How can one prove that on any given set $(a,b)\ ,\ a,b\in R$, there exists an algebraic operation $"*"$ that, together with the given interval, form a group? We know that $(a,b)$ is has an upper and lower bound.
Prove that any interval (a,b) with an algebraic operation "*" forms group
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abstract-algebra
group-theory
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1Any open interval is equinumerous to $\Bbb R$, and the real numbers have a natural group operation, to wit $(\Bbb R,+)$. – 2017-01-10
2 Answers
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Write a bijective function $f:(a,b)\to\Bbb R$ (a translated and scaled $\tan$ should do the work). Then define $$t*s:=f^{-1}(f(a)+f(b))$$
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0where $s,t \in (a,b)$, I suppose – 2017-01-10
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Hint: any open interval $(a,b)$ is homeomorphic to $\mathbb{R}$. Since $\mathbb{R}$ is a group under addition, use the homeomorphism to produce the group operation on $(a,b).$