# Tag Info

0

Rational numbers can have decimals because some of them can be written as fractions, so decimals like 0.1, 4.6666... (it repeats), and 4.7839583 are rational numbers because they can be written as (in order) $1\over 10$, $4$$2\over 3, and 4$$7,839,583\over 10,000,000$, which are pure fractions, but some other decimals aren't rational like $\pi$ ...

0

Let $q=m/n$ a rational. Then take the remainders $r_k$ of division of $10^k$ by $n$. As these powers are infinite and $r_k\in\{0,1,\ldots,n-1\}$, then there exist $k_1,k_2$, such that $r_{k_1}=r_{k_2}$, and hence $$n \mid 10^{k_2}-10^{k_1}=10^{k_2-k_1}(10^{k_1}-1),$$ and thus $10^{k_2-k_1}(10^{k_1}-1)=n\ell$, and hence $$... 1 When you divide numerator by denominator to find the decimal expression, there are three possibilities: The quotient is integer. The reaminder is eventually 0 after the comma. This gives a finite decimal representation. The interesting case: the remainder is never 0. Let n be the denominator (assume that it is positive). When you are finding decimal ... 0 anything that can be written as a fraction of two integers 23/99 or whatever is rational anything that cannot be written as the fraction of two integers like the sqare root of most numbers most infinite sums eg. pi. the logarithm, and most limits are not rational$$1.41421...=\sqrt{2} 1.100100001... = \sum_{n=0}^{\infty} 10^\left (-1 (n^2 )\right ...

1

$0.2\overline{34} = {1 \over 10} (2+0.\overline{34})$, and $0.\overline{34} = 100(0.\overline{34})-34$, so $99 (0.\overline{34}) = 34$, from which we get $0.2\overline{34} = {1 \over 10} (2+{34 \over 99}) = { 232 \over 990}$. Since $\gcd(232,990) = 2$, we get $0.2\overline{34} = { 116 \over 495}$.

3

1000 x = 234.3434... - 10 x = 2.3434... --------------------- 990 x = 232 x = 232/990 = 116/495 x can be represented as a quotient of two integers with denominator not equal to zero, so it is a rational number.

2

Every number which can be written in a decimal notation in such a way that from some point on, there exists a subsequence that repeats itself, is rational. For example, $1=1.0000000000000000000000...$ is rational as the zero repeats until infinity. $1/3=0.33333333333333333333...$ is also rational, as the three repeats on and on. Also rational is the number ...

15

I believe it is more interesting to prove a slighly more general result, which answers the question asked directly : Any number with a finite repeating pattern is rational. Proof : Say $X = a.u_1u_2u_3...u_n\overline{v_1v_2v_3...v_d}$ where $a$ is a whole number, $u_1,u_2,u_3...u_n, v_1v_2v_3...v_d$ are digits and $\overline{v_1v_2v_3...v_d}$ means that ...

6

We have $100x=23.434343\cdots$ and $10000x=2343.434343\cdots$ so $$9900x=2320$$ hence $x$ is rational and $$x=\frac{2320}{9900}=\frac{116}{495}$$

10

Let $y = .3434\ldots$. Then, $100y = 34.3434\ldots$. So $100y-y = 99y = 34$. Hence, $y = \frac{34}{99}$. Since $0.03434\ldots$ is just the above divided by $10$, we have $$0.2343434\ldots = 0.2 + \frac{34}{990}$$ which is certainly rational.

0

The question is not very clear, but: take as a trivial example $f(n)= \frac{\sqrt{2}}{n}+n+\frac{1}{2}$, $g(n)=n$. You have $f(n)\to+\infty$, $g(n)\to+\infty$ $f(n)-g(n)\to \frac{1}{2}$. However, f(n) is irrational for all $n$, and $g(n)$ is rational for all $n$. So, it's hard to imagine some sense in which both could defined as being "rational (or ...

