Proof by Induction Question including Rational Numbers











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I just recently covered 'rational numbers' in class and was assigned the following question to solve using induction for n, so that for all $q in mathbb{Q}$ {1}:



$$sum_{k=0}^n q^k = frac{q^{n+1}-1}{q-1}$$



I am not entirely sure on where to start, since up to this point I've only done proofs by induction involving natural numbers only. I've thought about leaving the variable q as it is, and doing



n = 1

then assuming statement is true for n, solve for n + 1


where $n in mathbb{N}$, but it leaves me with a dead end, since there are too many unknown variables involved.



I hope someone can help me with this question and explain to me what the best approach would be and why!
Thank you!!






induction rational-numbers







share|cite|improve this question


























    up vote
    3
    down vote

    favorite












    I just recently covered 'rational numbers' in class and was assigned the following question to solve using induction for n, so that for all $q in mathbb{Q}$ {1}:



    $$sum_{k=0}^n q^k = frac{q^{n+1}-1}{q-1}$$



    I am not entirely sure on where to start, since up to this point I've only done proofs by induction involving natural numbers only. I've thought about leaving the variable q as it is, and doing



    n = 1
    
then assuming statement is true for n, solve for n + 1


    where $n in mathbb{N}$, but it leaves me with a dead end, since there are too many unknown variables involved.



    I hope someone can help me with this question and explain to me what the best approach would be and why!
    Thank you!!






    induction rational-numbers







    share|cite|improve this question
























      up vote
      3
      down vote

      favorite









      up vote
      3
      down vote

      favorite











      I just recently covered 'rational numbers' in class and was assigned the following question to solve using induction for n, so that for all $q in mathbb{Q}$ {1}:



      $$sum_{k=0}^n q^k = frac{q^{n+1}-1}{q-1}$$



      I am not entirely sure on where to start, since up to this point I've only done proofs by induction involving natural numbers only. I've thought about leaving the variable q as it is, and doing



      n = 1
      
then assuming statement is true for n, solve for n + 1


      where $n in mathbb{N}$, but it leaves me with a dead end, since there are too many unknown variables involved.



      I hope someone can help me with this question and explain to me what the best approach would be and why!
      Thank you!!






      induction rational-numbers







      share|cite|improve this question













      I just recently covered 'rational numbers' in class and was assigned the following question to solve using induction for n, so that for all $q in mathbb{Q}$ {1}:



      $$sum_{k=0}^n q^k = frac{q^{n+1}-1}{q-1}$$



      I am not entirely sure on where to start, since up to this point I've only done proofs by induction involving natural numbers only. I've thought about leaving the variable q as it is, and doing



      n = 1
      
then assuming statement is true for n, solve for n + 1


      where $n in mathbb{N}$, but it leaves me with a dead end, since there are too many unknown variables involved.



      I hope someone can help me with this question and explain to me what the best approach would be and why!
      Thank you!!






      induction rational-numbers




      induction rational-numbers






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 28 at 21:18









      Rikk

      424




      424






















          3 Answers
          3






          active

          oldest

          votes

















          up vote
          1
          down vote



          accepted










          Your attempt is right! You do not have to change the $q$; leave it as it is and do the standard induction regarding to $n$.



          Therefore lets start with our basis case $n=0$:



          $$begin{align}
          sum_{k=0}^0q^k&=frac{q^{0+1}-1}{q-1}\
          1&=1
          end{align}$$



          Next consider the new values $n=m$ and $n=m+1$ which corresponde to the two equations



          $$begin{align}
          sum_{k=0}^{m}q^k&=frac{q^{m+1}-1}{q-1}\
          sum_{k=0}^{m+1}q^k&=frac{q^{m+2}-1}{q-1}
          end{align}$$



          Here the first one is our assumption the second one our hypothesis. Howsoever lets split up the second sum as the following



          $$sum_{k=0}^{m+1}q^k=underbrace{sum_{k=0}^{m}q^k}_{=text{assumption}}+q^{m+1}$$



          As we can see the remaining sum equals our assumption. Therefore we can replace it by the given fraction from above. This yields to



          $$sum_{k=0}^{m}q^k+q^{m+1}=frac{q^{m+1}-1}{q-1}+q^{m+1}=frac{q^{m+1}-1+q^{m+2}-q^{m+1}}{q-1}=frac{q^{m+2}-1}{q-1}$$



