2

We can't say what is the just next real number (or rational or irrational number) of a given real number (or rational or irrational number respectively), what is the actual fundamental reason behind it? Is it for the property of denseness of the sets? Please help me to sure about the answer of this question.

Martin Sleziak
  • 53,687
Satyajit
  • 31
  • 2
    “Give me answer with required proofs.” is not a polite way to ask for help. – k.stm Mar 02 '14 at 10:06
  • 2
    @k.stm It's obvious to me that the OP's lack of proficiency in english might hide any politeness he might intend. – Git Gud Mar 02 '14 at 10:10
  • 1
    @k.stm But yet the o.p. ends with "Please help me..." so I think the o.p. is being perfectly polite. Perhaps that sentence is just a poor translation, no? (I'm guessing English it not op's native language). To be more on topic: why do you think there should be a "next" real number? I think if you wanted to tell me there was one, you would need to prove it to me. Think about this: between any two real numbers, there is always another. – Ryan Sullivant Mar 02 '14 at 10:12
  • It wasn’t meant as a harsh critique. Just sayin’. – k.stm Mar 02 '14 at 10:14
  • 2
    @k.stm Well I just don't think that's any way to greet a new user. – Ryan Sullivant Mar 02 '14 at 10:17

4 Answers4

3

This is a property of the order of $\Bbb R$. We say that an order is dense if whenever $x<y$ there is some $z$ such that $x<z<y$. If $x$ is a real number, then if $y$ was "the next real number", we had some $z$ such that $x<z<y$, and so $z$ would be the next real number instead, but then we can do that again, ad infinitum.

However, one of the common assumptions in modern mathematics is "the axiom of choice", from which we can prove that there exists a well-order of any set, including $\Bbb R$. The fundamental property of a well-order is that every non-empty set has a minimum. It follows that every element which is not maximal, has an immediate successor.

So if we endow the set of real numbers with a well-ordering then there is a notion of "the next real number".

Some caveats:

  1. We cannot write an explicit formula which is guaranteed to produce a well-ordering of $\Bbb R$. We can prove its existence from the aforementioned axiom of choice. And so, since there is no natural well-ordering for $\Bbb R$ there is no deep sense to saying what is the next real number after $0$ (whereas in the natural numbers there is such sense for saying that $1$ is the successor of $0$).

    Using different well-orders would give us different notions for "the next real number".

  2. Any such well-ordering of $\Bbb R$ is guaranteed to be almost entirely incompatible with the natural order of $\Bbb R$. This is important to note, because we often expect the structure we give $\Bbb R$ would somehow respect any "naturally occurring structure" that $\Bbb R$ already carries, but this is very far from the truth.

  3. The above point stands in analogy to $\Bbb Q$ which is a countable set, and therefore can be written as $q_0,q_1,q_2,\ldots$, and that is in fact a well-ordering of $\Bbb Q$. Note that this indexing is very incompatible with the usual order of $\Bbb Q$. Although in the case of $\Bbb Q$ there are explicit enumerations of the set. So the analogy doesn't stretch beyond the second point.

Community
  • 1
Asaf Karagila
  • 393,674
1

You have to be precise with what does 'next' mean. It assumed some notion of order on the reals. If that notion of order is taken to be the usual one (whatever that means), then the fact that there is no 'next real' is a property of that particular order. In more detail, it is possible to construct the real numbers (in various essentially equivalent ways) from the rationals. Then one can define the ordering on the real numbers as a derived notion from the ordering of the rationals. Then one can prove that between any two real numbers there is a third, distinct one. This density property immediately implies that there is no 'next number' in the sense that if $x$ is a real number, then the set $\{y\in \mathbb R \mid x<y\}$ does not have a least element.

However, if one believes in the axiom of choice, then there exists an ordering on the reals which is a well-ordering, meaning that every real has a uniquely 'next real'. Explicitly exhibiting such an ordering is not possible though.

