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May 18th, 2019, 01:09 AM   #1
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Arrow Balance of arithmetic operations «Ghayna and Ayna»

Hello!
Balance of arithmetic operations «Ghayna and Ayna».pdf
https://drive.google.com/file/d/1DDn...ew?usp=sharing

Last edited by Edward Sky; May 18th, 2019 at 01:24 AM.
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May 18th, 2019, 07:50 AM   #2
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Math Focus: Algebraic Geometry
Not sure what this means. For $a,b \neq 0$ $\frac{b}{a} \times \frac{a}{b} = \frac{ba}{ba} = 1$ sure.

Also $\frac{a \times b}{b \times a} = 1$.

Also, $\frac{0}{0}$ is not zero as mentioned in the pdf.

Most of these points just show that the Commutative property works. I'm not sure what the purpose of this is, but if I'm missing something I'd love to hear it.

Last edited by skipjack; May 18th, 2019 at 05:18 PM. Reason: Typo
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May 18th, 2019, 04:30 PM   #3
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Ghayn and Ayn.ods (OpenOffice Calc)
https://drive.google.com/file/d/1hLh...02zQGF74N/view
Thanks for the comment.

Last edited by Edward Sky; May 18th, 2019 at 04:48 PM.
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May 27th, 2019, 11:01 AM   #4
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Balance of arithmetic operations «Alpha and Omega».pdf:
https://drive.google.com/file/d/1hqX...ew?usp=sharing
Alpha and Omega.ods (OpenOffice Calc):
https://drive.google.com/file/d/1znN...ew?usp=sharing
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May 27th, 2019, 03:29 PM   #5
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Math Focus: Algebraic Geometry
Your $\Omega=\frac{0}{0}$ does not exist. You cannot do any further manipulation with it.

$\frac{0}{0} = \Omega \implies 0 \times \Omega = 0$

This means $\Omega$ can be ANYTHING, and is therefore disallowed by convention. Allowing it leads to some issues. One such issue is displayed in the semi-famous $1=2$ proof, I'll leave out the deception to show the point.

$\Omega = 1$ in this case,

Let $a=b$ , then

$a^2 = ab$

$a^2 - b^2 = ab-b^2$

$(a+b)(a-b) = b(a-b)$ Keep in mind $(a-b)=0$, so we divide by $(a-b)$ here.

$a+b = b$

$2b=b$

$2=1$

Also, for your $0 \times \frac{10}{0}=0$, you can just let $\Omega = 1$ and cancel the $0$'s to find $10 = 0$.
Thanks from topsquark

Last edited by Greens; May 27th, 2019 at 03:36 PM.
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