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>> No.9431699 [View]
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9431699

>>9430943
Or maybe your moeshit circlejerks repel anyone with two brain cells to rub together.

>> No.9406825 [View]
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9406825

>>9406666
[math]\mathbb{B} \mathbb{R} \mathbb{A} \mathbb{I} \mathbb{N} [/math]
[math]\mathbb{E} \mathbb{X} \mathbb{P} \mathbb{A} \mathbb{N} \mathbb{D} \mathbb{I} \mathbb{N} \mathbb{G}[/math]

>> No.9353803 [View]
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9353803

>>9352860
[math]f(x) = \sqrt{1+x}[/math] is a contraction on [math][0, \infty)[/math], as f is closed on the same interval and differentiable with derivative [math]\frac{1}{2\sqrt{1+x}}[/math] bounded by 1/2. Then apply the Banach fixed point theorem.

>> No.9268198 [View]
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9268198

>>9268196
[math]\left \langle u, A u \right \rangle
=\left \langle A^\dagger u, u \right \rangle
=\left \langle A u, u \right \rangle
=\left \langle u, A u \right \rangle^\dagger[/math]
Must hurt to be such a brainlet as to be spooked by this.

>> No.9249030 [View]
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9249030

>>9248825
>double major in math
>first in class in the math department

>> No.9248033 [View]
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9248033

>Let S be the set of all totally ordered field extensions of Q.
>As Q is an element of S, S is nonempty.
>S is partially ordered under set inclusion, and every chain in S has an upper bound, namely the union of the chain (proof: think).
>Therefore by Zorn's lemma, S has a maximal element, say R.
>Assume there exists an upper bounded subset T of R with no least upper bound
>Construct a field extension R(t) with ordering induced by the ordering on R such that t ≤ any upper bound of T and ≥ any other element
>By assumption, this is a field extension of Q that is larger than R, contradiction
>Therefore R is Dedekind complete
should piss off both constructivelets and nonconstructivelets

>> No.9245592 [View]
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9245592

>posthuman
That's just the PHENOTYPE, stupid goy.

>> No.9240384 [View]
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9240384

>brainlets will get artificial PHENOTYPE enhancements
oh the horror
hope the titans form in time to stop this madness

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