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An infinite sum of rationals may be either rational or irrational. In fact, any real number, whether rational or not, is an infinite sum of rationals $$n+\frac{d_1}{10}+\frac{d_2}{100}+\frac{d_3}{1000}+\cdots\ :$$ this is exactly what its decimal representation means. The problem is that an "infinite sum" is not, strictly speaking, a sum. It is a sum ...

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Using your arguments, $e$ and $\pi$ are rational. Are they?

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The only rationals $r$ such that $\sin(\pi r)$ is rational are those for which $\sin(\pi r)$ is in $\{-1,-1/2,0,1/2,1\}$. This is because $2 \sin(\pi r)$ is an algebraic integer, and algebraic integers that are rational are ordinary integers.

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$$q=\frac{1}{3}\Rightarrow \sin \left( \frac{\pi}{2} \frac{1}{3}\right) =\frac{1}{2}$$

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No, that argument doesn't work. What it does establish is that $e+\pi$ can be written as an infinite sum of rational numbers - however, this is true of every number (since, for instance, $\pi=3+.1+.04+.001+\ldots$). However, this does not mean that $e+\pi$ is rational - in fact, your argument would necessarily also prove that $e$ and $\pi$ are also both ...

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You don't have all integer coefficients of your middle terms in the expanded polynomial. Multiplying through by three will get rid of fractional coefficients and give you the possibility of 3 being a root. The rational root theorem requires all integer coefficients.

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You are definitely mixing up contrapositive and contradiction. It is quite easy to do as a lot of proofs that can be done with one can be done with the other. Many students make this mistake early on in their careers. What you should say is that "Suppose $\dfrac{m}{nb}\in\Bbb Q$, then $b\in \Bbb Q$." The reason is that you want to prove "if $b\in \Bbb ... 2 By the duplication formula, $$\Gamma\left(\frac{2}{50}\right) = C\cdot \Gamma\left(\frac{1}{50}\right)\Gamma\left(\frac{13}{25}\right)$$ hence: $$\Gamma\left(\frac{1}{50}\right) = \frac{\Gamma\left(\frac{1}{25}\right)}{C\cdot \Gamma\left(\frac{13}{25}\right)}.$$ 2 That's as much as it can be simplified unless you wish to factor out a 2 in the denominator. 1 Since the numerator and denominator share no common factors, that is as far as you can simplify.$\frac{x+8}{2x+4}$is your answer. 1 This is it with additional constraints$x\not=\pm2$,also $$\frac{x+8}{2x+4}\not=\frac{x}{2x}+\frac{8}{4}$$ 1 It is tacitly assumed that the cut$\alpha$contains some positive rationals. There is an$a'$such that$a<a'<1$. We now need to localize the cut$\alpha$quite precisely. Starting from an arbitrary positive$x\in\alpha$and an$y\notin\alpha$you have to construct two numbers$c\in\alpha$and$d\notin\alpha$with$c\geq x$and $$a' d<c<d\ .$$ ... 1 If you are identifying$\mathbb{R}^2$with$\mathbb{C}$using the correspondence$(a, b) \mapsto a + bi$then$(1, 0)$is identified with$1$, and$(0, 1)$with$i$. The vectors$(1, 0)$and$(0, i)$don't lie in$\mathbb{C}$but$\mathbb{C}^2$. Now we can ask whether$\{1, i\}$is linearly independent in$\mathbb{C}$. The answer depends on how you are ... 1 (7,-4), (7,-4,-4), (7,-1,-4,-4),(7,-1,-1,-4,-4),(7,-1,-1,-1,-4,-4),(7,0,-1,-1,-1,-4,-4) You can duplicate numbers since you can take the mean of a 1 number subset and add it to the set. Then you are basically looking for a sum to zero Algorithmically, sort original array, look for the closest two numbers to zero, if they are multiples of each other you are ... 1 By the same reasoning, you could conclude that each of these series individually converges to a rational, which is false. It is indeed correct that a sum of rationals is always rational. The problem is that an infinite series is not a sum. This is a trap that many fall into. Even experienced mathematicians come to think of infinite series as a sums, but ... 1 Try to show that for every real number$a$and every positive integer$N$there exist$p,q$integers with$1 \le q \le N$such that$|qa - p| \le 1/(N+1)$. This can be shown just using the pigeonhole principle in a clever way. The result is called Dirchlet's approximation theorem. In this way you could easily find more detailed information in case it is ... 1 There might be simpler ways to prove this, but the only way I know how to do this is via the following theorem (which makes use of the pigeonhole principle): Theorem. Now show that given any positive integer$Q$we may find positive integers$p, q$with$1\leq q\leq Q$and such that$|x-p/q|\leq 1/(qQ)\leq 1/q^2$. Proof. Let$Q$be any positive integer. ... 1 Any subring of a field is a domain, so it's not surprising you can't apply the condition. If$S$is a multiplicatively closed subset of$\mathbb{Z}$(with$0\notin S$), then $$S^{-1}\mathbb{Z}=\left\{\frac{a}{b}\;\middle|\; a\in\mathbb{Z}, b\in S\right\}$$ is a subring of$\mathbb{Q}$containing$\mathbb{Z}$. A non trivial example of$S\$ is ...

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