Okay, I am more than aware that my logic has fallen through somewhere so please show me where.  But surely irrational numbers can't exist.  Okay, so an irrational number is infinitely long and never repeats, yeah?  But surely those things can't go hand in hand. Either it can be infinitely long, or never repeat, one would stop the other.   So each time you add a new number you reduce the chances, and there's an infinite number of numbers so you have a 1/infinity chance of getting the right number and remainder/no remainder. But you also have an infinite number of possibilities because you have an infinite number of numbers. Giving you an infinity/infinity chance. Which I'm pretty sure should come out to a probability of 1, or certain.  I know Infinity is a concept, not a number so you technically can't divide. But think about it an infinite number of tries to get something. So surely irrational numbers are impossible, just we can't generate them far enough to find the end/repeat.  Thank you in advance. Also, try to remember I'm 15. I'm good with concepts of stuff and will probably understand but you may have to say what some words mean. Sorry
120519
I realize the title of my question will probably cause the raising of some eyebrows, so let me explain. Not sure whether to file this under "math" or "philosophy". This also might be able to be explained away due to some possible misconceptions I have about irrational/transcendental numbers, so please correct me where I'm wrong, since I'm quite certain I am...but this is nagging me, so here it is:  Ok, so imagine a non-repeating decimal with an infinite number of digits in it. The numbers are random, but for the sake of argument, let's say the number starts with "0.2368210236..." and continues from there.  So I look at the first decimal digit - 2 - and think to myself "in the rest of this decimal, there are an infinite number of times that '2' shows up."  So I look at the first TWO decimal digits - 23 - and again, think to myself "due to the randomness and infinity of this decimal, there is statistically a 100% certainty that the numbers "23" show up, in that order, somewhere in the rest of this decimal.  Continuing this line of thought, I reach the conclusion that if I choose ANY number of digits (variable: x) that start off this infinite decimal number, then a) there is statistically a 100% certainty that particular sequence shows up somewhere else in the decimal, and b) that it shows up an infinite number of times.  If such is the case, then what happens as x approaches infinity? Would that not mean that, given enough length in the decimal, there would eventually come a point where the decimal starts repeating itself, ad infinitum? Would that not mean that what was originally thought of as an irrational number is actually rational? Would this line of thinking not apply to ALL irrational numbers?  I guess what I'm asking is this: could it be that there ARE no irrational numbers, just rational numbers whose scope and vastness outstrip our ability to see their rationality?  Alright, mathphiles; have at me!
31363
Followed my  ,  I still don't understand why the answers that were given are right.  By simply intuition and using the continuity of cosine, we get that we need to compute  $\lim_{n\to \infty}\cos (\pi n\sqrt{1-\frac{1}{n}})$ and because of that, as I see that, there are two clear limits: 1 and -1, and therefore the limit does not exist  Can someone please explain me why am I wrong? two answers that claimed that the limit is 0 got 25 votes together, so I must be mistaken, But still the answers are not satisfying me.  By the way, Wolfarmalpha  it does not have a limit.  Thank you.
76183
Okay, I'm not much of a mathematician (I'm an 8th grader in Algebra I), but I have a question about something that's been bugging me.  I know that $0.999 \cdots$ (repeating) = $1$. So wouldn't $1 - \frac{1}{\infty} = 1$ as well? Because $\frac{1}{\infty} $ would be infinitely close to $0$, perhaps as $1^{-\infty}$?  So $1 - 1^{-\infty}$, or $\frac{1}{\infty}$ would be equivalent to $0.999 \cdots$? Or am I missing something? Is infinity something that can even be used in this sort of mathematics?
26129
Can someone point me to a proof that the set of irrational numbers is uncountable? I know how to show that the set $\mathbb{Q}$ of rational numbers is countable, but how would you show that the irrationals are uncountable?
85824
Ok, so I've been playing around with radical graphs and such lately, and I discovered that if the   nth x = √(1st x √ 2nd x ... √nth x);   Then  $$\text{the "infinith" } x = x$$  Example:   $$\sqrt{4\sqrt{4\sqrt{4\sqrt{4\ldots}}}}=4$$ Try it yourself, type calc in Google search, hit  then a number, such as $4$, and repeat, ending with $4$, (or press the buttons instead).  I'm a math-head, not big enough though, I think this sequence is divergent or convergent or whatever, too lazy to search up the difference.  However, can this be explained to me? Like how the Pythagorean Theorem can be explained visually.
77444
There exists an infinite amount of rational and irrational numbers. But is there more irrational numbers than rational? And if so can a ratio of one to the other be calculated?
73502
Suppose $i_1$ and $i_2$ are distinct irrational numbers with $i_1 &lt; i_2$.  Is it necessarily the case that there is a rational number $r$ in the interval $[i_1, i_2]$?  How would you construct such a rational number?  [I posted this only so that the useful answers at  could be merged here before that question was deleted.]
47345
Just a curiosity. I just saw a video where pi was being calculated by the series  $$1-(1/3)+(1/5)-(1/7)+...$$  My question is how can $\pi$ be irrational if we can calculate it using rational numbers?
94366
Well, just another idea came up into my mind and i have no idea how to solve it :D Is there infinitely many prime numbers, which are not repunits and their inverse is also prime? (For example, inverse of 31 is 13 which is also prime. i didn't have any other name to describe the function!) P.S:I now know they are called s.