1

Okay, I think I now understand what you are saying. I'll enumerate my responses for readability. 1) You said "$x−y={a\over b}$ if and only if either $x$ and $y$ are both rational or irrational." This is not true. See the excellent answer here: The sum of irrationals is irrational? 2) The notion of growing 'irrationally' or 'rationally' is undefined. You ...

2

Clearly $f(n)$ constant does not work. Moreover for any non constant polinomial function it holds $\log n/ f(n)\to 0$, so no polinomial function $f$ satisfies $a^{b+c\log n} \sim a^{cf(n)}$.

1

Example. $f(x) = 1/\Gamma(x)$ is certainly a transcendental function. It has zeros, so it is not exp of an entire function. But your value $k$ is Euler's constant $\gamma$. Which has not yet been proved irrational... more If $c$ is any nonzero constant, then $\sin(cx)$ is a transcendental function. But your value $k$ is then $c \cot c$ if my ...

1

Well if $f(1) = 1$ and $f$ is differentiable at $1$, then $\lim (f(1+1/n))^n = \exp(f'(1))$. Now, there is absolutely no relationship whatsoever between differentiable functions whose derivative at $1$ is rational and "transcendental functions" (whatever that means), or wether $f$ is of the form $\exp(g)$ (or equivalently, functions with no zeros).

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$$f(x) = {e}^{-k(x-1)}$$ This function should be transcendental and satisfy the conditions you've shown, but $k$ can be rational.

