# Thread: Proof of Preserv. of order

1. ## Proof of Preserv. of order

I am stuck on this one!

$\displaystyle a>0, b>0 \implies ab>0$

2. Originally Posted by altave86
I am stuck on this one!

$\displaystyle a>0, b>0 \implies ab>0$
What axiomatic system do you consider for the order ,so that we can prove the above as a theorem??

3. Originally Posted by xalk
What axiomatic system do you consider for the order ,so that we can prove the above as a theorem??

-multiplicative identity

4. Originally Posted by altave86
-multiplicative identity
No ,no i mean what axioms for the order.For example one such system is the following:

1) $\displaystyle \forall x\forall y[ x<y\Longrightarrow \neg (y<x)]$

2) $\displaystyle \forall x\forall y\forall z[x<y\wedge y<z\Longrightarrow x<z]$

3)$\displaystyle \forall x\forall y[ x\neq y\Longrightarrow (x<y\vee y<x)]$

4)$\displaystyle \forall x\forall y\forall z[y<z\Longrightarrow (x+y<x+z)]$

5)$\displaystyle \forall x\forall y\forall z[0<x\wedge y<z\Longrightarrow x.y<x.z]$

Now in a such axiomatic system, if in axiom (5) you put:

x= a , y= 0 and z=b we have the following result:

$\displaystyle 0< a\wedge 0< b\Longrightarrow 0.a< a.b$ ,which is exactly the theorem you are trying to prove ,since 0.a = 0

However in another axiomatic system for order we can have a different proof

5. Ohh sorry but I am a bit confused about that. This is just the introduction o Number theory and we are taking a lookat the real numbers, especifically R as a field and a totally ordered set. We don't look upon that stuff or symbols, just basic things. Sorry.

Originally Posted by xalk
No ,no i mean what axioms for the order.For example one such system is the following:

1) $\displaystyle \forall x\forall y[ x<y\Longrightarrow \neg (y<x)]$

2) $\displaystyle \forall x\forall y\forall z[x<y\wedge y<z\Longrightarrow x<z]$

3)$\displaystyle \forall x\forall y[ x\neq y\Longrightarrow (x<y\vee y<x)]$

4)$\displaystyle \forall x\forall y\forall z[y<z\Longrightarrow (x+y<x+z)]$

5)$\displaystyle \forall x\forall y\forall z[0<x\wedge y<z\Longrightarrow x.y<x.z]$

Now in a such axiomatic system, if in axiom (5) you put:

x= a , y= 0 and z=b we have the following result:

$\displaystyle 0< a\wedge 0< b\Longrightarrow 0.a< a.b$ ,which is exactly the theorem you are trying to prove ,since 0.a = 0

However in another axiomatic system for order we can have a different proof

6. Originally Posted by altave86
Ohh sorry but I am a bit confused about that. This is just the introduction o Number theory and we are taking a lookat the real numbers, especifically R as a field and a totally ordered set. We don't look upon that stuff or symbols, just basic things. Sorry.
Here is a translation of the axioms written in a formal language.

x<y implies $\displaystyle \neg (y<x )$ for all x,y

x<y and y< z implies x< z for all x,y,z ( this is the transitive law)

$\displaystyle x\neq y$ implies x< y or y< x ,for all x,y ( this is another form of the trichotomy law)

y< z implies y+x < z+x ,for all x,y,z

0< x and y< z implies x.y < x.z ,for all x,y ,z ,
.

And if in this axiom we put x = a , y =0, z = b we have:

0< a and 0< b $\displaystyle \Longrightarrow$ 0< a.b .

Which the theorem you want to prove.

However in another axiomatic system your theorem could be considered as an axiom