Is there a real number which comes just after a particular real number?

Question

This is like when we say that the integer which comes just after $2$ is $3$.

Is there a real number which comes just after a particular real number?

For example: Is there a number which comes just after $0.5$?

If any one didn't get me, read Asaf's answer because that explains my question better.

Answer 1

If you are talking about a rational number or a real number after $\frac{1}{2}$, it is not possible. For suppose $x$ were such a number. Then $\frac{\frac{1}{2}+x}{2}$ is a number after $\frac{1}{2}$, but closer than $x$.

Answer 2

"Just after 0.5", well, in what context?

In the natural numbers the number that comes "just after $2$" is indeed $3$. But in the rational numbers, or the real numbers, what is the next real number after $2$? Is it $3$? Or is it $2.5$? Or is it $2.25$? Or so on and so forth.

Not every ordering comes with a well-defined notion of "just after", like the integers and the natural numbers have. And as luck would have it, the rational numbers and the real numbers are both examples for ordered sets which do not posses this property.

So as a real number, or even as a rational number, with the standard ordering, there is no "just after" any number.

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If one no longer desires to use the standard order, then it is not hard to come up with all manner of alternative orders, where the notion of "next real number" makes a lot of sense.

For example, we can prove that there is a bijection between $\Bbb R$ and $\Bbb{R\times Z}$. Then any structure we can give the set $\Bbb{R\times Z}$ can be translated to a structure on $\Bbb R$.

In particular the lexicographic order, $(r,k)\preceq (s,m)$ if and only if $r<s$ or, $r=s$ and $k\leq m$. Namely, we replace each real number with a copy of $\Bbb Z$. Now given any point on this order, $(r,k)$ it has a unique "next number" which is $(r,k+1)$.

Of course the translation from $\Bbb{R\times Z}$ to $\Bbb R$ is not in any way canonical or unique. It just exists, and so there's no way to just say what is the next real number after $0.5$ because that would greatly depend on this translation.

Answer 3

This is a great question! You stumbled upon an important problem:

"I know how to tell if a real number is bigger than another one, but apparently (even if there is an "order" between them) I cannot tell which one comes after $\frac 12$"

And you are right; there is no way to meaningfully define the "next" real number.

To understand why, let's think about natural numbers; they have a lot of properties, and among them, is the fact that you can order them. First you have $1$, then $2$, and so on.

So, which sets can be ordered? Of course any finite set can. What about infinite ones? Can $\mathbb Q$ be ordered?

The idea is that if you can find a bijection between your set $X$ and $\mathbb N$, then you can "copy" the order of $\mathbb N$ (because $1$ would correspond to an element of your set $X$, $2$ to another one and so on)

All the set that can be put in a such a relationship with $\mathbb N$ are called countable and all countable set can be ordered. (Note however that this does not ensure that the "next" element of $X$ is bigger; in fact the way you define bigger in $X$ may have nothing to do with the order in which its elements are arranged. )

For example, $\mathbb Q$ is (very un-intuitively) countable, so you can order the fractions and you can tell which (rational) number comes after $\frac 12$.

On the other hand, $\mathbb R$ is not countable, so you can't "automatically" do that. There is, as @Asaf Karagila points out in the comments, the concept of "well ordered" but I admit I know nothing about that. Refer to Asaf's comment for more info! :-)

The proof of both these fact were given by Cantor; look here, for example

Answer 4

Let $a$ and $b$ be positive reals, and $a \neq b$

To say that $b$ is the number just after $a$ it would mean that there is not any real number between $a$ and $b$. But there is! For example, we can average them: $$a < \frac{a+b}{2} < b$$

Well, maybe the number $\frac{a+b}{2}$ is the inmeadiate number after $a$, right? No. Because we can average again: $$a < \frac{a+ \frac{a+b}{2}}{2} < \frac{a+b}{2}$$

We can actually keep doing this forever. This is because the real numbers are dense, meaning that between any two distinct real numbers there is an infinite number of reals. So, there can't be an inmediate next real number to any real number (as Asaf says, in the context of the reals, of course.)

Answer 5

This is not possible because, between every two distinct points on a line, there is always a third one.

If it were possible, we couldn't say a line segment to have infinitely many points between its two end points.

Answer 6

Before we can answer whether there is a real number just after a real number $x$, we have to define what that means in mathematical terms.

I assume what you mean is the smallest $y$ such that $y > x$. As others have pointed out, such a $y$ cannot exist because $\frac{x+y}2$ will be between $x$ and $y$.

If we change the question slightly, we could ask for the smallest $y$ such that $y \geq x$. So far it doesn't sound particular interesting because it is quite obvious that then $y = x$.

But we can change the question a bit more and ask if we don't have just one $x$ but rather an infinite set $X$ of real numbers, does there exist a $y$ just after that set. In other words is there a smallest $y$ such that $\forall x \in X: y \geq x$.

It turns out that if any $y$ greater than all of $X$ exists, then there is also a smallest such $y$. This is known as the least upper bound of the set $X$. This is slightly different from what you asked for, but it is a very important property of the real numbers.

To see why this property is not true for all ordered fields consider $X = \{x \in \mathbb{Q}: x^2 < 2\}$. The least upper bound of this set is $\sqrt 2$ which exists in $\mathbb{R}$ but not in $\mathbb{Q}$.