*I know there are similar questions asked but but I want to know if it works the way I wrote it.
We are given the set $\Bbb Z_2 = \{0,1\}$ , the binary digits with arithmetic modulo 2. That is the usual set of rules we are accustomed to with the exception $1+1=1$
Proof by contradiction,
Let us suppose $\Bbb Z_2$ is an ordered field, then, we know, if $\Bbb F$ is an ordered field and $a,b,c \in \Bbb F$
$a<b \iff a+c<b+c$
Since $a<b$ we get, $a=0$ and $b=1$
For $c$ there can be two cases:
When $c=0$, we have $0+0<1+0 \implies 0 \lt1$
When $c=1$, we have $0+1<1+1 \implies 1\lt0$
Since $1\lt0$ is a contradiction, it follows that $Z_2$ is not an ordered field