# Tag Info

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Several. f(x) = x - x = 0 f(x) = x / x = 1, for x in R - {0} f(x) = round(x, 5), where round() rounds a number to the specified quantity of decimal places. The Dirac delta. f(x) = 1 if x in Q, 0 if x not in Q f(x) =The average of the first 10^6 digits of x (in base 10). And so on...

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According to a famous theorem about trascendental numbers for all $\alpha$ real algebraic with $1\ne\alpha\ne0$ the number $\ln \alpha$ is trascendental. Consider the function $f(x)=e^x$ which is a bijection of $\mathbb R$ onto $]0,\infty[$; take now un $r\in$ $\mathbb Q^+$, for instance $r=7$; there exists x such that $f(x)=7$ but $f(x)=e^x=7\iff x=\ln 7$ ...

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Hint: Use strengthenings of Bertrand's Postulate (please see Wikipedia), in particular Nagura's result that for $n\ge 25$ there is always a prime between $n$ and $n\left(1+\frac{1}{5}\right)$.

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I don’t understand your question, since it seems to be insufficiently quantified. Do you mean, “Is there a function that takes every irrational number to a rational?”? Or do you mean, “Given an irrational number, is there a function that takes it to a rational image?”? Your example seems to support the second interpretation, in which case I’d suggest the ...

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A well-known example. Is $m:=\sqrt{2}^{\sqrt{2}}$ rational? If yes, that is your answer. If no, then $m^\sqrt{2}$ is your answer: $m$ and $\sqrt{2}$ are irrational, but $$m^\sqrt{2} = \left(\sqrt{2}^\sqrt{2}\right)^\sqrt{2} =\sqrt{2}^{(\sqrt{2}\cdot\sqrt{2})} = \sqrt{2}^2 = 2$$ is rational.

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You can construct required function from any real function, rounding the result to finite number of decimals.

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What about $f(x)=x^2$? One has $f(\sqrt2)=2$.

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Sure. Choose any constant function $f : \mathbb R \setminus \mathbb Q \to \mathbb Q$, for example $$f(x) = 1 \quad\text{ for all } x \in \mathbb R \setminus \mathbb Q$$ If this answer seems flippant, you may want to sharpen the question.

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The cosine of $\pi$ is -1, a rational number.

