# ordering the complex numbersWatch

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#1
f : R^2 -> R
f(a,b) < f(p,q) => (a,b) < (p,q)

why can you not use this to put an order on the complex numbers?
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14 years ago
#2
Because the complex numbers simply aren't an ordered field. Consider, for example, a=1+i, b=2+i and p=8+i, q=6+i. Define f by f(r,s)= r-s. So f(a,b)=-1, and f(p,q)=2. -1<2 but it doesn't mean anything to say the ordered pair (1+i, 2+i)<(8+i,6+i).
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#3
(Original post by J.F.N)
Because the complex numbers simply aren't an ordered field. Consider, for example, a=1+i, b=2+i and p=8+i, q=6+i. Define f by f(r,s)= r-s. So f(a,b)=-1, and f(p,q)=2. -1<2 but it doesn't mean anything to say the ordered pair (1+i, 2+i)<(8+i,6+i).
sorry i meant to say f was bijective, and (a,b) is just a+bi, i understand it wouldn't say anything physical, but would it not be mathematically consistent to apply this definition anyway?
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14 years ago
#4
Ah, ignore that. It was rash of me. Its an ordered pair of reals. Let me think about it a bit more..
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14 years ago
#5
(Original post by fishpaste)
f : R^2 -> R
f(a,b) < f(p,q) => (a,b) < (p,q)

why can you not use this to put an order on the complex numbers?
If f is a bijection then the above may be used to order R^2 or C.

Any set can be totally ordered, it's just that C can't be ordered in a way that is invariant under translations. Such an order has to satisfy the further requirement that:

if a >= b then a+c >= b+c

as well as

if a >= b and b >= a then a = b
if a >= b and b >= c then a >= c
for any a,b then a >= b or b>= a
0
14 years ago
#6
The modulus function |z| is the basic way of looking at the size of a complex number.
In this way , u can say |a+bi|<|c+di|, or whatever it is.
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#7
(Original post by cms271828)
The modulus function |z| is the basic way of looking at the size of a complex number.
In this way , u can say |a+bi|<|c+di|, or whatever it is.
but this doesn't say so much

|1 + 3i| = |3 + i| =/=> 1 + 3i = 3 + i
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#8
(Original post by RichE)
If f is a bijection then the above may be used to order R^2 or C.

Any set can be totally ordered, it's just that C can't be ordered in a way that is invariant under translations. Such an order has to satisfy the further requirement that:

if a >= b then a+c >= b+c

as well as

if a >= b and b >= a then a = b
if a >= b and b >= c then a >= c
for any a,b then a >= b or b>= a
ahha, so it's not transitive? because there isn't a continuous bijection?
0
14 years ago
#9
The order you gave, with that bijection f, was a total order and so was transitive. It would satisfy

if a >= b and b >= a then a = b
if a >= b and b >= c then a >= c
for any a,b then a >= b or b>= a

but not

if a >= b then a+c >= b+c.
0
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