1. ## Proof Help!!

I'm not sure on how to prove statements with "if and only if"
Prove these using only field axioms, order axioms and the fact that 1>0 and x0=0
a>0 iff a^-1>0
a>1 iff 0<a^-1<1

2. Hello amm345
Originally Posted by amm345
I'm not sure on how to prove statements with "if and only if"
Prove these using only field axioms, order axioms and the fact that 1>0 and x0=0
a>0 iff a^-1>0
a>1 iff 0<a^-1<1
We need to establish first that $\Big((a>0) \land (b<0)\Big) \Rightarrow (ab < 0)$.

Now one of the axioms of an ordered field is that $\Big((a>0) \land (b>0)\Big) \Rightarrow (ab >0)$; and it is well established that $a(-b)=-(ab)$ (see here for instance.)

So $(a>0) \land (b<0)$

$\Rightarrow (a>0) \land ((-b)>0)$

$\Rightarrow a(-b)>0$

$\Rightarrow -(ab)>0$

$\Rightarrow ab < 0$, since $-(ab) +ab = 0$

We can now prove the first of the required statements by contradiction,as follows:

First consider the case where $a>0$ and assume that $a^{-1} < 0$. Then:

$\Big((a >0) \land (a^{-1}<0)\Big) \Rightarrow (a.a^{-1} <0)$ using the result above.

But $a.a^{-1} = 1>0$. Contradiction. Therefore $(a>0) \Rightarrow (a^{-1} >0)$.

Then prove it the other way around, and consider the case where $a^{-1} >0$, and assume that $a<0$.

$\Big((a^{-1} >0) \land (a<0\Big)) \Rightarrow (a^{-1}.a<0)$, again using the same result.

But $a^{-1}.a=1>0$. Contradiction. Therefore $(a^{-1}>0) \Rightarrow (a>0)$

We have now proved the implication in both directions. Hence $(a^{-1}>0) \iff (a>0)$

To prove the second of your two statements, establish first that $\Big((a>1)\land(b>1)\Big) \Rightarrow (ab >1)$. Hint:
$(b>1) \Rightarrow (b=1+c)$ for some $c>0$. Then use the distributive law on $a.(1+c)$.

Then use a contradiction method as I did for the first part, to show that $(a>1)\Rightarrow (a^{-1}<1)$. Hence, using the result above that $(0.

Finally, prove the result in the other direction to show that $(a>1) \iff (0.