Proof that (ab) mod n = (a mod n) (b mod n ) | Abstract Algebra

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Update: Check the comments for the mistake in this video.
Another fun proof on modular arithmetic.
I'll do another video with examples. I just realised that would be a lot more helpful in understanding the concept.
Ask me in the comment section if you have any doubts.
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(ab) mod n = (a mod n) (b mod n) (mod n) !

catalinliviugherghe
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4:30 No. If r1*r2 > n, then (a*b) mod n < r1*r2. Try, for example, n = 7, a = 12, and b = 10. Then a mod n = 5 = r1; b mod n = 3 = r2; (a*b) mod n = 120 mod 7 = 1. But r1*r2 = 15. All you can say is that (a*b) mod n = (r1*r2) mod n.

MrKarpovy