3

Say you constructed a number by counting out the positive odd numbers, in sequence:

$\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55...\}$

$135791113151719212325272931333537394143454749515355575961636567697173757779...$

This number would clearly be infinitely long, as there are an infinite number of odd numbers to choose from. Would this number be considered a natural number?

Follow-up:

If this number is not natural, what specifically disqualifies it? It certainly appears on the number line, just infinitely far from the origin, and it is a whole number. Moreover, which of the basic sets is this number a member of, and why?

  • No. What would it mean? – Henno Brandsma Feb 19 '16 at 16:34
  • If you consider analysis non-standard this number maybe some kind of hypernatural but it is not a natural number. –  Feb 19 '16 at 16:41
  • 5
    No, it doesn't "appear on the number line." The number line is infinite, but two points on the number line are always a finite distance apart. And $\pi$ appears on the number line, but it isn't an integer. (Though the above isn't even a real number, it is just a string of digits with no inherent meaning until we give it meaning. There is no useful meaning for it.) – Thomas Andrews Feb 19 '16 at 16:44
  • @ThomasAndrews If you were to stand on a physical representation of the number line (any finiteness of the physical universe aside) and look out in the positive direction, you would be looking in the direction of this number. The number may be infinitely far away, but it is there, and you're looking at it. – TheEnvironmentalist Feb 19 '16 at 16:46
  • 1
    First of all,, you are acting as if this thing "exists" on the number line, but that is an intuition based not in fact nor in definition. Notation is given meaning by definitions, and you haven't even defined what an infinite sequence of digits means. You just assume it means something. It doesn't. – Thomas Andrews Feb 19 '16 at 16:48

4 Answers4

12

No, it is not an integer. Nor a real number, nor anything meaningful, until you give it a meaning.

Answer this: If that is an integer $n$, what is $n+1$? Or what is $10\times n$? Looks like $10\times n=n$.

On any line, we have the concept of "between-ness." If $x,y,z$ are three distinct points on a line, then one of the points is between the other two.

Now, if your thing represents a number, let $x=0,y=123\dots,z=135\dots$.

You might first say, "Obviously, $y$ is between $x$ and $z$." But wait, is $z_1=0135\dots$ the same as $z$? If so, then it looks like $z$ should be between $0$ and $y$! So now, you are stuck trying to give meaning to the above expression, where $z_1=0135\dots$ is a different thing that $z_2=135\dots$.

Again, notation is given meaning by definitions. Just because we can write something down and it sort-of looks like something we've seen before doesn't mean it has meaning.


The natural numbers (positive integers) have a lot of representations. Base $10$ is a notation system for representing natural numbers, and we use it, and other notations (like Roman numerals, or base $2$, etc.) because the most primitive way to write a number $n$ is as a repetition of $n$ symbols, which is impractical. So we might write $10$ as $IIIIIIIIII$. That gets horrible fast, and it is very hard to write out positive integers in this way, so we find other notation to let us represent this, for clarity and practicality reasons.

Roman numerals are instructive for this reason, because $I,II,III$ are $1,2,3$. But then, somebody realized writing all those $I$s got tedious, and started "shortening" larger numbers in a predictable way.

But every positive integer is necessarily representable by a sequence of one symbol. It is unclear, even if you allowed infinite sequences of of that symbol, how you would distinguish $1357\dots$ from $1111\dots$ as different infinite sequences of dots.


Notation does not exist in a vacuum. Notation has meaning because we define it to have meaning. Escher can draw pictures of things that cannot exist, and you can write symbols that sort-of look somewhat like normal notation but aren't, but which has no useful meaning.


There does exist a type of number that has infinite digits, called $g$-adic numbers, were $g$ is a base. But these numbers start at the right. For example, there is a $10$-adic number:

$$...131197531$$

The $g$-adic numbers are not ordered, so we can't put them on a line. When $g$ is divisible by two or more distinct primes, they even have zero divisors - there are $a,b\neq 0$ with $ab=0$.

But $g$-adic numbers, while containing all the integers, also contain non-integers, and have very foreign behaviors.

When $g$ is prime, though, they become very useful and interesting for solving certain types of problems.

Thomas Andrews
  • 186,215
4

No it is not an integer. If you accepted all the infinite sequences of natural numbers as "natural numbers", then $\Bbb N$ would be uncountable (by Cantor's diagonal argument, for instance), which is impossible.

Watson
  • 24,404
  • I don't think so. Any natural number can be put into a mapping with itself. You wouldn't be able to construct a number not on the list, like with irrational and real numbers. – Saikat Feb 20 '16 at 01:20
  • 1
    @user230452 : Even the set of sequences containing only $0$'s and $1$'s (starting with $1$, say) is uncountable. If you had an enumeration: $$x_1 = 110101... \ x_2 = 100110... \ x_3 = 101101... $$ then define the sequence $y$ by $y_1=1$, $y_i = 1 - (x_{i})_i$ for $i≥2$. You won't be able to find $y$ in the $x_n$'s, contradiction. – Watson Feb 20 '16 at 10:11
3

No, because it's not a well-defined natural number. All members of $\mathbb{N}$ have to be finite; a construction like this doesn't make sense in $\mathbb{N}$.

DylanSp
  • 1,737
-2

"This is the follow-up:

"If this number is not natural, what specifically disqualifies it?

It's infinite. All real numbers are finite in value.

"It certainly appears on the number line, just infinitely far from the origin,

Which means it doesn't appear on the number line. The number line is the collection of all points that are zero or finite distance from the origin. This number is not so it doesn't appear on then number line.

"and it is a whole number."

No, it isn't. Now you are just making stuff up.

"Moreover, which of the basic sets is this number a member of, and why?"

Well, it's a member of the set of divergent series. X = $\sum_{k = 0}^{\infty} b_k*10^k$ which diverges...

blargen
  • 311
fleablood
  • 130,341