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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?

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  1. ganeshie8
    • one year ago
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    \(a\equiv b\pmod{c} \implies c\mid (a-b) \implies c\mid d(a-b)\\~\\ \implies d(a-b)\equiv 0 \pmod{c} \implies ad\equiv bd\pmod{c}\)

  2. ganeshie8
    • one year ago
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    do you see any restrictions on \(d\) ?

  3. anonymous
    • one year ago
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    No, I don't see any restriction on d. So it's true?

  4. ganeshie8
    • one year ago
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    Neither do I. So yes it is true for all \(d\in\mathbb Z\)

  5. anonymous
    • one year ago
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    how did you get d(a-b) ≡ 0 (mod c) though?

  6. ganeshie8
    • one year ago
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    That is the definition of congruence : \[n\mid m \iff m\equiv 0 \pmod{n}\]

  7. anonymous
    • one year ago
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    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)

  8. ganeshie8
    • one year ago
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    I don't see why you want to have that intermediate step of subtracting 0

  9. ganeshie8
    • one year ago
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    unless you're taking the pattern seriously : \[x\equiv y\pmod{n} \iff n\mid(x-y)\]

  10. anonymous
    • one year ago
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    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

  11. anonymous
    • one year ago
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    Yeah, that pattern ^^

  12. ganeshie8
    • one year ago
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    if it helps, then yes you may do that :)

  13. ganeshie8
    • one year ago
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    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

  14. anonymous
    • one year ago
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    right. It's true if gcd(c,d) = 1. The proof was provided in the book :)

  15. ganeshie8
    • one year ago
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    Yes, more generally, below holds : \[ad\equiv bd\pmod{c} \implies a\equiv b\pmod{\frac{c}{\gcd(d,c)}}\]

  16. anonymous
    • one year ago
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    :O I didn't know that. Thank you for the info!

  17. ganeshie8
    • one year ago
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    np

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