Ittay Weiss
  • 79,840
  • 7
  • 141
  • 236
0

The reason is if you say this so ans so number is the next number then I can immediately give you another in between both of them. Say x and y are rationals then (x+y)/2 is the real in between them. Suppose instead you have irrationals then see the digit where they differ and take rational approximation. And similarly do other cases

Martin Sleziak
  • 53,687
happymath
  • 6,148
0

Denseness is rather a relation between two sets than a propriety of a single one. That is, it has no meaning saying that "a set is dense". That between any two rational numbers there is another one distinct from both is independent from being $\mathbb Q$ a subset of $\mathbb R$. And it follows from the definition of $\mathbb Q$ and the proprieties of an ordering relation.

alex
  • 795
  • Please see my comment in user129901's answer. – Git Gud Mar 02 '14 at 10:18
  • I guess you are referring to the well ordering principle, which we must accept if we accept the axiom of choice. I agree, but I also think this goes a bit further than the kind of simpler answer the op was expecting. I suppose he has in mind the usual order on $\mathbb Q$. – alex Mar 02 '14 at 10:25
  • I add that in the "definition of $\mathbb Q$" I include also the order we choose. – alex Mar 02 '14 at 10:26
  • So, my doubt was: is it actually impossible to construct a well ordering on $\mathbb R$, though we know one must exist (that is, if we accept the axiom of choice, or Zorn's Lemma)? And, more to the point, what does this impossibility really mean? It is clear what it means that the roots of a 5th degree algebraic equation can't in general be found through algebraic expressions, though they certainly exist. But this impossibility about constructing a well ordering is more obscure, to me, and so far. – alex Mar 02 '14 at 10:44
  • It is impossible, yes. Why I do not know. I've never seen it proved. I'm afraid I can't answer your questions. Try asking Asaf. – Git Gud Mar 02 '14 at 10:53
  • alex, have you taken a course in Galois theory? What about set theory, now what about a second course in axiomatic set theory and a third course about forcing? Fields are much more accessible, so the first course on set theory is like the first course where you are introduced to fields (often linear algebra); the second course is the course about algebraic structures (usually groups and rings) which prepares the ground for understanding; and the third course is the course in Galois theory. If you tell this to a freshmen, they won't see why 5th degree equations don't have expressible radicals. – Asaf Karagila Mar 02 '14 at 14:12
  • [...] And similarly, people who didn't walk these three courses and many hours of thinking with set theory are not likely to immediately see why you cannot construct a well-ordering of $\Bbb R$ in the language of set theory (and I'm skipping the part where constructing, defining a formula, etc. etc. are all ill-defined, because that would require an additional course about logic and basic model theory). – Asaf Karagila Mar 02 '14 at 14:13
  • @Asaf Karagila I'm not sure why you are telling me these things. You write about seeing "why 5th degree equations don't have expressible radicals". Well, I once saw Galois Theory, but I can't say I remember it clearly. But what I actually said is that it is clear what it means that the roots are not expressible, even if I don't remember why in details any more. I might be wrong, but I suppose the average student can understand what it means, without knowing why, before even hearing about Galois Theory. – alex Mar 02 '14 at 14:30
  • @Asaf Karagila On the other hand, I wrote that it is not the same clear to me what it means that a well-ordering can't be constructed on $\mathbb R$, not why. I suppose it means it is not constructible within ZFS theory though. – alex Mar 02 '14 at 14:33
  • I am saying that because you wrote above that you don't see why it is impossible to construct a well order of the real numbers. To explain the phrase it's easy. There is no formula in the language of set theory which is provably a relation whose domain is all the real numbers and it defines a well-ordering. But do you see why that is the case? Now, polynomials and radicals are being taught in high school. If you would teach mathematical logic in high school, a freshman would understand what it means that it is impossible to define, or construct, a well-ordering of the real numbers. – Asaf Karagila Mar 02 '14 at 14:35
  • @Asaf Karagila After all, to understand the statement that $x^n+y^n=z^n$ with $n\in\mathbb N$ has no integer roots for $n>2$ is quite simple, but how many in the world do actually know why? – alex Mar 02 '14 at 14:36
  • @Asaf Karagila I'm sorry, I never wrote that, though it is of course very true. I'm not sure it is so easy to explain the phrase, 'cause one has to know what "the language of (axiomatic) set theory" is, and that's not that simple. That's what I meant. Now, about teaching mathematical logic in high school, here I couldn't agree more with you! – alex Mar 02 '14 at 14:43
  • Sorry, I must have misread. :-) – Asaf Karagila Mar 02 '14 at 14:44
  • No problems at all.:-) – alex Mar 02 '14 at 14:46
  • "Dense" is also a word applied to individual (partially) ordered sets: a densely ordered (po)set satisfies the property that for any $x<y$ there is a $z$ such that $x<z<y$. –  Mar 18 '14 at 07:51