Well you are on the right track, but p is not the irrational number right? The trick is best illuminated by considering the special case and generalizing.

If the square root of two were rational, we could find integers and such that . Furthermore we can choose these integers to be relatively prime, so in particular not both integers are even. Then similarly to your argument, we see that , so and thus is even. Then is divisible by , so is also even thus is even which is in contradiction with our choice of and .

This argument generalizes well. Keep in mind that our claims about the relationships between integers and their squares above comes from uniqueness of prime factorization.

I don't understand your question. What you've given is the definition of the infimum of a set of real numbers. From what I could follow of your answer, I think you are trying to prove that given every bounded above set of real numbers has a supremum, every bounded below set has a infimum as defined above which is unique. Then yes, your answer seems correct but it is not very clear.(2)I'm supposed to prove the theorem

If a nonemepty set S of real numbers is bounded below, then inf S is the unique real number such that

(a)for all x inS

(b) if , there is an in S such that

They gave me the set T = {x|-x \exists S}

Everything revolves around applying negatives to our established inequalities from the supremum as you were considering, but it seems like you were going about it kind of backwards. Half the battle in proof writing is understanding the answer, and the other half is making your answer understandable to others. Here is how it should start:

Let be a nonempty set of real numbers which is bounded below. Then there exists a real number such that for all . Then for all . Then for , for all . Then is bounded above and obviously nonempty, so there exists a supremum...

For your third question you had, there is no need for cases! You have it more or less exactly. One tricky aspect is that for the generality your question statement has, we need the extended real number system (real numbers with plus and minus infinity) and the definition that the supremum of an unbounded above set is positive infinity and likewise for the infimum. Regardless, you have the right idea: Let . Then so the desired inequality holds. If they are equal, then by the trichotomy of inequality each element of is equal as well thus it contains exactly one element. Likewise if it contains exactly one element then of course the infimum and supremum are the same.