Proof: a+bi=-a+bi

[uncool]

-a² isn't a complex quantity, and neither is 1/2 or 2. And how do you know if they mean that or not?

You are still using complex numbers - the square root itself is complex. And that's what the page is referring to.

 

And because you are taking the square root of a square and not getting the same number. You're effectively saying that i = -i because i = sqrt(-1) = sqrt(-1*-1*-1) = sqrt(-1)^3 = -i, which is blatantly false. Basically, every number has two square roots; if you restrict yourself to the positive real numbers, you can choose one unambiguously (namely, the positive square root). However, if you add in the negative reals and the complex numbers, you get ambiguity that you can't get rid of. That's how the math works. People have answered your question repeatedly now.

[John]

-a² isn't a complex quantity, and neither is 1/2 or 2. And how do you know if they mean that or not?

 

You're dealing with a fraction containing [math]i[/math], an imaginary number, in the denominator. You rewrite [math]i[/math] in the form [math](-1)^{\frac{1}{2}}[/math], but at the end of the day, you're still applying a power identity in a context where it isn't necessarily valid.

 

But even if I'm misreading or wrong about that, we know your result is "manifestly wrong" because, as shown in previous posts, it results in a logical contradiction. The fact that [math]a + bi \neq -a + bi[/math] for nonzero [math]a[/math] is extremely easy to demonstrate, and I'm not sure why you're clinging to this supposed "proof" of the contrary.

[Endercreeper01]

 

You're dealing with a fraction containing [math]i[/math], an imaginary number, in the denominator. You rewrite [math]i[/math] in the form [math](-1)^{\frac{1}{2}}[/math], but at the end of the day, you're still applying a power identity in a context where it isn't necessarily valid.

 

But even if I'm misreading or wrong about that, we know your result is "manifestly wrong" because, as shown in previous posts, it results in a logical contradiction. The fact that [math]a + bi \neq -a + bi[/math] for nonzero [math]a[/math] is extremely easy to demonstrate, and I'm not sure why you're clinging to this supposed "proof" of the contrary.

Well I know that now, but -1^1/2 equals i because its the same as doing I^1/2

You are still using complex numbers - the square root itself is complex. And that's what the page is referring to.

 

And because you are taking the square root of a square and not getting the same number. You're effectively saying that i = -i because i = sqrt(-1) = sqrt(-1*-1*-1) = sqrt(-1)^3 = -i, which is blatantly false. Basically, every number has two square roots; if you restrict yourself to the positive real numbers, you can choose one unambiguously (namely, the positive square root). However, if you add in the negative reals and the complex numbers, you get ambiguity that you can't get rid of. That's how the math works. People have answered your question repeatedly now.

root(-1*-1*-1) is root(-1³), not (root(-1))³

[ydoaPs]

Well I know that now, but -1^1/2 equals i because its the same as doing I^1/2

 

root(-1*-1*-1) is root(-1³), not (root(-1))³

(a^x)^y=a^xy=a^yx=(a^y)^x, no?

[Endercreeper01]

(a^x)^y=a^xy=a^yx=(a^y)^x, no?

Well actually, yes, but where did I say that I=-I?

[ydoaPs]

Well actually, yes

So, root(-1³)=root(-1)³, no?

[HalfWit]

So, root(-1³)=root(-1)³, no?

No, -1^3 = 1(1^3) since exponentiation has precedence over negation. One of the many persistent errors in this thread. Hello, people, two things please:

 

1) (-2)^2 = 2^2 does not imply that -2 = 2.

 

2) -1^48 = -1. (-1)^48 = 1.

[Endercreeper01]

No, -1^3 = 1(1^3) since exponentiation has precedence over negation. One of the many persistent errors in this thread. Hello, people, two things please:

 

1) (-2)^2 = 2^2 does not imply that -2 = 2.

 

2) -1^48 = -1. (-1)^48 = 1.

I agree. What are some more errors?

Edited October 16, 2013 by Endercreeper01

[uncool]

No, -1^3 = 1(1^3) since exponentiation has precedence over negation. One of the many persistent errors in this thread. Hello, people, two things please:

 

1) (-2)^2 = 2^2 does not imply that -2 = 2.

