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- This edit created a new page (also see list of new pages)
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19 April 2024
| m 04:35 | Pi is Irrational diffhist +50 Robkahn131 talk contribs | ||||
| m 04:33 | Pi Squared is Irrational diffhist +41 Robkahn131 talk contribs | ||||
| 04:27 | Pi Squared is Irrational/Proof 1 diffhist −4,346 Robkahn131 talk contribs | ||||
18 April 2024
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06:32 | Pi Squared is Irrational/Proof 1 2 changes history +4,913 [Robkahn131; Prime.mover] | |||
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06:32 (cur | prev) +285 Prime.mover talk contribs | ||||
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00:34 (cur | prev) +4,628 Robkahn131 talk contribs (filled in a few clarifying details) | |||
17 April 2024
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17:44 | Ideal Contained in Finite Union of Prime Ideals 6 changes history +823 [Prime.mover; Hbghlyj (5×)] | |||
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17:44 (cur | prev) +154 Prime.mover talk contribs | ||||
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13:02 (cur | prev) −6 Hbghlyj talk contribs | |||
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13:00 (cur | prev) −6 Hbghlyj talk contribs | |||
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13:00 (cur | prev) +661 Hbghlyj talk contribs (Add proof) | ||||
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12:50 (cur | prev) +28 Hbghlyj talk contribs (fixed {{explain|this needs to cite some results about summations}}) | ||||
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12:19 (cur | prev) −8 Hbghlyj talk contribs (fix typo "suppose suppose that") | |||
16 April 2024
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N 06:53 | Pi Squared is Irrational/Proof 2 3 changes history +3,770 [Simcha Waldman; Prime.mover (2×)] | |||
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06:53 (cur | prev) −314 Simcha Waldman talk contribs (Re-editing and explaining) | ||||
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06:00 (cur | prev) +1,126 Prime.mover talk contribs | ||||
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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)...") | |||
| N 05:05 | Pi Squared is Irrational/Proof 1 diffhist +2,341 Prime.mover talk contribs (Created page with "== Theorem == {{:Pi Squared is Irrational}} == Proof == <onlyinclude> {{AimForCont}} $\pi^2$ is rational. Then $\pi^2 = \dfrac p q$ where $p$ and $q$ are integers and $q \ne 0$. Note that $\paren {q \pi}^2 = q^2 \paren {\dfrac p q} = p q$ is an integer. Now let: :$\ds A_n = \frac {q^n} {n!} \int_0^\pi \paren {x \paren {\pi - x} }^n \sin x \rd x $ Integration by Parts twice gives: {{b...") | ||||
| 05:05 | Pi Squared is Irrational diffhist −4,858 Prime.mover talk contribs | ||||
15 April 2024
| 23:49 | Pi Squared is Irrational diffhist +2,783 Simcha Waldman talk contribs (Adding another proof) | ||||