9
$\begingroup$

I'm in ninth grade and I've been thinking about this for a while. It's related to a question that came to my mind, namely which is the highest number you can divide 11 by. Is "infinite" a number and is it the highest number by which 11 can divide it?

$\endgroup$
  • 17
    $\begingroup$ Infinity is not a number. $\endgroup$ – Sean Roberson Apr 12 at 14:38
  • 2
    $\begingroup$ @SeanRoberson What is a number for you to be saying that? $\endgroup$ – user647486 Apr 12 at 14:41
  • 2
    $\begingroup$ @user647486 Care to justify your statement? $\endgroup$ – Q.P. Apr 12 at 14:46
  • 8
    $\begingroup$ There is an arithmetic for the extended real numbers (real numbers with $\pm\infty.$) You can see its arithmetic here: en.wikipedia.org/wiki/…. $\endgroup$ – Adrian Keister Apr 12 at 14:50
  • 8
    $\begingroup$ Your title asks about dividing infinity by a number, then in your question you start out asking about dividing eleven by something, then you go back to dividing something by eleven. Dividing something by eleven and dividing eleven by something are very different, and you need to more clearly distinguish between the two. $\endgroup$ – Acccumulation Apr 12 at 16:05
13
$\begingroup$

Thinking and playing with infinity is a popular thing to do, especially for people your age. To do so correctly however, you need to understand what we actually mean by infinity and what we mean by a "number."

Note: For the remainder of this post, by "division" and "divided by" I will be referring to the usual division operator rather than the related concepts from number theory.

Infinity is not a "real number" (that is not to say it doesn't exist in some contexts, but it does not belong to the set of numbers known as 'the real numbers') so we are not even allowed to refer to operations involving infinity. It just flat out doesn't make sense in this context. There is no largest "real number" and so there is no "highest (real) number" that $11$ can be divided by.

Since you are talking about "dividing by infinity" then you are probably working in the extended real numbers rather than the real numbers. In such a context, yes infinity is the "largest" extended real number and $11$ can indeed be divided by it. In fact, every real number when treated as an extended real number instead can be divided by infinty and the result would be zero. In that context, yes... you are correct.

The extended real numbers however is not the usual context to be working in. If the question was asked "does there exist a largest number that $11$ can be divided by" without any additional clarification about context then the answer would be no.

(Note: infinity and negative infinity can not be divided by infinity)

$\endgroup$
6
$\begingroup$

This was getting too long for a comment, but it isn't really an answer - more an encouragement to hang in there with the kind of question you are asking.

The question "what is a number?" has caused many mathematicians to reflect on the basic material they work with and how it is defined. It leads to rich and fundamental questions about the foundation of mathematics.

Combined with geometric insights we can extend familiar numbers/systems to obtain projective spaces or the Riemann Sphere, both of which accommodate notions of infinity.

Algebraically it leads to reflection on what properties we expect our numbers to have, and the kinds of mathematical objects which have those properties.

And you might want to look up Cantor's diagonal argument and reflection on how infinite sets don't always behave in accord with our intuition. (An infinite set can be put in one-to-one correspondence with a proper subset of itself - we can match the integers with the even integers by pairing $n$ with $2n$)

It is possible to create systems of numbers in which the existence of infinite numbers is consistent with arithmetical operations - JH Conway's "Surreal Numbers" are an example.


To return a bit to the question:

If you are working with non-negative integers, then there is no infinite integer, and you can't divide by something which doesn't exist.

If you were working within the rational numbers (fractions), you could divide by any rational number (including any integer) except for zero. That would give you a valid fraction as an answer.

The non-negative integers and the rational numbers are systems of numbers which are carefully constructed to obey clearly stated rules - the need for careful definition arises because of the kind of question you are asking.

$\endgroup$
  • 6
    $\begingroup$ +1. Your answer has the flavor of an encouragement. If you want to encourage the asker then you may also consider "liking" the question. [I am always amazed when several people give very long answers to a question that they do not "like." So far, I am the only person who has "liked" (+1d) the question.] $\endgroup$ – Michael Apr 12 at 15:36
  • $\begingroup$ PS: In the comment trail of the question, my son wrote his comment using my account name "Michael." I encouraged him to look at the question because I thought he would appreciate it. $\endgroup$ – Michael Apr 12 at 15:56
5
$\begingroup$

The number $11$ can be divided by anything - real or complex. If you mean what is the greatest integer - $x$ - such that $$11\equiv 0\mod{x}$$ In other words, $11$ divided by $x$ leaves no remainder or fractional part, then the solution would be $x=11$.

