An easy one

Complex numbers with non zero imaginary part can never be compared.

All of the given numbers were complex in nature, and $complex$ numbers can never be compared.

It's simple to prove that it is impossible to define a total ordering of the complex numbers that satisfies all the properties of the ordering of the reals. The properties as I was taught them are:

• For any $x$ and $y$ , exactly one of these is true: $x < y$ , $x = y$ , $y < x$ .
• If $x < y$ and $y < z$ , then $x < z$ .
• If $x < y$ then $x + a < y + a$ .
• If $x < y$ and $0 < a$ , then $ax < ay$ .
• $0 \neq 1$ .

From these it can be proven that

• $0 < 1$ .
• If $x < y$ then $-y < -x$ .
• If $x < y$ and $a < 0$ , then $ay < ax$ .

Now suppose there is a total ordering of the complex numbers that meets this property. Then either

• $0 < i$ , in which case multiplying through by $i$ gives $0 < -1$ .
• $0 = i$ , in which case squaring both sides of the equation gives $0 = -1$ .
• $i < 0$ , in which case multiplying through by $i$ again gives $0 < -1$ .

each of which is nonsense.

It is, however, possible to define a partial ordering of the complex numbers that is an extension of that of the reals and satisfies all of the properties except for the total ordering one. If $\Im(z) = \Im(w)$ , then the numbers are ordered according to their real parts. Otherwise, neither $z < w$ nor $z > w$ . I have previously proven that it impossible even to define it even for some $z$ and $w$ with different imaginary parts without losing another of the properties.