1
$\begingroup$

Someone once told me that numbers such as $\sqrt{2}$ have a unique mathematical meaning:

$$\sqrt{2}=\lim\{1,1.4,1.41, ... \}$$

While I understand that this might be sufficient for a formal mathematical definition, it still fails to provide me with intuition. The numbers within the curly braces seem to be approaching a number that has a fixed, definite value, but the string of digits still goes on forever. Additionally, because there is no way of knowing all of the digits of $\sqrt{2}$, it seems strange that we can reason with this number so easily. Is there a more intuitive interpretation of what irrational numbers mean?

$\endgroup$
10
  • 1
    $\begingroup$ Who said there is no way of knowing all of the digits of $\sqrt2$? See here and here. $\endgroup$
    – user239203
    Apr 26, 2020 at 8:57
  • $\begingroup$ @Gae.S. Thank you for responding. Although I appreciate that there is a definite way of calculating the next digit of $\sqrt{2}$, my problem is that there is no way of working out all of the digits in the real world. This might seem trivial in mathematics, but it is problematic for me because it seems we do so much work with $\sqrt{2}$ without ever understanding 'what' it really is. $\endgroup$
    – Joe
    Apr 26, 2020 at 9:04
  • 3
    $\begingroup$ Geometric interpretation of $\sqrt2$: diagonal of unit square $\endgroup$
    – J. W. Tanner
    Apr 26, 2020 at 9:57
  • 1
    $\begingroup$ I should point out, the definition you provided is not really a definition. There's no clear pattern to the sequence of digits, especially from the first three terms. I know what you're going for here: the sequence of truncations of decimal expansions of $\sqrt{2}$, but this makes the definition impredicative: we are using $\sqrt{2}$ in its own definition! It may seem like an overly pedantic point, but I think it is part of the root cause of this lack of intuition. There's no intuitive pattern to this sequence of digits other than representing this number. $\endgroup$
    – user771918
    Apr 26, 2020 at 11:47
  • 1
    $\begingroup$ @Joe The definition of $\sqrt{2}$ is exactly what you wrote. You're right that it requires proof, but proving depends on how you're defining the real numbers. If you define $\Bbb{R}$ in terms of Dedekind cuts, we can define $\sqrt{2}$ to be the pair of sets $\{\{q \in \Bbb{Q} : q^2 < 2\}, \{q \in \Bbb{Q} : q^2 > 2\}\}$, for example. $\endgroup$
    – user771918
    Apr 26, 2020 at 15:23

1 Answer 1

Reset to default
1
$\begingroup$

The statement you've written down is a theorem, which states that given any $x \in \mathbb R \setminus \mathbb Q$, there exists a sequence of rational numbers $(a_n)_{n=1}^\infty$ such that: $\lim_{n\to \infty}=x$.

If one wants to talk about the intuition behind that statement, there's a (beautiful) observation to be made first: if $x \in \mathbb Q$, then the decimal expansion of $x$ will eventually become periodic, or terminate - if there is a "point after which" the expansion is composed entirely of zeros. This is a consequence of Dirichlet's pigeonhole principe.

Moreover, if $x \in \mathbb R \setminus \mathbb Q$, then the decimal expansion of $x$ will not terminate, and will never become periodic.

Now,an intuition for irrational numbers, based on this theorem, could go on as follows:

If $x$ is an irrational number, its decimal expansion does not terminate and never becomes periodic. However, there exists a sequence of rational numbers $(a_n)_{n=1}^\infty$ that converges to $x$ - and the decimal expansion of every individual term in that sequence is an approximation to the decimal expansion of $x$, which gets better and better the "further" you go along this sequence.

This is a soft way of saying that, as $n$ gets larger and larger, the approximation gets closer and closer to the actual decimal expansion of $x$, and the precision with which this happens can be as good as you want it to be - exact to within $1,000$ digits after the dot, or $1,000,000$, or $100,000,000$ and so on. Hope this helps.

$\endgroup$
4
  • $\begingroup$ Thank you for responding. This does seem like a nice way of providing intuition, as it demonstrates that $\sqrt{2}$ has a fixed value, and that it can be approximated using rational numbers. I have one question: is there a definition for the decimals in $\sqrt{2}$? I believe $0.\overline9$ means the limit of $0.9+0.99+0.999+...$ Is there a similar definition for non-terminating decimals such as $\sqrt{2}? (It might be the one that I provided, but user771918 said that was not sufficient.) $\endgroup$
    – Joe
    Apr 26, 2020 at 23:52
  • $\begingroup$ When we write $0. \overline 9$ we mean: $0.999999..$. Notice that $0.9+0.99+0.99 \neq 0. \overline 9$. Instead, a correct statement very much in the same spirit is that: $0. \overline 9 = 0.9+0.09+0.009+0.0009+..$. $\endgroup$
    – user764256
    Apr 27, 2020 at 8:12
  • $\begingroup$ About your definition for $\sqrt 2$ - it's not very clear in the sense of asking some deeper questions - what if I'd wanted to know the $n^{th}$ digit of that expansion, using your statement? I would intuitively take the $n^{th}$ term in your sequence - however I have no closed expression for that sequence, and this makes my mission quite hard even with the axiom of choice. Now, a very rich an active place in NT (and dynamical systems) is Diophantine approximation - which is concerned with finding the best rational approximations to real numbers - I will comment once more for lack of space. $\endgroup$
    – user764256
    Apr 27, 2020 at 8:57
  • $\begingroup$ As I was saying, some rational approximations converge faster than others - a particular sequence might give $2$, or $6$ or some other number of correct (new) digits after the dot with each term. This means that defining the decimals in $\sqrt 2$ is kind of unnecessary - after we've defined the real numbers, we've defined $\sqrt 2$ and hence it's expansion too. A very interesting, open and well known problem about the decimals of $\sqrt 2$ is to determine whether it's a normal number, which means that every digit 0-9 (in base 10) has a chance of $\frac{1}{10}$ to appear next at any point. $\endgroup$
    – user764256
    Apr 27, 2020 at 9:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .