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aaaa202
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When i is defined by an equation which has 2 solutions, how does it make sense to do algebra with complex numbers?
It doesn't! We don't define "i" by "[itex]i^2= -1[/itex]"- although elementary treatments may introduce it that way.aaaa202 said:When i is defined by an equation which has 2 solutions, how does it make sense to do algebra with complex numbers?
aaaa202 said:Okay so maybe you can help me understand where it goes wrong in the following (which motivated this thread for a start):
1. 1/i = sqrt(1)/sqrt(-1)
2. 1/i * i/i = sqrt(1/-1)
3. i / -1 = sqrt(-1)
4. -i = i
What is the problem by simply defining i=sqrt(-1) and where does it go wrong in the above when doing so?
aaaa202 said:I feel kind of stupid, but I don't understand this argument that if everything is true for i it will be for -i too. Could you give some examples? I feel it's kind of the same as choosing a right or lefthanded coordinate system but then again I have not understood that either.
I see. But what exactly is the difference then between defining i=(0,1) and i=√(-1). Is the only problem that the last one could make you mistakingly think that you can apply usual rules for square roots? And noting that this is not the case, are the 2 definitions the equivalent?HallsofIvy said:It doesn't! We don't define "i" by "[itex]i^2= -1[/itex]"- although elementary treatments may introduce it that way.
Rather, we define the complex numbers as pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (a, b)(c, d)= (ac- bd, ad+ bc). That way (a, 0)+ (c, 0)= (a+ c, 0) and (a, 0)(c, 0)= (ac, 0) sp we can associate the "complex number" (a, 0) with the real number, a. Also, if we define i= (0, 1) then we have [itex]i^2= (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0)[/itex] so that "[itex]i^2= -1[/itex]". We can then say (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1)= a(1)+ b(i)= a+ bi.
The point is that the way a mathematical concept is introduced to the student is not necessarily the way it is formally defined by mathematicians.
All you've done is arbitrary chosen (0,1) to be i instead of (0, -1). If (0,1) = i, then (0,-1) = -i, and then we're back where we started: -what's the difference?-HallsofIvy said:It doesn't! We don't define "i" by "[itex]i^2= -1[/itex]"- although elementary treatments may introduce it that way.
Rather, we define the complex numbers as pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (a, b)(c, d)= (ac- bd, ad+ bc). That way (a, 0)+ (c, 0)= (a+ c, 0) and (a, 0)(c, 0)= (ac, 0) sp we can associate the "complex number" (a, 0) with the real number, a. Also, if we define i= (0, 1) then we have [itex]i^2= (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0)[/itex] so that "[itex]i^2= -1[/itex]". We can then say (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1)= a(1)+ b(i)= a+ bi.
The point is that the way a mathematical concept is introduced to the student is not necessarily the way it is formally defined by mathematicians.
1MileCrash said:All you've done is arbitrary chosen (0,1) to be i instead of (0, -1). If (0,1) = i, then (0,-1) = -i, and then we're back where we started: -what's the difference?-
rbj said:originally the concept of "imaginary" numbers and then "complex" numbers did come from, i believe, the solution to quadratics (or higher-order polynomials) set to 0 when there were no "real" numbers (the set of rational and irrational numbers that are ordered on the "real" number line) that satisfy those equations.
rbj said:there's a reason i parenthetically inserted "or higher-order polynomials".
but, from a pedagogical POV, i don't understand why it's necessary to go above quadratics to introduce someone conceptually to imaginary and complex numbers. you don't need to go to cubic or quartic to see the need for imaginary and complex numbers.
Explain please how that is not about history. You explicitly used the phrase "originally...did come from". How am I supposed to parse that as a pedagogical point?originally the concept of "imaginary" numbers and then "complex" numbers did come from...
lostcauses10x said:Folks is this arguing helping?
pwsnafu said:About 1Mile's point about i and -i being equals.
As Halls pointed out complex numbers are formally defined as pairs, with ##i = (0,1)##. This space of numbers is of course ##\mathbb{C}##.
rbj said:they are qualitatively equivalent (both have equal claim to squaring to -1), but not quantitatively, since they are not zero and are negatives of each other.
[itex]\mathbb{C}[/itex] is not [itex]\mathbb{R}^2[/itex]. Halls was complete when he stated the rules of addition and multiplication of these pairs. my point is that the "formally defined as pairs" is not the original meaning (in case you're wondering, i mean the original pedagogical meaning), but can be made equivalent as long as you define the rules of addition (trivial, about the only way you can) and multiplication (less trivial, there are other ways to define multiplication of 2-vectors which are not compatible with complex numbers). i really don't see any point to it or advantage over the common pedagogy (geez, i have to be careful to remind snafu, lest the words be misconstrued).
R136a1 said:Getting all argumentative is getting us nowhere, rbj. I think all of pwsnafu's replies have been very accurate.
does define the rules of the complex system, not the number i.Rather, we define the complex numbers as pairs of real numbers
lostcauses10x said:The set of complex numbers is not reals squared. True.
And no one has said this.
The statement is ordered pairs of real parts.
Yet by leaving out i it can confuse some, and is easier for others.
rbj said:but the sets are one-to-one with each other. (what did they call that? "entire"? can't remember.)
When i is defined by an equation which has 2 solutions, how does it make sense to do algebra with complex numbers?
jbriggs444 said:bijection? There are several obvious bijections between the two entities, as long as they are viewed as simple sets.
isomorphism? If you tack on the appropriate addition and multiplication operations, there are two obvious isomorphisms between the two enties when viewed as fields.
automorphism? There is one non-trivial automorphism between [itex]\mathbb{C}[/itex] and itself. The mapping is complex conjugation (invert the imaginary part).
jbriggs444 said:bijection? There are several obvious bijections between the two entities, as long as they are viewed as simple sets.
isomorphism? If you tack on the appropriate addition and multiplication operations, there are two obvious isomorphisms between the two enties when viewed as fields.
automorphism? There is one non-trivial automorphism between [itex]\mathbb{C}[/itex] and itself. The mapping is complex conjugation (invert the imaginary part).
lostcauses10x said: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.
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.
lostcauses10x said: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 have had to adapt to all of this myself.
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.lostcauses10x said:Back to the original question.
Well how does it make sense to use the reals when the square root of 1 has two answers?
lostcauses10x said:(-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.
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.lostcauses10x said: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.
lostcauses10x said:I have had to adapt to all of this myself.
lostcauses10x said:"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.
lostcauses10x said: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.
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.