Given the definition:
A real number $x$ is said to be positive if $0\lt x$ and negative if $x\lt 0$, and the Axioms of order (Trichotomy, Transitivity, Monotonicity of addition and multiplication), prove that $0\lt 1$.
The approach that I have come across is as follows:
By the axiom of trichotomy, if $0$ is not less than $1$, then either $0=1$ or $1\lt 0$. But $0=1$ is false because the multiplicative identity axiom asserts that $0$ and $1$ are different numbers. If we assume that $1\lt 0$, then we form the following argument:
$1\lt 0$
$1+(-1)\lt 0+(-1)$
Since $1+(-1)=0$ and $0+(-1)=-1$, we have
$0\lt -1$ which by the definition above shows that $-1$ is positive. So,
$0(-1)\lt -1(-1)$
Since $0(-1)=0$ and $-1(-1)=1$, we have
$0\lt 1$
Here is where I am trying to understand how this proof works. If $p$ is the statement, $1\lt 0$ and $q$ is the statement, $0\lt 1$, we have shown that $p$ implies $q$. However, by the axiom of trichotomy and the fact that $0=1$ is false, $p$ and $q$ have opposite truth values, and the only way the conditional of $p$ and $q$ can be true, having $p$ and $q$ not equal to each other, is for $p$ to be false and $q$ to be true. Hence, $1\lt 0$ is false, $0=1$ is false, and we have $0\lt 1$.
By the way, I have also seen the proof which uses the fact that all nonzero real numbers have positive squares to show that $1\gt 0$, but I am wondering if my explanation is correct.
