Writing proofs with modular arithmetic

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I am enrolled in Discrete Mathematics 2 and I am having trouble understand a lot of the material. For the particular problems I need help with I need to:



Prove each of the given statements, assuming that
$a,b,c,d$, and $n$ are integers with $n>1$ and that $aequiv cbmod n$
and $bequiv dbmod n$.



These are the statements:




  1. a. $a+bequiv c+dbmod n$

    b. $a-bequiv c-dbmod n$

  2. $a^2equiv c^2bmod n$

  3. $a^mequiv c^mbmod n$ for all integers $mge1$ (Use mathematical
    induction on $m$).



I am in pretty bad shape as far as approaching and completing these problems and I need this to be explained to me as simply as possible (as if I am a 3 year old). Thank you all for your help in advanced.






discrete-mathematics proof-writing induction modular-arithmetic







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  • 2




    What does it mean for two numbers to be congruent mod n?
    – John Brevik
    Oct 30 '15 at 22:19










  • $a equiv b pmodn iff n ,vert , (a-b)$
    – JVV
    Oct 30 '15 at 22:23











  • ethz.ch/en/news-and-events/eth-news/news/2018/08/…
    – Piquito
    Aug 2 at 1:02














up vote
3
down vote

favorite
1












I am enrolled in Discrete Mathematics 2 and I am having trouble understand a lot of the material. For the particular problems I need help with I need to:



Prove each of the given statements, assuming that
$a,b,c,d$, and $n$ are integers with $n>1$ and that $aequiv cbmod n$
and $bequiv dbmod n$.



These are the statements:




  1. a. $a+bequiv c+dbmod n$

    b. $a-bequiv c-dbmod n$

  2. $a^2equiv c^2bmod n$

  3. $a^mequiv c^mbmod n$ for all integers $mge1$ (Use mathematical
    induction on $m$).



I am in pretty bad shape as far as approaching and completing these problems and I need this to be explained to me as simply as possible (as if I am a 3 year old). Thank you all for your help in advanced.






discrete-mathematics proof-writing induction modular-arithmetic







share|cite|improve this question



















  • 2




    What does it mean for two numbers to be congruent mod n?
    – John Brevik
    Oct 30 '15 at 22:19










  • $a equiv b pmodn iff n ,vert , (a-b)$
    – JVV
    Oct 30 '15 at 22:23











  • ethz.ch/en/news-and-events/eth-news/news/2018/08/…
    – Piquito
    Aug 2 at 1:02












up vote
3
down vote

favorite
1









up vote
3
down vote

favorite
1






1





I am enrolled in Discrete Mathematics 2 and I am having trouble understand a lot of the material. For the particular problems I need help with I need to:



Prove each of the given statements, assuming that
$a,b,c,d$, and $n$ are integers with $n>1$ and that $aequiv cbmod n$
and $bequiv dbmod n$.



These are the statements:




  1. a. $a+bequiv c+dbmod n$

    b. $a-bequiv c-dbmod n$

  2. $a^2equiv c^2bmod n$

  3. $a^mequiv c^mbmod n$ for all integers $mge1$ (Use mathematical
    induction on $m$).



I am in pretty bad shape as far as approaching and completing these problems and I need this to be explained to me as simply as possible (as if I am a 3 year old). Thank you all for your help in advanced.






discrete-mathematics proof-writing induction modular-arithmetic







share|cite|improve this question















I am enrolled in Discrete Mathematics 2 and I am having trouble understand a lot of the material. For the particular problems I need help with I need to:



Prove each of the given statements, assuming that
$a,b,c,d$, and $n$ are integers with $n>1$ and that $aequiv cbmod n$
and $bequiv dbmod n$.



These are the statements:




  1. a. $a+bequiv c+dbmod n$

    b. $a-bequiv c-dbmod n$

  2. $a^2equiv c^2bmod n$

  3. $a^mequiv c^mbmod n$ for all integers $mge1$ (Use mathematical
    induction on $m$).



I am in pretty bad shape as far as approaching and completing these problems and I need this to be explained to me as simply as possible (as if I am a 3 year old). Thank you all for your help in advanced.






discrete-mathematics proof-writing induction modular-arithmetic




discrete-mathematics proof-writing induction modular-arithmetic






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edited Feb 22 at 3:37









Parcly Taxel

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asked Oct 30 '15 at 22:15









user285507

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  • 2




    What does it mean for two numbers to be congruent mod n?
    – John Brevik
    Oct 30 '15 at 22:19










  • $a equiv b pmodn iff n ,vert , (a-b)$
    – JVV
    Oct 30 '15 at 22:23











  • ethz.ch/en/news-and-events/eth-news/news/2018/08/…
    – Piquito
    Aug 2 at 1:02












  • 2




    What does it mean for two numbers to be congruent mod n?
    – John Brevik
    Oct 30 '15 at 22:19










  • $a equiv b pmodn iff n ,vert , (a-b)$
    – JVV
    Oct 30 '15 at 22:23











  • ethz.ch/en/news-and-events/eth-news/news/2018/08/…
    – Piquito
    Aug 2 at 1:02







2




2




What does it mean for two numbers to be congruent mod n?
– John Brevik
Oct 30 '15 at 22:19




What does it mean for two numbers to be congruent mod n?
– John Brevik
Oct 30 '15 at 22:19












$a equiv b pmodn iff n ,vert , (a-b)$
– JVV
Oct 30 '15 at 22:23





$a equiv b pmodn iff n ,vert , (a-b)$
– JVV
Oct 30 '15 at 22:23













ethz.ch/en/news-and-events/eth-news/news/2018/08/…
– Piquito
Aug 2 at 1:02




ethz.ch/en/news-and-events/eth-news/news/2018/08/…
– Piquito
Aug 2 at 1:02










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I'll do the first one for you. The same logic follows for the rest. $aequiv c pmod n$ means $a=c+mn$, where $minmathbbZ$ and similarly for $b=d$ mod$n$. So $(c+d)=(a+m_1n)+(b+m_2n)=(a+b)+(m_1+m_2)nequiv(a+b) pmod n$






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    I'll do the first one for you. The same logic follows for the rest. $aequiv c pmod n$ means $a=c+mn$, where $minmathbbZ$ and similarly for $b=d$ mod$n$. So $(c+d)=(a+m_1n)+(b+m_2n)=(a+b)+(m_1+m_2)nequiv(a+b) pmod n$






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      up vote
      0
      down vote













      I'll do the first one for you. The same logic follows for the rest. $aequiv c pmod n$ means $a=c+mn$, where $minmathbbZ$ and similarly for $b=d$ mod$n$. So $(c+d)=(a+m_1n)+(b+m_2n)=(a+b)+(m_1+m_2)nequiv(a+b) pmod n$






      share|cite|improve this answer
























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        up vote
        0
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        I'll do the first one for you. The same logic follows for the rest. $aequiv c pmod n$ means $a=c+mn$, where $minmathbbZ$ and similarly for $b=d$ mod$n$. So $(c+d)=(a+m_1n)+(b+m_2n)=(a+b)+(m_1+m_2)nequiv(a+b) pmod n$






        share|cite|improve this answer














        I'll do the first one for you. The same logic follows for the rest. $aequiv c pmod n$ means $a=c+mn$, where $minmathbbZ$ and similarly for $b=d$ mod$n$. So $(c+d)=(a+m_1n)+(b+m_2n)=(a+b)+(m_1+m_2)nequiv(a+b) pmod n$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 21 at 18:50









        man on laptop

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        answered Oct 30 '15 at 22:27









        user153582

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