# Math Help - Proof of Preserv. of order

1. ## Proof of Preserv. of order

I am stuck on this one!

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

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

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

-Communtativity of addition and multip
-distributivity of add and mult
-associativity of add and mult
-multiplicative identity
-additive inverse (-c and 1/c)

4. Originally Posted by altave86
-Communtativity of addition and multip
-distributivity of add and mult
-associativity of add and mult
-multiplicative identity
-additive inverse (-c and 1/c)
No ,no i mean what axioms for the order.For example one such system is the following:

1) $\forall x\forall y[ x

2) $\forall x\forall y\forall z[x

3) $\forall x\forall y[ x\neq y\Longrightarrow (x

4) $\forall x\forall y\forall z[y

5) $\forall x\forall y\forall z[0

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

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

$0< a\wedge 0< b\Longrightarrow 0.a< a.b$ ,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) $\forall x\forall y[ x

2) $\forall x\forall y\forall z[x

3) $\forall x\forall y[ x\neq y\Longrightarrow (x

4) $\forall x\forall y\forall z[y

5) $\forall x\forall y\forall z[0

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

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

$0< a\wedge 0< b\Longrightarrow 0.a< a.b$ ,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 $\neg (y for all x,y

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

$x\neq y$ 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 $\Longrightarrow$ 0< a.b .

Which the theorem you want to prove.

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