In the ring $mathbb Z_5$ (class of modulo 5) compute $[3]^{-4}$.












0














$[3]$ means $x equiv 3mod(5)$. I know how to compute $[3]^{4}$ that is positive powers but I have not been in a situation where I need to take something modulo(n) to a negative power.
the book says $[3]^{-1}=[2]$ but why is this? what are the steps I need to take to find this?






abstract-algebra ring-theory







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  • Can you calculate $;3^4pmod5;$? Then take the inverse of that...
    – DonAntonio
    Nov 19 at 23:23
















0














$[3]$ means $x equiv 3mod(5)$. I know how to compute $[3]^{4}$ that is positive powers but I have not been in a situation where I need to take something modulo(n) to a negative power.
the book says $[3]^{-1}=[2]$ but why is this? what are the steps I need to take to find this?






abstract-algebra ring-theory







share|cite|improve this question
























  • Can you calculate $;3^4pmod5;$? Then take the inverse of that...
    – DonAntonio
    Nov 19 at 23:23














0












0








0







$[3]$ means $x equiv 3mod(5)$. I know how to compute $[3]^{4}$ that is positive powers but I have not been in a situation where I need to take something modulo(n) to a negative power.
the book says $[3]^{-1}=[2]$ but why is this? what are the steps I need to take to find this?






abstract-algebra ring-theory







share|cite|improve this question















$[3]$ means $x equiv 3mod(5)$. I know how to compute $[3]^{4}$ that is positive powers but I have not been in a situation where I need to take something modulo(n) to a negative power.
the book says $[3]^{-1}=[2]$ but why is this? what are the steps I need to take to find this?






abstract-algebra ring-theory




abstract-algebra ring-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 20 at 0:18









amWhy

191k28224439




191k28224439










asked Nov 19 at 23:22









Collin Roberts

32




32












  • Can you calculate $;3^4pmod5;$? Then take the inverse of that...
    – DonAntonio
    Nov 19 at 23:23


















  • Can you calculate $;3^4pmod5;$? Then take the inverse of that...
    – DonAntonio
    Nov 19 at 23:23
















Can you calculate $;3^4pmod5;$? Then take the inverse of that...
– DonAntonio
Nov 19 at 23:23




Can you calculate $;3^4pmod5;$? Then take the inverse of that...
– DonAntonio
Nov 19 at 23:23










3 Answers
3






active

oldest

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0














When doint arithmetic modulo $;p;$ , you get that $;x^{-1}=apmod piff xa=1pmod p;$ . In this case,



$$3^{-1}=2pmod 5;;text{since};;3cdot2=6=1pmod 5$$



You can do the same modulo any number $;p;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.






share|cite|improve this answer





























    0














    By little Fermat $,3^{-4}equiv (3^{-1})^4equiv 1,$ by $,3^{-1}notequiv 0 $ (or brute-force: $ (3^{-1})^4!equiv 2^4equiv 1)$



    Alternatively use $ underbrace{3^{4}}_{largeequiv 1}, 3^{-4}equiv 1$






    share|cite|improve this answer































      0














      By definition $3^{-4}$ is the equivalence class $x$ where $xcdot3^4 equiv 1 pmod 5$ (if any).



      We could calculate that $3^4 =81equiv 1 pmod 5$ and to so $xcdot 3^4 equiv x cdot 1equiv x equiv 1 pmod 5$ gives us ... $x equiv 1 pmod 5$.



      So $3^{-4}equiv 1 pmod 5$



      But that'd pretty ineffectual and ignores many things we should have at our finger tips.



      If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.



      If $a^{-1}$ so that $a^{-1}cdot a equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3aequiv 1pmod 5$ via $3a equiv 6pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a equiv 2 pmod 5$ and $3^{-4}equiv (3^{-1})^4 equiv 2^4equiv 16equiv 1$.



