New answers tagged

2

Every real number can be approximated arbitrarily closely by rational numbers; this is the basis for decimal notation. Now for an open set contained in the complement of $\Bbb Q$, its elements by definition cannot be approximated arbitrarily closely by rational numbers. So such a set is empty QED.


2

In any topological space $X$, for any subset $A$ of $X$, the interior of $A$ is the union of all open subsets contained in $A$, and the closure of $A$ is the intersection of all closed subsets that contain $A$. This is not the definition of interior and closure, but these characterizations follow easily from the definitions. (And each of the two propositions ...


5

How does it follow from this that the closure of $\mathbb{Q}$ is $\mathbb{R}$? Take $x \in \mathbb{R}$. Since $\mathbb{R}\ - \mathbb{Q}$ has empty interior, it follows that $x \notin int(\mathbb{R} - \mathbb{Q})$. It means that, for all $r > 0$, the ball $B(x,r)$ centred in $x$ with radius $r$ is not contained in $\mathbb{R}\ - \mathbb{Q}$. So, ...


1

a neighborhood of a point $x$ in $\mathbb{R}$ is an open interval, $(x- \epsilon _1 , x+ \epsilon _2 )$. if $x$ is an interior point of a set $S \subseteq \mathbb{R}$, then x has a neighborhood of that flavor, which is contained in $S$. if a set doesn't have any interior points, then, given any point in the set, you can't find an open interval $(x-\epsilon ...


4

This depends largely on the topology of the space that you're in. As copper.hat noted above, a set in a topological space has empty interior if it contains no open set. Recall that on $\mathbb{R}$ the open sets are generated by open intervals of the form $(a,b) = \{x \in \mathbb{R} \; | \; a <x < b\}$. So the complement of the rationals, which we ...


0

let $a=r+ir$ and $b=r-ir$, $r$ is a rational number and $ir$ is irrational , $a^2+b^2$ will be irrational if $ir^2$ is irrational now if the question was other way around , i.e $a+b$ irrational $a^2+b^2$ cannot be rational , that's also wrong as we can take both of them as square roots of an integer


6


2

This is not the answer to the question of irrationality. This a piece of information about an interesting closed form of the series. $$q_n=\sum_{k=0}^n 10^k = \frac{10^{n+1}-1}{9}$$ $q_0=1\quad;\quad q_1=11\quad;\quad q_2=111\quad...$ $\sum_{n=0}^m \frac{1}{q_n}=\frac{1}{1}+\frac{1}{11}+\frac{1}{111}+... \quad$ ($m+1$ terms). $$\sum_{n=0}^m ...


18

If we define the "Lambert series" $$ f(x) = \sum_{n=1}^\infty \frac{x^n}{1-x^n} = \sum_{n=1}^\infty \frac{1}{(1/x)^n-1}, $$ then since $q_n = (10^{n+1}-1)/9$, your number is $$ \sum_{n=1}^\infty \frac1{q_n} = 9f(\tfrac1{10})-1. $$ Chowla proved in 1947 that $f(\frac1{10})$ is irrational, and hence so is your number. You can find a proof in this 1948 paper of ...


1

Neither argument is complete. In each case, it's not enough to provide a listing of some of the set you want to show is countable; you also need to show that this list contains all the elements of that set. In the former case (the rationals), this isn't too hard; but technically it takes an argument. In the latter case (sets of naturals), this is in fact ...


2

The first argument is perfectly valid. The second ignores infinite subsets of the natural numbers: any infinite subset (e.g., the set of all odd natural numbers) doesn't correspond to the binary expansion of any natural number, so all the argument shows is that there are countably infinitely many finite subsets of the natural numbers.


0

$$\frac{3x^2}{3x-1} = \frac{0+3x^2}{3x-1}=\frac{(\frac{1}{3}-\frac{1}{3})+3x^2}{3x-1}=\frac{\frac{1}{3}+(-\frac{1}{3}+3x^2)}{3x-1}=\dots$$ Noting that $$x = \frac{3x}{3}=\frac{3x+0}{3}=\frac{3x+(-1+1)}{3}=\frac{(3x-1)+1}{3}$$ Then we have $$A = \frac{3x^2-\frac{1}{3}}{3x-1} = ...