$\endgroup$
  • 4
    $\begingroup$ small suggestion : you may want to rewrite 11 ≡ 0 mod x, OP may not understand that notation (at least I wouldn't on 9th grade) $\endgroup$ – Rod Apr 12 at 17:18
1
$\begingroup$

It all depends on the defintions of $\infty$, "number" and "divides". I will assume that you're asked what is the greatest integer number dividing 11 then basically in any sensible interpretation of words $\infty$, "integer number" and "divides" the answer to your question will be "11". As far as I know, $\infty$ is always interpreted distinct from the usual numbers (by usual I mean real numbers) (including integers), that is, you have an axiom (or a theorem)

$$\forall x \in \mathbb{Z} , x \neq \infty.$$ Hence $\infty$ could not possibly be the greatest integer number dividing 11 since it then would be an integer contradicting this axiom (or a theorem, whatever you wanna call it)

$\endgroup$
1
$\begingroup$

Julia thinks ∞ is evenly divisible by everything. Julia got ∞ by Π ℙ = ∞ (that is, the product of all prime numbers).

Anna thinks ∞ is evenly divisible by only by 1. Anna got ∞ by Π ℙ + 1 = ∞ (that is, one more than the product of all prime numbers).

∞ is not a number. Trying to treat it like one will eventually cause a headache.

(If you're having trouble getting the reference, there was this famous question I cannot find anymore that had five students giving five different answers to a "what is the next number in this sequence" question using five different fourth order polynomials.)

$\endgroup$
-1
$\begingroup$

It is really great to be thinking of stuff like this.

You will find people all around you who will say infinity is not a number and they will stop there, but there are a set of math classes in your future (Calculus) that are all about dealing with numbers that are virtually infinite--they call them limits.

All those people who say that it's absolutely impossible to deal with infinity will find it perfectly fine to deal with "x as x approaches infinity" and you get the same results you would intuit for infinity (Division by infinity is still a bit tricky if I recall correctly)

Anyway, if you have ever noticed that math is often about describing things, calculus is how you describe strange (arbitrary) curves and such--Exactly how much water fits into your strangely shaped sink, or how long will it take to drain said water since the speed it drains is slower as there is less water in the sink.

This kind of thing is calculated by slicing up curves into infinitely (Approaching infinite anyway) thin slices and adding the slices together to get an exact result.

(PS, I haven't done much calculus in decades, if this is factually wrong please feel free to fix it rather than just blindly downvoting. Thanks)

$\endgroup$
  • $\begingroup$ I don't think it's a great idea to compare limits to infinity. Limits are (usually) real numbers with fairly easily definable properties, unlike infinity. $\endgroup$ – Jam Apr 16 at 21:27
  • $\begingroup$ That's what confuses me. Why are people so reluctant to compare the two. How is a limit as it goes to infinity unable to simulate what would happen if you actually were able to act on infinity? Intuitively the results of limit calculations always seemed to match what would happen if you were able to drop infinity in it's place. There is always such a violent reaction to the concept of comparing them and it confuses the heck out of me why. Maybe I need to post a question about it :) $\endgroup$ – Bill K Apr 17 at 16:52
  • $\begingroup$ @Bill_K Fwiw, my previous comment and downvote weren't intended as a violent response, simply an opinion of the pedagogical value of your answer OP's question. Downvotes are a dime a dozen on this site so please don't feel discouraged or devalued :). Per your question, there's a few different types of 'infinity' that are often muddled together. The first (and probably easiest) type of infinity is the infinity of calculus, which represents boundless processes and sequences. For instance limn→∞ represents what happens to the sequence x1,x2,x3,…,xn as n gets larger (1/4). $\endgroup$ – Jam Apr 17 at 18:38
  • $\begingroup$ (cont.) This 'boundless sequence' infinity is the same one that is represented by $\int^\infty$, $\sum^\infty$ and in derivatives. which are a essentially special case of limits with $\lim_{n→\infty}\frac{f(x+1/n)−f(x)}{1/n}$. Clearly we can't do arithmetic on this infinity, since it's a process, not a value. Let's consider the second type of infinity, ostensibly represented by your definition $\lim_{n→\infty}x_n=L$. When we pick apart what this means with a rigorous definition of a limit, we see that it means that (2/4) $\endgroup$ – Jam Apr 17 at 18:40
  • $\begingroup$ (cont.) We can make the terms $x_n$ as close as we want to $L$, as long as we go beyond a sufficiently high $n$. This is due to the definition of a limit. Let's assume (falsely) that at some point, our $x_n$ get close to a limit $L=\infty$. Then (again by definition), we should be able to get even closer to $L=\infty$ if we continue further into the sequence. But how can we be closer to $\infty$? Is $10$ closer to $\infty$ than $100$ is? No - any number we pick will be infinitely far away from $\infty$. Hence we've contradicted ourselves, so we can't get "closer" to $\infty$ (3/4) $\endgroup$ – Jam Apr 17 at 18:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.