Analysis problem: x>o --> 1/x > 0

[b0it0i]

1. The problem statement, all variables and given/known data
Prove

If x>0 --> 1/x > 0


2. Relevant equations

ordered field axioms

closure, associativity, commutativity, identity, inverses, distributive law, trichotomy law, transitive law, preservation

x+z = y+z --> x = y
x.0 = 0
-1.x = -x
xy=0 iff x=0 or y=0
x<y iff -y<-x
x<y and z<0 then xz > yz


3. The attempt at a solution

i've tried this problem several times, and always hit a dead end

i tried a direct proof

assume x>0
therefore x does not equal 0

by existence of inverse

there exists a unique 1/x such that x (1/x) = 1

after that point, i get no where in my attempts

any suggestions?

you can user other "theorems" but you must also prove them

[morphism]

Assume 1/x<0. Can you derive a contradiction using one of the properties you listed? (Hint: which one deals with things <0?)

[b0it0i]

hey thanks a lot

prove: x > 0 --> 1/x > 0

assume 0 < x and assume 1/x less than or equal to 0, x cannot equal 0. then there exists a unique 1/x s.t. x(1/x) = 1. since 1/x is unique, 1/x cannot equal zero.
therefore 1/x < 0

since 0 < x and 1/x < 0

0. 1/x > x . 1/x
0 > 1

which contradicts 0 < 1

i would have to prove that 1 > 0, but i already done this.

thanks