How to make the nth root of a product act the same as simple multiplication in regard to parity?












0














I don't have any experience working with radicals, but I'm working on a function that requires products of nth roots to be positive or negative, depending on the number of negative factors.



I've done some initial research, and reviews these Stack questions: Square roots — positive and negative and The Product Rule of Square Roots with Negative Numbers but I couldn't find the information I was seeking (or am not fully understanding the answers.)



Are the following expressions true? If not, how can I produce the those results?



$sqrt[2]{1*-1} = -1$



$sqrt[3]{1*1*-1} = -1$



$sqrt[3]{1*-1*-1} = 1$




[update] This is what the function does:



$sqrt[n]{overline{Delta_1}*overline{Delta_2} *...*overline{Delta_n}} text{ }*text{ }
frac{overline{Delta_1}*overline{Delta_2} *...*overline{Delta_n}}{Delta_1*Delta_2*...*Delta_n}$



such that if there are an odd number of negative factors, the product is negative, otherwise positive.




  • Is there a more compact way to express this?


also, any tips on notation are appreciated.






notation roots radicals







share|cite|improve this question
























  • $sqrt{-1} =i$, so the first statement is wrong.
    – Larry
    Nov 26 at 18:04


















0














I don't have any experience working with radicals, but I'm working on a function that requires products of nth roots to be positive or negative, depending on the number of negative factors.



I've done some initial research, and reviews these Stack questions: Square roots — positive and negative and The Product Rule of Square Roots with Negative Numbers but I couldn't find the information I was seeking (or am not fully understanding the answers.)



Are the following expressions true? If not, how can I produce the those results?



$sqrt[2]{1*-1} = -1$



$sqrt[3]{1*1*-1} = -1$



$sqrt[3]{1*-1*-1} = 1$




[update] This is what the function does:



$sqrt[n]{overline{Delta_1}*overline{Delta_2} *...*overline{Delta_n}} text{ }*text{ }
frac{overline{Delta_1}*overline{Delta_2} *...*overline{Delta_n}}{Delta_1*Delta_2*...*Delta_n}$



such that if there are an odd number of negative factors, the product is negative, otherwise positive.




  • Is there a more compact way to express this?


also, any tips on notation are appreciated.






notation roots radicals







share|cite|improve this question
























  • $sqrt{-1} =i$, so the first statement is wrong.
    – Larry
    Nov 26 at 18:04
















0












0








0


1





I don't have any experience working with radicals, but I'm working on a function that requires products of nth roots to be positive or negative, depending on the number of negative factors.



I've done some initial research, and reviews these Stack questions: Square roots — positive and negative and The Product Rule of Square Roots with Negative Numbers but I couldn't find the information I was seeking (or am not fully understanding the answers.)



Are the following expressions true? If not, how can I produce the those results?



$sqrt[2]{1*-1} = -1$



$sqrt[3]{1*1*-1} = -1$



$sqrt[3]{1*-1*-1} = 1$




[update] This is what the function does:



$sqrt[n]{overline{Delta_1}*overline{Delta_2} *...*overline{Delta_n}} text{ }*text{ }
frac{overline{Delta_1}*overline{Delta_2} *...*overline{Delta_n}}{Delta_1*Delta_2*...*Delta_n}$



such that if there are an odd number of negative factors, the product is negative, otherwise positive.




  • Is there a more compact way to express this?


also, any tips on notation are appreciated.






notation roots radicals







share|cite|improve this question















I don't have any experience working with radicals, but I'm working on a function that requires products of nth roots to be positive or negative, depending on the number of negative factors.



I've done some initial research, and reviews these Stack questions: Square roots — positive and negative and The Product Rule of Square Roots with Negative Numbers but I couldn't find the information I was seeking (or am not fully understanding the answers.)



Are the following expressions true? If not, how can I produce the those results?



$sqrt[2]{1*-1} = -1$



$sqrt[3]{1*1*-1} = -1$



$sqrt[3]{1*-1*-1} = 1$




[update] This is what the function does:



$sqrt[n]{overline{Delta_1}*overline{Delta_2} *...*overline{Delta_n}} text{ }*text{ }
frac{overline{Delta_1}*overline{Delta_2} *...*overline{Delta_n}}{Delta_1*Delta_2*...*Delta_n}$



such that if there are an odd number of negative factors, the product is negative, otherwise positive.




