Complex numbers...

achyut joshi

does complex number have anything like consicutive numbers? . Like 2,3,4 are consicutive integers...

HallsofIvy

No. Both the rational and real numbers have the property that between any two rational (or real) numbers there exist at least one rational (or real) number and that property extends to the comple number- there cannot be a "next" complex number.

disregardthat

Sure they do. Every set can be well-ordered.

Give the complex numbers a well-ordering. For each complex number z, consider the minimal element of the set of complex numbers larger than z. This will be the least number larger than z, in other words, a consecutive element.

Now this of course applies to R and Q as well, but note that this cannot correspond to their usual order. The complex numbers however don't have any natural ordering (which satisfies e.g. a < b --> a+c < b+c), so I don't see how the order properties of the real numbers extends to the complex numbers.

ramsey2879

There is one similarity between natural numbers and complex numbers which is that two concecutive integers multiplied together then multiplied by 4 is one less than a perfect square. In other words if n is an integer then 4n(n+1)+1 is a perfect square. Likewise if z is a Gausian integer then 4z(z+1) + 1 is a perfect complex square. This leads me to define two consecutive Gausian Integers to be z and z plus 1.

TylerH

I think Halls was referring to the property of being dense(http://en.wikipedia.org/wiki/Dense_set).

Deveno

no, which is a bit frustrating. we can't "totally order" the complex numbers in a way that corresponds to our notion of "bigger".

there is, however, a "partial" order, where we can say the "size" of a complex number is how far away it is from 0 = 0+0i.

this partitions the complex plane into a set of concentric circles, all the points on any circle are "all just as big".

for Gaussian integers, that is complex numbers of the form m+ni, where m,n are integers, we can order them (in a similar way) by their norm:

N(m+ni) = m²+n² = |m+ni|².

numbers with the same gaussian norm, also lie "on the same circle". but we have no way of telling whether 3+4i should "come before" 4+3i, or not, because both have the same norm.

robert2734

suppose for numbers with the same norm, we arbitrarily list numbers counter clockwise from zero degrees. Then we can definitively say that one complex number comes before or after another in our catalog.

micromass

But you can never make an "ordered field" out of them: http://en.wikipedia.org/wiki/Ordered_field

lostcauses10x

http://planetmath.org/encyclopedia/OrderedField.html
A different explanation of ordered field.

Deveno

the trouble with this, is that even though we keep going "bigger" (counter-clockwise), we eventually "come full circle" (as we approach 360 degrees). this is the same kind of trouble with have with ordering finite fields: since p-1 = -1, is this number < 0 or > 0 (apparently it's BOTH).

if we just consider what happens on the unit circle, we can have z > z₀, w > z₀, but zw < z₀.

so this "<" doesn't behave well with respect to complex multiplication.

there are lots of possible orders. we could define a+bi < c+di as:

a < c, or if a = c, b < d.

but this "<" doesn't behave well with respect to complex addition.

the trouble is (more specifically), the field C doesn't have a "positive cone". with addition, we might be tempted to select the first quadrant, but multiplication can rotate, which nixes that idea.

robert2734

Consider the integers. We have -1< 0, -2< 0 and (-1)*(-2)>0. So the integers aren't an "orderred field" either but they do have an order to them. So back to OP's question: Can we order the complex numbers? yes. Does that order mean anything? I don't know.

lavinia

http://en.wikipedia.org/wiki/Ordered_field

The complex numbers can not be an ordered field in the sense that there is no total ordering, x < y, of them that respects the algebraic structure.

technically this means (see article) that if a < b then a + c < b + c for all numbers c
and if 0 < a and 0 <b then 0 < ab

Notice that since this is a total ordering that any number is either less than or greater than any other.

Following the Wikipedia article, i^2 must be positive - the square of any number is positive i.e. greater than zero in the ordering -so the complex numbers can not be ordered.

The Well Ordering Principle says that any set can be well ordered. A well ordering is stringer then a total ordering since every subset of a well ordered set must have a smallest element. The reals are not well ordered in the usual ordering so a well ordering of the complex numbers can not extend the ordering of the reals.

Deveno

yes but you are using numbers less than 0.

a "positive cone" refers to a subset of a field, considered its "positive elements". it must be closed under addition and multiplication. the positive cone of the real numbers derives from the fact the the natural numbers (including 0) form a sub-semi-ring of the semi-ring of integers (which is in fact, a full-fledged ring).

with the complex numbers, we do have an analog of the integers, the Gaussian integers Z[i], so one might hope that we could order THOSE, and then "fill in the gaps". but we don't have a suitable sub-semi-ring that qualifies as being dignified with the name "positive Gaussian integers".

geometrically, you can think of a total ordering as establishing a line where every number gets a unique place in line. for a given kind of "summing" and "multiplying" to be compatible with this order, we want addition and multiplication to be "order preserving on positive elements":

that is, if we write f(x) = c+x, and g(x) = cx:

0 < a < b --> f(a) < f(b)
0 < a < b --> g(a) < g(c)

well, both "stacking" (addition) and "stretching" along a given direction on a line do this. but on a plane, adding isn't "linear stacking" anymore, it's "head to tail" (triangle summing). and multiplying isn't just "stretching" (or dilation, as physicists are wont to call it), it's "stretching and rotating" (this turns out to be true even for the 4-dimensional quaternions, which is as many dimensions as we can get before things get really wacky).

that one extra degree of freedom of motion, really puts a monkey wrench in things. it makes you think in terms of circles (or should i say disks?), instead of segments.

in particular, the concepts "least upper bound" and "greatest lower bound" lose their utility as ideas when dealing with complex numbers, and induction loses a lot of power, as well. that's one of the reasons topology has become so important to mathematics....the complex numbers have some very desirable properties (including being able to solve problems that don't seem to even involve complex numbers), but we've had to devise a better framework of "closer" and "farther" than "smaller" and "bigger".

Ashu2912

No, I guess not because there is no inequality or comparision in case of different complex numbers. We can compare their magnitudes, but never the complex numbers themselves...

mdemonion

Complex numbers are useful abstract quantities that can be used in calculations and result in physically meaningful solutions..The sum and difference of two complex numbers are defined by adding or subtracting