Which is greater?

[Ikerous]

Which is greater: i or 2i?

I've asked like six people.. no one has helped any.

 

[Revisedsoul]

if i is positive then 2i is greater. if i is negitive then i is greater. if your meaning i to be a variable

 

[Ikerous]

i is sqrt(-1)

(Sorry, i woulda said that but i figured it was implied)

 

[Death.Trap]

No, i stands for an imaginary number. I would say that 2i is greater.

 

[Locust]

i may be negative though. and if it is, then it would be 2*square root of -1. I believe this would result in negative so it would be a greater negative number than the square root of -1

 

[Ikerous]

No, i stands for an imaginary number. I would say that 2i is greater.

Why..? (Some sort of reasoning would be greatly appreciated, although 2i is the more popular answer, i cant actually prove it)

i may be negative though. and if it is, then it would be 2*square root of -1. I believe this would result in negative so it would be a greater negative number than the square root of -1

Exactly >_> Which is why i have no clue as to the answer..

 

[Hazar]

I think it's 2i.

Since i^1 is always an imaginary number, and can't be comprehended, then 2 times i must be twice as big as i.

 

[Ikerous]

I think it's 2i.

Since i^1 is always an imaginary number, and can't be comprehended, then 2 times i must be twice as big as i.

But as abc pointed out 2 times a number isnt necessarily bigger...

 

[Revisedsoul]

ok, but its imposible to get the square root of a negitive number. so they are both equal

 

[Ikerous]

ok, but its imposible to get the square root of a negitive number. so they are both equal

Its not exactly impossible; its just imaginary.
And if 2i = i then 2 = 1.. and thats not true :-/

 

[Revisedsoul]

Its not exactly impossible; its just imaginary.
And if 2i = i then 2 = 1.. and thats not true :-/

how is it imaginary. by defenition square root of something is a number times by itself equaling the number being rooted, if i explained that right. and no number multiplied by itself can equal a negitive

 

[Ikerous]

how is it imaginary. by defenition square root of something is a number times by itself equaling the number being rooted, if i explained that right. and no number multiplied by itself can equal a negitive

http://dictionary.reference.com/search?q=imaginary unit
(Note the symbol is i)

 

[Revisedsoul]

oh ok, my guess is then 2i is greater, beacause if i is correcponding to the poini (0,1) then 2(0,1) = (0,2). yet again that is mixing match forumlas a little

 

[Ikerous]

oh ok, my guess is then 2i is greater, beacause if i is correcponding to the poini (0,1) then 2(0,1) = (0,2). yet again that is mixing match forumlas a little

Um, you dont know a great deal about imaginary numbers do ya...?
And i dont think you can just multiply like that o.o
Anyone actualy know?
I asked a few people from my calc class but they didn't know :-/

 

[Revisedsoul]

not really, i belive thats in calc, and i didnt get the chance to take it. and no you cant mutliply locations like that, i was just having fun

 

[SpuD]

Technically it would be 2, but there is no real way to prove, since i can be anything

 

[Ikerous]

Technically it would be 2, but there is no real way to prove, since i can be anything

What reasoning did you use to get to 2?
And i cant be anything :-/
i is a constant, not a variable

 

[nofx]

Im taking algebra 2 right now and were doing theese gay little things I believe 2i should be greater because if you did (i+i) it would be 2i
but if it was just like i^2 = -1 and i^3 = -i and i^4 = 1 so w/e eh you can probably figure out fro this info

 

[Ikerous]

Im taking algebra 2 right now and were doing theese gay little things I believe 2i should be greater because if you did (i+i) it would be 2i
but if it was just like i^2 = -1 and i^3 = -i and i^4 = 1 so w/e eh you can probably figure out fro this info

Um.. so how does that make 2i bigger..?

 

[nofx]

I dont have my math folder with me right now but im going to make a conjecture and think spud is right here there's really no way to prove really which would be greater

 

[Ikerous]

I dont have my math folder with me right now but im going to make a conjecture and think spud is right here there's really no way to prove really which would be greater

Yea, it most likely is impossible, but i havent heard a reason why it's impossible.
Spud said that "i can be anything" which it cant...
C'mon, someones gotta actually know for sure :-/

 

[nofx]

I talked to my friend hes becoming an engineer but he wasnt so sure about the answer himself but we got maybe a little closer to a more defined reason... ok think of it this way i = sqrt -1 then 2i = sqrt -4 so 2 would be equal to -2 so i guess its lesser than i i being = -1

 

[Ikerous]

I talked to my friend hes becoming an engineer but he wasnt so sure about the answer himself but we got maybe a little closer to a more defined reason... ok think of it this way i = sqrt -1 then 2i = sqrt -4 so 2 would be equal to -2 so i guess its lesser than i i being = -1

Okay.. lemme follow this..
i = sqrt(-1) Yea, i know that
2i = sqrt(-4) Yup, still gotcha
"so 2 would be equal to -2" Wha...?
"i being = -1" no :-/

Thanks for asking someone though ^_^
Just help me understand what you're explainin a lil better...

