My Math Forum Balance of arithmetic operations «Ghayna and Ayna»

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 May 18th, 2019, 01:09 AM #1 Newbie   Joined: May 2019 From: Russian Federation Posts: 3 Thanks: 0 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.
 May 18th, 2019, 07:50 AM #2 Member     Joined: Oct 2018 From: USA Posts: 98 Thanks: 70 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
 May 18th, 2019, 04:30 PM #3 Newbie   Joined: May 2019 From: Russian Federation Posts: 3 Thanks: 0 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.
 May 27th, 2019, 11:01 AM #4 Newbie   Joined: May 2019 From: Russian Federation Posts: 3 Thanks: 0 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
 May 27th, 2019, 03:29 PM #5 Member     Joined: Oct 2018 From: USA Posts: 98 Thanks: 70 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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