Algebra with Complex Numbers & Imaginary Unit

[lostcauses10x]

"b a real and positive integer"

b a real integer yes.

Not a problem in that statement. Yet it can be confusing due to it is a part of i. Simply put if b =1, and A= 1 then b still is not equal to b. a is only equal to b at (0,0)
Yet the complex number is two sides, which is why I see the process of teaching in the ordered pair coordinates to be of great use.
Since the simplicity of a+bi can and is for every number of the complex, it can be taken to a+b, and of course to (a,b) easily, as long as (a,b) is defined as being of the complex system, with both a and b being numbers of the reals set, of course stating that a is not equal to b except at (0,0)

The complex number is two dimensions, even though I also shorthand to just i or reals at times.
Not a problem to adapt.
I even see a lot of uses for this.

 

[Mark44]

Back to the original question.


Well how does it make sense to use the reals when the square root of 1 has two answers?

We get this question a lot here at PF. The expression $\sqrt{1}$ has one answer: 1. I will grant you that 1 has two square roots, a positive one and a negative one, but the symbol $\sqrt{1}$ represents the principal, or positive, square root of 1.

(-1) squared= ?, and (1) squared equals what?
Yet both -1 and 1 are two separate numbers.

Oh and the thinking -i is the same as i is also a mistake.
Yes squared they become the same answer, yet -i times i= 1, not -1.

 

[Mark44]

First to have the ordered pairs, takes two sets of reals; not just one: with the rules as stated in this thread. The intersection of coarse is at (0,0) with one set vertical to the other.
End result is that "i" is equivalent to (0,1)
Which is why I referred to the link I posted.

The problem with boards and such talks is that to state all involved is not practical. Just to get into a talk about a term or terms can lead a topic way off.

And I learned the old way myself. I have always had a bit of trouble when they say (a+bi) that b is a real.

I don't understand your difficulty. Complex numbers work in a way very similar to two-dimensional vectors. If you have two vectors that don't point in the same direction or in opposite direction, any vector in the plane can be written as a linear combination of your two vectors (that is, as the sum of scalar multiples of the two vectors). In the case of complex numbers, the vectors are "1", a vector that points to the right, and "i", a vector that points straight up. The complex number a + bi is a*1 + b*i, where a and b are real numbers.

I have had to adapt to all of this myself.

 

[rbj]

"b a real and positive integer"

b a real integer yes.

Not a problem in that statement.

so then we extend the concept to negative integers (what that would mean is $-bi + bi = 0$) then extend it to rational $b$, and then hand wave to every real $b$.

Yet it can be confusing due to it is a part of i. Simply put if b =1, and A= 1 then b still is not equal to b.

makes zero sense to me.

a is only equal to b at (0,0)

i guess. but this doesn't really say anything.

Yet the complex number is two sides, which is why I see the process of teaching in the ordered pair coordinates to be of great use.

my objection is to teach complex numbers (which i have done in the context of electrical engineering classes) solely as ordered pairs.

Since the simplicity of a+bi can and is for every number of the complex, it can be taken to a+b,

how? $i$ is not 1.

and of course to (a,b) easily, as long as (a,b) is defined as being of the complex system, with both a and b being numbers of the reals set, of course stating that a is not equal to b except at (0,0)

this almost sounds like you're making sense, but you're not. at least not to me.

The complex number is two dimensions, even though I also shorthand to just i or reals at times.
Not a problem to adapt.
I even see a lot of uses for this.

the problem with teaching complex numbers without $i$ (and without $i^{\ 2} = -1$) is that you must present as an axiom that

(a,b) \times (x,y) \triangleq (ax-by, bx+ay)

besides the addition axiom

(a,b) + (x,y) \triangleq (a+x, b+y)

the latter which is pedagogically trivial. but the former is not. the first thing students will ask is "why the he11 define multiplication like that?" how do you answer that? you might say, "that's the only way we can make the distributive property work" and then proceed to prove it, but it's far more complicated than starting out with $i$ such that $i^{\ 2} = -1$.

 

[1MileCrash]

Back to the original question.


Well how does it make sense to use the reals when the square root of 1 has two answers?
(-1) squared= ?, and (1) squared equals what?
Yet both -1 and 1 are two separate numbers.

1 is not defined to be the square root of 1.

Suppose we inverted the negative and positive imaginary numbers, and left all arithmetic the same. Is everything that was true before, still true? I say yes.

