How can J be equal to two different things?

Klaynos...

 

... not what i have ever done...

And your multiplication of the brackets is STILL wrong.

It wasn't... one of the J's where meant to be J'.

 

[math]J=a+bi[/math]

[math]J=a-bi[/math]

 

Gives the form:

 

 

Have you not?

 

ah ok, just read your above post

Klaynos...

 

How is the multiplication wrong?

And your multiplication of the brackets is STILL wrong.

 

Still stands though.

Yes, again. One of them was meant to have an expression J'

Klaynos...

 

How is the multiplication wrong?

 

Compare it to how I multiply out brackets earlier in the thread, for starters you don't have a²

And again... how is it wrong? I can easily plug real values into a and b, and the equation turns out right

 

You;ve lost me

[math]

(a-\sqrt{b})(a+\sqrt{b})=a-bi^{2}=a-b(-1)=a+b=x

[/math]

 

Is just not true.

 

You're wrong, it's simple mathematics, let me explain:

 

 

[math]

(c-d)(e+f)= ce + cf - de - df

[/math]

 

If we subsitute c = e = a and d = f = sqrt(b)

[math]

ce + cf - de - df=(a-\sqrt{b})(a+\sqrt{b})=aa+a\sqrt{b} - \sqrt{b}a -\sqrt{b}\sqrt{b}= a^2 - \sqrt{b}\sqrt{b} = a^2 \mp b

[/math]

 

Which part of this do you disagree with?

 

 

Is how it's done.

It isn't? Well, lets plug some values into a and b.

 

[math](1 + i\sqrt{2})(1 − i\sqrt{2}) = 1 − 2i² = 1 + 2 = 3[/math]

 

where a=1

and b=2

 

for some reason it won't show... so bare with me while i type it out

Grav, are you sure you know how? It doesn't look like it. Here's a trick I learned in middle school. It's called FOIL. FOIL stands for "First Outside Inside Last." It means you take the sum of the products of the first term in each binomial, the product of the first term of the first binomial and the last of the second(the outside ones), the product of the last term of the first binomial and the first of the second(the inside ones), and the product of the last term in both binomials.

 

(a+b)(c+d)=ac+ad+bc+bd

(1 + i√2)(1 − i√2) = 1 − 2i² = 1 + 2 = 3

 

Of course i know how to calculate the yourdad. I wouldn't say i cud if i couldn't

who cares about numbers? You're not using numbers though, you're using letters, put in a=5 and b =16 and see what you get.... Because I sure as hell get a different answer than what your forumla gives.

 

And where have you're i's come from all of a sudden?

 

a=1 is a special case, as 1*1 = 1

{i} was necessery, because i was talking about time dimensions. I only added the values because you said the equation was wrong, and i have shown it isn't, from my technical view.

 

Besides, i am assuming a and b have values; i just don't know what kind of value.

{i} was necessery, because i was talking about time dimensions. I only added the values because you said the equation was wrong, and i have shown it isn't, from my technical view.

 

Besides, i am assuming a and b have values; i just don't know what kind of value.

 

Yes they have values fine, but 1 and 2 are NOT valid to show that the equation is true in all cases, as 1^2 = 1.

 

You have a technical view? What training have you had, did this cover expanding brackets? Try putting in the values I suggest 5 and 16, and tell me if I'm right or wrong that the original equation doesn't give the same answer as the one you've come up with (and I wouldn't like to speculate how).

You mean a=5 and b=16? Well of course it will give different answers.

 

But using 1 and 2 was only an example.

 

God only knows what the values of a and b are, because as far as i can tell, they would mean something to the universe, but very little to us.

 

You see, this is why i like posting here. If i do potentially get anything wrong, it will be noticed for sure.

no no no

 

a=5 and b=16

 

Will not give a differnt answer to a=1 and b=2 but the two parts of your equation:

 

[math]

(a-\sqrt{b})(a+\sqrt{b})=a-bi^{2}=a-b(-1)=a+b=x

[/math]

 

[math]

(5-\sqrt{16})(5+\sqrt{16})=9

[/math]

 

[math]

5-16i^{2}=21

[/math]

 

Therefore your equation MUST be wrong. It's quite simple.

Wait a minute...

 

That's what i meant. Just because you calculate 21, doesn't proove the equation wrong. That's why i keep values (not by choice though), out of a and b.

 

The equation is right. You are but adding different values to a and b, in an attempt to show it is wrong. How can it be wrong, when essentially, we don't know the value of x?

And you're writing the Algebra GUT?

Yes

 

Klayos' arguement is wrong. He is seeing it from the wrong angle.

Wait a minute...

 

That's what i meant. Just because you calculate 21, doesn't proove the equation wrong. That's why i keep values (not by choice though), out of a and b.

 

The equation is right. You are but adding different values to a and b, in an attempt to show it is wrong. How can it be wrong, when essentially, we don't know the value of x?

 

Because the first and second bits of your derivation do not correlate with each other, you cannot show one to be the other, surely if it was correct just plugging in equations to both sides of the equation should ALWAYS yield the same answer as each other.

 

How am I seeing it wrong? Your derivation does not follow from the first statement.

Oh wait.. i have it wrong with Klayos... i now see what he means. What i mean, is that a and b cannot be determined with any values, but we can suppose [math](a-i\sqrt{b})(a-i\sqrt{b})[/math] is a-bi^2. Klayos must remain with the sequence i have given, to reach x, or he splits the equation, and gets two different answers, with the values given. This was why i used 1 and 2 as values. They seem appropriate.

 

Gives the form:

 

[math](a-\sqrt{b})(a+\sqrt{b})=a-bi^{2}=a-b(-1)=a+b=x[/math]

 

Look at this equation you've written. How can the first term in the first brackets, [math]a[/math], be multiplied by the first term in the second brackets, [math]a[/math], and NOT result in [math]a^2[/math] ?!?

 

Unless you are redefining mathematics, why do you keep defending this mistake?

 

What's really weird that you get the second term right.... you get [math]\sqrt{b}[/math] times [math]\sqrt{b}[/math] as [math]b[/math]. Why don't you see the mistake in the first part?!?