# Thread: Completeness Axiom in R (Real numbers)

1. ## Completeness Axiom in R (Real numbers)

So I am going over the Completeness Axiom, and I have a question, First I will explain what I think my answer is and my logic , and I am waiting for feedback thank you.

So the question is about the Examples that the lecturer gave,

As you can see in the picture it says for integers, not dense but without gaps and for rationals it says dense but with gaps.

I am having trouble with understanding this,

I understand the fact that rationals have gaps, because there are numbers like square root of something or number e or pi, numbers that can't be represented with rationals, therefore they are not rationals so there are a lot of gaps on the number line.

But how come Z is without gaps and not dense. I mean I understand the fact that it's not dense because you can't put a number between 1 and 2 or any two integers.
But why is it without gaps? Is it because numbers like 1/2 are not defined in Z? According to the definition that they gave us an integer is in the form of n-m when n and m is naturals, so you can't really talk about fractions in Z because it has no meaning, I guess. Is my logic correct?

Thank you.

2. ## Re: Completeness Axiom in R (Real numbers)

Originally Posted by davidciprut
So I am going over the Completeness Axiom,
So the question is about the Examples that the lecturer gave,
As you can see in the picture it says for integers, not dense but without gaps and for rationals it says dense but with gaps.
I am having trouble with understanding this,
I understand the fact that rationals have gaps, because there are numbers like square root of something or number e or pi, numbers that can't be represented with rationals, therefore they are not rationals so there are a lot of gaps on the number line.
But how come Z is without gaps and not dense. I mean I understand the fact that it's not dense because you can't put a number between 1 and 2 or any two integers.
But why is it without gaps? Is it because numbers like 1/2 are not defined in Z? According to the definition that they gave us an integer is in the form of n-m when n and m is naturals, so you can't really talk about fractions in Z because it has no meaning, I guess. Is my logic correct? .
Have you studied Dedekind cuts ?
That is what your attachment suggests you are working with. I have a small complaint about the handout. It should made it clear that in all cases $c\in S$
From that web page in the link that $S=\{r\in\mathbb{Q}|~r<\sqrt{2}\text{ or }r>\sqrt{2}\}=\mathbb{Q}$.

Note that in the example, we can define $L~\&~U$ but there is no greatest or least element in one or the the other set. Hence a gap.
Between any two rational there is a third rational number, density.
But you have shown that there is no integer between $n~\&~n+1$ so $\mathbb{Z}$ cannot be dense. By the same logic it cannot have gaps.

Does that help?

3. ## Re: Completeness Axiom in R (Real numbers)

Originally Posted by Plato
Note that in the example, we can define $L~\&~U$ but there is no greatest or least element in one or the the other set. Hence a gap.
That depends on our set though, I mean If we define L=[-infinity,2] and U=[2,infinity], this case 2 is the greatest , the least element, respectively in U and L, no? At this point it's not a dedekind cut, but still, in the definition that the lecturer gave it doesn't specify it as a dedekind cut, and that is confusing. Maybe there is a point that I am missing here? I understood that we are working with dedekind cuts but the teacher didn't even define what a dedekind is and he is using it, I get so confused sometimes by the lecturer notes...

Originally Posted by Plato
But you have shown that there is no integer between n~\&~n+1 so \mathbb{Z} cannot be dense. By the same logic it cannot have gaps.
What do you mean by ''the same logic''? It's still not clear why it doesn't have gaps, can you explain?

Thank you.

4. ## Re: Completeness Axiom in R (Real numbers)

Originally Posted by davidciprut
What do you mean by ''the same logic''? It's still not clear why it doesn't have gaps, can you explain?
It is really simple: $S=\{r\in\mathbb{Z}|~n<0.5\text{ or }n>0.5\}=\mathbb{Z}$.

$L=\{n\in\mathbb{Z}: n<0.5\}\text{ and }U=\{n: n>0.5\}$

$0=\sup(L)~\&~0\in L$ and $1=\inf(U)~\&~1\in U$ but as you proved $(0,1)\cap\mathbb{Z}=\emptyset$ so no gaps.