0

Although there are various ways to solve this problem, especially since you're given the expression $\frac{1/3}{3x-1}+A$ for $\frac{3x^2}{3x-1}$ and so methods like solving for $A$ by subtraction are apparent, it seems most natural to me to interpret this as an exercise in dividing polynomials. The "best" approach is always a rather subjective assessment, ...


2

You can solve the equation for A and get $A = \frac {3x^2-\frac 13}{3x-1}$. In the numerator 3 can be factored out. $A = \frac {3(x^2-\frac 19)}{3x-1}$ $x^2-\frac 19$ is equivalent to the third binomial formula $a^2-b^2=(a-b)\cdot (a+b)$. Therfore $ x^2-\frac 19=(x-\frac 13)\cdot (x+\frac 13)$ $A= \frac {3\cdot (x-\frac 13)\cdot (x+\frac ...


1

One may write $$ \frac {3x^2}{3x-1}=\frac {\frac13\left(9x^2-1\right)+\frac13}{3x-1}=\frac {\frac13}{3x-1}+\frac {\frac13\left(3x-1\right)(3x+1)}{3x-1}=\frac {\frac13}{3x-1}+x+\frac13. $$


1

$\frac{3x^2}{3x-1} = \frac{3x^2-x+x}{3x-1} = \frac{x(3x-1)}{3x-1}+\frac{x}{3x-1} = x+\frac{x}{3x-1} = x+\frac{x-\frac{1}{3}+\frac{1}{3}}{3x-1}$ $=x+\frac{\frac{1}{3}(3x-1)}{3x-1} + \frac{\frac{1}{3}}{3x-1} = x+\frac{1}{3}+\frac{\frac{1}{3}}{3x-1}$ At each step, I generally did one of two things: I added and subtracted the same amount (equivalent to adding ...


1

You have $$\frac{3x^2}{3x-1}=\frac{1/3}{3x-1}+A$$ so $$3x^2=\frac{1}{3}+A(3x-1)$$ $$A=\frac{3x^2-1/3}{3x-1}=\frac{1}{3}\cdot\frac{9x^2-1}{3x-1}=\frac {1}{3}\frac{(3x-1)(3x+1)}{3x-1}$$ etc.


6

HINT From $\frac {3x^2}{3x-1} = \frac {\frac 13}{3x-1}+A $ you get $A = \frac {3x^2}{3x-1} - \frac {\frac 13}{3x-1}$.


2

The question makes no sense, really. Take any convergent series; take its limit $l$, and if the limit is not rational, add something just to the first term so the new limit is rational. (For example, add $-l$ to the first term to make the sum of the series equal to $0$.) Did you mean something else, such as "all the terms are rational"? And perhaps also ...


0

One could say that $\sum0=0$, it grows slower than both of your conditions and it is a rational number. Also, $\sum\frac{[\ln(2)]^n}{n!}=2$, it is both a rational number, and it grows slower than a geometric sequence. I'm not quite sure if I know any that grow slower than your factorial problem, but if it did, it could easily be tweaked to produce a ...


2

Is $\pi$ rational or irrational? It is irrational. Here is the number with its first several decimals $$\pi = 3.141592653589793238... $$ Can $\pi$ not be represented as $22/7$? No, this fraction is only an approximation of the true value. No fraction of integers will perfectly represent $\pi$ (exactly because it is irrational). It actually only ...


1

$\frac{22}{7}$ and $3.14$ are rational. $\pi=3.14...$, a non-terminating decimal expansion, is irrational (it is even transcendental) and is only approximately equal to them.


1

$\pi$ is irrational It was prooved. See there : proof $\pi$ is irrational


2

$f(x)=\frac{1}{x^2+1}$ is analytic but not a polynomial.


