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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.

Sufyan Naeem
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Man_Of_Wisdom
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6 Answers6

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"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.

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.

Asaf Karagila
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  • Sorry if I misunderstand, I may lack of basic knowledge on the subject. So finding a bijection from $\Bbb R\times \Bbb Z$ to $\Bbb R$ do gives to the real numbers a well-order relation (induced from $(\Bbb R\times \Bbb Z,\preceq)$ )? I mean, every bijection we can find gives us a different well-order and a differend "successor"? – MphLee May 30 '15 at 19:35
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    No, this is not at all a well-ordering. It has no minimal element, and no $\Bbb Z$-interval has any minimal element either. – Asaf Karagila May 30 '15 at 19:36
  • Oh yeah...damn... I see it now... Z intervals goes to negative infinity... – MphLee May 30 '15 at 19:38
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    But even if you would use $\Bbb N$. The entire set of real numbers still has no minimum. – Asaf Karagila May 30 '15 at 19:39
  • Yea yea, I got it too late, cauz $\preceq$ first orders the pairs using the usual reals order using the first term of the pair. – MphLee May 30 '15 at 19:41
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    Alternatively, there's a really simple bijection between $\mathbb{R}$ and $[0,1) \times \mathbb{Z}$, by taking fractional and integer parts. Using lexicographic order here, we have for instance $1.4 \prec 2.4 \prec 3.4 \prec 1.5$ etc, and the next number after 2.4 is 3.4. – Nate Eldredge May 31 '15 at 02:14
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    @Nate: Very clever! – Asaf Karagila May 31 '15 at 08:08
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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 $x$.

CPM
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    Do you want to say "rational number or irrational number"? – Man_Of_Wisdom May 30 '15 at 18:13
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    The rationals and irrationals make up the real numbers. I am saying if you are talking about (1) the number after $\frac{1}{2}$ among $\mathbb{Q}$ (the rationals) or (2) the number after $\frac{1}{2}$ among $\mathbb{R}$ (the real numbers), then over either of these sets, your question has no answer. There is no immediate successor.element like there is over the integers, $\mathbb{Z}$. – CPM May 30 '15 at 18:16
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    Alright guy!... – Man_Of_Wisdom May 30 '15 at 18:23
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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

Ant
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    It is important to note that this ordering of $\mathbb Q$ is very different from the usual order, as in $\frac13 \ge_\ast \frac12$ with the usual enumeration. – AlexR May 30 '15 at 18:24
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    @AlexR Yes, I just added a note, realizing that there was the possibility of a misunderstanding (after I say that all countable sets can be ordered).. Thanks ;-) – Ant May 30 '15 at 18:27
  • So which rational number comes after 1/2? I don't think you can tell me that – Peter May 30 '15 at 18:50
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    @Peter After $\frac 12$, in the bijection proposed by Cantor, comes $\frac 13$. And after that $3$. You can find more here: http://www.homeschoolmath.net/teaching/rational-numbers-countable.php – Ant May 30 '15 at 18:54
  • Just because something is uncountable doesn't mean that you cannot order it in a way that makes "the next number" make sense. – Asaf Karagila May 30 '15 at 18:59
  • @AsafKaragila Ah, I thought so! can you expand on that comment? :-) – Ant May 30 '15 at 19:01
  • This is just the notion of well-orders (and generally, scattered orders). The question whether or not every set can be well-ordered, is of course, equivalent to the axiom of choice and even in the case of the real numbers there's no way to prove (or disprove) the existence of such linear order from the axioms of ZF. But by all means, countable well-orders which are very "long" and very complicated exist all the same, as well as uncountable well-orders. – Asaf Karagila May 30 '15 at 19:04
  • @AsafKaragila I see.. thank you for the clarification :-) – Ant May 30 '15 at 19:23
  • @Ant may be, you should have written "which comes just after $\frac{1}{2}$" and not "which comes after $\frac{1}{2}$" in your second para. :-) – Man_Of_Wisdom Jun 02 '15 at 17:32
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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.)

Felipe Gavilan
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  • Strictly speaking, wouldn't you say the (usual) order on the reals is dense? (Saying a set is dense usually means something different from what you are saying.) –  May 31 '15 at 05:14
  • I understand the density of a set as it is defined in general topology. I used the word "dense" because, as I understand, some teachers used to label the property of there being an infinite number of reals between any two distinct reals. Perhaps I should have said something like "the real line is dense because there is no gaps in it". – Felipe Gavilan May 31 '15 at 11:44
  • @FelipeGavilan The notion "has no gaps" is a bit misleading, though, because this usually corresponds to connectedness which $\Bbb R$ has and $\Bbb Q$ doesn't (even though the order on $\Bbb Q$ is dense). – Mario Carneiro Jun 01 '15 at 04:53
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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.

Sufyan Naeem
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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 $\mathbb{Q}$.

kasperd
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