Doubt on rational and real numbers 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.
 A: 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.
A: 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 $\mathbb{Q}$.
A: 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.
A: 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.
A: 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.
A: 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 $\{n,m\in \Bbb N:\, n+2m>1\}$ has the least element $(1,1)$.
