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## New answers tagged rational-numbers

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If m and n are integers, then $\frac{m}{n}$ can be represented as a terminating fraction in base B if and only if $\frac{B^N}{n}$ is an integer for some integer $N$. Even then, it may not be represented exactly on a computer if it doesn't fit into the finite number of bits allowed. $5764607523034235/576460752303423488 - 0.01$ is about $2 \times 10^{-19}$. ...

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For the above problem there are four sets of solutions (this is intuitive: for a, b, c, & d). In the case of positive rational r and any odd number n we can eliminate all but one of the solutions: $d^n = 5 \wedge c^n = 1 \wedge a^n + b^n = 30 \wedge r = 5 \wedge a^n \in Z$ In the case of any odd number n≥3 we refer to the generating function: $a^{2 n ... 0 Your proof (after the edit, with the separate even and odd sequences of non-repeating and repeating decimals) seems correct, except for the claim (just before the edit) about one "extra" rational number in the interval$[0,1]$. (Remember how you can always book one more guest in the Hilbert Hotel.) The part that you added to the proof to handle the ... 2 Firstly, you should not say "infinite integers" if you mean "infinitely many integers". If one admits such a thing as an infinite integer, and$n$is an infinite integer, then$n$and$n+1$are infinite integers, but they are not infinitely many, since there are only two of them. This is standard usage in mathematics, regardless of what usages may prevail ... 0 Integers can be broken up into 3 subsets:$\{ 0 \}$,${+,\mathbb{N}}$and$\{-, \mathbb{N}\}$, i.e. 0 and two instances of the set of naturals, one for positive integers and the other for negatives. The integer$0 $corresponds with the rational$0$. If we can construct a correspondence between the positive rationals and the naturals, then we can use it ... 1 One can prove that $$Trd (\mathbb{Q}(S)) \aleph_0 + 1 \le \dim (\mathbb{Q}(S)) \le (Trd (\mathbb{Q}(S)) +1) \aleph_0 .$$ In particular$\dim (\mathbb{Q}(S)) = Trd (\mathbb{Q}(S)) \aleph_0$if$Trd (\mathbb{Q}(S))$is non-zero. Let$S' \subset S$be a transcendence basis of$\mathbb{Q}(S)$then$S'$is algebraically indepenent over the rationals and ... 4 First note that:$x\to \sin(x)$is locally a bijection on the interval you have given , so it's rational for an infinite number of arguments, in particular whenever it's evaluated on$\arcsin(r)$for rational$r$Another way of solving your problem is assuming that$\sin(x)$and$\cos(x)$are together rationals and here we will come up with the Pythagorean ... 3 I don't know if this is what you are looking for, but if we work on the unit circle,$sinx$is the y-coordinate of any point$(x,y)$. Then$a^2+b^2=1$, where$a=cosx, b=sinx$. Then $$b=\sqrt{1-a^2} \in \mathbb Q$$ iff$(1-a^2)= \frac {p^2}{q^2}$, so$y$as the coordinate of a point is rational iff the x coordinate is of the form$ \sqrt { 1-p^2/q^2}$... 3 The only nice$x$in this range for which$\sin x$is rational are$0$,$\pi/6$, and$\pi/2$. Every other preimage of a rational number is neither rational nor a rational multiple of$\pi$. (In fact I'm almost sure that if$\sin x$is rational and not in$\{-1,-\frac12,0,\frac12,1\}$, then$x$is algebraically independent of$\pi$). 0$\sin(x)=$opposite/hypotenuse.$y^2+x^2=1$If$\frac{y}{y^2+x^2}$is rational. So we know$y$must be rational and consider:$y^2+x^2$? what conditions would make$y^2+x^2$rational? (fixed errors) 3 What you want to prove is: given an irrational number$b$, then one of the following numbers : $$r^{\frac{1}{b}}\ \ \ \ \ \ r\in \Bbb Q$$ is irrational, this seems to be an attainable result for what we know nowadays about the transcendence of numbers, we have for example the following theorem: Six Exponentials Theorem:Let$(x_1,x_2)$and ... 0 Suppose we have already proven that given$s\in L_x$it is always possible to find$q\in\mathbb Q$such that $$s<q^2<x$$ Then we can factor$s$as$s=\frac sq\cdot q$and check that$\frac sq,q\in L_{\sqrt x}$. Let us show how to find$q\in\mathbb Q$with$s<q^2<x$. Fix$t^2>x$and consider some$q^2<x$. Then$q<t$and for ... 2 Let$r \in L_2$, i.e.$0 < r < 2$. We want to show that there exist$r_1, r_2 \in L_\sqrt2$such that$r = r_1 \cdot r_2$. Since$\dfrac 2r > 1$we can choose$v \in \mathbb N$such that $$1 + \frac 1v < \frac{2}{r} \, . \tag 1$$ Then define $$u = \max \{ x \in \mathbb N \mid \frac{x^2}{v^2} < 2 \} \tag 2$$ so that $$... 3 Set \,x=\sqrt{\dfrac{4-2\sqrt 3\strut}7}, and eliminate progressively the radicals. You'll obtain the equation:$$49x^4-56x^2+4=0$$If x is rational, say x=\dfrac pq, p and q coprime, we know p is a divisor of 4, and q a divisor of 49, whence the statement. 2 I am not quite sure where Fubini factors in, but your question can be answered by a slightly stronger statement. Proposition Let A, B\subset \mathbb{R} be Lebesgue measurable sets of positive measure, then the difference set A - B contains an interval. To wit, if A - B contains an interval, it must contain a rational number, which would ... 0 Hint: Let N=\{n\in\Bbb N: x^n\notin\alpha\}. This is non-empty, because x>1, so let n+1=\min N. 1 According to the Wikipedia article on Pisot–Vijayaraghavan numbers, this is true. In fact, if x is algebraic (not necessarily rational) then it has to be a Pisot number, and in particular an algebraic integer. A rational integer is of course a bona fide integer. 0 The length of the period is given by the multiplicative order of 10 \pmod q, where q is your quotient. It is closely related to the discrete logarithm. Wikipedia lists several algorithms that are faster than going through all the powers of ten, which is relevant if you're dealing with very large numbers (hundreds of digits). 0 Let \dfrac{x}{y} be a normalized fraction such as \gcd(x, y) = 1. We want to prove that adding an integer to a fraction with the naive algorithm produces another normalized fraction (a fraction whose greatest common divisor equals 1). The algorithm produces the following fraction: \dfrac{x}{y} + n = \dfrac{x + n·y}{y} One of the properties of the ... 1 As shown in this answer, n can be written as the sum of two squares if and only if, in the prime factorization of n, each prime that is \equiv3\pmod4 appears with even exponent. If x^z+y^2=3z^2, then 3 appears with odd exponent. Thus, there are no rational solutions of$$ \left(\frac xz\right)^2+\left(\frac yz\right)^2=3\tag{1}$$As noted, ... 1 For$N$an integer, the general result is that if$x^2+y^2=N$has rational solutions, then it has at least one integer solution. 5$a)$it amounts to solving in$\mathbb{Z}: x^2+y^2=3z^2$. You have that$x^2+y^2 = 0 \pmod 3 \to x = y = 0 \pmod 3$, and you get back the original one using descending method, and this proves$x = y = z = 0$, but this means the first equation$x^2+y^2 = 3\$ has no rational solutions.

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