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Hey all,

Great forum here! I am working through "Elementary Analysis" by Kenneth Ross in preparation for an intro to analysis course I am taking in the fall. I was wondering if the following is a valid proof? Included are the axioms I used in the proof, but if you need the complete list I will post them. I'm still learning to use LaTeX so my appologies for not using it here.

Key

<= "is less than or equal to (or > for greater)"

!= "does not equal"

Problem 3.4: Prove 0 < 1

The following properties hold:

...

M4. For each a != 0, there is an element a^(-1) such that aa^(1) = 1

02. If a <= b and b <= a, then a = b.

...

The following are consequences of an ordered field:

...

(iv) 0 <= a^2 for all a

...

Proof by contradiction: Suppose that 0>=1. By (iv), we also have that 0 <= 1^2 = 1. Therefore 0 <= 1 and 1 <= 0, and hence 0 = 1, by O2. But there is an element 1^(-1) = 1 such that (1)(1)^(-1) = 1, and 0 != 1, by M4. Thus 0 = 1 and 0 != 1, a contradiction. Therefore 0 >= 1 is false, and hence 0 < 1.

I am not sure about using M4 to prove 0 != 1, I am pretty shaky on my logic when it comes to quantifiers.

Thanks for any input!

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# Valid proof?

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