  • Is there a more compact way to express this?


also, any tips on notation are appreciated.






notation roots radicals




notation roots radicals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 at 3:42

























asked Nov 26 at 17:57









mnp

62




62












  • $sqrt{-1} =i$, so the first statement is wrong.
    – Larry
    Nov 26 at 18:04




















  • $sqrt{-1} =i$, so the first statement is wrong.
    – Larry
    Nov 26 at 18:04


















$sqrt{-1} =i$, so the first statement is wrong.
– Larry
Nov 26 at 18:04






$sqrt{-1} =i$, so the first statement is wrong.
– Larry
Nov 26 at 18:04












2 Answers
2






active

oldest

votes


















1














We have that





  • $sqrt[2]{1times(-1)} = sqrt{-1} neq -1quad color{red}checkmark$ indeed $-1times -1=1$


  • $sqrt[3]{1times1times(-1)} = sqrt[3]{-1} = -1quad color{green}checkmark$ indeed $-1times -1times -1=-1$


  • $sqrt[3]{1times(-1)times(-1)} = sqrt[3]{1}=1quad color{green}checkmark$ indeed $1times 1times 1=1$


As a general rule




  • for $nin mathbb{N}$ even and $age 0$ we have


$$sqrt[n] a=b iff bge 0 quad b^n=a$$




  • for $nin mathbb{N}$ odd and we have


$$sqrt[n] a=b iff b^n=a$$






share|cite|improve this answer































    1














    I am not sure if this is the method you want.



    We have Euler's formula
    $$e^{itheta} = isintheta+costheta$$
    We can take $n^{th}$ root of both sides to obtain
    $$sqrt[n]{e^{itheta}}=sqrt[n]{isintheta+costheta}tag{1}$$



    It seems like you are only asking for the cases for $1$ and $-1$, so let's do the following.



    It is clear that $sqrt[n]{1}$ is $1$ or $-1$ regardless of whether $n$ is odd or even. The problem is how to figure out $sqrt[n]{-1}$. For $n$ is odd, $sqrt[n]{-1}=-1$.



    For $n$ is even, let's suppose $theta = pi$. Then, from $(1)$ we have
    $$sqrt[n]{e^{ipi}}= sqrt[n]{-1}$$
    $$e^{ifrac{pi}{n}} = sqrt[n]{-1}$$
    $$sqrt[n]{-1}= isinfrac{pi}{n}+cosfrac{pi}{n}$$
    I apologize if this is not what you are looking for.






    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









      1














      We have that





      • $sqrt[2]{1times(-1)} = sqrt{-1} neq -1quad color{red}checkmark$ indeed $-1times -1=1$


      • $sqrt[3]{1times1times(-1)} = sqrt[3]{-1} = -1quad color{green}checkmark$ indeed $-1times -1times -1=-1$


      • $sqrt[3]{1times(-1)times(-1)} = sqrt[3]{1}=1quad color{green}checkmark$ indeed $1times 1times 1=1$


      As a general rule




      • for $nin mathbb{N}$ even and $age 0$ we have


      $$sqrt[n] a=b iff bge 0 quad b^n=a$$




      • for $nin mathbb{N}$ odd and we have


      $$sqrt[n] a=b iff b^n=a$$






      share|cite|improve this answer




























        1














        We have that





        • $sqrt[2]{1times(-1)} = sqrt{-1} neq -1quad color{red}checkmark$ indeed $-1times -1=1$


        • $sqrt[3]{1times1times(-1)} = sqrt[3]{-1} = -1quad color{green}checkmark$ indeed $-1times -1times -1=-1$


        • $sqrt[3]{1times(-1)times(-1)} = sqrt[3]{1}=1quad color{green}checkmark$ indeed $1times 1times 1=1$


        As a general rule




        • for $nin mathbb{N}$ even and $age 0$ we have


        $$sqrt[n] a=b iff bge 0 quad b^n=a$$




        • for $nin mathbb{N}$ odd and we have


        $$sqrt[n] a=b iff b^n=a$$






        share|cite|improve this answer


























          1












          1








          1






          We have that





          • $sqrt[2]{1times(-1)} = sqrt{-1} neq -1quad color{red}checkmark$ indeed $-1times -1=1$


          • $sqrt[3]{1times1times(-1)} = sqrt[3]{-1} = -1quad color{green}checkmark$ indeed $-1times -1times -1=-1$


          • $sqrt[3]{1times(-1)times(-1)} = sqrt[3]{1}=1quad color{green}checkmark$ indeed $1times 1times 1=1$


          As a general rule




          • for $nin mathbb{N}$ even and $age 0$ we have


          $$sqrt[n] a=b iff bge 0 quad b^n=a$$




          • for $nin mathbb{N}$ odd and we have


          $$sqrt[n] a=b iff b^n=a$$






          share|cite|improve this answer














          We have that





          • $sqrt[2]{1times(-1)} = sqrt{-1} neq -1quad color{red}checkmark$ indeed $-1times -1=1$


          • $sqrt[3]{1times1times(-1)} = sqrt[3]{-1} = -1quad color{green}checkmark$ indeed $-1times -1times -1=-1$


          • $sqrt[3]{1times(-1)times(-1)} = sqrt[3]{1}=1quad color{green}checkmark$ indeed $1times 1times 1=1$


          As a general rule




          • for $nin mathbb{N}$ even and $age 0$ we have


          $$sqrt[n] a=b iff bge 0 quad b^n=a$$




          • for $nin mathbb{N}$ odd and we have


          $$sqrt[n] a=b iff b^n=a$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 26 at 18:10

























          answered Nov 26 at 18:04









          gimusi

          1




          1























              1














              I am not sure if this is the method you want.



