Formula q_711 (Q542): Difference between revisions

Latest revision as of 13:17, 20 April 2020

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English	Formula q_711	No description defined

q_711

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\left(\bigcup _{k\in \mathbb {N} }(0+p_{k}\mathbb {Z} )\right)\cup (-1+\ell _{1}\mathbb {Z} )\cup (1+\ell _{2}\mathbb {Z} )=\mathbb {Z}

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Revision as of 13:17, 20 April 2020 SchuBot (talk \| contribs) SchuBot, Bots, Bureaucrats, flow-bot, Suppressors, Administrators 4,665 edits ‎Created claim: arqmath formula id (P8): q_711 ← Older edit	Latest revision as of 13:17, 20 April 2020 SchuBot (talk \| contribs) SchuBot, Bots, Bureaucrats, flow-bot, Suppressors, Administrators 4,665 edits ‎Created claim: Defining formula (P1): \left(\bigcup_{k\in \mathbb{N}}(0+p_k\mathbb{Z})\right)\cup (-1+\ell_1 \mathbb{Z})\cup (1+\ell_2\mathbb{Z})=\mathbb{Z}
	Property / Defining formula
		$\left(\bigcup _{k\in \mathbb {N} }(0+p_{k}\mathbb {Z} )\right)\cup (-1+\ell _{1}\mathbb {Z} )\cup (1+\ell _{2}\mathbb {Z} )=\mathbb {Z}$ `\left(\bigcup_{k\in \mathbb{N}}(0+p_k\mathbb{Z})\right)\cup (-1+\ell_1 \mathbb{Z})\cup (1+\ell_2\mathbb{Z})=\mathbb{Z}`
	Property / Defining formula: $\left(\bigcup _{k\in \mathbb {N} }(0+p_{k}\mathbb {Z} )\right)\cup (-1+\ell _{1}\mathbb {Z} )\cup (1+\ell _{2}\mathbb {Z} )=\mathbb {Z}$ / rank
		Normal rank