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Yes we know that two complex numbers can be equal to one another , but why can't we say that a complex number is greater/lesser from another complex number ?

Jalaj
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    Which order relation do you suggest? – Did Feb 01 '16 at 09:23
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    We can, just not in a way that either gives a partial order or, alternately, in a way that respects the arithmetic operations. – Derek Elkins left SE Feb 01 '16 at 09:23
  • Related: http://math.stackexchange.com/questions/487997/total-ordering-on-complex-numbers, http://math.stackexchange.com/questions/310931/comparing-complex-numbers. – Martin R Feb 01 '16 at 09:24

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You can easily arrange the complex numbers in some total order. For example, you can define that $$ a+i\cdot b \leq c+i\cdot d \quad\Leftrightarrow\quad (a \leq c) \lor \big((a=b) \land (b \leq d) \big), $$ i.e. you sort the numbers first by their real part, and then by their imaginary part.

You won't, however, find an order that is compatible with multiplication and addition in the same way the the usual order on the real numbers is. Every ordering of the complex numbers will violate at least of the the requirements

  • If $u \leq v$ and $w \leq z$ then $u + w \leq v + z$
  • If $u \geq 0$ and $v \geq 0$ then $u\cdot v \geq 0$.

The order from above satisfies the first requirement, but violates the second, because e.g. $1+i\cdot 2 \geq 0$ but $(1+i\cdot 2)^2 = -3 + i\cdot 4 < 0$.

fgp
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    Here's a hint on how to prove that no order on the complex numbers can satisfy both of those requirements. Suppose there is an order that satisfies them. First, prove that $a\ne0$ implies $a^2>0$. Next, prove that $a^2+b^2=0$ implies $a=b=0$. Finally, note that $i^2+1^2=0$, but $i\ne0$ and $1\ne0$; contradiction. – Akiva Weinberger Feb 01 '16 at 13:25
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"Order" is a defined concept which is included in the definition of real numbers, an ordered field. Complex numbers are a field which are not defined with this property. If you want to tell me that one complex number is greater than another, you can, it's just that the field isn't defined to be that way and doing so leads to a contradiction. To avoid said contradiction, we don't define order on the complex field and hence the field has no order.

See: Proof of Non-Ordering of Complex Field

Community
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user311790
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