# Thread: 2 ordered field theorems that need proving

1. ## 2 ordered field theorems that need proving

My textbook has kindly decided to present twotheorems without proof, meanwhile all others have proofs. I don't want to come up with a flawed proof, so I need some help.

The first of the two involved proving that 0 =/= 1
I did this one, with the help of mhf.

The remaining theorem will probably incorporate the latter into its proof...

Def. a < b <=> a<=b and a =/=b

Thm. For any ordered field, 0 < 1.
Proof:

I'm not really sure where to start, contradiction seems like it's the way to go...

Suppose that for some ordered field, 1 <= 0
0 =/= 1, so 1 < 0
by another theorem that i've proved already, (-0) < (-1)
but (-0) = 0 < (-1)

Q.E.D.

2. Originally Posted by Noxide
My textbook has kindly decided to present twotheorems without proof, meanwhile all others have proofs. I don't want to come up with a flawed proof, so I need some help.

The first of the two involved proving that 0 =/= 1
I did this one, with the help of mhf.

The remaining theorem will probably incorporate the latter into its proof...

Def. a < b <=> a<=b and a =/=b

Thm. For any ordered field, 0 < 1.
Proof:

I'm not really sure where to start, contradiction seems like it's the way to go...

Suppose that for some ordered field, 1 <= 0
0 =/= 1, so 1 < 0
by another theorem that i've proved already, (-0) < (-1)
but (-0) = 0 < (-1)
If a< b and 0< c then ac< bc.
Use that with a= 0, b= -1 and c= -1.

Q.E.D.

3. Thm. For any ordered field, 0 < 1.
Proof:

Suppose that for some ordered field, 1 <= 0
(-0) <= (-1)
(-0) = 0 <= (-1)
Using a<= b and 0 <= c implies ac <= bc, 0 * (-1) = 0 <= 0 = 0*(-1)
but 0=/= 1, so 0 < 0
but 0 = 0
Q.E.D.

Thanks