Proof that if $a > 0$, then $\frac{1}{a} > 0$ using just field and order axioms This is my attempt to prove that, if $a > 0$, then $\frac{1}{a} > 0$, using just field and order axioms.
$$a > 0 \implies a \cdot \left(\frac{1}{a}\right)^2 > 0 \cdot \left(\frac{1}{a}\right)^2 \implies a \cdot \left(\frac{1}{a}\right)^2 > 0 \implies a \cdot \left(\left(\frac{1}{a}\right) \cdot \left(\frac{1}{a}\right)\right) > 0$$
By associativity of multiplication:
$$\left(a \cdot \left(\frac{1}{a}\right)\right) \cdot \left(\frac{1}{a}\right) > 0$$
We know that $a \cdot \left(\frac{1}{a}\right) = 1$, because one is the multiplication inverse of the other. 
So we have:
$$1 \cdot \left(\frac{1}{a}\right) > 0 \implies \frac{1}{a} > 0 $$
Which is what we wanted to prove. 
My question are: 


*

*In the field axioms, there's no axiom that says that $0$ times another number is $0$, but I have used this property. How would you translate it to field and order axioms?

*I have assumed that $\left(\frac{1}{a}\right)^2$ is positive, have you noticed? Note that I need a positive number in order not to change the $>$ sign. But this is not present in the field or order axioms. How would you explain this?
 A: *

*$x\cdot 0 = x\cdot (0+0) = x\cdot 0 + x\cdot 0$, now add $-(x\cdot 0)$ on both sides

*If $x>0$ then $x\cdot x>x\cdot 0=0$. If $x<0$, then $0<-x$, hence $x\cdot x=(-x)\cdot(-x)>(-x)\cdot 0=0$ (I assume you already know $(-x)(-y)=xy$). Only if $x=0$, we do not have $x\cdot x>0$. So you need to show $\frac1a\ne 0$.

You can simplify your overall proof anyway: Given $a>0$, assume than not $\frac 1a>0$.
Then either $\frac 1a=0$, hence $1=a\cdot \frac1a=a\cdot 0=0$, contradiction.
Or $\frac1a<0$, then $0<-\frac 1a$, hence $-1=a\cdot(-\frac1a)>0$, then also $1=(-1)(-1)>0$, leading to $0=1+(-1)>0$, contradiction. Of course this argument also relies on similar basic "standard" results about fields.
A: *

*If you want to prove that $x\cdot 0=0$ for any real number $x$ using only the field axioms (and without the addition property of equality yet) you need to start with the fact that $0$ as the additive identity is unique in the set of real numbers, that is, if $y$ is a real number such that $y+a=a$ for any $a\in\mathbb{R}$, then $y=0$. This is easy to see using only field axioms:
$$y=y+0=y+[a+(-a)]=(y+a)+(-a)=a+(-a)=0.$$
Now, let's look at $x+x\cdot 0$. For any $x\in\mathbb{R}$
$$x+x\cdot 0 = x\cdot1 + x\cdot 0 = x(1+0) = x(1) = x.$$
By uniqueness of the additive identity, $x\cdot 0=0$.


*For this, I will assume that by $1/a$ you mean the multiplicative inverse of a nonzero real number $a$, that is, $$a\cdot 1/a = 1,$$
and that the order relation $<$ is defined as follow: $a<b$ if $b-a$ is positive; and that you already have the notion of the Trichotomy property, that is, for any real number $a$, only one holds: $a>0$ (positive), $a=0$, or $-a>0$ (additive inverse of $a$ is positive).
We first show that if $a\neq 0$ is a real number, then $a^2>0$. By Trichotomy property, either $a>0$ ($a$ is positive) or $-a>0$ (negative $a$ is positive w/c is the same as $a<0$). By the property of positive numbers (the set of positive numbers is closed under multiplication)

*

*If $a>0$, then $a\cdot a=a^2$ is positive.

*If $a<0$, then $(-a)(-a)=a^2$ is positive.

In either case, $a^2>0$. That $(1/a)^2>0$ immediately follows since $1/a$ is a nonzero real number.