      But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a notequiv 0$ $a^4 equiv 1 pmod 5$ so we can figure either:



      $3^{-4}equiv 3^{4-4} equiv 3^0 equiv 1 pmod 5$ or



      $3^{-4}equiv (3^{-1})^4 equiv 1 pmod 5$.






      share|cite|improve this answer





















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        3 Answers
        3






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        3 Answers
        3






        active

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        active

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        active

        oldest

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        0














        When doint arithmetic modulo $;p;$ , you get that $;x^{-1}=apmod piff xa=1pmod p;$ . In this case,



        $$3^{-1}=2pmod 5;;text{since};;3cdot2=6=1pmod 5$$



        You can do the same modulo any number $;p;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.






        share|cite|improve this answer


























          0














          When doint arithmetic modulo $;p;$ , you get that $;x^{-1}=apmod piff xa=1pmod p;$ . In this case,



          $$3^{-1}=2pmod 5;;text{since};;3cdot2=6=1pmod 5$$



          You can do the same modulo any number $;p;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.






          share|cite|improve this answer
























            0












            0








            0






            When doint arithmetic modulo $;p;$ , you get that $;x^{-1}=apmod piff xa=1pmod p;$ . In this case,



            $$3^{-1}=2pmod 5;;text{since};;3cdot2=6=1pmod 5$$



            You can do the same modulo any number $;p;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.






            share|cite|improve this answer












            When doint arithmetic modulo $;p;$ , you get that $;x^{-1}=apmod piff xa=1pmod p;$ . In this case,



            $$3^{-1}=2pmod 5;;text{since};;3cdot2=6=1pmod 5$$



            You can do the same modulo any number $;p;$ ...under the condition, of course, that you have an invertible number. If the modulo is prime then any non-multiple of that prime is invertible.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 19 at 23:25









            DonAntonio

            176k1491225




            176k1491225























                0














                By little Fermat $,3^{-4}equiv (3^{-1})^4equiv 1,$ by $,3^{-1}notequiv 0 $ (or brute-force: $ (3^{-1})^4!equiv 2^4equiv 1)$



                Alternatively use $ underbrace{3^{4}}_{largeequiv 1}, 3^{-4}equiv 1$






                share|cite|improve this answer




























                  0














                  By little Fermat $,3^{-4}equiv (3^{-1})^4equiv 1,$ by $,3^{-1}notequiv 0 $ (or brute-force: $ (3^{-1})^4!equiv 2^4equiv 1)$



                  Alternatively use $ underbrace{3^{4}}_{largeequiv 1}, 3^{-4}equiv 1$






                  share|cite|improve this answer


























                    0












                    0








                    0






                    By little Fermat $,3^{-4}equiv (3^{-1})^4equiv 1,$ by $,3^{-1}notequiv 0 $ (or brute-force: $ (3^{-1})^4!equiv 2^4equiv 1)$



                    Alternatively use $ underbrace{3^{4}}_{largeequiv 1}, 3^{-4}equiv 1$






                    share|cite|improve this answer














                    By little Fermat $,3^{-4}equiv (3^{-1})^4equiv 1,$ by $,3^{-1}notequiv 0 $ (or brute-force: $ (3^{-1})^4!equiv 2^4equiv 1)$



                    Alternatively use $ underbrace{3^{4}}_{largeequiv 1}, 3^{-4}equiv 1$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 20 at 0:21

























                    answered Nov 20 at 0:12









                    Bill Dubuque

                    208k29190627




                    208k29190627























                        0














                        By definition $3^{-4}$ is the equivalence class $x$ where $xcdot3^4 equiv 1 pmod 5$ (if any).



                        We could calculate that $3^4 =81equiv 1 pmod 5$ and to so $xcdot 3^4 equiv x cdot 1equiv x equiv 1 pmod 5$ gives us ... $x equiv 1 pmod 5$.



                        So $3^{-4}equiv 1 pmod 5$



                        But that'd pretty ineffectual and ignores many things we should have at our finger tips.



                        If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.



                        If $a^{-1}$ so that $a^{-1}cdot a equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3aequiv 1pmod 5$ via $3a equiv 6pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a equiv 2 pmod 5$ and $3^{-4}equiv (3^{-1})^4 equiv 2^4equiv 16equiv 1$.



