Proof with Supremum, Infimum, and Well ordering principle

[magic88]

Homework Statement

We just started learning about supremums and infimums in my math proofs class. I am having trouble with the following question:

Let x, y be real numbers with y - x > 1. Prove that there exists an integer n such that x < n < y. Hint--use the well ordering principle.

Homework Equations

Well-ordering principle: a nonempty set S of real numbers is said to be well-ordered if every nonempty subset of S has a least element. This is proved using induction.

Upper bound: a real number m is called an upper bound of S if for every s in S, s <= m.
Lower bound: a real number m is called an lower bound of S if for every s in S, m <= s.
Maximum: If m is an upper bound of S and m is in S, then we call m the maximum of S.
Minimum: If m is a lower bound of S and m is in S, then we call m the minimum of S.

Supremum: least upper bound
Infimum: greatest lower bound

The Attempt at a Solution

I am unsure of how to incorporate the well ordering principle into my proof. I was thinking of writing something like, "y - x > 1 \Leftrightarrow y > x + 1 \Leftrightarrow x < x + 1 < y. So there is an integer z such that x \leq z < x + 1 or x < z \leq x + 1." And then show the rest with ceiling and floor... but I'm sure this is incorrect, as we are supposed to use the well ordering principle.

Please let me know if extra clarification is needed.

Any help would be GREATLY appreciated!

Thanks in advance,

Kendra

 

[micromass]

Your solution is correct, but you need to work this out:

So there is an integer z such that x \leq z < x + 1 or x < z \leq x + 1."

I know it is an obvious truth, but you need to prove it. You will need the well-ordering principle for that.

 

[magic88]

Thanks for your reply, micromass.

I'm drawing a blank on how to incorporate the well ordering principle into proving it. It seems like I could just use the ceiling or floor function to show x <= z <= x+1. Do I let the let the ceiling of x be the least element of a set of z integers in the closed interval [x, x+1], and then say that all subsets of that set have a least element?

 

[micromass]

Yes, that's the way you do it. You take the set A of all integers greater then x, thus

A=\{z\in \mathbb{Z}~\vert~x\leq z\}

You show that A is nonempty and then you apply the wellordering principle on A. You will obtain an integer z. Then you only need to show z\leq x+1.

 

[magic88]

Thank you, micromass! I understand now. Thanks for getting me on track :)

 

[mohdha]

Sorry but how to prove A non empty