I am just curious if 0 is a real number. The definition of a real number is all rational and irrational numbers. And the def definition for rational number is that &quot;$\mathbb{Q}={a\div b|a,b\in\mathbb{Z}}$&quot;. But in done say that $0$ is a whole number and some day that it is not. In some websites some say that $0$ is a real number, and in some not. Please help me. Thanks in advance
120430
There are several things that I don't understand. I know that zero is a real number, but I'm confused on the how and why aspects. What defines a real number, compared to a number that's considered non-real? For what reason is zero real, and how is it that a value such as infinity isn't when zero is?  Also, what purpose does zero even have? It's obvious that it's used as a place holder, but what value does it have alone? What makes it any different from other numbers? I realize that it is typically interchanged with nothing, but that would technically mean that zero is nothing—and that's not true. People have told me already so many times that zero does not necessarily mean nothing, and I figured it out a long while ago anyway. So, what I want to know is the actual value that can be associated with zero.  However, my question brings up another topic that has been argued and disputed about and disagreed on for years. Zero divided by zero, or 0/0. There are a lot of varying answers about this, but what's been constant throughout everything is that people claim 0/0 is not a real number. So, if I may ask, why is it that zero is considered a real number, and not 0/0? And why is it considered non-real in the first place? True, there is no defined value, but the expression is anything but meaningless; if it was, then there wouldn't be so many people trying to understand it. I guess what I'm trying to figure out is the reason behind the expression's classification as non-real—especially when the sole number of zero is real, despite not having an evident value.  This covers a few different things, so I was uncertain about what to make the title of these questions. I apologize for that. If anyone can answer, though, I'd be extremely grateful!
109545
The sum of a finite number of rational numbers is of course a rational number, but the sum of an infinite number of rational numbers might be an irrational number.  Can someone give me some intuition why this sum might be irrational?  I just "don't feel it."
10637
A well-known mathematical fact is that the rational numbers are countable, i.e. there is a bijective function  $$f:\mathbb{N}\rightarrow \mathbb{Q}$$  I am interesting in making a list of all explicit such bijections since each one that I know have a different philosophy behind it.   This is one of the more counterintuitive facts about infinite, at least when one enters inside set theory. When I explain this in the first time to anyone, he/she is surprised. Thus I think  it will be useful for showing the fact in a more clear manner or as possible exercises to show alternatives ways with respect to the standard one.
61757
I'm being asked to prove that the set of irrational number is dense in the real numbers.  While I do understand the general idea of the proof:   Given an interval $(x,y)$, choose a positive rational number (say) $z=\sqrt{2}$.  By density of rationals, there exists a rational number $p$ in the interval $(x/z, y/z)$, which essentially means that $$\frac{x}{\sqrt{2}} &lt; p &lt; \frac{y}{\sqrt{2}}.$$  I find that $pz$ is irrational, since it is the product of a rational and irrational number. However, my instructions besides writing that proof down is to specifically verify that $y=xz$ is irrational.  What does this have to do with anything and how does it prove denseness of irrationals?  Regardless, assume that $x$ is a nonzero rational number and that $z$ is irrational. For the sake of contradiction, assume that $y=xz$ is rational. This should mean that $y/x$ rational as well, and therefore that $z$ is rational, a contradiction.
74671
Show that   $\log_7 n$ is either an integer or an irrational number where n is a positive number.   I assumed that it is rational and tried to get a contradiction for $\log_7 n = a/b$, where b does not divide a, but how can I show that $7^{a/b}$ is not an integer to achieve a contradiction since n is an integer ? If I can exclude rational numbers from the range of log function then it is either integer or irrational.  Or do you suggest other methods ?
48880
In my understanding of Cantor's diagonal argument, we start by representing each of a set of real numbers as an infinite bit string.  My question is: why can't we begin by representing each natural number as an infinite bit string? So that 0 = 00000000000..., 9 = 1001000000..., 255 = 111111110000000...., and so on.  If we could, then the diagonal argument would imply that there is a natural number not in the natural numbers, which is a contradiction.