              We have Euler's formula
              $$e^{itheta} = isintheta+costheta$$
              We can take $n^{th}$ root of both sides to obtain
              $$sqrt[n]{e^{itheta}}=sqrt[n]{isintheta+costheta}tag{1}$$



              It seems like you are only asking for the cases for $1$ and $-1$, so let's do the following.



              It is clear that $sqrt[n]{1}$ is $1$ or $-1$ regardless of whether $n$ is odd or even. The problem is how to figure out $sqrt[n]{-1}$. For $n$ is odd, $sqrt[n]{-1}=-1$.



              For $n$ is even, let's suppose $theta = pi$. Then, from $(1)$ we have
              $$sqrt[n]{e^{ipi}}= sqrt[n]{-1}$$
              $$e^{ifrac{pi}{n}} = sqrt[n]{-1}$$
              $$sqrt[n]{-1}= isinfrac{pi}{n}+cosfrac{pi}{n}$$
              I apologize if this is not what you are looking for.






              share|cite|improve this answer


























                1














                I am not sure if this is the method you want.



                We have Euler's formula
                $$e^{itheta} = isintheta+costheta$$
                We can take $n^{th}$ root of both sides to obtain
                $$sqrt[n]{e^{itheta}}=sqrt[n]{isintheta+costheta}tag{1}$$



                It seems like you are only asking for the cases for $1$ and $-1$, so let's do the following.



                It is clear that $sqrt[n]{1}$ is $1$ or $-1$ regardless of whether $n$ is odd or even. The problem is how to figure out $sqrt[n]{-1}$. For $n$ is odd, $sqrt[n]{-1}=-1$.



                For $n$ is even, let's suppose $theta = pi$. Then, from $(1)$ we have
                $$sqrt[n]{e^{ipi}}= sqrt[n]{-1}$$
                $$e^{ifrac{pi}{n}} = sqrt[n]{-1}$$
                $$sqrt[n]{-1}= isinfrac{pi}{n}+cosfrac{pi}{n}$$
                I apologize if this is not what you are looking for.






                share|cite|improve this answer
























                  1












                  1








                  1






                  I am not sure if this is the method you want.



                  We have Euler's formula
                  $$e^{itheta} = isintheta+costheta$$
                  We can take $n^{th}$ root of both sides to obtain
                  $$sqrt[n]{e^{itheta}}=sqrt[n]{isintheta+costheta}tag{1}$$



                  It seems like you are only asking for the cases for $1$ and $-1$, so let's do the following.



                  It is clear that $sqrt[n]{1}$ is $1$ or $-1$ regardless of whether $n$ is odd or even. The problem is how to figure out $sqrt[n]{-1}$. For $n$ is odd, $sqrt[n]{-1}=-1$.



                  For $n$ is even, let's suppose $theta = pi$. Then, from $(1)$ we have
                  $$sqrt[n]{e^{ipi}}= sqrt[n]{-1}$$
                  $$e^{ifrac{pi}{n}} = sqrt[n]{-1}$$
                  $$sqrt[n]{-1}= isinfrac{pi}{n}+cosfrac{pi}{n}$$
                  I apologize if this is not what you are looking for.






                  share|cite|improve this answer












                  I am not sure if this is the method you want.



                  We have Euler's formula
                  $$e^{itheta} = isintheta+costheta$$
                  We can take $n^{th}$ root of both sides to obtain
                  $$sqrt[n]{e^{itheta}}=sqrt[n]{isintheta+costheta}tag{1}$$



                  It seems like you are only asking for the cases for $1$ and $-1$, so let's do the following.



                  It is clear that $sqrt[n]{1}$ is $1$ or $-1$ regardless of whether $n$ is odd or even. The problem is how to figure out $sqrt[n]{-1}$. For $n$ is odd, $sqrt[n]{-1}=-1$.



                  For $n$ is even, let's suppose $theta = pi$. Then, from $(1)$ we have
                  $$sqrt[n]{e^{ipi}}= sqrt[n]{-1}$$
                  $$e^{ifrac{pi}{n}} = sqrt[n]{-1}$$
                  $$sqrt[n]{-1}= isinfrac{pi}{n}+cosfrac{pi}{n}$$
                  I apologize if this is not what you are looking for.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 26 at 18:43









                  Larry

                  1,6082722




                  1,6082722






























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