1

Suppose the resulting sequence is periodic with period $T$. Consider a shift of $2T$: being even, it puts odd elements to odd positions, that is, works as a shift on our initial subsequence $a_1, a_3, a_5, \dots$, which consequently must itself be periodic. If it is not, then the resulting sequence can't be either. The countability argument by @almagest is ...


1

To elaborate on the discussion in the comments: suppose that $x+y=S$ and $xy=k$ are given. We claim that this data specifies the pair $(x,y)$ up to order. Indeed, declaring $x$ to be the larger of the two we easily see that $$2x=\sqrt {S^2-4k}+S$$. Just to emphasize, if $(x',y')$ was a another pair with $x'≥y',x'+y'=S,x'y'=k$ then the same algebra would ...


2

The subgroups of $(\mathbb{R},+)$ are up to isomorphism the torsion-free abelian groups of rank $\alpha$ for every cardinal $\alpha\leq 2^{\aleph_0}$, because $(\mathbb{R},+)$ is isomorphic to $(\mathbb{Q}^{(2^{\aleph_0})},+)$ (weak direct product) as a $\mathbb{Q}$-vector space and thus also as a group. The torsion-free abelian groups of rank $2$ already ...


3

I've read that Euler ranked the harmoniousness of musical intervals, and thus the simplicity of rational numbers, in a way that appears to correspond to this function: $f(p) = p-1$, if $p$ is prime $f(ab) = f(a/b) = f(a)+f(b)$ assuming, of course, that the argument is expressed in lowest terms.


2

$f(\frac{a}{b})=\frac{1}{|a|+|b|}$ For the following conditions on $x$, $f(x)$ is either zero or not defined: $x$ irrational $x=0$ Higher output values implies high purity.


3

In the same spirit as Rob Arthan's answer, you could use the first time a rational number appears in the Calfin-Wilf sequence. Define a sequence $a_n$ with $a_0 = 0$, $a_1 = 1$ obeying the recurrences $a_{2n} = a_n$ and $a_{2n + 1} = a_n + a_{n+1}$. We get $$0, 1, 1, 2, 1, 3, 2, 3, 1, 4, 3, 5, \ldots$$ The Calfin-Wilf sequence is $c_n = a_n / a_{n+1}$: ...


4

A natural pureness measure is the level of the Stern-Brocot tree at which the fraction occurs. Since each run of consecutive left or right steps in the tree corresponds to a continued fraction term equal to the length of the run, the pureness may be defined as the sum of the continued fraction terms.


11

I have often used the sum of the terms in the Continued Fraction. This is finite for rational numbers the same for $x$ and $\frac1x$ for $0\lt x\lt 1$, the same for $x$ and $1-x$ $$ 1=(1)\to1 $$ $$ \frac12=(0,2)\to2\quad\text{and}\quad2=(2)\to2 $$ $$ \frac23=(0,1,2)\to3 $$ $$ \frac{53}{41}=(1,3,2,2,2)\to10 $$ etc.


8

The order of the Farey sequence where the (fractional part) of the rational number first occurs is a measure that should be of interest to you. As you will see from the link, the Farey sequences have many fascinating properties.


22

You can take, $$\frac{a}{b}\mapsto \frac{a+b}{\gcd(a,b)},$$ Note that this is independent of the choice of representative since $\gcd(na,nb)=n\gcd(a,b)$ for non-negative integers $n$. For your examples, $$\frac{1}{1}\mapsto 2,\quad \frac{1}{2}\mapsto 3,\quad\frac{2}{3}\mapsto 5,\quad\frac{53}{41}\mapsto 94.$$ Another possibility is $$\frac{a}{b}\mapsto ...


3

How about $f(\frac{p}{q})=pq$ for rational numbers and $f(x)=0$ when it's irrational?


2

Maybe you can get inspiration from the Thomae's function.