6

Suppose $\sin a$ is transcendental. Then from the addition and subtraction formulas for sine, the two values $\sin(a+b),\ \sin(a-b)$ are both rational iff each of $r=\sin a \cos b$ and $s=\cos a \sin b$ are rational. Supposing $r$ is rational, from the first of these $\cos b=r/\sin a.$ Then $$s=\cos a \sin b = (\pm \sqrt{1-\sin^2 a})(\pm \sqrt{1-r^2/\sin^2 ... 1 It is useless to attempt to determine the nature of such a limit simply from the nature of the respective sequences. If both p_i and q_i converge, then \lim_{i\to\infty}\frac {p_i}{q_i} = \frac{\lim p_i}{\lim q_i} is simply the quotient of two real numbers, because any convergent sequence of rational numbers (ie any Cauchy sequence) uniquely defines ... 1$$ \frac{\sin x -(e^x-1)}{x^2} \to - \frac12 $$4 A more boring example: Let f be a transcendental function. Then$$\frac{(f(x)+x)-f(x)}{x}$$works. 4 If f_3(0)\ne0 then$$ \lim_{x\to 0} \frac{f_1(x)-f_2(x)}{f_3(x)}=0\in\mathbb{Q}. $$The above example was given before the OP added the condition f_3(0)=0. If you want something less trivial$$ \lim_{x\to0}\frac{e^{2x}-e^x}{x}=1\in\mathbb{Q}. $$5 The axiom of choice is necessary for producing a Hamel basis for \Bbb R over \Bbb Q, meaning we can prove that without the axiom of choice we cannot prove the existence of a Hamel basis; and of course it is not equivalent to the existence of such basis. Let me start with the easy part. The second part. There is no statement which is limited to a ... 3 No. To show the existence of such a basis one uses the Axiom of Choice. And anything that uses this axiom explicitly and necessarily defies explicit description. (If one could give an explicit description, one would make choices that can be described in finitely many words). For a deeper insight into this, check out Asaf's answer that will occur here in a ... 1 You are looking for the Hamel Basis. There is no way to write this basis explicitly, since its existence is dependent on the Axiom of Choice. 1 Here is a definition that might help you to write out (1) in a formal way. Definition: Let F be a field, a vector space over F is a set V together with a binary operation + on V under which V is an abelian group, an action of F on V (that is, a map F\times V \rightarrow V) denoted by rv, for all r\in F and for all v\in V which ... 1 Another proof for point 2: \mathbb{Q}^2 is countable, \mathbb{R} is not, so \mathbb{Q}^2 can't be a vector space over \mathbb{R}. 1 This set is open in relation to \mathbb{Q}. Remember that a set A\subset\mathbb{R} is open in \mathbb{Q} iff there is an open set B\subset \mathbb{R} such that A=\mathbb{Q}\cap B. Given that every open interval is an open set of \mathbb{R}, (\sqrt2,\sqrt3) is open, which means that (\sqrt2,\sqrt3)\cap\mathbb{Q} is open in \mathbb{Q}. 6 Note that in {\bf Q}, V\cap {\bf Q} is open when V is open in {\bf R}. So let V=(\sqrt{2},\sqrt{3}). So V\cap {\bf Q} is open in {\bf Q} 1 Another way of knowing if \frac{a}{b} is greater than \frac{c}{d} is by converting these fractions to fractions with a common denominator. You have to get the LCM of the denominators and use it as the LCD of the two fractions. Convert each fraction to an equivalent fraction whose denominator is the LCD. When comparing two fractions with like ... 0 Since only things related to multiplication and not addition are involved in this problem, looking at the prime factorization of things should prove useful. (we can extend prime factorization to rational numbers: e.g. 4/6 has prime factorization 2 \cdot 3^{-1}) 0 Imagine we have \frac{m^2}{n^2}=x,\quad x,m,n\in\mathbb{N},n\neq 1, GCD(m,n)=1. Now we know there is a prime p so that it appears in the factorization of x with odd exponent a - else x would be a perfect square. Let us say x=p^\alpha r. Now we know that m^2=xn=p^{a}rn^2. From this we get p|m (why?). Since GCD(m,n)=1, we also know that ... 0 The easiest way to do this is to take a rational, written in lowest terms, and square it. In particular,$$\left(\frac{p}{q}\right)^2=\frac{p^2}{q^2}$$where the fraction on the right is also in lowest terms, since \gcd(p^2,q^2)=\gcd(p,q)^2=1. However, any integer must have a denominator of 1, thus its square root must have denominator q satisfying ... 3 There is a classic proof on the irrationality of \sqrt{2}. You should be able to modify it to achieve this result. Give it a shot and maybe you can answer your own question. 0 For Q1 do long division. Either you reach a remainder 0 and the division terminates, or you don't. If you don't, the number of possible remainders is finite and you must repeat. For Q2, if the repeating decimal has n digits, multiply by 10^n and subtract the original. If it terminates with n digits after the decimal point, just multiply by 10^n ... 