Proof: a+bi=-a+bi

[Endercreeper01]

Here is proof that a+bi=-a+bi:
Using the distributive proprty, we can write a+bi as i(a/i+b). a/i is also equal to ai, so then it becomes i(ai+b). That equals ai²+bi, and i²=-1, so therefore it makes -a+bi.

 

Equation form of proof:

a+bi=i(a/i+b)=i(ai+b)=ai²+bi=-a+bi

 

Proof that a/i=ai:

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.

 

What do you guys think?

[John Cuthber]

I think that you have forgotten that taking the square root of something is ambiguous.

 

Root(4) has two perfectly acceptable answers, 2 and -2

I think that you have used that as the basis for saying

2=-2 because they are both root(4).

Edited October 13, 2013 by John Cuthber

[HalfWit]

a/i is also equal to ai,

 

How do you figure that? Multiplying both sides by i gives a = -a, which is your mistake.

[ydoaPs]

a/i is also equal to ai

a/i=-ai

[studiot]

If I write something I know is wrong like 2=3, I work at it and worry at it until I find my mistake (and I always eventually do).

What I don't do is to present it to others as a proof that i have discovered something new.

 

In fact i have made some small advances in my time, but always it has then been in accord with what I already know and extended my knowledge.

[Endercreeper01]

I think that you have forgotten that taking the square root of something is ambiguous.

 

Root(4) has two perfectly acceptable answers, 2 and -2

I think that you have used that as the basis for saying

2=-2 because they are both root(4).

If you raise to the 1/2 power, would it still be ambiguous?

How do you figure that? Multiplying both sides by i gives a = -a, which is your mistake.

See bottom of 1st post

 

If I write something I know is wrong like 2=3, I work at it and worry at it until I find my mistake (and I always eventually do).

What I don't do is to present it to others as a proof that i have discovered something new.

 

In fact i have made some small advances in my time, but always it has then been in accord with what I already know and extended my knowledge.

All I am doing is just showing you something and asking what you think.

Here is proof that a+bi=-a+bi:

Using the distributive proprty, we can write a+bi as i(a/i+b). a/i is also equal to ai, so then it becomes i(ai+b). That equals ai²+bi, and i²=-1, so therefore it makes -a+bi.

 

Equation form of proof:

a+bi=i(a/i+b)=i(ai+b)=ai²+bi=-a+bi

 

Proof that a/i=ai:

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.

 

What do you guys think?

[ydoaPs]

-a^2/2=a^2/2i, or just ai.

 

What do you guys think?

That's not right. You can't get from -1(a^2/2) to ai.

 

As I said before, a/i=-ai. Multiply a/i by i/i, and you get -ai, not ai. And -a^2/2=-a.

All I am doing is just showing you something and asking what you think.

All we're doing is showing you why you're wrong.

[Endercreeper01]

That's not right. You can't get from -1(a^2/2) to ai.

 

As I said before, a/i=-ai. Multiply a/i by i/i, and you get -ai, not ai. And -a^2/2=-a.

Sorry, that was a typo. I meant (-a²)^1/2.

if a/i=-ai, then that means i/i=1, and you are just multiplying a/i by one (accoding to a/i=-ai), and that gives you a/i.

Edited October 13, 2013 by Endercreeper01

[ydoaPs]

Sorry, that was a typo. I meant (-a²)^1/2.

That's also wrong. You can't get there from where you started.

 

if a/i=-ai, then that means i/i=1, and you are just multiplying a/i by one (accoding to a/i=-ai), and that gives you a/i.

Yeah, it's multiplying by one to get the I out of the bottom like you're supposed to.

[Endercreeper01]

That's also wrong. You can't get there from where you started.

 

Yes you can. You raise (a^2/i^2=a^2/-1=-a^2) to the power of 1/2. Edited October 14, 2013 by Endercreeper01

[John Cuthber]

Yes you can. You raise (a^2/i^2=a^2/-1=-a^2) to the power of 1/2.

Raising to the power half is taking the square root, and, as I said that's ambiguous.