          And thus we have shown that from our assumption the hypothesis follows and therefore we are done.






          share|cite|improve this answer





















          • Thank you @mrtaurho! I'm glad my attempt was going in the right direction! But is there a formal reason for why the basis case is n = 0? Although, it is a fair step, considering n = 0 does work perfectly when looking at the statement.
            – Rikk
            Nov 28 at 21:40












          • @Rikk I general you try to find the smallest possible natural number for which it works out. In your case it is given that $ninmathbb N$ - or at least you assumed that it has to be so - and $n=0$ is the first possible value of this set. In case it does not works out for $0$, one may consider $n=1$ and so on. Take a look at this for an example where you cannot start by setting $n=0$.
            – mrtaurho
            Nov 28 at 21:43












          • Fair! I am so used to using n = 1, that I have forgotten about using 0 as the first possible value of natural numbers. Thank you a lot for your clarification, regardless!
            – Rikk
            Nov 28 at 21:46










          • No problem. I am happy that I could clear your doubts :)
            – mrtaurho
            Nov 28 at 21:47


















          up vote
          0
          down vote













          Well ...



          Write down that assertion when $n=1$ and verify that it's true.



          Assume that it's true for some particular (unspecified) value of $n$. Write down what that tells you - I recommend doing that with an ellipsis
          $$
          1 + q + q^2 + cdots + q^n = ldots
          $$
          Then add $q^{n+1}$ and see if you can turn that into what you want.



          Having said that, I wish you didn't have to use induction. It's much easier just expanding with the distributive law:
          $$
          (q-1)(1 + q + q^2 + cdots + q^n) .
          $$
          Induction should be saved for problems where it really helps.






          share|cite|improve this answer




























            up vote
            0
            down vote













            Hint:



            The set in which the arithmetic operations take place has no importance. You indeed have use induction on the exponent.



            The simplest, from my point of vizw, would be to prove by induction the factorisation formula:
            $$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+dots+x+1).$$






            share|cite|improve this answer





















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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              1
              down vote



              accepted










              Your attempt is right! You do not have to change the $q$; leave it as it is and do the standard induction regarding to $n$.



              Therefore lets start with our basis case $n=0$:



              $$begin{align}
              sum_{k=0}^0q^k&=frac{q^{0+1}-1}{q-1}\
              1&=1
              end{align}$$



              Next consider the new values $n=m$ and $n=m+1$ which corresponde to the two equations



              $$begin{align}
              sum_{k=0}^{m}q^k&=frac{q^{m+1}-1}{q-1}\
              sum_{k=0}^{m+1}q^k&=frac{q^{m+2}-1}{q-1}
              end{align}$$



              Here the first one is our assumption the second one our hypothesis. Howsoever lets split up the second sum as the following



              $$sum_{k=0}^{m+1}q^k=underbrace{sum_{k=0}^{m}q^k}_{=text{assumption}}+q^{m+1}$$



              As we can see the remaining sum equals our assumption. Therefore we can replace it by the given fraction from above. This yields to



              $$sum_{k=0}^{m}q^k+q^{m+1}=frac{q^{m+1}-1}{q-1}+q^{m+1}=frac{q^{m+1}-1+q^{m+2}-q^{m+1}}{q-1}=frac{q^{m+2}-1}{q-1}$$



              And thus we have shown that from our assumption the hypothesis follows and therefore we are done.






              share|cite|improve this answer





















              • Thank you @mrtaurho! I'm glad my attempt was going in the right direction! But is there a formal reason for why the basis case is n = 0? Although, it is a fair step, considering n = 0 does work perfectly when looking at the statement.
                – Rikk
                Nov 28 at 21:40












              • @Rikk I general you try to find the smallest possible natural number for which it works out. In your case it is given that $ninmathbb N$ - or at least you assumed that it has to be so - and $n=0$ is the first possible value of this set. In case it does not works out for $0$, one may consider $n=1$ and so on. Take a look at this for an example where you cannot start by setting $n=0$.
                – mrtaurho
                Nov 28 at 21:43












              • Fair! I am so used to using n = 1, that I have forgotten about using 0 as the first possible value of natural numbers. Thank you a lot for your clarification, regardless!
                – Rikk
                Nov 28 at 21:46










              • No problem. I am happy that I could clear your doubts :)
                – mrtaurho
                Nov 28 at 21:47















              up vote
              1
              down vote



              accepted










              Your attempt is right! You do not have to change the $q$; leave it as it is and do the standard induction regarding to $n$.



