# Thread: more proofs in analysis

1. U guys are amazing, I dont see how you can just come up with these answers so quickly. I consider myself good at math, but I am not so good at proofs, and it seems as if you have no weaknesses.....do you teach anywere?

2. Originally Posted by ThePerfectHacker
The way I proved it for homework, is by contradiction. I assumed that there are only finitely many rational numbers. And I let the set S represent all the rationals, which is finite. Then I show that leads to a contradiction.
Can you write out the proof please. Plato has a problem with mine, and he's a smart fellow, so maybe there is a problem. I don't think the general logic is wrong, but maybe the proof doesn't explain itself clearly.

3. Originally Posted by Jhevon
Read the proof again, i showed there is at least one m that satisfies nx < m < ny. But there are infinitely many n's that can work. Thus, for infinitely many n's i can find at least one m for each of them, so i end up with infinitely many m/n numbers
Well you may well think that! But it does not work.
In your original proof there is a quantification error. You select n and then find m. That means that m depends on n. Each new n will possibly produce a different m and there is no way to be sure that they distinct.

TPH, has a similar way as I gave. It will work.

4. Originally Posted by luckyc1423
U guys are amazing, I dont see how you can just come up with these answers so quickly. I consider myself good at math, but I am not so good at proofs, and it seems as if you have no weaknesses.....do you teach anywere?
On the contrary, when it comes to proofs i am very weak. It's just that i am currently taking a proofs class and we did questions very similar to these. However, i believe the perfecthacker has no weaknesses. i'm pretty sure he can come up with proofs for these quickly

5. Originally Posted by Plato
Well you may well think that! But it does not work.
In your original proof there is a quantification error. You select n and then find m. That means that m depends on n. Each new n will possibly produce a different m and there is no way to be sure that they distinct.

TPH, has a similar way as I gave. It will work.
It doesn't really matter if they are not distinct. n/n is as much a rational number as m/n, provided n is not zero

6. Originally Posted by ThePerfectHacker
I do not see how that is a problem. Since "sup S" is used, does it not mean the definition is "well-defined", i.e. it is unique. Hence there is nothing to prove.
TPH, you can find this problem in many basic analysis textbooks. The idea of "well-defined" as in a formal logic is not used in such texts.

7. Originally Posted by Plato
Well you may well think that! But it does not work.
In your original proof there is a quantification error. You select n and then find m. That means that m depends on n. Each new n will possibly produce a different m and there is no way to be sure that they distinct.

TPH, has a similar way as I gave. It will work.
Do you still have a problem with my proof Plato? Can you write out your way please. I am very interested in other ways to do it. TPH uses contradiction, your way seems to require induction.

8. Originally Posted by Plato
TPH, you can find this problem in many basic analysis textbooks. The idea of "well-defined" as in a formal logic is not used in such texts.
One thing what bothered my is that my text defines the limit as follows....

"The limit of a sequence a_n is L iff |a_n-L|<e for ..... "
It bothered me because we never shown that if the limit exists it must be unique. And hence there is nothing wrong with the definition. But just saying the definition like that is wrong, you cannot introduce such a definition. Neither the author nor professor mentioned that there is something wrong with the definition. I believe is because they do not want to leave the realm of analysis and ented the world of logic.

9. Originally Posted by ThePerfectHacker
One thing what bothered my is that my text defines the limit as follows....

"The limit of a sequence a_n is L iff |a_n-L|<e for ..... "
It bothered me because we never shown that if the limit exists it must be unique. And hence there is nothing wrong with the definition. But just saying the definition like that is wrong, you cannot introduce such a definition. Neither the author nor professor mentioned that there is something wrong with the definition. I believe is because they do not want to leave the realm of analysis and ented the world of logic.
As for the definition of limit you are talking about, what page is it on, i don't recall that definition. However, i do believe we addressed that if a convergent sequence has a limit, the limit is unique in class

10. Originally Posted by Jhevon
As for the definition of limit you are talking about, what page is it on, i don't recall that definition. However, i do believe we addressed that if a convergent sequence has a limit, the limit is unique in class
That is true. But we need to first show that the limit is unique before we can make that definition. But we first made the definition and then we showed it is unique.

A correct way is to say,
"There exists a number L such that |a_n-L|<e for ...."

Then we show that if L exists then it is unique.

And then we can define what lim a_n means. Because otherwise it is not well-defined.

11. I remember that you are using Ken Ross’s book. That definition is on page 25. And it does not have ‘iff’ in it. It just says that for each c>0 there exists a number N such that if n>N implies |s_n – s |<c.

The implies is one way. On page 27 it is proved that the limit is unique if it exists. Next we can show that any convergent sequence is a Cauchy sequence.
Then you show that if a sequence is a Cauchy sequence then it has a limit.

12. Originally Posted by Plato
I remember that you are using Ken Ross’s book. That definition is on page 25. And it does not have ‘iff’ in it. It just says that for each c>0 there exists a number N such that if n>N implies |s_n – s |<c.

The implies is one way. On page 27 it is proved that the limit is unique if it exists. Next we can show that any convergent sequence is a Cauchy sequence.
Then you show that if a sequence is a Cauchy sequence then it has a limit.

Oh, you have our book Plato? We have a different edition though. The pages are 33 and 35 for us. My proof for the infinite rationals thingy was similar to the proof for the denseness of Q in the book. I see the problem you had with it. It really doesn't show infinite rationals between real numbers, it shows there is at least one. The proof of infinite rationals has to go a bit further, using your or TPH proof

13. Originally Posted by Plato
And it does not have ‘iff’ in it.
Let me tell you something that used to bother me for a long time.

Definition: An integer n>1 is prime if it has no nontrivial proper divisors.

Now, look at the logic.

If it has no non-trivial proper divisors then it is prime.

But what is an integer is prime, what can we say?

I was finally satisfied when the first page of John Fraleigh's book on Algebra says "..each definition if an if and only if statement.." Meaning, it is automatically assumed that it goes both ways.
Since, I do not like to write it the standard way because, it used to before me, I used "iff".

Also, note when I define things I tried to remove the conditionals from the definition. Over here it is a little strange looking if you removed the conditionals, rather I used iff.

14. Let’s clear that proof up. (You both realize that this was like a retest for the student. You did the work for him/her. You should get the extra points.) My copy of Ross is 1991 edition.

Start with the fact that between any two numbers there is a rational number. There is a rational number, r_1, such that x<r_1<y, There is a rational number, r-2, x<r_2<r_1, there is a rational number, r_3, x<r_3<r_2<r_1<y. Then by induction we define a sequence {r_n} of distinct rationals between x & y. I use this approach because it foreshadows the proof that any real number is the limit of a sequence of rational numbers.

For TPH’s proof. Suppose that the set of rationals in the open interval (x,y) is finite. Any finite set contains its first term; call it r. Now x<r, but there is a rational, s, x<s<r<y. That is a contradiction.

P.S. Here is the definition of prime many of us use.
A positive integer is prime if it has exactly two divisors.
Using that why is 1 not prime.

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