1

Option 1 Writing them out gives:$${1, \frac{1}{2}, 2, \frac{1}{3}, 3, \frac{1}{4}, \frac{2}{3}, \frac{3}{2}, 4, \frac{1}{5}, 5, \frac{1}{6}, \frac{2}{5}, \frac{3}{4}, \frac{4}{3}, \frac{5}{2}, 6, \frac{1}{7}, \frac{3}{5}, \frac{5}{3}, 7, \frac{1}{8}, \frac{2}{7}, \frac{4}{5}, \frac{5}{4}, \frac{7}{2}, 8, \frac{1}{9}, \frac{3}{7}, \frac{7}{3}, 9, ...


3

Given any sort of object $x$ for which it makes sense to multiply $x$ by rational numbers, you can call the set in question simply "the set of rational multiples of $x$", or sometimes, "the $\mathbb{Q}$-span of $x$". However, it should be pointed out that no matter what (non-zero) natural number $n$ you choose, the set of rational multiples of $n$ is ...


2

An uncountable number of counterexamples: If $x \in \mathbb R \backslash \mathbb Q$, then $\dfrac1{x} \in \mathbb R \backslash \mathbb Q$, but $x\cdot \dfrac1{x} =1 \not \in \mathbb R \backslash \mathbb Q$


2

Yes, it is formal enough. The statement implies a $ for all $. The denial of that $for all$ element, a proposition $p"$is true, is that $exists$ a case that does not fulfill. You have shown that there is a case for $a$ and $b$ that does not fulfill.


2

Since $S = \emptyset$ does not satisfy (3), $S$ is nonempty, say $x \in S$. Now (2) implies that $1 = x/x \in S$. It then follows that $S$ is closed under reciprocals, and hence also under multiplication. So it suffices to show that $1$ can be written as $y + z$ for some $y,z \in S$, because then each $x = xy + xz$ is a sum of elements of $S$. Suppose for ...


1

First notice that we only need to find one pair $s_1$, $s_2 \in S$ such that $s_1+s_2 \in S$. Then for any $x \in S$, we have $$x=x.1=x\Bigg(\frac{s_1}{s_1+s_2}+\frac{s_2}{s_1+s_2}\Bigg) = x\frac{ s_1}{s_1+s_2}+x\frac{s_2}{s_1+s_2} $$ and we may then take $y = x\frac{ s_1}{s_1+s_2}$, $z = x\frac{s_2}{s_1+s_2}$. Now $S$ is a subgroup of $\mathbb{Q}^{*}$, ...


1

The distinction here is the difference between finite sums and infinite sums. You correctly observe that finite sums of rational numbers are again rational numbers. However, infinite sums of rational numbers need not be rational. This is because, in some sense, real numbers are what happens when you take rational numbers and add infinity. Real numbers are ...


1

Indeed, it is not true that an infinite sum of rational numbers needs to be a rational number; you're example with $3+.1+.04+\cdots$ proves this. While it is true that any finite sum of rationals is again a rational, this is false for the infinite case. Remember that an infinite sum is just like an infinite sequence of partial sums. So we can consider ...


2

It is a common error to use induction to prove something about infinity. Using repeated applications of the rule $a, b \in \mathbb{Q} \Rightarrow a + b \in \mathbb{Q}$ you can only prove that any finite sum of rational numbers is again rational, but nothing about infinite series.


3

$\mathbb{Q}$ is closed under the addition of two elements, and so is closed under the addition of a finite number of elements But it is not necessarily closed under the addition of a countably infinite terms, even when the partial sums are increasing and bounded above, as shown by your example In other words, the rationals are not complete, and Cauchy ...


1

First of all, you have an if and only if statement, so you need to prove both directions. One of them should be trivial. If $A,B$ are perfect squares, then it shuold be easy to show that $\sqrt x$ is rational. Now, for the other direction, follow the same steps you are used to: Assume that $x$ is rational. Therefore, $x=\frac pq$ for some $p,q$ such that ...



Top 50 recent answers are included