Does modular arithmetic have different rules for different cases?

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The discussion revolves around proving the equation 2((a+b)mod 2π) = (2a+2b)mod 4π, with a and b being nonnegative and less than 2π. Participants suggest breaking the problem into cases based on the value of (a+b), specifically considering intervals 0 to 2π and 2π to 4π. The original poster seeks assistance with modular arithmetic identities to aid in the proof. The conversation emphasizes the importance of understanding modular arithmetic rules in different scenarios. Overall, the thread highlights a common challenge in applying modular arithmetic to isomorphism problems.
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Homework Statement


I'm trying to prove the following (as part of an isomorphism problem).
The problem is I don't know much about modular arithmetic, so would appreciate some help.

2((a+b)mod \ 2π) = (2a+2b)mod \ 4π.

given that a,b are both nonnegative and less than 2π.

Homework Equations


The Attempt at a Solution


I believe there are certain identities for modular arithmetic that I could use, but do not know. Can anyone tell me some or give me a link to some modular arithmetic identities? Thanks.

BiP
 
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Bipolarity said:

Homework Statement


I'm trying to prove the following (as part of an isomorphism problem).
The problem is I don't know much about modular arithmetic, so would appreciate some help.

2((a+b)mod \ 2π) = (2a+2b)mod \ 4π.

given that a,b are both nonnegative and less than 2π.


Homework Equations





The Attempt at a Solution


I believe there are certain identities for modular arithmetic that I could use, but do not know. Can anyone tell me some or give me a link to some modular arithmetic identities? Thanks.

BiP

Break it into cases i) 0<=(a+b)<2pi and ii) 2pi<=(a+b)<4pi.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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