Euler's theorem with fractions












2












$begingroup$


Suppose I have this:



$frac{6^{666}}{2^{6}}$ (mod $125$)



I saw it is possible to reduce only the numerator's power modulo Euler's phi function. Can someone explain why is that possible?



It is essentially this:



$frac{6^{666 space (mod phi(125))}}{2^{6}}$ (mod $125$)






elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$












  • $begingroup$
    What's the question? There's no need to reduce the exponent in $2^6$ as it is already small.
    $endgroup$
    – lulu
    Jan 28 at 17:06






  • 1




    $begingroup$
    Can you clarify your question? It really isn't clear what you are asking.
    $endgroup$
    – lulu
    Jan 28 at 17:23










  • $begingroup$
    My question is why it works?
    $endgroup$
    – Michael Munta
    Jan 28 at 17:25






  • 1




    $begingroup$
    Why what works?
    $endgroup$
    – lulu
    Jan 28 at 17:27






  • 1




    $begingroup$
    @MichaelMunta If, say, you had $frac {6^{666}}{2^{501}}$ you could use Euler to write that as $frac {6^{66}}{2^1}pmod {125}$.
    $endgroup$
    – lulu
    Jan 28 at 17:41


















2












$begingroup$


Suppose I have this:



$frac{6^{666}}{2^{6}}$ (mod $125$)



I saw it is possible to reduce only the numerator's power modulo Euler's phi function. Can someone explain why is that possible?



It is essentially this:



$frac{6^{666 space (mod phi(125))}}{2^{6}}$ (mod $125$)






elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$












  • $begingroup$
    What's the question? There's no need to reduce the exponent in $2^6$ as it is already small.
    $endgroup$
    – lulu
    Jan 28 at 17:06






  • 1




    $begingroup$
    Can you clarify your question? It really isn't clear what you are asking.
    $endgroup$
    – lulu
    Jan 28 at 17:23










  • $begingroup$
    My question is why it works?
    $endgroup$
    – Michael Munta
    Jan 28 at 17:25






  • 1




    $begingroup$
    Why what works?
    $endgroup$
    – lulu
    Jan 28 at 17:27






  • 1




    $begingroup$
    @MichaelMunta If, say, you had $frac {6^{666}}{2^{501}}$ you could use Euler to write that as $frac {6^{66}}{2^1}pmod {125}$.
    $endgroup$
    – lulu
    Jan 28 at 17:41
















2












2








2


2



$begingroup$


Suppose I have this:



$frac{6^{666}}{2^{6}}$ (mod $125$)



I saw it is possible to reduce only the numerator's power modulo Euler's phi function. Can someone explain why is that possible?



It is essentially this:



$frac{6^{666 space (mod phi(125))}}{2^{6}}$ (mod $125$)






elementary-number-theory modular-arithmetic







share|cite|improve this question









$endgroup$




Suppose I have this:



$frac{6^{666}}{2^{6}}$ (mod $125$)



I saw it is possible to reduce only the numerator's power modulo Euler's phi function. Can someone explain why is that possible?



It is essentially this:



$frac{6^{666 space (mod phi(125))}}{2^{6}}$ (mod $125$)






elementary-number-theory modular-arithmetic




elementary-number-theory modular-arithmetic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 28 at 17:05









Michael MuntaMichael Munta

106111




106111












  • $begingroup$
    What's the question? There's no need to reduce the exponent in $2^6$ as it is already small.
    $endgroup$
    – lulu
    Jan 28 at 17:06






  • 1




    $begingroup$
    Can you clarify your question? It really isn't clear what you are asking.
    $endgroup$
    – lulu
    Jan 28 at 17:23










  • $begingroup$
    My question is why it works?
    $endgroup$
    – Michael Munta
    Jan 28 at 17:25






