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#1
Nov 8 2015 06:51pm
Let a and b be integers, and m be a positive integer.
Prove that:
ab ≡ [ (a (mod m)) · (b (mod m)) ] (mod m).

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#2
Nov 8 2015 07:11pm
maybe i misread it? we know ab (mod m) = a (mod m) * b (mod m), so doesn't that just become:
ab ≡ [ (a (mod m)) · (b (mod m)) ] (mod m)
ab ≡ [ (ab (mod m)) ] (mod m)
ab ≡ ab (mod m)
?
Member
Posts: 10,049
Joined: Aug 11 2014
Gold: 540.00
#3
Nov 8 2015 07:20pm
Quote (carteblanche @ Nov 8 2015 09:11pm)
maybe i misread it? we know ab (mod m) = a (mod m) * b (mod m), so doesn't that just become:
ab ≡ [ (a (mod m)) · (b (mod m)) ] (mod m)
ab ≡ [ (ab (mod m)) ] (mod m)
ab ≡ ab (mod m)
?


how?
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