 

[qckbeam]

i is the positive square root of -1. It is not an integer, and it is not real. It's not something you can represent using normal numbers. Conventional mathematics has no idea what i actually is. Still it is possible to define the "<" relationship between two complex numbers. Here is the rule for it:

(a+ib) < (c+id) provided either a < c or a=c and b < d

In your case:

(0+1i) < (0+2i)

Therefore, since a (0) is equal to c (0) and b (1) is less than d (2) the expression 0+1i is less than 0+2i

It's not very useful however. It doesn't hold in all cases.

 

[Ames]

In order to solve this problem, we have to go into an imaginary plane of existence. As we know already i = sqrt(-1). So adding in your post, 2i would be equal to sqrt(-4). However, this would mean that 2 would be equal to -2 as you point out.

Now what did I just do? Absof*ckinglutely nothing.

Anyways 2i > i. They can be added and subtracted like algebraic terms I believe.

 

[Viperidae]

I'm pretty sure you only (practically) use imaginary numbers for the square roots of negative integers correct?

Then 2i should be smaller, because multiplying a negative by two usually gives you a bigger negative.

But that doesn't make sense, because i is neither negative nor positive, that's why we made it up in the first place!

ArGh, brain meltage!

I suck at math.

 

[Ikerous]

i is the positive square root of -1. It is not an integer, and it is not real. It's not something you can represent using normal numbers. Conventional mathematics has no idea what i actually is. Still it is possible to define the "<" relationship between two complex numbers. Here is the rule for it:

(a+ib) < (c+id) provided either a < c or a=c and b < d

In your case:

(0+1i) < (0+2i)

Therefore, since a (0) is equal to c (0) and b (1) is less than d (2) the expression 0+1i is less than 0+2i

It's not very useful however. It doesn't hold in all cases.

Wow.. someone answered with actual math! W00t!
Okay.. i entirely understand everything you said
However, where did you get that 'rule' from?
I just cant pull a rule outta thin air w/o proving it..

Wait.. how doyou know a and c are 0?
Because i didn't define them?
If i did define them itd change the answer
Yet i dont need to define them
Since i'm not graphing, im just comparing..

But that doesn't make sense, because i is neither negative nor positive

Thats what i've been thinking! And since its neither negative nor postive you cant tell how itd react when multiplied by an integer...

 

[Ames]

You can graphically represent imaginary numbers at a right angle to the normal number line. Both number lines are known as a complex plane which can be shown in a xy coordinate system.

Damn math.

 

[Viperidae]

Damn math.

Indeed.

 

[Dulrough]

2i = 1

That or my math teacher is a lying bastard.

 

[Ikerous]

2i = 1

That or my math teacher is a lying bastard.

lol XD Sorry to tell ya... but he is.
cuz 2i = sqrt(-4) which clearly isnt 1..
i^4 is 1.. so is e^(2i*pi)..

 

[Teta_Bonita]

they are both undefined.
....right?

 

[oldagerocker]

this thread sucks, i can't do math, so il go along with it anyway.

i + p = pi
so,
2i = (pi - p)*2 = errm.. i dont know but pi is 3.154

i win. thanks.

 

[Letters]

Is it really that hard to say "I dunno" and move on?

 

[oldagerocker]

you got to have a go man! I had a good 5 minutes bullsh*tting that one

 

[Letters]

Heh, I wasn't really talking about your post oldagerocker. I meant the people actually trying.

 

[Locust]

I^2=-1
2*I^2=-2
I=a negative number
2*I=a negative number
I<2I
Or some bullshit like that.

 

[Ikerous]

I^2=-1
2*I^2=-2
I=a negative number
2*I=a negative number
I<2I
Or some bullshit like that.

I isnt a negative number..
I squared is negative
Which means thats not right :-/
I still cant figure it out
Cept by graphing it
Which id prefer not doing
Someone help >_>

 

[Seeky]

Here's what I came up with. I have doubts about the concreteness of this, because the real number line is separate from the imaginary number line.

Let i = sqrt(-1).

So, assume i^2 = i^2.

Thus, i^2 = -1.
i^2 = i * i, so i = (-1 / i).

i = (-1 / i), so 2i = (-2 / i).

Assume i > 2i. Thus
(-1 / i) > (-2 / i).

Multiply by i on both sides. Thus, -1 > -2. Therefore, i > 2i.

I'm not sure what happens to the inequality when you multiply by an imaginary number. If it does not flip the sign, the above should be sound.

 

[mortiz]

I'd just like to ask...what is the point of imaginary numbers? Did some dood just make up some numbers that don't make sense in normal maths just to get rich?