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Math Help - Real analysis questions ( elementary)

  1. #1
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    Real analysis questions ( elementary)

    (1) I'm supposed to prove that \sqrt{p} is irrational if p is prime.

    I did the following
    p = \frac{a^2}{b^2}
    b^2p=a^2
    Which means a^2 and a are factors of p but p is irrational.

    Is that enough for the proof ?

    (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 \alpha such that

    (a) x\geq\alpha for all x inS
    (b) if \epsilon>0 , there is an x_{0}in S such that x_{0}<\alpha +\epsilon

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

    MY answer
    I said that if T is bounded from above then x\leq\beta
    and x_{0}>\beta+\epsilon
    so supT = \beta

    If T is bounded from above then S must be bounded from below because T = {x|-x \exists S}

    Let x'=-x \exists S
    x'\geq-\beta
    \epsilon>0,  x'_{0}< -\beta +\epsilon

    Suppose there is an \alpha such that -\beta<\alpha which is a lower bound for S

    Taking \epsilon=\alpha-(-\beta)
    x'_{0}<-\beta+\alpha-(-\beta) means that
    x'_{0}<\alpha which means that \alpha is not lower bound of S.


    Therefore theorem1.1.8 is true. And there is no number greater than beta that satisfies the conditions

    Is this enough ?

    (3) It says to show that inf S \leq supS
    If S is a nonempty set of real number, and give sufficient conditions for equality

    I said......................

    If inf S exists  \forall x \exists S
    x \geq \alpha=infS
    AND
    if sup exists then \forall x \exists S
    x\leq\alpha=supS

    Case1
    x>\alpha and x<\beta \forall x
    then according to the transitive property
    \alpha<beta which implies infS<supS

    Case2
    x=\alpha and x<\beta \forall x
    then it follows again for the transitivity of real numbers that
    \alpha<\beta

    Case3
    x= \alpha and x= \beta \forall x
    then it follows that
    \alpha = \beta

    So the inequality

     infS \leq supS is satisfied.







    MY major problem is i'm not sure if this suffices as proofs or not.
    Last edited by justchillin; August 10th 2009 at 08:27 AM.
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  2. #2
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    Quote Originally Posted by justchillin View Post
    (1) I'm supposed to prove that \sqrt{p} is irrational if p is prime.

    I did the following
    p = \frac{a^2}{b^2}
    b^2p=a^2
    Which means a^2 and a are factors of p but p is irrational.

    Is that enough for the proof ?
    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 p=2 and generalizing.

    If the square root of two were rational, we could find integers a and b\ne 0 such that \sqrt{2}=\dfrac{a}{b}. 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 2b^2=a^2, so a^2 and thus a is even. Then a^2 is divisible by 4, so b^2=a^2/2 is also even thus b is even which is in contradiction with our choice of a and b.

    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.

    (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 \alpha such that

    (a) x\geq\alpha for all x inS
    (b) if \epsilon>0 , there is an x_{0}in S such that x_{0}<\alpha +\epsilon

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

    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 S be a nonempty set of real numbers which is bounded below. Then there exists a real number M such that M\leq s for all s\in S. Then -M\geq -s for all s \in S. Then for t=\{-s:s\in S\}, t\leq -M for all t\in T. Then T 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 s\in S. Then \inf S \leq s \leq \sup S so the desired inequality holds. If they are equal, then by the trichotomy of inequality each element of S 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.
    Last edited by siclar; August 10th 2009 at 08:52 AM. Reason: To address third question
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  3. #3
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    Quote Originally Posted by siclar View Post
    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 p=2 and generalizing.

    If the square root of two were rational, we could find integers a and b\ne 0 such that \sqrt{2}=\dfrac{a}{b}. 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 2b^2=a^2, so a^2 and thus a is even. Then a^2 is divisible by 4, so b^2=a^2/2 is also even thus b is even which is in contradiction with our choice of a and b.

    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 made a mistake i meant to say that " but p is a prime number " and since a^2 and a are factors of p this goes against the definition of a prime number. I understand what you are saying, the previous question in the book i'm using actually asked to show that root 2 was irrational and i had already used the above method.
    Quote Originally Posted by siclar View Post
    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.

    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 S be a nonempty set of real numbers which is bounded below. Then there exists a real number M such that M\leq s for all s\in S. Then -M\geq -s for all s \in S. Then for t=\{-s:s\in S\}, t\leq -M for all t\in T. Then T is bounded above and obviously nonempty, so there exists a supremum...


    I was trying to prove that every bounded set of real numbers has an infimum. So i guess i did it right ?
    Quote Originally Posted by siclar View Post
    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 s\in S. Then \inf S \leq s \leq \sup S so the desired inequality holds. If they are equal, then by the trichotomy of inequality each element of S 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.
    Thanks a lot . I'm really new to analysis, so i am not very good at it yet. It says
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  4. #4
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    You definitely have the right ideas. I would just say that your proofs you have given need a bit more clarity.
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  5. #5
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    Thanks this is my first " real" math course. I have another thread about Dedekind's cut can you help me ? I guess now that i think about it i should have just posted the question here.
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  6. #6
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    Once you have written it up I will be more than happy to take a look at it for you.
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