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We have $$10!\cdot18!=10!\cdot17!\cdot6\cdot3$$ and $$12!\cdot17!=10!\cdot17!\cdot 6\cdot 22.$$ Thus considering the LCM of $3$ and $22$, we get that $$\text{LCM}(10!\cdot18!,12!\cdot17!)=10!\cdot17!\cdot 6\cdot 22\cdot3.$$ This can be rewritten as $$10!\cdot17!\cdot 6\cdot ... 0$$10!18!=10!17!\times 18=10!17!\times 6 \times 312!17!=10!17!\times 12\times11=10!17!\times 6 \times22\text{G.C.D}=10!17!\times6=\frac{10!18!\times3}{18}=\frac{10!18!}{3!}$$0$$(a,b,c)=d\Rightarrow d|a(a,b,c)=d\Rightarrow d|b$$Therefore$$d|(a,b)$$So$$d|1\Rightarrow d=13 For the sake of illustration, I have made a plot of the powers of 2 (those of 10 grow too fast). You see that they nicely align on a smooth and regular curve. Then it is a natural thing to define the powers for non-integer numbers too. And by continuity of the curve, for every number y, there will be some x such that y=2^x. How exactly we can ... 4 We have that a^n for some natural number n is equal to \underbrace{a\cdot a \cdots a}_{n\text{ times}} $$just as you said. This exponentiation operation fulfills a few relations, which I'm sure you've seen, such as a^n\cdot a^m = a^{n+m}, (a^n)^m = a^{nm} and a^n:a^m = a^{n-m} (but only for n>m at the moment). As long as n and m are ... 3 By convention, if the power is a fraction of the form \frac1n it means taking the n-th root. That is consistent with the rule, already true for integer exponents, that taking the x-th power of the y-th power is the same as taking the (x.y)-th power. By extension then, if the power is any fraction, it means ordinary exponentiation with the power of ... 2$$ \underbrace{10\cdot 10 \cdot 10 \cdots 10}_{x \text {times}} $$is how 10^x is defined for natural numbers x. Mathematicians like to extend definitions, and that has happened with exponentiation too. When doing these extensions, it's nice if they lead to pleasent results like well-known rules still applying. A first step is observing that ... 2 For any finite decimals number a=0.d_1d_2d_3\ldots d_n 000\ldots we have that a=\frac{d_1d_2d_3\ldots d_n}{10^{n+1}}. When we write x^a then what we mean is$$x^a=(x^{d_1d_2d_3\ldots d_n})^{1/10^{n+1}}$$That is we take x to the d_1d_2d_3\ldots d_nth power and then the 10^{n+1}th root of it. if b is with infinite decimals (and presumingly not ... 2 To show gcd(x-y,n)\neq n, we assume the opposite first for the sake of contradiction, then n|x-y contradicting x\not\equiv y\pmod{n}. To show gcd(x+y,n)\neq n, we assume the opposite first then n|x+y contradicting x\not\equiv -y\pmod{n}. To show gcd(x-y,n)\neq1, we assume the opposite, then since x^2-y^2 is a multiple of n and x-y,n are ... 0 TheGreatDuck, For n \in \mathbb{N}, x \in \mathbb{R} we have$$ \lfloor nx \rfloor = \sum\limits_{i=0}^{n-1} \lfloor x + \frac{i}{n} \rfloor $$0 If x is irrational, then$$ \sum_{k=1}^{n} \lfloor kx \rfloor = \sum_{k=1}^{n} kx - \sum_{k=1}^{n} \{ kx \} = \frac{n(n+1)x}{2} - n \cdot \frac{1}{n}\sum_{k=1}^{n} \{ kx \}. $$Since the sequence ( \{ kx \} ) is equidistributed, we know that$$ \lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^{n} \{ kx \} = \int_{0}^{1} x \, dx = \frac{1}{2}. $$Consequently ... 0 This equation is equal to: [\frac ni] mod  m = [\frac ni] - m[\frac {[\frac ni]}m] This function will be a degrading loop. It will equal 0 at every point [\frac ni] is evenly divisible by m. However, once i > n, the greatest integer will equal 0 on its own, and then the entire function will have reached its asymptote of y=0. It is a series ... 0 It all depends on the value of x. If x is a positive integer such that [x] = x, then the series will be equal to the series of kx, which I believe reduces to:$$ x \frac {k(k+1)}{2} $$As for the other possibilities, good luck. There is no known property involving multiplication inside of greatest integer (that I know of). I will try to look into ... 1 It comes from the formula$$x^n+y^n=(x+y)(x^{n-1}-x^{n-2}y+x^{n-3}y^2-...-xy^{n-2}+y^{n-1})$$whenever n is odd. 1 If a \mid c and if b \mid c, then c = ap and c = bq for some integers p,q; if in addition (a,b) = 1, then ap = bq implies b \mid p and a \mid q; but then we are done. 0 This is (up to the sign) the well known von Mangoldt function :$$\Lambda(n)= -\sum_{d|n} \mu(d) \log(d)$$2 Not an elementary solution. But you can solve it in integers. (10n-3)^2=5\cdot 2^{p+3}-31. Clearly p+3\ge 0. If p+3=3k, k\ge 0, then (5(10n-3))^2=\left(5\cdot 2^k\right)^3-775. But a^2=b^3-775 has 14 integral solutions (see http://oeis.org/A081120 and http://oeis.org/A081120/b081120.txt), which can be found with a program or these tables. ... 2 We need some proposition depending on n that we can prove by induction. In this case, it is$$P(n)\ \colon\ \sum_{i=1}^n i \geq \frac{n^2}{2}.$$First check the base case:$$P(1)\ \colon\ 1\geq \frac{1}{2}.$$This is true, so we have proved the base case P(1). Next suppose that P(n) is true, for some n\geq 1. We wish to prove that then P(n+1) is ... 0 Hint: \sum_{i =1}^n i \ge \frac {n^2} 2 iff 2\sum_{i =1}^n i = (1 + .... + n) + (1 + ..... + n) \ge n^2. What is (1 + ...... + n) + (1 + ...... +n)? Hint: addition is commutative and associative. 1 In any binary representation of 2n+1, we must have the zeroth power of 2 appearing (because the number 2n+1 is odd). It can't possibly have another 1 in its binary representation, because that would use up all the 1's we're allowed (that is, all two of them), leaving an odd number to be made as a sum of (even) powers of 2. Therefore, any odd number has ... 