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In high school, we are taught that we do not have $2i < 3i$, i.e., the complex number system is not an ordered field.

(Real number, for example, is an ordered field. For example, $2 < 3$).

Why?

My comment to this is because in the complex corrdinate, in $Re-Im$ coordinate, the concept of complex number is somewhat a rotation around the origin.

sleeve chen
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    That's more or less right. On the number line, which of two numbers is furthest to the right is a well-behaved notion (especially with respect to multiplication). In the complex plane, it isn't so well-behaved anymore. Not even the notion of positive and negative has a well-established generalization. – Arthur Jun 25 '15 at 09:04
  • See also http://math.stackexchange.com/questions/1184547/total-order-on-complex-numbers. – Dietrich Burde Jun 25 '15 at 09:12
  • see my answer to: http://math.stackexchange.com/questions/1264203/complex-number-inqualities/1264307#1264307 – Emilio Novati Jun 25 '15 at 09:17

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If you have an order in your field, you will want your order to be compatible with the field operations. This basically means that the sum and product of positives should be positive; and it has many consequences. One of them is that if $a <b $, then $b-a>0$. So if $2i <3i $, $$ 0 <3i-2i=i. $$ Now $i $ is positive, and by squaring we get $-1>0$. Probably not what you intended.

Martin Argerami
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