How do I divide a number that cannot be divided? Watch

King Leonidas
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Please help,

thanks
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Juichiro
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Is this a troll? It can't be divided.
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claret_n_blue
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I don't understand your question.

In what domain are we talking? Real, complex?

All real numbers can be divided...you just might not get a whole number.
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King Leonidas
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How do I divide .2% of the calculus theorem?
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interstitial
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Which theorem of calculus are you talking about? :holmes:

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I'm aware this is a joke, just thought I'd play along for a bit


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ThatPerson
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(Original post by King Leonidas)
Please help,

thanks
You need to use Symbolic Mathematics.
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King Leonidas
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(Original post by majmuh24)
Which theorem of calculus are you talking about? :holmes:

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I'm aware this is a joke, just thought I'd play along for a bit


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I'm being totally serious, my maths is really bad I want to create my own theorem.
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Ateo
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Define S as the set of reals that cannot be divided. Then to divide do as you please king, it will be vacuously true, just don't kick me down a hole.
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interstitial
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(Original post by ThatPerson)
You need to use Symbolic Mathematics.
Indeed! Which method would you suggest - exponential ladders, squaring the ratio, or the Dalekian algorithm?

(Original post by King Leonidas)
I'm being totally serious, my maths is really bad I want to create my own theorem.
Dalek's already done that.

If  \{ S : x \in S \implies \ \text{x \ cannot \ be \ divided} \} , and a \in S, we can use Kaluza-Klein theory along with the topological 42 dimensional fractal self similarity between \mathbb R \to S

\dfrac{a}{b} = \vartheta ^2 \cdot \mathcal{L} \left( \dfrac{a}{b} \right)

Where  \mathcal L denotes the Laplace Transform of \dfrac{a}{b} as a function of parameter t, which is given by \mathcal{L} \{ f(t) \} = \displaystyle\int_0^{\infty} e^{-st} \cdot f(t) where s is an arbitrary parameter.

Symbolic maths is the way to go
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Ateo
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(Original post by majmuh24)
Indeed! Which method would you suggest - exponential ladders, squaring the ratio, or the Dalekian algorithm?

Dalek's already done that.
Gauss may have been the prince, but Dalek is the god of mathematics, period.
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Kvothe the Arcane
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(Original post by majmuh24)
Indeed! Which method would you suggest - exponential ladders, squaring the ratio, or the Dalekian algorithm?



Dalek's already done that.

If  \{ S : x \in S \implies \ \text{x \ cannot \ be \ divided} \} , and a \in S, we can use Kaluza-Klein theory along with the topological 42 dimensional fractal self similarity between \mathbb R \to S

\dfrac{a}{b} = \vartheta ^2 \cdot \mathcal{L} \left( \dfrac{a}{b} \right)

Where  \mathcal L denotes the Laplace Transform of \dfrac{a}{b} as a function of parameter t, which is given by \mathcal{L} \{ f(t) \} = \displaystyle\int_0^{\infty} e^{-st} \cdot f(t) where s is an arbitrary parameter.

I've created my own branch of maths too
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I see. The scary thing is that this makes as much sense to me as certain real maths.
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Ari Ben Canaan
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I know a proof, but this page is too narrow to contain it. Soz.
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