              Therefore lets start with our basis case $n=0$:



              $$begin{align}
              sum_{k=0}^0q^k&=frac{q^{0+1}-1}{q-1}\
              1&=1
              end{align}$$



              Next consider the new values $n=m$ and $n=m+1$ which corresponde to the two equations



              $$begin{align}
              sum_{k=0}^{m}q^k&=frac{q^{m+1}-1}{q-1}\
              sum_{k=0}^{m+1}q^k&=frac{q^{m+2}-1}{q-1}
              end{align}$$



              Here the first one is our assumption the second one our hypothesis. Howsoever lets split up the second sum as the following



              $$sum_{k=0}^{m+1}q^k=underbrace{sum_{k=0}^{m}q^k}_{=text{assumption}}+q^{m+1}$$



              As we can see the remaining sum equals our assumption. Therefore we can replace it by the given fraction from above. This yields to



              $$sum_{k=0}^{m}q^k+q^{m+1}=frac{q^{m+1}-1}{q-1}+q^{m+1}=frac{q^{m+1}-1+q^{m+2}-q^{m+1}}{q-1}=frac{q^{m+2}-1}{q-1}$$



              And thus we have shown that from our assumption the hypothesis follows and therefore we are done.






              share|cite|improve this answer





















              • Thank you @mrtaurho! I'm glad my attempt was going in the right direction! But is there a formal reason for why the basis case is n = 0? Although, it is a fair step, considering n = 0 does work perfectly when looking at the statement.
                – Rikk
                Nov 28 at 21:40












              • @Rikk I general you try to find the smallest possible natural number for which it works out. In your case it is given that $ninmathbb N$ - or at least you assumed that it has to be so - and $n=0$ is the first possible value of this set. In case it does not works out for $0$, one may consider $n=1$ and so on. Take a look at this for an example where you cannot start by setting $n=0$.
                – mrtaurho
                Nov 28 at 21:43












              • Fair! I am so used to using n = 1, that I have forgotten about using 0 as the first possible value of natural numbers. Thank you a lot for your clarification, regardless!
                – Rikk
                Nov 28 at 21:46










              • No problem. I am happy that I could clear your doubts :)
                – mrtaurho
                Nov 28 at 21:47













              up vote
              1
              down vote



              accepted







              up vote
              1
              down vote



              accepted






              Your attempt is right! You do not have to change the $q$; leave it as it is and do the standard induction regarding to $n$.



              Therefore lets start with our basis case $n=0$:



              $$begin{align}
              sum_{k=0}^0q^k&=frac{q^{0+1}-1}{q-1}\
              1&=1
              end{align}$$



              Next consider the new values $n=m$ and $n=m+1$ which corresponde to the two equations



              $$begin{align}
              sum_{k=0}^{m}q^k&=frac{q^{m+1}-1}{q-1}\
              sum_{k=0}^{m+1}q^k&=frac{q^{m+2}-1}{q-1}
              end{align}$$



              Here the first one is our assumption the second one our hypothesis. Howsoever lets split up the second sum as the following



              $$sum_{k=0}^{m+1}q^k=underbrace{sum_{k=0}^{m}q^k}_{=text{assumption}}+q^{m+1}$$



              As we can see the remaining sum equals our assumption. Therefore we can replace it by the given fraction from above. This yields to



              $$sum_{k=0}^{m}q^k+q^{m+1}=frac{q^{m+1}-1}{q-1}+q^{m+1}=frac{q^{m+1}-1+q^{m+2}-q^{m+1}}{q-1}=frac{q^{m+2}-1}{q-1}$$



              And thus we have shown that from our assumption the hypothesis follows and therefore we are done.






              share|cite|improve this answer












              Your attempt is right! You do not have to change the $q$; leave it as it is and do the standard induction regarding to $n$.



