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anonymous
 one year ago
if a ≡ b (mod c), then is it true that ad ≡ bd (mod c) for all integer d?
anonymous
 one year ago
if a ≡ b (mod c), then is it true that ad ≡ bd (mod c) for all integer d?

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3\(a\equiv b\pmod{c} \implies c\mid (ab) \implies c\mid d(ab)\\~\\ \implies d(ab)\equiv 0 \pmod{c} \implies ad\equiv bd\pmod{c}\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3do you see any restrictions on \(d\) ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, I don't see any restriction on d. So it's true?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Neither do I. So yes it is true for all \(d\in\mathbb Z\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how did you get d(ab) ≡ 0 (mod c) though?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3That is the definition of congruence : \[n\mid m \iff m\equiv 0 \pmod{n}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh I see. Could you have done this though? c  d(ab) is the same as c  d(ab)  0, which by definition d(ab) ≡ 0 (mod c)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3I don't see why you want to have that intermediate step of subtracting 0

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3unless you're taking the pattern seriously : \[x\equiv y\pmod{n} \iff n\mid(xy)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I was thinking about the definition in terns of two things. x ≡ y (mod z) means z  (x  y) so by subtracting 0, i.e c  d(ab)  0, I was comparing d(ab) to x and 0 to y

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah, that pattern ^^

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3if it helps, then yes you may do that :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3btw the converse of the conditional in main question does not hold : \[ad\equiv bd\pmod{c} \implies a\equiv b\pmod{c}\] is false in general

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right. It's true if gcd(c,d) = 1. The proof was provided in the book :)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.3Yes, more generally, below holds : \[ad\equiv bd\pmod{c} \implies a\equiv b\pmod{\frac{c}{\gcd(d,c)}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0:O I didn't know that. Thank you for the info!
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