proof with multiplicative inverses mod n [duplicate]












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  • Multiplicative inverse questions

    2 answers




I am trying to write a proof for-



show that if [b] and [c] are both multiplicative inverses of [a] in Zn, then b congruent c (mod n)



I don't know a lot about multiplicative inverse proofs and any help will be appreciated.






proof-writing inverse







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marked as duplicate by rschwieb abstract-algebra
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 Jan 30 at 19:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    If you search the site you'll find more than one solution demonstrating that multiplicative inverses are unique. Even if they aren't exactly your situation, they will point out to you exactly why this is so (sometimes in a more general context.)
    $endgroup$
    – rschwieb
    Jan 30 at 19:14


















0












$begingroup$



This question already has an answer here:




  • Multiplicative inverse questions

    2 answers




I am trying to write a proof for-



show that if [b] and [c] are both multiplicative inverses of [a] in Zn, then b congruent c (mod n)



I don't know a lot about multiplicative inverse proofs and any help will be appreciated.






proof-writing inverse







share|cite|improve this question











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marked as duplicate by rschwieb abstract-algebra
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 Jan 30 at 19:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    If you search the site you'll find more than one solution demonstrating that multiplicative inverses are unique. Even if they aren't exactly your situation, they will point out to you exactly why this is so (sometimes in a more general context.)
    $endgroup$
    – rschwieb
    Jan 30 at 19:14
















0












0








0





$begingroup$



This question already has an answer here:




  • Multiplicative inverse questions

    2 answers




I am trying to write a proof for-



show that if [b] and [c] are both multiplicative inverses of [a] in Zn, then b congruent c (mod n)



I don't know a lot about multiplicative inverse proofs and any help will be appreciated.






proof-writing inverse







share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Multiplicative inverse questions

    2 answers




I am trying to write a proof for-



show that if [b] and [c] are both multiplicative inverses of [a] in Zn, then b congruent c (mod n)



I don't know a lot about multiplicative inverse proofs and any help will be appreciated.





This question already has an answer here:




  • Multiplicative inverse questions

    2 answers







proof-writing inverse




proof-writing inverse






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 30 at 19:29







Kerry Brennan

















asked Jan 30 at 19:00









Kerry BrennanKerry Brennan

1




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marked as duplicate by rschwieb abstract-algebra
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marked as duplicate by rschwieb abstract-algebra
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 Jan 30 at 19:13


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    If you search the site you'll find more than one solution demonstrating that multiplicative inverses are unique. Even if they aren't exactly your situation, they will point out to you exactly why this is so (sometimes in a more general context.)
    $endgroup$
    – rschwieb
    Jan 30 at 19:14




















  • $begingroup$
    If you search the site you'll find more than one solution demonstrating that multiplicative inverses are unique. Even if they aren't exactly your situation, they will point out to you exactly why this is so (sometimes in a more general context.)
    $endgroup$
    – rschwieb
    Jan 30 at 19:14


















$begingroup$
If you search the site you'll find more than one solution demonstrating that multiplicative inverses are unique. Even if they aren't exactly your situation, they will point out to you exactly why this is so (sometimes in a more general context.)
$endgroup$
– rschwieb
Jan 30 at 19:14






$begingroup$
If you search the site you'll find more than one solution demonstrating that multiplicative inverses are unique. Even if they aren't exactly your situation, they will point out to you exactly why this is so (sometimes in a more general context.)
$endgroup$
– rschwieb
Jan 30 at 19:14












1 Answer
1






active

oldest

votes


















1












$begingroup$

Multiplication is commutative and associative.
$$[b] = [b][1] = [b]([c][a]) = [c]([b][a]) = [c][1] = [c]$$






share|cite|improve this answer









$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Multiplication is commutative and associative.
    $$[b] = [b][1] = [b]([c][a]) = [c]([b][a]) = [c][1] = [c]$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Multiplication is commutative and associative.
      $$[b] = [b][1] = [b]([c][a]) = [c]([b][a]) = [c][1] = [c]$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Multiplication is commutative and associative.
        $$[b] = [b][1] = [b]([c][a]) = [c]([b][a]) = [c][1] = [c]$$






        share|cite|improve this answer









        $endgroup$



        Multiplication is commutative and associative.
        $$[b] = [b][1] = [b]([c][a]) = [c]([b][a]) = [c][1] = [c]$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 30 at 19:02









        Nathaniel MayerNathaniel Mayer

        1,863516




        1,863516















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