[studiot]

 

endercreeper post#1

Here is proof that a+bi=-a+bi:

 

 

 

endercreeper post#6

All I am doing is just showing you something and asking what you think

 

A tad more than just showing me something, I think.

[Endercreeper01]

A tad more than just showing me something, I think.

I'm asking what you think about something

Raising to the power half is taking the square root, and, as I said that's ambiguous.

When you take the square root of a negative, you have to decide on either -i or i to be the imaginary unit, and so there is no ambiguity.

 

The concept of principal square root cannot be extended to real negative numbers since the two square roots of a negative number cannot be distinguished until one of the two is defined as the imaginary unit, at which point and can then be distinguished. Since either choice is possible, there is no ambiguity in defining as "the" square root of .

http://mathworld.wolfram.com/PrincipalSquareRoot.html

Edited October 14, 2013 by Endercreeper01

[studiot]

 

I'm asking what you think about something

 

 

No, you are doing more. You are asserting something we all know to be false, and then asking what I think.

 

Why are you doing this?

[Endercreeper01]

 

No, you are doing more. You are asserting something we all know to be false, and then asking what I think.

 

Why are you doing this?

Why do you think I am false? You can't just say that I am asserting something false. You have to tell me why, just like others did.

[John]

http://mathworld.wolfram.com/PrincipalSquareRoot.html

 

The ambiguity lies in steps where you claim that [math]a = \sqrt{a^{2}}[/math]. In fact, [math]\sqrt{a^{2}} = \pm a[/math].

 

Anyway, to be clear, [math]\frac{a}{i} = \frac{a + 0i}{0 + i} = \left(\frac{a + 0i}{0 + i}\right) \left(\frac{0 - i}{0 - i}\right) = \frac{-ai}{-i^{2}} = \frac{-ai}{-(-1)} = -ai[/math], as ydoaPs said.

 

Here's a proof that [math]a + bi \neq -a + bi[/math] or [math]a = 0[/math]:

 

Assume [math]a + bi = -a + bi[/math] and [math]a \neq 0[/math]. Then [math]a = -a \implies a + a = 0 \implies 2a = 0 \implies a = 0[/math]. Contradiction!

[Endercreeper01]

 

The ambiguity lies in steps where you claim that [math]a = \sqrt{a^{2}}[/math]. In fact, [math]\sqrt{a^{2}} = \pm a[/math].

 

Anyway, to be clear, [math]\frac{a}{i} = \frac{a + 0i}{0 + i} = \left(\frac{a + 0i}{0 + i}\right) \left(\frac{0 - i}{0 - i}\right) = \frac{-ai}{-i^{2}} = \frac{-ai}{-(-1)} = -ai[/math], as ydoaPs said.

 

Here's a proof that [math]a + bi \neq -a + bi[/math] or [math]a = 0[/math]:

 

Assume [math]a + bi = -a + bi[/math] and [math]a \neq 0[/math]. Then [math]a = -a \implies a + a = 0 \implies 2a = 0 \implies a = 0[/math]. Contradiction!

By sq.root(a²), I meant a^2/2, which equals (a^1/2)² and (a²)^1/2. Because 2/2=1, and (a^b)^c=a^bc, then you are raising it to the power of 2/2, or 1, which equals a.

Edited October 15, 2013 by Endercreeper01

[John]

To amend my earlier post (with apologies--my excuse is I'm tired ), it's probably better to say [math]\sqrt{a^{2}} = |a|[/math].

 

Also, another problem is that the power identity [math]\frac{a^{x}}{b^{x}} = \left(\frac{a}{b}\right)^{x}[/math] is valid for [math]a,b \in \mathbb{R_{+}}[/math] and [math]x \in \mathbb{R}[/math] , but not necessarily for complex [math]a[/math] or [math]b[/math] unless [math]x[/math] is an integer.

Edited October 15, 2013 by John

[John Cuthber]

 

When you take the square root of a negative, you have to decide on either -i or i to be the imaginary unit, and so there is no ambiguity.