 

2) -1^48 = -1. (-1)^48 = 1.

I want to point out that that specific error started with EnderCreeper01, and ydoaps was using the (still incorrect) same notation. To correct what ydoaps said in terms of notation:

 

So, root((-1)^3) = (root(-1))^3, right?

[Endercreeper01]

I want to point out that that specific error started with EnderCreeper01, and ydoaps was using the (still incorrect) same notation. To correct what ydoaps said in terms of notation:

 

So, root((-1)^3) = (root(-1))^3, right?

What was that error?

[HalfWit]

I want to point out that that specific error started with EnderCreeper01, and ydoaps was using the (still incorrect) same notation. To correct what ydoaps said in terms of notation:

 

So, root((-1)^3) = (root(-1))^3, right?

No, because "root" is ambiguous. Every nonzero complex number has two distinct square roots. And there is no canonical way to distinguish them. So you have to say which of the two roots you mean in the equation above. All the problems in this thread come down to choosing one square root on the left side and the other square root on the right.

 

This is different from defining sqrt(2) as the positive of the two real numbers whose square is 2. In the real numbers, we can define a privileged subset of positive numbers. In the real numbers, 2 can be algebraically distinguished from -2.

 

But in the complex numbers, there are no positive numbers so that we can not distinguish between the two square roots of a number without explicitly saying which square root we're choosing.

 

There's a somewhat heavygoing discussion of this subject here ...

 

http://en.wikipedia.org/wiki/Branch_point

 

Basically they teach you this stuff in a math major class on complex analysis. The key point is that in the reals, sqrt can be defined unambiguously as the positive one. In the complex numbers, sqrt can not be defined unambiguously except by explicit saying which square root you mean.

[ydoaPs]

No, because "root" is ambiguous. Every nonzero complex number has two distinct square roots. And there is no canonical way to distinguish them. So you have to say which of the two roots you mean in the equation above. All the problems in this thread come down to choosing one square root on the left side and the other square root on the right.

 

This is different from defining sqrt(2) as the positive of the two real numbers whose square is 2. In the real numbers, we can define a privileged subset of positive numbers. In the real numbers, 2 can be algebraically distinguished from -2.

 

But in the complex numbers, there are no positive numbers so that we can not distinguish between the two square roots of a number without explicitly saying which square root we're choosing.

 

There's a somewhat heavygoing discussion of this subject here ...

 

http://en.wikipedia.org/wiki/Branch_point

 

Basically they teach you this stuff in a math major class on complex analysis. The key point is that in the reals, sqrt can be defined unambiguously as the positive one. In the complex numbers, sqrt can not be defined unambiguously except by explicit saying which square root you mean.

If you had been reading along instead of nitpicking notation, you'd know that was the point.

[uncool]

No, because "root" is ambiguous. Every nonzero complex number has two distinct square roots. And there is no canonical way to distinguish them. So you have to say which of the two roots you mean in the equation above. All the problems in this thread come down to choosing one square root on the left side and the other square root on the right.

 

This is different from defining sqrt(2) as the positive of the two real numbers whose square is 2. In the real numbers, we can define a privileged subset of positive numbers. In the real numbers, 2 can be algebraically distinguished from -2.

 

But in the complex numbers, there are no positive numbers so that we can not distinguish between the two square roots of a number without explicitly saying which square root we're choosing.

 

There's a somewhat heavygoing discussion of this subject here ...

 

http://en.wikipedia.org/wiki/Branch_point

 

Basically they teach you this stuff in a math major class on complex analysis. The key point is that in the reals, sqrt can be defined unambiguously as the positive one. In the complex numbers, sqrt can not be defined unambiguously except by explicit saying which square root you mean.

As ydoaPs said, that was the point - Endercreeper has been claiming that this is false, so ydoaPs was using his own logic to demonstrate that Endercreeper was wrong. My correction was along the same lines. Further, you have made an error yourself; namely, that sqrt cannot be defined as "the positive one" on the entire real line; it can only be so defined on the nonnegative reals.

What was that error?