              Therefore lets start with our basis case $n=0$:



              $$begin{align}
              sum_{k=0}^0q^k&=frac{q^{0+1}-1}{q-1}\
              1&=1
              end{align}$$



              Next consider the new values $n=m$ and $n=m+1$ which corresponde to the two equations



              $$begin{align}
              sum_{k=0}^{m}q^k&=frac{q^{m+1}-1}{q-1}\
              sum_{k=0}^{m+1}q^k&=frac{q^{m+2}-1}{q-1}
              end{align}$$



              Here the first one is our assumption the second one our hypothesis. Howsoever lets split up the second sum as the following



              $$sum_{k=0}^{m+1}q^k=underbrace{sum_{k=0}^{m}q^k}_{=text{assumption}}+q^{m+1}$$



              As we can see the remaining sum equals our assumption. Therefore we can replace it by the given fraction from above. This yields to



              $$sum_{k=0}^{m}q^k+q^{m+1}=frac{q^{m+1}-1}{q-1}+q^{m+1}=frac{q^{m+1}-1+q^{m+2}-q^{m+1}}{q-1}=frac{q^{m+2}-1}{q-1}$$



              And thus we have shown that from our assumption the hypothesis follows and therefore we are done.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 28 at 21:26









              mrtaurho

              3,0351930




              3,0351930












              • Thank you @mrtaurho! I'm glad my attempt was going in the right direction! But is there a formal reason for why the basis case is n = 0? Although, it is a fair step, considering n = 0 does work perfectly when looking at the statement.
                – Rikk
                Nov 28 at 21:40












              • @Rikk I general you try to find the smallest possible natural number for which it works out. In your case it is given that $ninmathbb N$ - or at least you assumed that it has to be so - and $n=0$ is the first possible value of this set. In case it does not works out for $0$, one may consider $n=1$ and so on. Take a look at this for an example where you cannot start by setting $n=0$.
                – mrtaurho
                Nov 28 at 21:43












              • Fair! I am so used to using n = 1, that I have forgotten about using 0 as the first possible value of natural numbers. Thank you a lot for your clarification, regardless!
                – Rikk
                Nov 28 at 21:46










              • No problem. I am happy that I could clear your doubts :)
                – mrtaurho
                Nov 28 at 21:47


















              • Thank you @mrtaurho! I'm glad my attempt was going in the right direction! But is there a formal reason for why the basis case is n = 0? Although, it is a fair step, considering n = 0 does work perfectly when looking at the statement.
                – Rikk
                Nov 28 at 21:40












              • @Rikk I general you try to find the smallest possible natural number for which it works out. In your case it is given that $ninmathbb N$ - or at least you assumed that it has to be so - and $n=0$ is the first possible value of this set. In case it does not works out for $0$, one may consider $n=1$ and so on. Take a look at this for an example where you cannot start by setting $n=0$.
                – mrtaurho
                Nov 28 at 21:43












              • Fair! I am so used to using n = 1, that I have forgotten about using 0 as the first possible value of natural numbers. Thank you a lot for your clarification, regardless!
                – Rikk
                Nov 28 at 21:46










              • No problem. I am happy that I could clear your doubts :)
                – mrtaurho
                Nov 28 at 21:47
















              Thank you @mrtaurho! I'm glad my attempt was going in the right direction! But is there a formal reason for why the basis case is n = 0? Although, it is a fair step, considering n = 0 does work perfectly when looking at the statement.
              – Rikk
              Nov 28 at 21:40






              Thank you @mrtaurho! I'm glad my attempt was going in the right direction! But is there a formal reason for why the basis case is n = 0? Although, it is a fair step, considering n = 0 does work perfectly when looking at the statement.
              – Rikk
              Nov 28 at 21:40














              @Rikk I general you try to find the smallest possible natural number for which it works out. In your case it is given that $ninmathbb N$ - or at least you assumed that it has to be so - and $n=0$ is the first possible value of this set. In case it does not works out for $0$, one may consider $n=1$ and so on. Take a look at this for an example where you cannot start by setting $n=0$.
              – mrtaurho
              Nov 28 at 21:43






              @Rikk I general you try to find the smallest possible natural number for which it works out. In your case it is given that $ninmathbb N$ - or at least you assumed that it has to be so - and $n=0$ is the first possible value of this set. In case it does not works out for $0$, one may consider $n=1$ and so on. Take a look at this for an example where you cannot start by setting $n=0$.
              – mrtaurho
              Nov 28 at 21:43














              Fair! I am so used to using n = 1, that I have forgotten about using 0 as the first possible value of natural numbers. Thank you a lot for your clarification, regardless!
              – Rikk
              Nov 28 at 21:46




              Fair! I am so used to using n = 1, that I have forgotten about using 0 as the first possible value of natural numbers. Thank you a lot for your clarification, regardless!
              – Rikk
              Nov 28 at 21:46












              No problem. I am happy that I could clear your doubts :)
              – mrtaurho
              Nov 28 at 21:47




              No problem. I am happy that I could clear your doubts :)
              – mrtaurho
              Nov 28 at 21:47










              up vote
              0
              down vote













              Well ...