123408
The proof that the set of real numbers is uncountably infinite is often concluded with a contradiction. In the following argument I use a similar proof by contradiction to show that the set of rational numbers can be shown to be uncontably infinite:  Let us assume that the set of rational numbers is countable. This implies that we can write all the rational numbers between 0 and 1 in a list. Let us say, it looks something like this:     $ 0.\color{red}0000000001000000...\\ 0.0\color{red}10000001110000...\\ 0.10\color{red}000001000000... \\ ....$   ans so on. Now, create a new number that picks $n^{th}$ digit from $n^{th}$ number and  changes it to $0$ if it is $1$ and vice versa.     $0.101.... $   This number, say $x$, is between 0 and 1, and so it must appear in the list. Suppose it appears at the $n^{th}$ position in the list. However, based on how we constructed $x$, we know that the $n^{th}$ digit of $x$ is different from the $n^{th}$ digit of the $n^{th}$ number in the list. So, $x$ should be different from the $n^{th}$ number in the list at the $n^{th}$ digit and hence, this is a contradiction. We can conclude that such a list of the rational numbers does not exist and hence, the set of rational numbers is uncontably infinite.  I know that there is another mapping between the set of rational numbers and natural numbers that can show that there is one-to-one mapping between the two sets (). However, the above proof in that sense is not really conclusive and that the set of real numbers might also be countably infinite and it is just that we are not yet aware of the mapping.  Please provide your comments if the above proof is somehow flawed for rational numbers or if I'm missing something.
26398
claims that every repeating decimal represents a rational number.  According to the following definition, how can we prove that fact?  Definition:  A number is rational if it can be written as $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0$.
4721
So I answered this question:  I believe my answer is correct, however one thing I couldn't explain fully, and which is bugging me, is why the rationals $\mathbb Q$, integers $\mathbb Z$ and naturals $\mathbb N$ all have cardinality $| \mathbb Q |=| \mathbb Z | = |\mathbb N| = \aleph_0$, when $\mathbb N \subsetneq \mathbb Z \subsetneq \mathbb Q$.  The basic proof of the other question, from my answer, is that for any $r$ in $\mathbb Q$ there is a single function $f(r)$ such that $f(r) \in \mathbb Z$, and the same holds between $\mathbb Z$ and $\mathbb N$. Because there is this 1:1 transformation possible, there must be the same number of numbers in the three sets, because otherwise there would be a number in one set for which the bijection could not produce a number of the other set, and this is not so.  Now, Belgi explained why $1&lt;2$ in the other question by defining the value 0 as the cardinality of the empty set $\emptyset$, 1 as the cardinality of a set of sets containing only the empty set, and 2 as the set of sets containing the empty set and a set containing the empty set, then proceeding as follows:     Now that we have defined the natural numbers we can define when one number   is smaller than another. The definition is $x&lt;y$ is $x\subset y$   and $x\neq y$ .      Clearly $\{\emptyset\}\neq\{\emptyset,\{\emptyset\}\}$ and $\{\emptyset\}\subset\neq\{\emptyset,\{\emptyset\}\}$   So this a proof, by definition, of why $1&lt;2$.   ... however, by the same definition, because $\mathbb N \subsetneq \mathbb Z \subsetneq \mathbb Q$, then $| \mathbb N | &lt; | \mathbb Z | &lt; |\mathbb Q|$ and thus a maximum of one of these quantities can be the ordinal quantity $\aleph_0$.
1939
Loosely, a number is describable if it can be unambiguously defined by a finite string over a finite alphabet. Numbers such as $\frac{1}{3}$, $\sum_{n=0}^\infty \frac{1}{n!}$, and "the ratio of circumference to diameter of a circle" are all describable numbers. One can  show that the set of all such numbers is countable.  Let $U$ be the set of indescribable real numbers. I "claim" $U$ is empty, and so every real number is describable. Suppose to the contrary. Place a well-order on $U$ (this can be done assuming the Axiom of Choice). The least element $u$ of $U$ admits the description "the least element of $U$ according to the specified well-order", contradicting the indescribability of $u$.  Obviously, something has gone awry here. I suspect it is some combination of my loose definition of describability and my self-referential "description" of $u$. My (slightly open-ended) question is "What's wrong here?".  P.S. There is a  addressing something rather like my question. As I understand it, the author's response is "the given description isn't really a description". If so, I'd enjoy some elaboration.