  • 1




    $begingroup$
    Why what works?
    $endgroup$
    – lulu
    Jan 28 at 17:27






  • 1




    $begingroup$
    @MichaelMunta If, say, you had $frac {6^{666}}{2^{501}}$ you could use Euler to write that as $frac {6^{66}}{2^1}pmod {125}$.
    $endgroup$
    – lulu
    Jan 28 at 17:41




















  • $begingroup$
    What's the question? There's no need to reduce the exponent in $2^6$ as it is already small.
    $endgroup$
    – lulu
    Jan 28 at 17:06






  • 1




    $begingroup$
    Can you clarify your question? It really isn't clear what you are asking.
    $endgroup$
    – lulu
    Jan 28 at 17:23










  • $begingroup$
    My question is why it works?
    $endgroup$
    – Michael Munta
    Jan 28 at 17:25






  • 1




    $begingroup$
    Why what works?
    $endgroup$
    – lulu
    Jan 28 at 17:27






  • 1




    $begingroup$
    @MichaelMunta If, say, you had $frac {6^{666}}{2^{501}}$ you could use Euler to write that as $frac {6^{66}}{2^1}pmod {125}$.
    $endgroup$
    – lulu
    Jan 28 at 17:41


















$begingroup$
What's the question? There's no need to reduce the exponent in $2^6$ as it is already small.
$endgroup$
– lulu
Jan 28 at 17:06




$begingroup$
What's the question? There's no need to reduce the exponent in $2^6$ as it is already small.
$endgroup$
– lulu
Jan 28 at 17:06




1




1




$begingroup$
Can you clarify your question? It really isn't clear what you are asking.
$endgroup$
– lulu
Jan 28 at 17:23




$begingroup$
Can you clarify your question? It really isn't clear what you are asking.
$endgroup$
– lulu
Jan 28 at 17:23












$begingroup$
My question is why it works?
$endgroup$
– Michael Munta
Jan 28 at 17:25




$begingroup$
My question is why it works?
$endgroup$
– Michael Munta
Jan 28 at 17:25




1




1




$begingroup$
Why what works?
$endgroup$
– lulu
Jan 28 at 17:27




$begingroup$
Why what works?
$endgroup$
– lulu
Jan 28 at 17:27




1




1




$begingroup$
@MichaelMunta If, say, you had $frac {6^{666}}{2^{501}}$ you could use Euler to write that as $frac {6^{66}}{2^1}pmod {125}$.
$endgroup$
– lulu
Jan 28 at 17:41






$begingroup$
@MichaelMunta If, say, you had $frac {6^{666}}{2^{501}}$ you could use Euler to write that as $frac {6^{66}}{2^1}pmod {125}$.
$endgroup$
– lulu
Jan 28 at 17:41












2 Answers
2






active

oldest

votes


















4












$begingroup$

It's valid to mod out the arguments of a fraction - just like it is for the arguments sums and products.



If $(B,n)= 1,$ then $bmod n!:, begin{align}Aequiv a\ Bequiv bend{align},Rightarrow, dfrac{A}B,overset{rm def}equiv Acdot B^{-1}equiv acdot b^{-1},overset{rm def}equiv, dfrac{a}b$



So the answer to your question as to "why it works" is that unwinding the definition of a fraction yields a composition of a product and inverse operation - and those operations are compatible with modular aithmetic hence so too is their composition (modular "division" by units = invertibles = integers $B$ coprime to the modulus).



See this answer for much further discussion.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So if $B equiv b$ then also $B^{-1} equiv b^{-1}$?
    $endgroup$
    – Michael Munta
    Jan 28 at 19:29










  • $begingroup$
    @Michael Yes, multiply $,Bequiv b $ by $ b^{-1}B^{-1} $ [assuming $,(B,n)=1,,$ so also $,(b,n)=(B,n)=1$] $ $
    $endgroup$
    – Bill Dubuque
    Jan 28 at 20:06












