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Hollywood_chrissy

  • 3 years ago

Prove that if a≡b mod n and c≡d mod n then a−c≡b−d mod n??

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  1. A.Avinash_Goutham
    • 3 years ago
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    u start like a=kn + b........

  2. FoolForMath
    • 3 years ago
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    \(a≡b \pmod n \implies a=n.k+b \) where \( k\in \mathbb{N} \) \(c≡d \pmod n \implies c=n.p+d \) where \( p\in \mathbb{N} \) \(a-c = nk+b-np-d = n(k-p) +(b-d)\), where \( (k-p)\in \mathbb{N} \). Hence \((a−c)≡(b−d) \pmod n \) QED!

  3. A.Avinash_Goutham
    • 3 years ago
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    c= pn+d a-c=n(k-p)+b-d :P......m late

  4. FoolForMath
    • 3 years ago
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    Similarly we can show that \( (a+c)≡(b+d) \pmod n \) and more generally \( (pa+qc)≡(pb+qd) \pmod n \)

  5. Hollywood_chrissy
    • 3 years ago
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    so basically we need to show that difference between a and c will be a multiple of n

  6. FoolForMath
    • 3 years ago
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    Yes.

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