101469
Suppose that $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous on $\mathbb{R}$ and that $f(r)=0$ for all $r \in \mathbb{Q}$. Prove that $f(x)=0$ for all $x \in \mathbb{R}$.   My attempt: Define a sequence $(x_n)$ where $x_n \in \mathbb{Q}$ for all $n \in \mathbb{N}$ and assume that $(x_n) \rightarrow a \not\in \mathbb{Q}$. Since $f$ is continuous, we have $\lim_n{f(x_n)}=f(a)=0$. Since $a$ is arbitrary irrational number, we have $f(a)=0$ for all $a \not\in \mathbb{Q}$. Hence, we proved the statement.  Is my proof valid? or is there any flaw ?
51931
Consider $\mathbb{Q}$ which is countable, we may enumerate $\mathbb{Q}=\{q_1, q_2, \dots\}$. For each rational number $q_k$, cover it by an open interval $I_k$ centered at $q_k$ with radius $\epsilon/2^k$.  The total length of the intervals is a geometric progression that sums up to $\epsilon$.  Each real number is arbitrarily close to a rational number since $\mathbb{Q}$ is dense in $\mathbb{R}$. Thus, each real number is in one of the open intervals.  Thus the entire real line is covered by the union of the $I_k$, thus $\mathbb{R}$ is a null set with measure zero.  Clearly there is something wrong in the above proof, however I am not sure where is it?  Thanks for any help.
101854
It is a common saying that there are more real numbers than integers, since you can not uniquely map integers to real numbers - even after using up all integers, there will still be at least one real number left that you haven't mapped to any integer.  So I applied the same argument to the two intervals in question. One obvious way to map number from $[0,1]$ to $[0,2]$ is $\times 2$ - $0$ is still $0$, $0.25$ becomes $0.5$, and $1$ becomes $2$. Good enough.  However, if we change the second interval to an open interval, namely $(0,2)$, then the number $0$ and $1$ in the original mapping is now the "remaining" number, while all numbers in $(0,2)$ have already been "used". Does that mean that $[0,1]$ have more real numbers than $(0,2)$? (even though $(0,2)$ seems to be bigger than $[0,1]$&hellip;)  Does my argument even make sense? I did some Googling but didn't found anything useful.
23130
Prove that Real Numbers and the interval $(0,\infty)$ have the same cardinality.   Attempt:  Consider the function $f(x) = e^x$.   The domain of this function is all real numbers.  The range of this function is from $0$ to infinity.   Let $e^a = e^b$.   Then $\ln(e^a) = \ln(e^b)$  Then $a\ln(e) = b\ln(e)$  This means that $a = b$  Hence, $f$ is injective.   Let $c &gt; 0$   Then $e^{ln(c)} = c$  Since $c &gt; 0, \ln(c)$ is defined, so $f(\ln(c)) = c$  Therefore, f is surjective.   Then f is bijective.   Hence, $\mathbb{R}$ and $(0, \infty)$ have the same cardinality.
111341
how to show that between any two distinct real numbers there are infinitely many rational numbers?  do we need to use the fact that a real number is an equivalent class of cauchy sequence of rational numbers?
24591
Claim. Let $a$ and $b$ be real numbers. If $a\geq b$  then $(a,b), [a,b)$ and $(a,b]$ are empty set.   My is question is that why?
5808
$$\Bbb{Q} = \left\{\frac ab \mid \text{$a$ and $b$ are integers and $b \ne 0$} \right\}$$ In other words, a rational number is a number that can be written as one integer over another. For an integer, the denominator is $1$ in that case. For example, $5$ can be written as $\dfrac 51$.  Is $5$ a rational number? Or is $\dfrac 51$  a rational number? I'm not able to figure out what the definition is actually saying. What are the numbers that cannot be written as one integer over another?  Irrational numbers are the numbers that cannot be written as one integer over another. Roots of numbers that are not perfect squares are examples of irrational numbers.  However, what is this then: $\dfrac {\sqrt 7} {2}$?