Can you claim $1+i>1$?

Question

The title says it all: I know you can claim $i+1>i$. But can you also claim $1+i >1$? If not why can't I?

Answer 1

As mentioned in the comments, there is no way to "order" $\mathbb{C}$ in a way that is compatible with the operations of addition and multiplication that $\mathbb{C}$ is equipped with - $\mathbb{C}$ is not an ordered field with the usual operations. What this means in practice is that we cannot compare the sizes of complex quantities in a meaningful way using $<$ or $>$.

One way we commonly get around this problem is to instead consider the relative magnitudes of complex quantities. For example, $|1+i|=\sqrt{2}>1=|i|$, and $|1+i|=\sqrt{2}>1=|1|$. This is perfectly valid, because we have moved away from $\mathbb{C}$ to $\mathbb{R}$, which is an ordered field and hence allows us to make these comparisons.

Answer 2

It's easy to extend $>$ to the complex numbers in such a way that the new relation is the same for the real numbers.

Define $a+bi>c+di$ to stand for $a>c$ or $a=c$ and $b>d$ (as customary, $a,b,c,d$ denote real numbers; “and” binds more strictly than “or”).

This can be easily proved to be a (strict) order relation: indeed it is false that $a+bi>a+bi$; transitivity is a simple case analysis.

With this relation it is true that $1+i>1$.

Unfortunately, you cannot claim that $z_1>z_2$ and $z_3>z_4$ implies $z_1+z_3>z_2+z_4$, nor that $z_1>z_2$ and $z>0$ implies $z_1z>z_2z$ (find counterexamples).

In more technical terms, $\mathbb{C}$ is not an ordered field with respect to this order relation. It is not an ordered field under any order relation. Indeed, in an ordered field one has $1>0$ and every nonzero square is $>0$. Since $i^2=-1$, we'd find $-1>0$, so $1-1>0+0$, which is a contradiction.

Answer 3

No, because that would imply that $i>0,$ which would imply that $i^2>0,$ or in other words that $$-1>0,$$ which is nonsense in our frame of ordering.