  • $begingroup$
    It's important to realize, as Bill Dubuque pointed out, that you must have $gcd(B, n) = 1$. But if so, if $B equiv b$ then $1equiv B*B^{-1} equiv b*B^{-1}$ and $b^{-1} equiv b^{-1}(b*B^{-1}) equiv B^{-1}$. Which isnt surprising as equivalence carries over multiplication and we are multiplying the inverses to get $1$
    $endgroup$
    – fleablood
    Feb 12 at 18:01



















0












$begingroup$

As $2$ is relatively prime to $125$ then $2$ is invertable so there is a $[frac 12]$ so that $2[frac 12]equiv 1 pmod {125}$ (just let $[frac 12] = 63$ but we don't actually care what $[frac 12]$ is; just that it exists) so for any $m = 2^jb$ then $frac m{2^j}equiv frac m{2^j}(2^j*[frac 12]^j) equiv m*[frac 12]^j$.



So $frac {6^{666}}{2^6} equiv 6^{666}[frac 12]^6 equiv 6^{phi (666)}[frac 12]^6pmod {125}$



For notation purposes it is pefectly acceptable to write $frac 12 equiv 63 pmod {125}$ and to use the fraction notation. (Although it's perhaps misleading to use the $2^{-1} equiv 63 pmod {125}$ notation instead.)



It's just important to realize that the residuce class is not ${frac 12 + 125k| k in mathbb Z}$ but ${m|2m equiv 1 pmod 125} = {m|exists kin mathbb Z; 2m = 1 + 125k}={m|2m = 1 + 125k$ for some odd $k} ={frac {1+125(2k+1)}2| k in mathbb Z}={63+ 125k|k in mathbb Z}$.



And it's important to realize that if $k$ and $n$ arent relatively prime there isn't and such $k^{-1} mod n$ and we can't use $frac 1k pmod n$






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    It's valid to mod out the arguments of a fraction - just like it is for the arguments sums and products.



    If $(B,n)= 1,$ then $bmod n!:, begin{align}Aequiv a\ Bequiv bend{align},Rightarrow, dfrac{A}B,overset{rm def}equiv Acdot B^{-1}equiv acdot b^{-1},overset{rm def}equiv, dfrac{a}b$



    So the answer to your question as to "why it works" is that unwinding the definition of a fraction yields a composition of a product and inverse operation - and those operations are compatible with modular aithmetic hence so too is their composition (modular "division" by units = invertibles = integers $B$ coprime to the modulus).



    See this answer for much further discussion.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So if $B equiv b$ then also $B^{-1} equiv b^{-1}$?
      $endgroup$
      – Michael Munta
      Jan 28 at 19:29










    • $begingroup$
      @Michael Yes, multiply $,Bequiv b $ by $ b^{-1}B^{-1} $ [assuming $,(B,n)=1,,$ so also $,(b,n)=(B,n)=1$] $ $
      $endgroup$
      – Bill Dubuque
      Jan 28 at 20:06












    • $begingroup$
      It's important to realize, as Bill Dubuque pointed out, that you must have $gcd(B, n) = 1$. But if so, if $B equiv b$ then $1equiv B*B^{-1} equiv b*B^{-1}$ and $b^{-1} equiv b^{-1}(b*B^{-1}) equiv B^{-1}$. Which isnt surprising as equivalence carries over multiplication and we are multiplying the inverses to get $1$
      $endgroup$
      – fleablood
      Feb 12 at 18:01
















    4












    $begingroup$

    It's valid to mod out the arguments of a fraction - just like it is for the arguments sums and products.



    If $(B,n)= 1,$ then $bmod n!:, begin{align}Aequiv a\ Bequiv bend{align},Rightarrow, dfrac{A}B,overset{rm def}equiv Acdot B^{-1}equiv acdot b^{-1},overset{rm def}equiv, dfrac{a}b$



    So the answer to your question as to "why it works" is that unwinding the definition of a fraction yields a composition of a product and inverse operation - and those operations are compatible with modular aithmetic hence so too is their composition (modular "division" by units = invertibles = integers $B$ coprime to the modulus).