              Write down that assertion when $n=1$ and verify that it's true.



              Assume that it's true for some particular (unspecified) value of $n$. Write down what that tells you - I recommend doing that with an ellipsis
              $$
              1 + q + q^2 + cdots + q^n = ldots
              $$
              Then add $q^{n+1}$ and see if you can turn that into what you want.



              Having said that, I wish you didn't have to use induction. It's much easier just expanding with the distributive law:
              $$
              (q-1)(1 + q + q^2 + cdots + q^n) .
              $$
              Induction should be saved for problems where it really helps.






              share|cite|improve this answer

























                up vote
                0
                down vote













                Well ...



                Write down that assertion when $n=1$ and verify that it's true.



                Assume that it's true for some particular (unspecified) value of $n$. Write down what that tells you - I recommend doing that with an ellipsis
                $$
                1 + q + q^2 + cdots + q^n = ldots
                $$
                Then add $q^{n+1}$ and see if you can turn that into what you want.



                Having said that, I wish you didn't have to use induction. It's much easier just expanding with the distributive law:
                $$
                (q-1)(1 + q + q^2 + cdots + q^n) .
                $$
                Induction should be saved for problems where it really helps.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  Well ...



                  Write down that assertion when $n=1$ and verify that it's true.



                  Assume that it's true for some particular (unspecified) value of $n$. Write down what that tells you - I recommend doing that with an ellipsis
                  $$
                  1 + q + q^2 + cdots + q^n = ldots
                  $$
                  Then add $q^{n+1}$ and see if you can turn that into what you want.



                  Having said that, I wish you didn't have to use induction. It's much easier just expanding with the distributive law:
                  $$
                  (q-1)(1 + q + q^2 + cdots + q^n) .
                  $$
                  Induction should be saved for problems where it really helps.






                  share|cite|improve this answer












                  Well ...



                  Write down that assertion when $n=1$ and verify that it's true.



                  Assume that it's true for some particular (unspecified) value of $n$. Write down what that tells you - I recommend doing that with an ellipsis
                  $$
                  1 + q + q^2 + cdots + q^n = ldots
                  $$
                  Then add $q^{n+1}$ and see if you can turn that into what you want.



                  Having said that, I wish you didn't have to use induction. It's much easier just expanding with the distributive law:
                  $$
                  (q-1)(1 + q + q^2 + cdots + q^n) .
                  $$
                  Induction should be saved for problems where it really helps.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 28 at 21:25









                  Ethan Bolker

                  40.7k546108




                  40.7k546108






















                      up vote
                      0
                      down vote













                      Hint:



                      The set in which the arithmetic operations take place has no importance. You indeed have use induction on the exponent.



                      The simplest, from my point of vizw, would be to prove by induction the factorisation formula:
                      $$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+dots+x+1).$$






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        Hint:



                        The set in which the arithmetic operations take place has no importance. You indeed have use induction on the exponent.



                        The simplest, from my point of vizw, would be to prove by induction the factorisation formula:
                        $$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+dots+x+1).$$






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          Hint:



                          The set in which the arithmetic operations take place has no importance. You indeed have use induction on the exponent.



                          The simplest, from my point of vizw, would be to prove by induction the factorisation formula:
                          $$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+dots+x+1).$$






                          share|cite|improve this answer












                          Hint:



                          The set in which the arithmetic operations take place has no importance. You indeed have use induction on the exponent.



                          The simplest, from my point of vizw, would be to prove by induction the factorisation formula:
                          $$x^{n}-1=(x-1)(x^{n-1}+x^{n-2}+dots+x+1).$$







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 28 at 21:26









                          Bernard

                          117k637110




                          117k637110






























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