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/sci/ - Science & Math

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>> No.11553887 [View]
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11553887

>>11553646
you forgot to mention it "coincidentally" happened RIGHT AFTER Trump got acquitted. Really tickles the noggin.

>> No.11536014 [View]
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11536014

>>11534699
>supremum

>> No.11460568 [View]
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11460568

>>11459409
Lmao what the fuck did you accidentally solve captcha and pasted that shit lol
The funniest thing are you's

>> No.11450156 [View]
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11450156

>>11446142
>10 million people die of cancer a year
I actually did not know that, fucking hell that is gnarly.

>> No.9163516 [View]
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9163516

>>9163511

How did you discover my secret?

>> No.9040582 [View]
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9040582

I just realized something. Since for any Field [math](\mathbb{K},+,\times)[/math] with corresponding identities [math]0,1[/math]
[math]a\times 0 = 0[/math] follows essentially from distributivity, one cannot extend the notion of a field to an
algebraic object that is distributive in a symmetric fashion consistently.

By that I mean that taking a set [math]\mathbb{K}[/math] such that [math]\mathbb{K}^+ = \mathbb{K}\backslash\{1\}[/math], [math]\mathbb{K}^\times = \mathbb{K}\backslash\{0\}[/math]
form abelian groups [math]\left((\mathbb{K}^+,+),(\mathbb{K}^\times,\times)\right )[/math] and enforce
[eqn]a\times(b+c) = (a\times b)+(a\times c)\\
a+(b\times c) = (a+b)\times(a+c)[/eqn]
as well as associativity you'll find that [math]1 + a = 1[/math] and [math]0 \times a = 0[/math] being well-defined,
but expressions like [math] 0\times 0[/math] becoming ill-defined for any case where [math]\mathbb{K}\ne\{1,0\}[/math]
(I did not check that one, however).

>> No.6637352 [View]
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6637352

>>6637258
Dear god

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