0 Every rational number not having a finite number of decimals is a repeating decimal because of the following reason. When you make the division of m by n you get, at most, n different remainders. So, if the division doesn't end, after, at most, n+1 steps, you get the same remainder you have got in a previous step. From this, you get a periodic ... 2 Since OP has asked for a solution that does not involve the theory of continued fractions, I present this alternative solution which involves no theory whatsoever. #include <stdio.h> int main(void) { unsigned n, d; for (d=1; 1; d++) { for (n=1; n<d; n++) { double q = (double)n/d; if (33.0/100 < ... 2 Let x be the desired quantity. We have$$\cfrac1{3+\cfrac1{33}} < x < \cfrac1{3+\cfrac1\infty}$$so the continued fraction expansion of x must be$$\cfrac1{3+\cfrac 1{n+y}}$$where 33<n+y<\infty. We get the smallest possible denominator by truncating the continued fraction after n (taking y=0) and taking n as small as possible, so ... 5 Since {\sqrt{a}}^{\sqrt{b}} is expressed as an algebraic number not equal to 0 or 1 raised to an irrational algebraic power, the result will be transcendental (and hence irrational) by the Gelfond–Schneider theorem. 1 Yet another answer: (credit goes 99% to lhf) Let \alpha = (\sqrt{2}+1)^{1/n}, \beta = (\sqrt{2}-1)^{1/n}. Since \alpha and \beta are algebraic integers, \alpha+\beta is an algebraic integer. Since a rational algebraic integer is an integer, it suffices to show that \alpha+\beta is not an integer. The problem is trivial for n=1, so we assume ... 1 Let \alpha = (1+\sqrt{2})^{1/n}, \beta = (-1+\sqrt{2})^{1/n}. Then \alpha\beta = 1, so \alpha and \beta are roots of the polynomial X^2-(\alpha+\beta)X+1. If \alpha+\beta\in\mathbb{Q}, then \alpha and \beta lie in a quadratic extension of \mathbb{Q}. Since \alpha^n = 1 + \sqrt{2}, we have \alpha\in\mathbb{Q}[\sqrt{2}], and since ... 0 Let B(x-\epsilon/2, \epsilon/2) and B(x+\epsilon/2, \epsilon/2) be open balls. By density of the rationals, there is a rational in each of these balls, and it is trivial to check that they are less than \epsilon apart. 0 You say in comments that you know that between any two real numbers there is a rational number. So apply that to the the two real numbers x and x+\dfrac\varepsilon2, and then again to the two real numbers x and x-\dfrac\varepsilon2. 2 If you are allowed to use the property that the rational numbers are dense in \mathbb{R}, and it sounds like you are from the comments, then: x - \frac{\epsilon}{2} and x + \frac{\epsilon}{2} are both real numbers, so find q, q' rational such that$$x - \frac{\epsilon}{2} < q < x < q' < x + \frac{\epsilon}{2}.$$Then |q' - q| < |x + ... 1 Let x=x_0,x_1x_2x_3... be the decimal representation of x>0. Given \varepsilon>0, take n such that \dfrac{2}{10^n}<\varepsilon. Than the rationals q=\sum_{k=0}^n\dfrac{x_k}{10^k}-\dfrac{1}{10^n} and q'=\sum_{k=0}^n\dfrac{x_k}{10^k}+\dfrac{1}{10^n}, satisfies the conditions desired. For x<0, make a similar aproach. 5 Use the fact that for every n there exists a polynomial P_n[x] monic of degree n with integral coefficients so that we have the identity: \begin{eqnarray*} t^n + \frac{1}{t^n} = P_n(t+ \frac{1}{t}) \end{eqnarray*} The P_n's are connected with the Chebyshev polynomial of first kind since for t = e^{i \theta} we get 2 \cos n \theta = P_n (2 \cos ... 12 Assume that for a certain positive integer n$$ \big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\in \mathbb Q. $$Clearly, \,\big(\sqrt{2}-1\big)^{1/n}\cdot\big(\sqrt{2}+1\big)^{1/n}=1, and thus$$ \left(\big(\sqrt{2}-1\big)^{1/n}+\big(\sqrt{2}+1\big)^{1/n}\right)^2= \big(\sqrt{2}-1\big)^{2/n}+2+\big(\sqrt{2}+1\big)^{2/n} \in \mathbb Q, $$and ... 4 In general, if x is irrational, then$$ a_n=\frac{\lfloor 2^nx\rfloor}{2^n}<x, $$is an increasing (a_n\le a_{n+1}) sequence of rationals converging to x. Here \lfloor a \rfloor is the integer part of a. Note. It is possible to extract a strictly increasing sequence, as all the terms of the sequence are strictly smaller than the limit. 8 One more example$$f(x) = \sin(2\pi\log_2x)f(2x) = \sin(2\pi\log_2(2x)) = \sin(2\pi(1 + \log_2x)) = \sin(2\pi\log_2x) = f(x)$$4 Consider the following function:$$ f(x)=\left\{\begin{array}{ccc} k & \text{if} & x= 2^mk,\,\,\, \text{for some}\,\,\,k,m\in \mathbb Z, \\ 0 & \text{otherwise}. \end{array} \right. $$Indeed, \,\,f(2x)=f(x). 2 The only continuous function satisfying the relation is the constant function. Indeed the given equality gives$$f(x)=f\left(\frac x{2^n}\right)\xrightarrow{n\to\infty}f(0)$$To construct a general function let S\subset \Bbb R and consider the set$$A=\{2^n s\;|\; n\in\Bbb Z,\;s\in S\}$$and define f by$$f(x)=\text{constant if}\; x\in A,\; ...

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