That (-1)^3 does not mean the same thing as -1^3; the first means -1*-1*-1 (as I had in my post); the second means -(1*1*1). They come out to the same quantity, but using that misses the point of the question.

[Endercreeper01]

As ydoaPs said, that was the point - Endercreeper has been claiming that this is false, so ydoaPs was using his own logic to demonstrate that Endercreeper was wrong. My correction was along the same lines. Further, you have made an error yourself; namely, that sqrt cannot be defined as "the positive one" on the entire real line; it can only be so defined on the nonnegative reals.

 

The principle square root, or the square root most people mean when they say square root, can be.

 

That (-1)^3 does not mean the same thing as -1^3; the first means -1*-1*-1 (as I had in my post); the second means -(1*1*1). They come out to the same quantity, but using that misses the point of the question.

I did not say that (-1)³ doesn't mean -1³. I said that root((-1)^3) # (root(-1))^3

 

 

 

[uncool]

The principle square root, or the square root most people mean when they say square root, can be.

No, it can't, since i is neither positive nor negative. You can't say "the positive one" except when dealing with real numbers directly, and the square root of a negative number is not real.

I did not say that (-1)³ doesn't mean -1³. I said that root((-1)^3) # (root(-1))^3

 

You did say (in effect) that (-1)^3 means the same thing as -1^3, by saying:

root(-1*-1*-1) is root(-1³)

Further, you haven't answered the question I've been pointing out. According to your logic, don't you have that:

 

root((-1)^3) = ((-1)^3)^(1/2) = (-1)^(3/2) = ((-1)^(1/2))^3 = (root(-1))^3?

=Uncool-

[Endercreeper01]

No, it can't, since i is neither positive nor negative. You can't say "the positive one" except when dealing with real numbers directly, and the square root of a negative number is not real.

You did say (in effect) that (-1)^3 means the same thing as -1^3, by saying:

 

Further, you haven't answered the question I've been pointing out. According to your logic, don't you have that:

 

root((-1)^3) = ((-1)^3)^(1/2) = (-1)^(3/2) = ((-1)^(1/2))^3 = (root(-1))^3?

=Uncool-

That's like saying 2=-2 just because 2² and (-2)² are the same

[ydoaPs]

That's like saying 2=-2 just because 2² and (-2)² are the same

And that's what you're doing.

[uncool]

That's like saying 2=-2 just because 2² and (-2)² are the same

I'm only using your logic.

[John Cuthber]

OK, lets have a look at the OP

"Because i²=-1, we can write a/i as a/-1^1/2. a is also sq.root(a²), or a^2/2, so it is also a^2/2/-1^1/2. Since a^1/2/b^1/2=(a/b)^1/2, we can write this as (a²/-1)^1/2, which is also equal to -a^2/2. -a^2/2 =a^2/2i, or just ai."

 

I'm just going to put it on a few lines and add notes to make it clearer

 

Because i²=-1, we can write a/i as a/-1^1/2.

OK, fair enough

a is also sq.root(a²), or a^2/2,

There's where it gets ambiguous a isn't the only possible answer to that, it could be -a

so it is also a^2/2/-1^1/2.

So -a is also = a^2/2/-1^1/2.

Since a^1/2/b^1/2=(a/b)^1/2, we can write this as (a²/-1)^1/2,

The same problem again. You chose just one of the two possible values.

Lets put in some numbers

a^1/2/b^1/2=(a/b)^1/2

Say a=64 and b =16

a/b is 4 unequivocally.

But 64 ^1/2 is just as much+8 as it is -8

Similarly b^1/2 is 4 or -4 depending on which choice you take

and

(a/b) ^1/2 is +2 or -2

So, the two sides of this

a^1/2/b^1/2=(a/b)^1/2

 

can both be +2 or -2

And if one of them is (arbitrarily chosen to be) +2 while the other is (also arbitrarily chosen to be)-2 then they are clearly not the same.

So you can not, in general, say that

a^1/2/b^1/2=(a/b)^1/2

 

At that point, your "proof" goes out of the window.

So this last bit

"which is also equal to -a^2/2. -a^2/2 =a^2/2i,

or just ai."

also doesn't hold.

 

 

Like I said at the outset, you took square roots without due care and attention.

