Zn under addition modulo n forms an abelian group.

[@murtazanabi5735]

Point to point explanation... Good job...

[@radhakantabal4575]

God level explanation

[@ashunegi_]

Your way of explaining is very good

[@mathematicalbond2725]

Thanks a lot

[@daryaumed6021]

Thanks 😊

[@akash1616]

Mam yeh samajh nahi aaya ki a +n ( b +n c ) = a +n ( b+c ) kaise hua isme b+n c (b+c ) kaise ho gaya

[@neerkumar1026]

Saved my day

[@Rsingh1]

Thanks!

[@sonasharma4118]

Can we prove this by letting n= any number

[@mathematicalbond2725]

I am here sharing the link of RUclips video of my channel in which the topic you require is explained
ruclips.net/video/vRs8OXS4sYg/видео.htmlsi=kxAnGCEh8UzSJWC9

[@MahendraKumar-fu7bp]

I like the way you explained. Just a point which I think can be added :When checking existence of inverse in Zn modulo n
Actually (n-a) belongs to( Zn ,+)only when a>0 but when a=0 , we have to write it seperately, isn't it?

[@mathematicalbond2725]

Yes. U r right. By chance it is left.

[@anarel8086]

No it is not necessary to write for 0 separately as in Zn; [0] is identified by [n] and n-a is n-n in that case.

[@bharatdan3980]

Thankyou ma'am

[@kritikagupta4957]

N ko n se divide kare ge toh 1 aaye ga

[@rahul_73]

Thanks ma'am

[@lochansharma1007]

Thanku mam

[@mathematicalbond2725]

Welcome Lochan

[@murtazanabi5735]

I recommend u use board and appear Profisinally....

[@Ragavi......]

Mam teach in English , my kind request.

[@mathematicalbond2725]

My students can't understand only English. So I have to use both the languages.