-1%29+%2F+i