    See this answer for much further discussion.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So if $B equiv b$ then also $B^{-1} equiv b^{-1}$?
      $endgroup$
      – Michael Munta
      Jan 28 at 19:29










    • $begingroup$
      @Michael Yes, multiply $,Bequiv b $ by $ b^{-1}B^{-1} $ [assuming $,(B,n)=1,,$ so also $,(b,n)=(B,n)=1$] $ $
      $endgroup$
      – Bill Dubuque
      Jan 28 at 20:06












    • $begingroup$
      It's important to realize, as Bill Dubuque pointed out, that you must have $gcd(B, n) = 1$. But if so, if $B equiv b$ then $1equiv B*B^{-1} equiv b*B^{-1}$ and $b^{-1} equiv b^{-1}(b*B^{-1}) equiv B^{-1}$. Which isnt surprising as equivalence carries over multiplication and we are multiplying the inverses to get $1$
      $endgroup$
      – fleablood
      Feb 12 at 18:01














    4












    4








    4





    $begingroup$

    It's valid to mod out the arguments of a fraction - just like it is for the arguments sums and products.



    If $(B,n)= 1,$ then $bmod n!:, begin{align}Aequiv a\ Bequiv bend{align},Rightarrow, dfrac{A}B,overset{rm def}equiv Acdot B^{-1}equiv acdot b^{-1},overset{rm def}equiv, dfrac{a}b$



    So the answer to your question as to "why it works" is that unwinding the definition of a fraction yields a composition of a product and inverse operation - and those operations are compatible with modular aithmetic hence so too is their composition (modular "division" by units = invertibles = integers $B$ coprime to the modulus).



    See this answer for much further discussion.






    share|cite|improve this answer











    $endgroup$



    It's valid to mod out the arguments of a fraction - just like it is for the arguments sums and products.



    If $(B,n)= 1,$ then $bmod n!:, begin{align}Aequiv a\ Bequiv bend{align},Rightarrow, dfrac{A}B,overset{rm def}equiv Acdot B^{-1}equiv acdot b^{-1},overset{rm def}equiv, dfrac{a}b$



    So the answer to your question as to "why it works" is that unwinding the definition of a fraction yields a composition of a product and inverse operation - and those operations are compatible with modular aithmetic hence so too is their composition (modular "division" by units = invertibles = integers $B$ coprime to the modulus).



    See this answer for much further discussion.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Feb 12 at 17:27

























    answered Jan 28 at 17:52









    Bill DubuqueBill Dubuque

    213k29195654




    213k29195654












    • $begingroup$
      So if $B equiv b$ then also $B^{-1} equiv b^{-1}$?
      $endgroup$
      – Michael Munta
      Jan 28 at 19:29










    • $begingroup$
      @Michael Yes, multiply $,Bequiv b $ by $ b^{-1}B^{-1} $ [assuming $,(B,n)=1,,$ so also $,(b,n)=(B,n)=1$] $ $
      $endgroup$
      – Bill Dubuque
      Jan 28 at 20:06












    • $begingroup$
      It's important to realize, as Bill Dubuque pointed out, that you must have $gcd(B, n) = 1$. But if so, if $B equiv b$ then $1equiv B*B^{-1} equiv b*B^{-1}$ and $b^{-1} equiv b^{-1}(b*B^{-1}) equiv B^{-1}$. Which isnt surprising as equivalence carries over multiplication and we are multiplying the inverses to get $1$
      $endgroup$
      – fleablood
      Feb 12 at 18:01


















    • $begingroup$
      So if $B equiv b$ then also $B^{-1} equiv b^{-1}$?
      $endgroup$
      – Michael Munta
      Jan 28 at 19:29










    • $begingroup$
      @Michael Yes, multiply $,Bequiv b $ by $ b^{-1}B^{-1} $ [assuming $,(B,n)=1,,$ so also $,(b,n)=(B,n)=1$] $ $
      $endgroup$
      – Bill Dubuque
      Jan 28 at 20:06












