anonymous
  • anonymous
if a ≡ b (mod c), then is it true that ad ≡ bd (mod c) for all integer d?
Mathematics
chestercat
  • chestercat
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ganeshie8
  • ganeshie8
\(a\equiv b\pmod{c} \implies c\mid (a-b) \implies c\mid d(a-b)\\~\\ \implies d(a-b)\equiv 0 \pmod{c} \implies ad\equiv bd\pmod{c}\)
ganeshie8
  • ganeshie8
do you see any restrictions on \(d\) ?
anonymous
  • anonymous
No, I don't see any restriction on d. So it's true?

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ganeshie8
  • ganeshie8
Neither do I. So yes it is true for all \(d\in\mathbb Z\)
anonymous
  • anonymous
how did you get d(a-b) ≡ 0 (mod c) though?
ganeshie8
  • ganeshie8
That is the definition of congruence : \[n\mid m \iff m\equiv 0 \pmod{n}\]
anonymous
  • anonymous
oh I see. Could you have done this though? c | d(a-b) is the same as c | d(a-b) - 0, which by definition d(a-b) ≡ 0 (mod c)
ganeshie8
  • ganeshie8
I don't see why you want to have that intermediate step of subtracting 0
ganeshie8
  • ganeshie8
unless you're taking the pattern seriously : \[x\equiv y\pmod{n} \iff n\mid(x-y)\]
anonymous
  • anonymous
I 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(a-b) - 0, I was comparing d(a-b) to x and 0 to y
anonymous
  • anonymous
Yeah, that pattern ^^
ganeshie8
  • ganeshie8
if it helps, then yes you may do that :)
ganeshie8
  • ganeshie8
btw 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
  • anonymous
right. It's true if gcd(c,d) = 1. The proof was provided in the book :)
ganeshie8
  • ganeshie8
Yes, more generally, below holds : \[ad\equiv bd\pmod{c} \implies a\equiv b\pmod{\frac{c}{\gcd(d,c)}}\]
anonymous
  • anonymous
:O I didn't know that. Thank you for the info!
ganeshie8
  • ganeshie8
np

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