Let $A = \{x \sin x : x \in \mathbb{Z}\} \subset \mathbb{R}$. Is $A$ a ring under the usual addition and multiplication operations of $\mathbb{R}$? It looks like it's not, but I can't find something concrete to justify this.
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2$\begingroup$ Under what operations ? The usual ones in $\;\Bbb R\;$ ? $\endgroup$– DonAntonioFeb 25, 2014 at 16:36
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3$\begingroup$ Is this even closed under negation? After all, $(-x)\sin(-x) = x\sin x$ because $\sin x = -\sin (-x)$... $\endgroup$– fgpFeb 25, 2014 at 16:49
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1$\begingroup$ It's not immediately clear why there isn't some $y\in\mathbb{Z}$ such that $y\sin{y}=-x\sin{x}$ though (I agree that this is unlikely). $\endgroup$– mdpFeb 25, 2014 at 16:55
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1$\begingroup$ To take things even further (and actually possibly point to a proof): is there even one non-trivial identity among some finite number of members of $A$? $\endgroup$– Steven StadnickiFeb 25, 2014 at 17:16
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2$\begingroup$ Another identity is that $f(1 \sin 1, 2 \sin 2) = 0$ where $$f(x,y) = y^2 - 8 x^2 (1-x^2) (1-2x^2)^2$$ $\endgroup$– user14972Feb 25, 2014 at 17:32
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2 Answers
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Suppose $A$ were a ring; then in particular, we have $2\sin 1=n\sin n$ for some $n$. Now, use the duplication formulas for $\sin$ to write $\sin n$ as $(\sin 1)\cdot P_n(\cos 1)$, where $P_n(x)$ is a polynomial of degree $n$ (this can be shown straightforwardly via induction; these polynomials are known as the Chebyshev polynomials of the second kind). This yields an identity of the form $nP_n(\cos 1)-2=0$, contradicting the fact that $\cos 1$ is transcendental. In fact, this shows that no non-trivial linear relation can hold among any finite number of members of $A$.
answered Feb 25, 2014 at 17:24
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$\begingroup$ Could you at least hint at a justification of the last sentence? $\endgroup$ Feb 25, 2014 at 17:28
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$\begingroup$ @MannyReyes I should correct that to 'additive identity', actually (I was thinking of $A$ as a group, not a ring). The proof in that case is almost identical: use the $\sin n$ identity to write each of the terms in the form $(\sin 1)\cdot P_n(\cos 1)$, sum, and derive a contradiction. $\endgroup$ Feb 25, 2014 at 17:31
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2$\begingroup$ Thanks, the "additive identity" helps a lot. So it seems that you're actually proving that $\{\sin(n) \mid n \in \mathbb{N}\}$ is a $\mathbb{Z}$-linearly independent subset of $\mathbb{R}$. Very nice! $\endgroup$ Feb 25, 2014 at 17:35
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$\begingroup$ (Note that similarly to a now-deleted answer from @MarcinŁoś, this uses the Lindemann-Weirstraß theorem to derive its conclusion - in my case, it's implicit in the assertion that $\cos 1$ is transcendental.) $\endgroup$ Feb 25, 2014 at 20:39
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There is no 1 (multiplicative identity) so it is not a ring. x sin(x) is either 0 (when x=0) or is a transcendental number due to Lindemann–Weierstrass theorem when x <> 0
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$\begingroup$ Depends on your definition of a ring. Rings are not usually required to have an identity though. $\endgroup$– the_foxMar 5, 2014 at 17:31
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$\begingroup$ True that is unclear from the question if they mean ring with identity or not. The other answer also assumes there is a 1 in the ring when it say there is an n such that 2 sin 1 = n sin n. Because the 2 here is an element 2 in the ring to multiply 1 sin 1 by. Hence it assumes there is a 1 in the ring to make the element 2 from. $\endgroup$ Mar 5, 2014 at 21:09
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$\begingroup$ Now I am curious if you assume that there is not a 1 in the ring if you can still prove the statement :-) $\endgroup$ Mar 5, 2014 at 21:15
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$\begingroup$ I think that Steven's answer tackles closure of addition, i.e. whether $1\sin 1 + 1 \sin 1 = 2\sin 1$ is an element of this set or not. $\endgroup$– the_foxMar 6, 2014 at 1:36
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$\begingroup$ Aha I had missed that point. Thanks for explaining! So given we are using the 1 from R here in regular way it seems like we must be assuming the ring has identity. $\endgroup$ Mar 6, 2014 at 10:48