    • $begingroup$
      It's important to realize, as Bill Dubuque pointed out, that you must have $gcd(B, n) = 1$. But if so, if $B equiv b$ then $1equiv B*B^{-1} equiv b*B^{-1}$ and $b^{-1} equiv b^{-1}(b*B^{-1}) equiv B^{-1}$. Which isnt surprising as equivalence carries over multiplication and we are multiplying the inverses to get $1$
      $endgroup$
      – fleablood
      Feb 12 at 18:01
















    $begingroup$
    So if $B equiv b$ then also $B^{-1} equiv b^{-1}$?
    $endgroup$
    – Michael Munta
    Jan 28 at 19:29




    $begingroup$
    So if $B equiv b$ then also $B^{-1} equiv b^{-1}$?
    $endgroup$
    – Michael Munta
    Jan 28 at 19:29












    $begingroup$
    @Michael Yes, multiply $,Bequiv b $ by $ b^{-1}B^{-1} $ [assuming $,(B,n)=1,,$ so also $,(b,n)=(B,n)=1$] $ $
    $endgroup$
    – Bill Dubuque
    Jan 28 at 20:06






    $begingroup$
    @Michael Yes, multiply $,Bequiv b $ by $ b^{-1}B^{-1} $ [assuming $,(B,n)=1,,$ so also $,(b,n)=(B,n)=1$] $ $
    $endgroup$
    – Bill Dubuque
    Jan 28 at 20:06














    $begingroup$
    It's important to realize, as Bill Dubuque pointed out, that you must have $gcd(B, n) = 1$. But if so, if $B equiv b$ then $1equiv B*B^{-1} equiv b*B^{-1}$ and $b^{-1} equiv b^{-1}(b*B^{-1}) equiv B^{-1}$. Which isnt surprising as equivalence carries over multiplication and we are multiplying the inverses to get $1$
    $endgroup$
    – fleablood
    Feb 12 at 18:01




    $begingroup$
    It's important to realize, as Bill Dubuque pointed out, that you must have $gcd(B, n) = 1$. But if so, if $B equiv b$ then $1equiv B*B^{-1} equiv b*B^{-1}$ and $b^{-1} equiv b^{-1}(b*B^{-1}) equiv B^{-1}$. Which isnt surprising as equivalence carries over multiplication and we are multiplying the inverses to get $1$
    $endgroup$
    – fleablood
    Feb 12 at 18:01











    0












    $begingroup$

    As $2$ is relatively prime to $125$ then $2$ is invertable so there is a $[frac 12]$ so that $2[frac 12]equiv 1 pmod {125}$ (just let $[frac 12] = 63$ but we don't actually care what $[frac 12]$ is; just that it exists) so for any $m = 2^jb$ then $frac m{2^j}equiv frac m{2^j}(2^j*[frac 12]^j) equiv m*[frac 12]^j$.



    So $frac {6^{666}}{2^6} equiv 6^{666}[frac 12]^6 equiv 6^{phi (666)}[frac 12]^6pmod {125}$



    For notation purposes it is pefectly acceptable to write $frac 12 equiv 63 pmod {125}$ and to use the fraction notation. (Although it's perhaps misleading to use the $2^{-1} equiv 63 pmod {125}$ notation instead.)



    It's just important to realize that the residuce class is not ${frac 12 + 125k| k in mathbb Z}$ but ${m|2m equiv 1 pmod 125} = {m|exists kin mathbb Z; 2m = 1 + 125k}={m|2m = 1 + 125k$ for some odd $k} ={frac {1+125(2k+1)}2| k in mathbb Z}={63+ 125k|k in mathbb Z}$.



    And it's important to realize that if $k$ and $n$ arent relatively prime there isn't and such $k^{-1} mod n$ and we can't use $frac 1k pmod n$






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      As $2$ is relatively prime to $125$ then $2$ is invertable so there is a $[frac 12]$ so that $2[frac 12]equiv 1 pmod {125}$ (just let $[frac 12] = 63$ but we don't actually care what $[frac 12]$ is; just that it exists) so for any $m = 2^jb$ then $frac m{2^j}equiv frac m{2^j}(2^j*[frac 12]^j) equiv m*[frac 12]^j$.



