# Thread: Real analysis questions ( elementary)

1. ## Real analysis questions ( elementary)

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

I did the following
$\displaystyle p = \frac{a^2}{b^2}$
$\displaystyle b^2p=a^2$
Which means $\displaystyle a^2$ 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 $\displaystyle \alpha$ such that

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

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

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

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

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

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

Taking $\displaystyle \epsilon=\alpha-(-\beta)$
$\displaystyle x'_{0}<-\beta+\alpha-(-\beta)$ means that
$\displaystyle x'_{0}<\alpha$ which means that $\displaystyle \alpha$ 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 $\displaystyle \leq$ supS
If S is a nonempty set of real number, and give sufficient conditions for equality

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

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

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

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

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

So the inequality

$\displaystyle infS \leq supS$ is satisfied.

MY major problem is i'm not sure if this suffices as proofs or not.

2. Originally Posted by justchillin
(1) I'm supposed to prove that$\displaystyle \sqrt{p}$ is irrational if p is prime.

I did the following
$\displaystyle p = \frac{a^2}{b^2}$
$\displaystyle b^2p=a^2$
Which means $\displaystyle a^2$ 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 $\displaystyle p=2$ and generalizing.

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

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 $\displaystyle \alpha$ such that

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

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 $\displaystyle S$ be a nonempty set of real numbers which is bounded below. Then there exists a real number $\displaystyle M$ such that $\displaystyle M\leq s$ for all $\displaystyle s\in S$. Then $\displaystyle -M\geq -s$ for all $\displaystyle s \in S$. Then for $\displaystyle t=\{-s:s\in S\}$, $\displaystyle t\leq -M$ for all $\displaystyle t\in T$. Then $\displaystyle T$ 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 $\displaystyle s\in S$. Then $\displaystyle \inf S \leq s \leq \sup S$ so the desired inequality holds. If they are equal, then by the trichotomy of inequality each element of $\displaystyle S$ 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.

3. Originally Posted by siclar
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 $\displaystyle p=2$ and generalizing.

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

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.
Originally Posted by siclar
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 $\displaystyle S$ be a nonempty set of real numbers which is bounded below. Then there exists a real number $\displaystyle M$ such that $\displaystyle M\leq s$ for all $\displaystyle s\in S$. Then $\displaystyle -M\geq -s$ for all $\displaystyle s \in S$. Then for $\displaystyle t=\{-s:s\in S\}$, $\displaystyle t\leq -M$ for all $\displaystyle t\in T$. Then $\displaystyle T$ 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 ?
Originally Posted by siclar
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 $\displaystyle s\in S$. Then $\displaystyle \inf S \leq s \leq \sup S$ so the desired inequality holds. If they are equal, then by the trichotomy of inequality each element of $\displaystyle S$ 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

4. You definitely have the right ideas. I would just say that your proofs you have given need a bit more clarity.

5. 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.

6. Once you have written it up I will be more than happy to take a look at it for you.