Dealing with Infinity Rational/Irrational Numbers

158 Views Asked by At

I have two questions which I think both concern the same problem I am having. Is $...121212.0$ a rational number and is $....12121212....$ a rational number? The reason I was thinking it could be a number is when you take the number $x=0.9999...$, then $10x=9.999...$ . Therefore, we conclude $9x=9$ which means $x=1$. Why could or couldn't you do the same thing and divide the first number in similar fashion by defining it as $x$ and then taking $x/100$?

3

There are 3 best solutions below

1
On BEST ANSWER

You write:

I thought the real numbers were defined as the numbers on the number line.

This isn't really a definition of the real numbers, since "number line" is a bit vague, but: a key fact about the number line as generally understood is that the distance between any two points is finite. If I imagine a number with digits stretching infinitely far to the left, such a number is infinitely large, that is, infinitely far away from zero; and these don't have a place on the number line as generally understood.

This is not to say that we can't give a mathematically precise meaning to such objects! Indeed, one particular formalization of them - the $p$-adic numbers - plays an important role in number theory and algebraic geometry (and they allow manipulations such as that in your last comment). However, it's important to note that these are not, in fact, real numbers.

Put another way, while you can manipulate these expressions in an interesting way (e.g. conclude that $...99999=-1$), that does not in any way mean that they correspond to something in the particular number system "the real numbers"; rather, it merely suggests that they may be interesting objects in their own right. There are lots of very interesting objects (besides the $p$-adics mentioned above) that we can make sense of, which are not real numbers:

0
On

The "numbers" you have written are not real numbers at all, so they are not rational or irrational. The decimal expansion of a real number cannot continue infinitely to the left. Why? Well, intuitively, such a number would be "infinitely large" and there are no infinitely large real numbers. More precisely, an expression like $...121212.0$ would denote the sum of the series $$\sum_{n=0}^{\infty}a_n\cdot 10^n$$ where $a_n=2$ if $n$ is even and $a_n=1$ if $n$ is odd. But this series diverges: the partial sums get larger and larger without bound, since you keep adding larger powers of $10$. So there is no real number that is the sum of the series.

0
On

$0.a_0a_1a_2..... = \sum\limits_{k=0}^{\infty} a_k*10^{-k} = \lim\limits_{n\rightarrow \infty }\sum\limits_{k=0}^{n} a_k*10^{-k}$. This limit exists. For one thing the terms $a_k*10^{-k}$ get small and approach infinitesimal. (But more importantly, the difference between the finite sums becomes infinitesimal.)

So this is a valid number. It may or may not be rational.

$.....a_3a_2a_1.0$ if it were to mean anything would have to mean $\sum\limits_{k=0}^{\infty} a_k*10^{k} = \lim\limits_{n\rightarrow \infty}\sum\limits_{k=0}^{n} a_k*10^{k}$. This limit does NOT exist. The terms $a*10^k$ get large infinitely and these do not converge to any finite number.

So this is not a number of any sort or any meaningful concept.

When numbers get infinitely small they approach $0$ and it is possible (but it doesn't always happen) that we can add them infinitely and have the limit exist (but it is important to realize there are exceptions). (Decimals, however, can be added infinitely and converge. I won't go into details.)

When number get infinitely large they .... blow up. They do not converge to anything but increase to infinite. We can not ever add them infinitely and have the limit exist.