                        But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a notequiv 0$ $a^4 equiv 1 pmod 5$ so we can figure either:



                        $3^{-4}equiv 3^{4-4} equiv 3^0 equiv 1 pmod 5$ or



                        $3^{-4}equiv (3^{-1})^4 equiv 1 pmod 5$.






                        share|cite|improve this answer


























                          0














                          By definition $3^{-4}$ is the equivalence class $x$ where $xcdot3^4 equiv 1 pmod 5$ (if any).



                          We could calculate that $3^4 =81equiv 1 pmod 5$ and to so $xcdot 3^4 equiv x cdot 1equiv x equiv 1 pmod 5$ gives us ... $x equiv 1 pmod 5$.



                          So $3^{-4}equiv 1 pmod 5$



                          But that'd pretty ineffectual and ignores many things we should have at our finger tips.



                          If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.



                          If $a^{-1}$ so that $a^{-1}cdot a equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3aequiv 1pmod 5$ via $3a equiv 6pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a equiv 2 pmod 5$ and $3^{-4}equiv (3^{-1})^4 equiv 2^4equiv 16equiv 1$.



                          But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a notequiv 0$ $a^4 equiv 1 pmod 5$ so we can figure either:



                          $3^{-4}equiv 3^{4-4} equiv 3^0 equiv 1 pmod 5$ or



                          $3^{-4}equiv (3^{-1})^4 equiv 1 pmod 5$.






                          share|cite|improve this answer
























                            0












                            0








                            0






                            By definition $3^{-4}$ is the equivalence class $x$ where $xcdot3^4 equiv 1 pmod 5$ (if any).



                            We could calculate that $3^4 =81equiv 1 pmod 5$ and to so $xcdot 3^4 equiv x cdot 1equiv x equiv 1 pmod 5$ gives us ... $x equiv 1 pmod 5$.



                            So $3^{-4}equiv 1 pmod 5$



                            But that'd pretty ineffectual and ignores many things we should have at our finger tips.



                            If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.



                            If $a^{-1}$ so that $a^{-1}cdot a equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3aequiv 1pmod 5$ via $3a equiv 6pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a equiv 2 pmod 5$ and $3^{-4}equiv (3^{-1})^4 equiv 2^4equiv 16equiv 1$.



                            But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a notequiv 0$ $a^4 equiv 1 pmod 5$ so we can figure either:



                            $3^{-4}equiv 3^{4-4} equiv 3^0 equiv 1 pmod 5$ or



                            $3^{-4}equiv (3^{-1})^4 equiv 1 pmod 5$.






                            share|cite|improve this answer












                            By definition $3^{-4}$ is the equivalence class $x$ where $xcdot3^4 equiv 1 pmod 5$ (if any).



                            We could calculate that $3^4 =81equiv 1 pmod 5$ and to so $xcdot 3^4 equiv x cdot 1equiv x equiv 1 pmod 5$ gives us ... $x equiv 1 pmod 5$.



                            So $3^{-4}equiv 1 pmod 5$



                            But that'd pretty ineffectual and ignores many things we should have at our finger tips.



                            If $p$ is prime $Z_p$ is a field but if not then $Z_p$ has some zero divisors.



                            If $a^{-1}$ so that $a^{-1}cdot a equiv 1$ then $a^{-k} = (a^{-1})^k$ so we could attempt to find $3^{-1}$ by solve $3aequiv 1pmod 5$ via $3a equiv 6pmod 5$ so (as $3,5$ are relatively prime) then $3^{-1}= a equiv 2 pmod 5$ and $3^{-4}equiv (3^{-1})^4 equiv 2^4equiv 16equiv 1$.



                            But most importantly, we were ignoring that $Z_5$ is a fields so $3^{-1}$ exists and we don't need to figure out what it is, it's enough to no it exists and that by Fermat's Little Theorem for all $a notequiv 0$ $a^4 equiv 1 pmod 5$ so we can figure either:



                            $3^{-4}equiv 3^{4-4} equiv 3^0 equiv 1 pmod 5$ or



                            $3^{-4}equiv (3^{-1})^4 equiv 1 pmod 5$.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 20 at 0:35









                            fleablood

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