anonymous
  • anonymous
Is this problem an application of the cancelation law of congruences? Prove that if bd ≡ bd' (mod p), where p is a prime, and p does not divide b, then d ≡ d' (mod p)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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ganeshie8
  • ganeshie8
That follows trivially from euclid lemma : \(p\mid mn \implies p\mid m~~\text{or}~~p\mid n\)
ganeshie8
  • ganeshie8
\[bd\equiv bd'\pmod{p} \implies p\mid b (d-d')\] since \(p\nmid b\), it must be the case that \(p\mid (d-d')\) which is equivalent to saying \(d\equiv d'\pmod{p}\)
anonymous
  • anonymous
ah.... I see. Is gcd(b,p) = 1 though?

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ganeshie8
  • ganeshie8
\(p\nmid b ~~\iff~~\gcd(p,b)=1 \)
anonymous
  • anonymous
so in general, m does not divide n if and only if gcd(m,n)=1?
ganeshie8
  • ganeshie8
that is correct only if \(m\) is a prime
ganeshie8
  • ganeshie8
consider an example : \(m = 4,~ n = 6\) \(4\nmid 6\) but \(\gcd(4,6)\ne 1\)
ganeshie8
  • ganeshie8
btw, in this thread \(p\) is assumed to be a prime.
anonymous
  • anonymous
Ah... I see. Thank you :')
anonymous
  • anonymous
So there are two ways to do the problem above. Euclid lemma or cancelation law (since gcd(b,p) = 1)
ganeshie8
  • ganeshie8
pretty sure there are many other ways to prove it, as you can see the proof is trivial
anonymous
  • anonymous
right. Thank you :)

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