0 After reducing the zeroes, what you have is 2^{35}3^{22}7^{8}\cdots \equiv 8 \cdot 9 \cdots \equiv 2 \pmod {10}. 0 For t>0 and n\ne0 we have \exp (\pi t n^2)>1/0!+\pi t n^2/1!>\pi t n^2 so 0<\exp (-\pi t n^2)<1/\pi t n^2. 1 After a while the intuition you described is enough. But if you want some detail, we can separate out the term n=0. The rest is twice \sum_1^\infty e^{-\pi t n^2}. Now observe that for positive n we have e^{-\pi tn^2}\le e^{-\pi tn}. The series \sum_1^\infty e^{-\pi t n} is a geometric series with positive common ratio less than 1. 1 Because of gcd(9,19)=1 and 200\equiv 2\ (\ mod\ 18\ ), we have$$9^{200}\equiv 9^2\equiv 5\ (\ mod\ 19\ )$$2 In 2012 Feng and Wu showed that$$\limsup_{n} \delta_n \frac{\log \gamma_n}{2\pi}\geq 2.7327.$$Note however that the quantity$$\mu=\liminf_n\delta_n \frac{\log \gamma_n}{2\pi}$$is far more interesting. Unconditionally, the best bound is \mu<0.525396 due to this recent paper of Preobrazhenskii, however even under the Riemann Hypothesis this cannot be ... 1 There isn't going to be a 'nice' closed form, and as amcalde mentions, proving that seems hard. For the infinite series, we can write$$\sum_{n=1}^\infty r^n\sqrt{a+nd}=\sqrt{d}\Phi\left(r,-\frac{1}{2},\frac{a}{d}\right)-\sqrt{a},$$where \Phi is the Lerch Transcendent. 3 Add up the following: The number of strings with  1 zero and 11 ones is \dbinom{12}{ 1}= 12 The number of strings with  2 zeros and 10 ones is \dbinom{12}{ 2}= 66 The number of strings with  3 zeros and  9 ones is \dbinom{12}{ 3}=220 The number of strings with  4 zeros and  8 ones is \dbinom{12}{ 4}=495 The number of strings with  ... 0 Let n=\frac{p}{q} with gcd(p,q)=1. Then we want \dfrac{9p^2+30pq-9q^2}{q^3} to be an integer. This implies that q must divide 9p^2, hence q divides 9. If q=3, then we want p^2+10p-9 to be divisible by 3, which happens for any p=2 mod 3. If q=9, then we want 81 to divide p^2+30p and hence divide p+30, so this happens for p=6 (mod ... 0 You are correct. Possible values for (-1)^r where r is irrational are all complex (not necessarily imaginary) and none of them lie on the real axis. (Below, k is an integer, so (2k+1)r can never be an integer by irrationality of r.)$$(-1)^r=\exp ((2k+1)r\pi i)$$6 Write n=mk and suppose that x^n+y^n=z^n. Then (x^k)^m + (y^k)^m = (z^k)^m, so (x^k,y^k,z^k) is a solution to x^m+y^m=z^m. 0 First observe that \nu_3(n) answers the question “When written in base 3, how many zeroes does n end with?” Then it is easy to see that for a\neq b, \nu_3(a-b) is even if and only if a and b, when written in base 3, differ in the final digit, or end in the same two (but not the same three) digits, or end in the same four (but not five) ... 4 As you mentioned on Chat, Kap and I wrote a note on forms of different discriminants (but positive definite forms, meaning negative discriminants). This was corrected and extended by John Voight, now at Dartmouth; also published. The best known examples are the pair x^2 + xy + y^2 and x^2 + 3 y^2. The proof that these represent the same numbers is some ... 1 This is a very basic question that follows trivially via summation by parts. We have that $\sum_{n \leq x} \frac{d_{sf}(n)}{n} = \frac{1}{x} \sum_{n \leq x} d_{sf}(n) + \int_{1}^{x} \frac{1}{t^2} \sum_{n \leq t} d_{sf}(n) \, dt.$ Using the expression $\sum_{n \leq x} d_{sf}(n) = A x \log x + B x + E(x),$ where the error term E(x) satisfies E(x) = ... 1 Since we know how the old components of a base change B are given by v_{new}=B^{-1}v_{old} we use it on the quadratic form v\mapsto v^{\top}Qv. But let us note that$$v^{\top}Qv=(BB^{-1}v)^{\top}Q(BB^{-1}w)=(B^{-1}v)^{\top}(B^{\top}QB)(B^{-1}w)=v'^{\top}\overline Q v'.$$Then \overline Q=B^{\top}QB would be the matrix in the new coordinates. ... 2 This is a well-studied problem, a key-word is "Linnik's theorem." Let us denote the least prime itself by p(a,b), so p(a,b) = b + n(a,b)a. It is more common to express results in that way, and one can pass from on to the other easily. Then Linnik proved p(a,b) \le c a^L for some constants c and L. Meanwhile the best constant L for which ... 2 Let x=4n-1. Then the given equation translates to: x^2+7=2^{p+3}, which asks for solutions of the form 4k-1 to the Ramanujan-Nagell equation. The solution to this might have been discussed elsewhere on this site. 3 Let's say we have some partition A = \{a_1, a_2,\ldots,a_N\}, B = \{b_1, b_2,\ldots,b_N\} with a_i<a_{i+1} and b_i > b_{i+1} for all i, and set$$ S = \sum_{i = 1}^N |a_i-b_i| $$Because the a_i are increasing, and the b_i are decreasing, there is a number k such that a_i<b_i for all i \leq k and a_i>b_i for all i > k. ... 0 I too found the answer to be: x ≡ -1 (mod 60). My solution: 2 It follows from the hypothesis that 3,4,5,6 divides (x+1)  and therefore x+1 is a common multiple of these 4 numbers. 1 We have$$x \equiv 2 \pmod {3}x \equiv 3 \pmod {4}x \equiv 4 \pmod {5}$$in the form x\equiv a_i\pmod{m_i} and M=m_1m_2m_3=3\cdot4\cdot5 = 60 Then using chinese remainder theorem we have to find b_i such that b_i\dfrac{M}{m_i} = 1\pmod{m_i} Then$$b_1 \cdot\frac{60}{3}\equiv 1 \pmod {3} \implies b_1 \equiv 20^{-1} \pmod {3} \implies b_1 = ...

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HINT: Get a contradiction by showing that at some point, 7 must divide one of the numbers. Look at what happens to remainders mod 7 under $a \to a^2+4$ and $(a,b) \to ab+4$.

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