Edited October 17, 2013 by John Cuthber

[Endercreeper01]

OK, lets have a look at the OP

"Because i²=-1, we can write a/i as a/-1^1/2. a is also sq.root(a²), or a^2/2, so it is also a^2/2/-1^1/2. Since a^1/2/b^1/2=(a/b)^1/2, we can write this as (a²/-1)^1/2, which is also equal to -a^2/2. -a^2/2 =a^2/2i, or just ai."

 

I'm just going to put it on a few lines and add notes to make it clearer

 

Because i²=-1, we can write a/i as a/-1^1/2.

OK, fair enough

a is also sq.root(a²), or a^2/2,

There's where it gets ambiguous a isn't the only possible answer to that, it could be -a

so it is also a^2/2/-1^1/2.

So -a is also = a^2/2/-1^1/2.

Since a^1/2/b^1/2=(a/b)^1/2, we can write this as (a²/-1)^1/2,

The same problem again. You chose just one of the two possible values.

Lets put in some numbers

a^1/2/b^1/2=(a/b)^1/2

Say a=64 and b =16

a/b is 4 unequivocally.

But 64 ^1/2 is just as much+8 as it is -8

Similarly b^1/2 is 4 or -4 depending on which choice you take

and

(a/b) ^1/2 is +2 or -2

So, the two sides of this

a^1/2/b^1/2=(a/b)^1/2

 

can both be +2 or -2

And if one of them is (arbitrarily chosen to be) +2 while the other is (also arbitrarily chosen to be)-2 then they are clearly not the same.

So you can not, in general, say that

a^1/2/b^1/2=(a/b)^1/2

 

At that point, your "proof" goes out of the window.

So this last bit

"which is also equal to -a^2/2. -a^2/2 =a^2/2i,

or just ai."

also doesn't hold.

 

 

Like I said at the outset, you took square roots without due care and attention.

So therefore, a/I is both ai and -ai.

 

And that's what you're doing.

How?

[uncool]

So therefore, a/I is both ai and -ai.

You completely ignored what he wrote in blue. No, a/i is not both ai and -ai; it is -ai, period. Edited October 18, 2013 by uncool

[HalfWit]

Somebody make it stop.

[Endercreeper01]

You completely ignored what he wrote in blue. No, a/i is not both ai and -ai; it is -ai, period.

He said there was ambiguity

 

 

 

There's where it gets ambiguous a isn't the only possible answer to that, it could be -a

So -a is also = a^2/2/-1^1/2.

Since a^1/2/b^1/2=(a/b)^1/2, we can write this as (a²/-1)^1/2,

The same problem again. You chose just one of the two possible values.

Lets put in some numbers

a^1/2/b^1/2=(a/b)^1/2

Say a=64 and b =16

a/b is 4 unequivocally.

But 64 ^1/2 is just as much+8 as it is -8

Similarly b^1/2 is 4 or -4 depending on which choice you take

and

(a/b) ^1/2 is +2 or -2

So, the two sides of this

a^1/2/b^1/2=(a/b)^1/2

 

can both be +2 or -2

And if one of them is (arbitrarily chosen to be) +2 while the other is (also arbitrarily chosen to be)-2 then they are clearly not the same.

So you can not, in general, say that

a^1/2/b^1/2=(a/b)^1/2

 

 

In fact, you yourself said so.

Basically, every number has two square roots; if you restrict yourself to the positive real numbers, you can choose one unambiguously (namely, the positive square root). However, if you add in the negative reals and the complex numbers, you get ambiguity that you can't get rid of. That's how the math works. People have answered your question repeatedly now.

Edited October 18, 2013 by Endercreeper01

[uncool]

He said there was ambiguity

 

 

 

 

 

 

 

 

In fact, you yourself said so.

Yes, and that's the point - you can't simply say x^(1/2), since that is ambiguous. You can't just say (-1)^(1/2) must be i.

=Uncool-

[Endercreeper01]

Yes, and that's the point - you can't simply say x^(1/2), since that is ambiguous. You can't just say (-1)^(1/2) must be i.

=Uncool-

But that means a/i equals ai and -ai