Inequality and complex number.

[gotjrgkr]

hi, while trying to study complex analysis, i have a few problems.
i already know that in complex number system, it's impossible for any order relation to exist.
but i was confused to this fact when i saw the proof of triangle inequality.
;
Let z,w be complex numbers. Then, triangle inequality states that absolute value of z+w is less than or equal to the sum of each absolute value of z and w.

i think the absolute value of a complex number also belongs to the complex field because real number is identified with special ordered pair whose second coordinate is zero. There is strange thing here. triangle inequality use inequality even though the absolute value is a special complex number. how is this possible?
please let me know if you know about this.
thanks!

 

[nicksauce]

But... the absolute value of a complex number is a real number, and orders are obviously well defined for real numbers.

 

[gamma5772]

You are correct in stating that the reals are, in some sense, a special case of the complexes. However, saying there is no order relation on the complexes does not imply there cannot be an order relation on a subset of the complexes.

Something that is true of a set does not imply it is true of a subset. A trivial example is the set {0,1,2} has 3 elements. The subset {0,1} does not have 3 elements.

 

[gotjrgkr]

please see my thought below.
Let R = {(a,0): a is a real number} where the element a is derived from some subset (cut) of rational number, and order relation < on R = { ((p,0),(q,0)) : p<q where p and q are real numbers, and are derived from cut} and write (p,0)<(q,0) if ((p,0), (q,0)) belongs to <.

Do you mean the order relation I've suggested above exists?
i have also tried to think about it like above.
i , however , couldn't find some statements related with above order relation in any textbook. could you tell me some textbooks explaing the existence of such an order relation?

 

[gamma5772]

please see my thought below.
Let R = {(a,0): a is a real number} where the element a is derived from some subset (cut) of rational number, and order relation < on R = { ((p,0),(q,0)) : p<q where p and q are real numbers, and are derived from cut} and write (p,0)<(q,0) if ((p,0), (q,0)) belongs to <.

I do not fully understand what you are trying to do with subsets of the rationals, but I might understand the thrust of your argument.

Identify C with pairs of reals. Let R = {(a,0): a is real}. Let O be the set { ((p,0),(q,0)) : p<q with p and q real}. Write (p,0)<(q,0) if ((p,0), (q,0)) belongs to O.

O is not an order relation on C. It violates trichotomy -- that for all x, y, exactly one of the following holds: x<y, x>y, or x=y. To see this, consider (0,1) and (0,2). Neither ((0,1),(0,2)) nor ((0,2),(0,1)) is in O, and (0,1) is not equal to (0,2).

However, O is an order relation on R.

i , however , couldn't find some statements related with above order relation in any textbook. could you tell me some textbooks explaing the existence of such an order relation?

Rudin, Principles of Mathematical Analysis gives definitions and some basic properties of ordered fields. You may find this useful. (In fact, Ch. 1, Problem 8 asks you to prove the complexes cannot be ordered).

 

[gotjrgkr]

I do not fully understand what you are trying to do with subsets of the rationals, but I might understand the thrust of your argument.

Identify C with pairs of reals. Let R = {(a,0): a is real}. Let O be the set { ((p,0),(q,0)) : p<q with p and q real}. Write (p,0)<(q,0) if ((p,0), (q,0)) belongs to O.

O is not an order relation on C. It violates trichotomy -- that for all x, y, exactly one of the following holds: x<y, x>y, or x=y. To see this, consider (0,1) and (0,2). Neither ((0,1),(0,2)) nor ((0,2),(0,1)) is in O, and (0,1) is not equal to (0,2).

However, O is an order relation on R.



Rudin, Principles of Mathematical Analysis gives definitions and some basic properties of ordered fields. You may find this useful. (In fact, Ch. 1, Problem 8 asks you to prove the complexes cannot be ordered).

Thank you for replying my question.
^^