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Show new changes starting from 11:11, 20 April 2024

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20 April 2024

04:13

Projection from Product Category‎ diffhist +52‎ Leigh.Samphier talk contribs

19 April 2024

m 13:33

Definition:Differential Equation‎ diffhist +10‎ Prime.mover talk contribs

18 April 2024

08:59

Functor Category is Category‎ diffhist +17‎ Leigh.Samphier talk contribs

08:50

Universal Property of Quotient Ring‎ diffhist +17‎ Leigh.Samphier talk contribs

08:44

Snake Lemma‎ diffhist +50‎ Leigh.Samphier talk contribs

17 April 2024

			17:44		Ideal Contained in Finite Union of Prime Ideals‎‎ 6 changes history +823‎ [Prime.mover‎; Hbghlyj‎ (5×)]
					17:44 (cur \| prev) +154‎ Prime.mover talk contribs
			m		13:02 (cur \| prev) −6‎ Hbghlyj talk contribs
			m		13:00 (cur \| prev) −6‎ Hbghlyj talk contribs
					13:00 (cur \| prev) +661‎ Hbghlyj talk contribs (Add proof)
					12:50 (cur \| prev) +28‎ Hbghlyj talk contribs (fixed {{explain\|this needs to cite some results about summations}})
			m		12:19 (cur \| prev) −8‎ Hbghlyj talk contribs (fix typo "suppose suppose that")

16 April 2024

			N 06:53		Pi Squared is Irrational/Proof 2‎‎ 3 changes history +3,770‎ [Simcha Waldman‎; Prime.mover‎ (2×)]
					06:53 (cur \| prev) −314‎ Simcha Waldman talk contribs (Re-editing and explaining)
					06:00 (cur \| prev) +1,126‎ Prime.mover talk contribs
			N		05:06 (cur \| prev) +2,958‎ Prime.mover talk contribs (Created page with "== Theorem == {{:Pi Squared is Irrational}} == Proof == {{tidy}} {{MissingLinks}} <onlyinclude> {{AimForCont}} $\pi^2$ is rational. Then $\pi^2 = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$. Let us define a polynomial: :$\ds \map f x = \frac{(1-x^2)^n}{n!} = \sum_{m\,=\,n}^{2n}\frac{c_m}{n!}x^m,\quad:c_m\in\Z$ $\map f x = f(-x)$ and so we get {{begin-eqn}} {{eqn \| l = f^{(k)}(x) = (-1)^kf^{(k)}(-x)...")

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			17:44		Ideal Contained in Finite Union of Prime Ideals‎‎ 6 changes history +823‎ [Prime.mover‎; Hbghlyj‎ (5×)]
					17:44 (cur \| prev) +154‎ Prime.mover talk contribs
			m		13:02 (cur \| prev) −6‎ Hbghlyj talk contribs
			m		13:00 (cur \| prev) −6‎ Hbghlyj talk contribs
					13:00 (cur \| prev) +661‎ Hbghlyj talk contribs (Add proof)
					12:50 (cur \| prev) +28‎ Hbghlyj talk contribs (fixed {{explain\|this needs to cite some results about summations}})
			m		12:19 (cur \| prev) −8‎ Hbghlyj talk contribs (fix typo "suppose suppose that")

			N 06:53		Pi Squared is Irrational/Proof 2‎‎ 3 changes history +3,770‎ [Simcha Waldman‎; Prime.mover‎ (2×)]
					06:53 (cur \| prev) −314‎ Simcha Waldman talk contribs (Re-editing and explaining)
					06:00 (cur \| prev) +1,126‎ Prime.mover talk contribs
			N		05:06 (cur \| prev) +2,958‎ Prime.mover talk contribs (Created page with "== Theorem == {{:Pi Squared is Irrational}} == Proof == {{tidy}} {{MissingLinks}} <onlyinclude> {{AimForCont}} $\pi^2$ is rational. Then $\pi^2 = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$. Let us define a polynomial: :$\ds \map f x = \frac{(1-x^2)^n}{n!} = \sum_{m\,=\,n}^{2n}\frac{c_m}{n!}x^m,\quad:c_m\in\Z$ $\map f x = f(-x)$ and so we get {{begin-eqn}} {{eqn \| l = f^{(k)}(x) = (-1)^kf^{(k)}(-x)...")