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First of all, we know that the set of real numbers is a subset of the set of complex numbers: $\mathbb R \subset \mathbb C$.
Also, we know that we cannot compare two complex numbers. For example if we have ${a+bi}$ and ${c+di}$ complex numbers, we cannot state which one of them is greater. On the other hand, $\mathbb R \subset \mathbb C$ implying that real numbers as a subset of complex numbers cannot be compared. However, we have defined (we know) how to compare real numbers. Does this have a mathematical explanation?

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    $\begingroup$ We can compare the absolute values of complex numbers, but there cannot be an order on complex numbers. Consider that $i>0$ implies $i^2>0$ , which is false because of $-1<0$. $i<0$ also leads to a contradiction because multiplying with $i$ leads to $i^2>0$ $\endgroup$ – Peter Feb 20 '17 at 12:19
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    $\begingroup$ How do you come to the conclusion : "If there is no ordering on some set $S$, there is no ordering on any subset of $S$ as well" ? $\endgroup$ – Peter Feb 20 '17 at 12:25
  • $\begingroup$ With similar reasoning on the order of complex numbers, we will come to contradiction again. We know how to order real numbers. $\endgroup$ – Armen Gabrielyan Feb 20 '17 at 12:26
  • $\begingroup$ Can you provide an counter-example other than this case with real and complex numbers? $\endgroup$ – Armen Gabrielyan Feb 20 '17 at 12:29
  • $\begingroup$ It just means that elements in $\Bbb C \setminus \Bbb R$ can not be compared as well. That is elements in $\Bbb R$ can be compared. then we extended this set to $\Bbb C$ where the extra elements can't be compared within themselves as well as with elements of $\Bbb R$. $\endgroup$ – Error 404 Feb 20 '17 at 12:40
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Your assertion

we cannot compare two complex numbers

is incorrect as stated, as every set can be well-ordered.

The correct statement is that

$\mathbb{C}$ is not an ordered field,

that is, it does not admit an order compatible with the field structure.

$\mathbb{R}$, instead, is an ordered field. This shows that an extension of an ordered field need not be an ordered field.

Note that the fields that can be ordered are the formally real fields, that is, those fields where $-1$ cannot be written as a sum of squares. This shows that the $\mathbb{C}$ vs. $\mathbb{R}$ situation is somewhat typical.

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  • $\begingroup$ But how does this prove that "an extension of an ordered field need not be an ordered field?" How could we prove this statement? $\endgroup$ – Armen Gabrielyan Feb 20 '17 at 12:41
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    $\begingroup$ To prove such a statement, an example is enough. Now $\mathbb{R}$ is an ordered field, and its extension $\mathbb{C}$ is not. $\endgroup$ – Andreas Caranti Feb 20 '17 at 12:44
  • $\begingroup$ But maybe there exist some $A$ and $B$ sets ($\mathbb A \subset \mathbb B$), for which this statement does not hold true. How could we be sure? $\endgroup$ – Armen Gabrielyan Feb 20 '17 at 12:49
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    $\begingroup$ @ArmenGabrielyan Perhaps you need some training in understanding of 'proofs' and 'counter-examples'. As you say there might exist such $A \subset B$ so what! Even if they exist it doesn't matter. But, point here is that if we have one counter-example then we can't say this implies that. Observe the word 'need not be' in your first comment. 'need not be $\neq$ implies'. $\endgroup$ – Error 404 Feb 20 '17 at 12:57
  • $\begingroup$ Now I got your point. But I'd like you to further elaborate on the following: we have that $\mathbb{C}$ is not an ordered field meaning that we cannot order the elements of $\mathbb{C}$. Let's take $1$ and $2$ complex numbers. As they are from complex numbers set, they cannot be ordered. Now let's take $1$ and $2$ real numbers. As this time they are real numbers, we can order them. This implies that at one time $1$ and $2$ as complex numbers cannot be ordered while at another time $1$ and $2$ as real numbers can be ordered. $\endgroup$ – Armen Gabrielyan Feb 20 '17 at 13:11
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You can define a "least common denominator" of two numbers $p,q\in\mathbb Q$, but you cannot do the same for two real numbers, even though $\mathbb Q\subset\mathbb R$.

So, there is nothing strange or "paradoxical" in seeing a property of a subset of $X$ that is not a property of $X$ itself.

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From your comments I think you are concerned with the property of comparing two real numbers getting lost when we transit to complex numbers.

Think of this:

1)When we transit from natural numbers to integers there is a loss of a property. In naturals, suppose we have $a \lt b$, then $ca \lt cb \; \; \forall c \in \Bbb N$. But when we enter in integers, this property is not valid since if we multiply by a negative number both sides then we get $ca \gt cb$ and $c$ is a negative number.

2)When we transit from integers to rationals, there is again a loss of a property. In integers, you know which number is next provided you are given an initial number. But in rationals, if I give you $\frac 23$, can you tell me next rational number? So set of rationals is dense but set of integers is not.

3)When we transit from rationals to reals, another property is lost. Every rational number can be written in the form $\frac pq$ where $p,q \in \Bbb Z$. But can every element of $\Bbb R$ be expressed like that? Another loss would be that rationals are countable but reals are not.

4)When we transit from reals to complex, the loss is what you have observed. Even though we know two real numbers can be compared by 'usual' 'less than' and 'greater than' orders, but in complex numbers these exact same orders fail.

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  • $\begingroup$ If you ask me whether $1$ and $2$ as complex numbers can be compared using 'less than' and 'greater than' orderings of reals then I would say yes. Because the reason is they are real numbers. $\endgroup$ – Error 404 Feb 22 '17 at 10:23

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