Why $(\mathbb R^*, \cdot)$ isn't an ordered group?

Question

I don't understand why $(\mathbb R^*, \cdot)$ (group with multiplication) is not ordered ?

Thank you.

Answer 1

Suppose $\preceq$ was an order on $(\Bbb R^*,\cdot)$. Now let's pose the question - which is larger with respect to $\preceq$, $-1$ or $1$? If $-1\prec1$, then by multiplying both sides by $-1$ we should have $(-1)\cdot(-1)\prec1\cdot(-1)$, so $1\prec -1$, which is impossible. Similarly we cannot have $1\prec -1$, so this group cannot be totally ordered.

Answer 2

I believe it's because multiplication by a negative is order reversing. The operation is not strictly order preserving. But if you just restrict to the positive real numbers $\Bbb R^{+}$ then it is an ordered group under multiplication.