 

http://mathworld.wolfram.com/PrincipalSquareRoot.html

Since you have to choose, there is obviously an ambiguity: and, since you got the wrong answer, you obviously chose the wrong one.

[studiot]

 

Why do you think I am false? You can't just say that I am asserting something false. You have to tell me why, just like others did.

 

 

I don't have to tell you anything.

 

But since you asked so pretty please nicely:

 

You asserted

 

 

Here is proof that a+bi=-a+bi:

 

 

You have asserted that a particular complex number Z1 is equal to another complex number Z2

Equality is the most fundamental property definition of a complex number after the definition of the complex numbers themselves.

 

Definition:

A complex number Z1 is equal to another complex number Z2 IFF ( if and only if) the real and imaginary parts are separately equal.

That is Re(Z1) = Re(Z2) and Im(Z1) = Im(Z2).

 

Your numbers fail to meet this criterion.

 

end of.

 

Incidentally in you playing with complex square roots are you aware that not all complex numbers have square roots?

 

Amaton has started a thead to ask about this, you should perhaps follow it.

[math]{Z^{\frac{1}{2}}}[/math]

 

Is only defined on the cut Z plane which excludes the negative real axis?

http://www.scienceforums.net/topic/79290-the-12-power-and-square-roots/

Edited October 15, 2013 by studiot

[HalfWit]

 

Incidentally in you playing with complex square roots are you aware that not all complex numbers have square roots?

 

 

I haven't been following this (silly IMO) thread in any detail. But I'm not aware of the above either. Please explain.

[Endercreeper01]

Also, another problem is that the power identity [math]\frac{a^{x}}{b^{x}} = \left(\frac{a}{b}\right)^{x}[/math] is valid for [math]a,b \in \mathbb{R_{+}}[/math] and [math]x \in \mathbb{R}[/math] , but not necessarily for complex [math]a[/math] or [math]b[/math] unless [math]x[/math] is an integer.

http://mathworld.wolfram.com/ExponentLaws.html says that x doesn't have to be an integer, but it has to be a real number. 1/2 and 2 are real numbers

 

 

I don't have to tell you anything.

 

But since you asked so pretty please nicely:

 

You asserted

 

 

You have asserted that a particular complex number Z1 is equal to another complex number Z2

Equality is the most fundamental property definition of a complex number after the definition of the complex numbers themselves.

 

Definition:

A complex number Z1 is equal to another complex number Z2 IFF ( if and only if) the real and imaginary parts are separately equal.

That is Re(Z1) = Re(Z2) and Im(Z1) = Im(Z2).

 

Your numbers fail to meet this criterion.

 

end of.

 

I know that, but what about my process of finding my equation? What would be wrong there?

 

 

I don't have to tell you anything.

 

To back up your statement then you would need to

 

 

Incidentally in you playing with complex square roots are you aware that not all complex numbers have square roots?

 

Amaton has started a thead to ask about this, you should perhaps follow it.

[math]{Z^{\frac{1}{2}}}[/math]

 

Is only defined on the cut Z plane which excludes the negative real axis?

http://www.scienceforums.net/topic/79290-the-12-power-and-square-roots/

Why not?

[studiot]

 

I haven't been following this (silly IMO) thread in any detail. But I'm not aware of the above either. Please explain

 

That's a very good question. I will have to rethink that part.

[John]

http://mathworld.wolfram.com/ExponentLaws.html says that x doesn't have to be an integer, but it has to be a real number. 1/2 and 2 are real numbers

 

You didn't read the entire page. See near the bottom:

 

Note that these rules apply in general only to real quantities, and can give manifestly wrong results if they are blindly applied to complex quantities.

 

Note that one of said "manifestly wrong results" is the idea that a + bi = -a + bi for nonzero a.

Edited October 16, 2013 by John

[Endercreeper01]

 

You didn't read the entire page. See near the bottom:

 

 

Note that one of said "manifestly wrong results" is the idea that a + bi = -a + bi for nonzero a.

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

Edited October 16, 2013 by Endercreeper01