# Thread: Simple proof -1<0<-1 using axioms

1. ## Simple proof -1<0<-1 using axioms

This is probably a 2 or 3 line proof but just can't seem to get it. (mistyped in the title)

Prove -1<0<1 using Axioms 1-10 and a>0 => -a<0

Thanks!

2. Originally Posted by economanc
Prove -1<0<1 using
Axioms 1-10 and a>0 => -a<0
If you want any help you must give us the list of Axioms 1-10.

3. Originally Posted by Plato
If you want any help you must give us the list of Axioms 1-10.
oops ok:
A3: every integer has an identity with respect to addition
A4: every integer has an inverse with respect to addition
A5: Multiplication is associative
A6: Multiplication is commutative
A7: every integer has an identity with respect to multiplication
A8: Distributive Law: x(y+z)=xy+xz
A9: Closure Property: a set A is closed with respect to addition and multiplication if x+y is in A and xy is in A
A10: Trichotomy Law: For every integer x, exactly one of the following is true: x is either in the set of negative integers, set of positive integers, or x=0

You can also use the following properties that I have already proved:

P1: a+b=a+c => b=c
P2: a0=0a=0
P3: (-a)b=a(-b)=-(ab)
P4: -(-a)=a
P5: (-a)(-b)=ab
P6: a(b-c)=ab-ac
P7: (-1)a=-a
P8: (-1)(-1)=1

4. There seems to be a missing definition.
You should have: $\displaystyle a>b$ if and only if $\displaystyle a-b$ is a positive integer.

Having that we get $\displaystyle 1-0=1$ which is a positive integer so $\displaystyle 1>0$.

By the given you get $\displaystyle a > 0\; \Rightarrow \; - a < 0$.
There you have it.

5. you are correct, left off that definition.

Thank you! So simple, I should have known