Now suppose we inverted the negative and positive real numbers, and left all arithmetic the same. Then statements do not remain true. For example, if this behaved like switching the imaginary signs, the true statement 3^(1) = 3 would be changed into (-3)^(-1) = (-3), but that is false.

But note that for switching sign of the imaginaries:
3^(i) = A + Bi for the appropriate reals A, B
and
3(-i) = A - Bi = A + B(-i)

So the statement 3^(i) = A + Bi remains true when the sign of i is changed.

Oh and the thinking -i is the same as i is also a mistake.
Yes squared they become the same answer, yet -i times i= 1, not -1.

I don't see your point.

(-i)(i) = 1 says equal things about (-i) and (i). This doesn't show any difference between the two (by "difference," I mean some statement that is true for one and false for the other). The only difference is "call one negative, call one positive." That's why in my example above, changing which ones we call negative and positive has no effect on the truth value of any statement.

The only difference between i and -i is "relative to each other." They are truly two sides of a coin.

 

[rbj]

1mile,

i think the wikipedia article is clear about this. i don't know what the problem they have with it.

$-1$ and $+1$ are not equivalent. one is the multiplicative identity and the other is not.

in contrast, $-i$ and $i$ are qualitatively equivalent. there is not one single property that one has that the other does not. but they are not zero, so being negatives of each other, they cannot be equal.

 

[lostcauses10x]

Stuff to catch up on.
A quick scan:

my objection is to teach complex numbers (which i have done in the context of electrical engineering classes) solely as ordered pairs.

I agree.

 

[lostcauses10x]

1MileCrash
I do not know what to say to you. I really don't.
I have encountered this before, and for some reason it just does not get across.
(0+1i) does not = (0-1i).
How do you come to the conclusion you have?

Edited to remove shorthand.

 

[rbj]

I have encountered this before, and for some reason it just does not get across.

there's always hope. he might be able to get it across to you eventually.

(0+1i) does not = (0-1i).
How do you come to the conclusion you have?

please list (using words) a single property that $0 + 1i$ has and that $0 - 1i$ does not have. or vise versa.

can you respond to the Wikipedia article pointed to several times?

 

[R136a1]

please list (using words) a single property that $0 + 1i$ has and that $0 - 1i$ does not have. or vise versa.

That they have the "same" property, doesn't mean that they are equal. That's all he said. Same as a category has the "same" properties as its dual, doesn't mean that they're equal.

 

[HallsofIvy]

they are qualitatively different. only one of those two numbers are the multiplicative identity.



it is no mistake to think of -i and i as qualitatively the same. every property -i has, +i also has.

No, -i has the property that it is equal to -1 times i. i does not have that property!

 

[rbj]

No, -i has the property that it is equal to -1 times i. i does not have that property!

sorry, Halls. epic fail.

replace every occurrence of $-i$ with $i$ (which has the consequence that every occurrence of $i$ is replaced by $-i$) and you will see your mistake.

 

[R136a1]

sorry, Halls. epic fail.

Can we please have a civil discussion without immature stuff like "epic fail". This is not a contest.

 

[rbj]

$-1$ and $+1$ are not equivalent. one is the multiplicative identity and the other is not.

in contrast, $-i$ and $i$ are qualitatively equivalent. there is not one single property that one has that the other does not. but they are not zero, so being negatives of each other, they cannot be equal.

That they have the "same" property,

all of their properties are the same.

doesn't mean that they are equal.

please respond to what i said, not to what i didn't say (in fact what i actually explicitly denied).

That's all he said. Same as a category has the "same" properties as its dual, doesn't mean that they're equal.

i didn't say "equal". i said being that "they are not zero [and] being negatives of each other, they cannot be equal."

but they are equivalent. $i$ and $-i$ are interchangeable (and every single theorem in every single textbook and journal article would continue to be just as valid). and that cannot be said of $1$ and $-1$.

 

[Mark44]

No, -i has the property that it is equal to -1 times i. i does not have that property!

sorry, Halls. epic fail.

What HallsOfIvy wrote is obviously true, so it's difficult to see why you think this is an "epic fail."

replace every occurrence of $-i$ with $i$ (which has the consequence that every occurrence of $i$ is replaced by $-i$) and you will see your mistake.

Since the OP hasn't been back for quite some time, it seems to me that this thread has run its course. I am closing it.