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I am going through the numbers system from an analysis book. It is written that:

1) there is no rational number $p \ ( > 0)$ which satisfies $p^2=2$.

2) The set $\{p: p^2 < 2\}$ does not have a greatest element and the set $\{p: p^2 > 2\}$ does not have a smallest element.

Now it is written that the above two imply that rational number system has certain gaps. But, only the first is enough to show that the there are some numbers which are not rational i.e. rational number does not completely describe the number system i.e it has certain gaps. What is the implication of the second?

My basic confusion is what is meant by "gap/hole"? why $\Bbb Q$ is said to have hole whereas $\Bbb R$ is said not having holes.

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The first alone doesn't guarantee gaps, the set $\mathbb{N}$ hasn't got a supremum in $\mathbb{Q}$ either. –  vonbrand Apr 13 '13 at 2:03
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Incidentally, I believe $p$ supposed to be restricted to the positive numbers. –  Hurkyl Apr 13 '13 at 2:19
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(2) is all you need for a Dedekind cut. –  oldrinb Apr 13 '13 at 3:18
    
Thanks for the comment, I updated it in the problem –  sosha Apr 13 '13 at 3:18
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I think that you meant $\{p : p^2>2\text{ and } p>0\}$, otherwise most negative numbers have the property that $p^2>2$, so clearly there is no smallest element. –  Asaf Karagila Apr 13 '13 at 11:01

7 Answers 7

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Think of the natural numbers, in which are no numbers n and m that satisfy n+2m=1 (assuming 0 is not a natural number). However, the set {n,m:n+2m>1} has the least element (1,1).

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Well, you'd need an ordering on the pairs, but yes. –  Thomas Andrews Apr 13 '13 at 2:20

What the text may be trying to assert is that the rational numbers can be split into two "separate" pieces. The second claim shows two such pieces, and notes that no point of either piece is "infinitely close" to the other piece (e.g. we would say that $0$ is infinitely close to the set of positive numbers, since we can find positive numbers arbitrarily close to $0$). To show that their union is all of $\mathbb{Q}$, we need the first claim, that the element we have "left out" (that is $\sqrt{2}$), is not in $\mathbb{Q}$.

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thanks for the reply. It is not clear to me the significance of "no greatest / smallest" element. –  sosha Apr 13 '13 at 2:31
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@prasenjit When we say a set $A$ has a greatest element, it means there exists $a\in A$ for which $x\leq a$ whenever $x\in A$. That is, every element of $a$ is smaller than $a$, or equal to it. However, there might be sets, such as $(0,1)$, that have no greatest element. For any $0<\ell <1$ we have, $\ell <\frac{\ell+1}2<1$, so whatever element you give me, I will find one which is stricly larger and still in that set. We say in such a case there is no greatest or last element in said set. –  Pedro Tamaroff Apr 13 '13 at 2:34
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@PeterTamaroff: Thanks for your reply. I did not ask the definition of greatest /smallest element ? I am interested in knowing whether having a greatest/ smallest element has any role in the statement " the set of rational numbers has certain gaps". Why it is proved that both the set mentioned in point 2 does not have greatest and smallest element respectively before stating that rational number has certain gaps –  sosha Apr 13 '13 at 3:09

You could think about it this way: there's no real number $x$ such that $x^2 = -1$, but that doesn't imply $\mathbb{R}$ has holes. It just means that there are things beyond $\mathbb{R}$. The point of the second assertion is to show that not only $\mathbb{Q}$ isn't complete (in an everyday sense, in the same way that $\mathbb{R}$ isn't complete because you can extend it to $\mathbb{C}$), but in some sense you can split the whole of it into two parts and leave a gap in between.

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Does the textbook talk about the Intermediate Value Theorem anywhere nearby? It may be trying to show that it does not hold in $\mathbb{Q}$.

Let $f(x) = x^2$. It is continuous, even in $\mathbb{Q}$. Since $2$ is between $f(1)$ and $f(2)$, it would seem intuitive that there exists a $c \in [1, 2]$ such that $f(c) = 2$. However, unlike in $\mathbb{R}$, this is not true.

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What is a gap in a set of numbers? It is a number which is greater than some of them and lesser than others, so that it is between or among them in some way, yet which is not one of them!

Note that we can divide rational numbers into three sets: those less than $1/2$, then $1/2$ itself, and lastly those greater than $1/2$. This clearly represents a three-way partition which includes all of the rationals. Moreover, we can remove the $1/2$ and we then have a gap between the remaining ones. The $1/2$ sits between the other two sets. We know that no rationals are missing between either set and $1/2$ because both sets include rationals which are arbitrarily close to $1/2$.

The second argument shows that in making this kind of three-way partition, we can replace $1/2$ by a number which is not a rational, such as $\sqrt 2$, and yet it still works the same way. We then have a situation in which the upper and lower partition contain all of the rationals, separated by a number betwen them which isn't a rational. Just like a removed $1/2$, it represents a gap: except that since it isn't a rational, it doesn't have to be removed: it simply is that gap.

The second argument is needed because it is not enough to know that $\sqrt 2$ isn't a rational, but also that it sits among them, in the same continuum.

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Thanks. Very nice explanation. In case of $R$ we cannot partition like that as there is no real number which has has a negative square. One question, what is the significance of the two sets having no greatest/smallest element here ? –  sosha Apr 13 '13 at 5:33
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Well, no! That is just the first argument: the square root of a negative isn't a real number, it's some other number. The second part of the argument is: this non-real number does not sit among the reals. It cannot divide a set of reals such that some are bigger than it and some are smaller! –  Kaz Apr 13 '13 at 5:34
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In the case of the rationals, the partitioning works in spite of the pivotal number not being a rational! An irrational number is being used to form the endpoint of an open interval of rationals! –  Kaz Apr 13 '13 at 5:37

There is no rational number $q$ such that $q^2=-1$, but the lack of such a rational number does not imply a "gap" in the rational numbers - there are no perfect squares of rationals immediately to either side of $-1$ either.

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The set of irrational numbers is dense in $\mathbb R$. In other words, there are a lot irrational numbers in $\mathbb R$.

ADDEd: Suppose there is smallest element $x$ such that $p^2>2$. As $\mathbb R\setminus \mathbb Q$ is dense in $\mathbb R$, then the open set $(2^{\frac 1 2}, x)$ will intersects the set $\mathbb R\setminus \mathbb Q$, then there is a irrational (a rational, as $\mathbb Q$ is also dense in $\mathbb R$), which is bigger $2^{\frac 1 2}$ and smaller $x$, a contradiction!

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What are you answering here? –  Pedro Tamaroff Apr 13 '13 at 2:07
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$\mathbb{Q}$ is also dense in $\mathbb{R}$. –  Javier Badia Apr 13 '13 at 2:09
    
What is the implication of the second ? –  Paul Apr 13 '13 at 2:09
    
I know what "dense" means. I am asking what you're aiming to convey by enunciating such thing. How does it answer the OP's questions? –  Pedro Tamaroff Apr 13 '13 at 2:09
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Then by that logic, there are also a lot of rational numbers. I'm saying that yoru statement doesn't contain a lot of information particular to the irrationals. –  Javier Badia Apr 13 '13 at 2:11

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