59768
I have often heard it said (by professional cosmologists) that if the universe is infinite, then there necessarily exist infinitely many copies of me repeated throughout.  The reasoning seems to be that any finite volume of space can contain at most finitely many fundamental particles existing in finitely many configurations.  From here it is implied that all possible (permissible) configurations must both exist and occur infinitely often.  This does not sound like a sound argument to me, so I have to believe that I have misunderstood the argument.    Is this unsettling implication true in an infinite universe, and if so how is it justified.
69633
Let's say you tell me to produce randomly a number from 1-100, and I choose the number 47. Can it be said that there is a specific reason I chose the number 47, and that it is not completely random? By random I mean that there is absolutely no reason this number in particular was chosen.    It seems odd to me, since it seems like since I have applied conscious effort to produce or choose a number within that range, or even without range, that it can't be truly random, but there might be a reason why I chose that number amongst the many others I could have chosen.  I guess the general question, which I'm interested to know if any philosophers in the past have discussed, is whether our thoughts are necessarily based on prior thoughts, or if thoughts can be truly spontaneous and random?
90289
This question is me trying to understand (again) why there can be no one-to-one correspondence between the sets of natural and real numbers.  The source of confusion is this: if we abstract completely from the properties and operations on numbers and just think of them as strings of digits, it appears to be possible to list all the real numbers between $0$ and $1$ and all positive integers having the same digits in the same (or reverse) order, for example:  $$0.1 \to 1$$  $$0.01 \to 10$$  $$0.0101 \to 1010$$  $$0.110101 \to 101011$$  I use binary expansion for simplicity.  Since irrational numbers have infinite expansions, we would have to consider some kind of 'infinite natural numbers', like:  $$101110001\dots$$  or  $$\dots101110001$$  I think the main (and only) reason why such numbers don't exist is this requirement:   We need the sets of real and natural numbers to be ordered. We need to be able to do arithmetics on real and natural numbers.   Obviously, if we consider any of two types of 'infinite natural numbers' and the usual absolute value, we find that it's impossible to order them, or do arithmetics with them.  The only way is to change the norm (to $p-$adic norm for example).  While with the real numbers it's still possible to order them, and do arithmetics with them when they have infinite amount of digits in their expansion (or at least in principle).     Is my reasoning correct? Are the two requirements (ordering and arithmetics) enough to explain why natural numbers with infinite amount of digits do not exist, while real numbers do?   This confusion is not restricted to me alone, see  for example
11305
How many integers from 1-999 do not have any repeated digits?  The answer is explained in this , but why is the last set 9*9*8? Why not 9*9*9?
20567
There are infinitely many real numbers between any two real numbers, therefore there are infinitely many real numbers in the range $[0,1]$ as there are in $[1, \infty)$. In a mathematical sense, are there more numbers in one range as opposed to the other?  What about $[0,1]$ and $[0, \infty)$ ?  I'd assume this is akin to how limits work in that the limit as $x\to\infty$ of $\frac{x^x}{\log(x)}$ is $\infty$, even though both the numerator and denominator approach infinity.
10186
I am trying to figure out how to prove that all rational numbers are either terminating decimal or repeating decimal numerals, but I am having a great difficulty in doing so. Any help will be greatly appreciated.
40663
I realize this is possibly a copy of another question (The Axiom of Choice and definability) but I would appreciate an explanation with less set theoretical explanations if that is at all possible.  I recently took the opportunity to look up the axiom of choice (AC). Wikipedia has both a formal and an informal definition.  Based on my shortcomings in set theory I have to rely on the informal version: “given any collection of bins, each containing at least one object, it is possible to make a selection of exactly one object from each bin”.  It seems that this can be used to define an endless series of decimals such as an arbitrary number x between 0 and 1, in term corresponding to an arbitrary positive real number through such a function as f(x)=1/x-1, i.e. it can be used to define any real number (assuming a real number is an infinite decimal series).  A beauty spot is the cases of 0.99999… and 1.00000… - discussed in a number of questions on the site. It points to several issues:  I argue, albeit in a roundabout way, that for a choice involving the AC to have any value then the number received through the choice must also allow for existing mathematical operations, i.e. 2*0.11111…=0.22222…  as discussed in another question. (“You're also assuming that you can multiply 0.9999... by 10 and get 9.9999... (or, rather, that arithmetic with infinite decimals works normally), which is not at all unreasonable to assume. – Isaac Jul 20 '10 at 20:59”) i)  0.99999…must be equal to 1.00000… for the same reason. ii) The fact that 0.99999…=1.00000…can challenge the one-to-one correlation between the real numbers, as we view as being without these doublets. However it can be disregarded as   I was assured in a recent question, in that removing a countable set such as 0,99999…, between the endless decimal series and 0.099999…and 0.0099999…  etc. from a uncountable set would not change the one-to-one correlation to numbers without this anomaly. (I was referred to i.a. the Cantor–Bernstein–Schroeder theorem).  Reasoning in this manner I perceive the AC to be identical of postulating - or at least allowing for the definition of – real numbers. My question is whether or not this is the case.  Edit much later: I should add that I primarily am looking at the problems of defing invividual real numbers, so called pointwise definition, rather than definition of the whole set of reals. I didn't realize at the time that these are two different issues.
37277
Every infinite continued fraction is irrational.  But can every number, in particular those that are not the root of a polynomial with rational coefficients, be expressed as a continued fraction?