      So $frac {6^{666}}{2^6} equiv 6^{666}[frac 12]^6 equiv 6^{phi (666)}[frac 12]^6pmod {125}$



      For notation purposes it is pefectly acceptable to write $frac 12 equiv 63 pmod {125}$ and to use the fraction notation. (Although it's perhaps misleading to use the $2^{-1} equiv 63 pmod {125}$ notation instead.)



      It's just important to realize that the residuce class is not ${frac 12 + 125k| k in mathbb Z}$ but ${m|2m equiv 1 pmod 125} = {m|exists kin mathbb Z; 2m = 1 + 125k}={m|2m = 1 + 125k$ for some odd $k} ={frac {1+125(2k+1)}2| k in mathbb Z}={63+ 125k|k in mathbb Z}$.



      And it's important to realize that if $k$ and $n$ arent relatively prime there isn't and such $k^{-1} mod n$ and we can't use $frac 1k pmod n$






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        As $2$ is relatively prime to $125$ then $2$ is invertable so there is a $[frac 12]$ so that $2[frac 12]equiv 1 pmod {125}$ (just let $[frac 12] = 63$ but we don't actually care what $[frac 12]$ is; just that it exists) so for any $m = 2^jb$ then $frac m{2^j}equiv frac m{2^j}(2^j*[frac 12]^j) equiv m*[frac 12]^j$.



        So $frac {6^{666}}{2^6} equiv 6^{666}[frac 12]^6 equiv 6^{phi (666)}[frac 12]^6pmod {125}$



        For notation purposes it is pefectly acceptable to write $frac 12 equiv 63 pmod {125}$ and to use the fraction notation. (Although it's perhaps misleading to use the $2^{-1} equiv 63 pmod {125}$ notation instead.)



        It's just important to realize that the residuce class is not ${frac 12 + 125k| k in mathbb Z}$ but ${m|2m equiv 1 pmod 125} = {m|exists kin mathbb Z; 2m = 1 + 125k}={m|2m = 1 + 125k$ for some odd $k} ={frac {1+125(2k+1)}2| k in mathbb Z}={63+ 125k|k in mathbb Z}$.



        And it's important to realize that if $k$ and $n$ arent relatively prime there isn't and such $k^{-1} mod n$ and we can't use $frac 1k pmod n$






        share|cite|improve this answer









        $endgroup$



        As $2$ is relatively prime to $125$ then $2$ is invertable so there is a $[frac 12]$ so that $2[frac 12]equiv 1 pmod {125}$ (just let $[frac 12] = 63$ but we don't actually care what $[frac 12]$ is; just that it exists) so for any $m = 2^jb$ then $frac m{2^j}equiv frac m{2^j}(2^j*[frac 12]^j) equiv m*[frac 12]^j$.



        So $frac {6^{666}}{2^6} equiv 6^{666}[frac 12]^6 equiv 6^{phi (666)}[frac 12]^6pmod {125}$



        For notation purposes it is pefectly acceptable to write $frac 12 equiv 63 pmod {125}$ and to use the fraction notation. (Although it's perhaps misleading to use the $2^{-1} equiv 63 pmod {125}$ notation instead.)



        It's just important to realize that the residuce class is not ${frac 12 + 125k| k in mathbb Z}$ but ${m|2m equiv 1 pmod 125} = {m|exists kin mathbb Z; 2m = 1 + 125k}={m|2m = 1 + 125k$ for some odd $k} ={frac {1+125(2k+1)}2| k in mathbb Z}={63+ 125k|k in mathbb Z}$.



        And it's important to realize that if $k$ and $n$ arent relatively prime there isn't and such $k^{-1} mod n$ and we can't use $frac 1k pmod n$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 12 at 17:54









        fleabloodfleablood

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