Clarification of Answers given in Spivak's Calculus, Chapter 1-24












1












$begingroup$


I am self-studying Spivak's Calculus.



In Chapter 1, Question 24, $a_1 + ldots + a_n$ is defined as $a_1 + (a_2 + (a_3 + ldots + (a_{n-2} + (a_{n-1} + a_n))) ldots )$.



In part (a), we are asked to proof that $(a_1 + ldots + a_k) + a_{k+1} = a_1 + ldots + a_{k+1}$. The hint given was to use induction.



In the answers, it is written that:



begin{align} (a_1 + ldots + a_{k+1}) + a_{k+2} &= [(a_1 + ldots + a_{k}) + a_{k+1}] + a_{k+2} \ &= (a_1 + ldots + a_{k}) + (a_{k+1} + a_{k+2}) \ &= a_1 + ldots + a_{k} + (a_{k+1} + a_{k+2}) \ &= a_1 ldots + a_{k+2} end{align}



For the first and third equality, it is true because the equation holds for $k$. For the second equality, it is true because $a + (b + c) = (a + b) + c$. However, for the last equality, it is written that it is true by the definition of $a_1 + ldots + a_{k+2}$.



I do not understand how the last inequality is derived from the definition. Any help will be appreciated.






elementary-number-theory induction arithmetic







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    To make it more clear, let say for the moment that $b=a_{k+1}+a_{k+2}$. Then on your third line you have $a_1+cdots +a_k+b$ which is by definition is $a_1+left(a_2+cdots left(a_{k-1}+(a_k+b)right)right)$ which is the same as $a_1+left(a_2+cdots left(a_{k-1}+(a_k+(a_{k+1}+a_{k+2})right)right)$ but this is the very definition of $a_1+cdots +a_{k+2}$.
    $endgroup$
    – user9077
    Dec 10 '18 at 16:03












  • $begingroup$
    Thank you for the reply. I would then ask why does $a_1 + ldots + a_n$ have to be added from the back and not, say, from the front? Or should the logic be that is could be added in both ways?
    $endgroup$
    – Joshua
    Dec 11 '18 at 0:35










  • $begingroup$
    Ok I see it now, treating $b = a_{k+1} + a_{k+2}$, applying Spivak's definition, and finally expanding $b$ would result in the equality. Thanks!
    $endgroup$
    – Joshua
    Dec 11 '18 at 0:42
















1












$begingroup$


I am self-studying Spivak's Calculus.



In Chapter 1, Question 24, $a_1 + ldots + a_n$ is defined as $a_1 + (a_2 + (a_3 + ldots + (a_{n-2} + (a_{n-1} + a_n))) ldots )$.



In part (a), we are asked to proof that $(a_1 + ldots + a_k) + a_{k+1} = a_1 + ldots + a_{k+1}$. The hint given was to use induction.



In the answers, it is written that:



begin{align} (a_1 + ldots + a_{k+1}) + a_{k+2} &= [(a_1 + ldots + a_{k}) + a_{k+1}] + a_{k+2} \ &= (a_1 + ldots + a_{k}) + (a_{k+1} + a_{k+2}) \ &= a_1 + ldots + a_{k} + (a_{k+1} + a_{k+2}) \ &= a_1 ldots + a_{k+2} end{align}



For the first and third equality, it is true because the equation holds for $k$. For the second equality, it is true because $a + (b + c) = (a + b) + c$. However, for the last equality, it is written that it is true by the definition of $a_1 + ldots + a_{k+2}$.



I do not understand how the last inequality is derived from the definition. Any help will be appreciated.






elementary-number-theory induction arithmetic







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    To make it more clear, let say for the moment that $b=a_{k+1}+a_{k+2}$. Then on your third line you have $a_1+cdots +a_k+b$ which is by definition is $a_1+left(a_2+cdots left(a_{k-1}+(a_k+b)right)right)$ which is the same as $a_1+left(a_2+cdots left(a_{k-1}+(a_k+(a_{k+1}+a_{k+2})right)right)$ but this is the very definition of $a_1+cdots +a_{k+2}$.
    $endgroup$
    – user9077
    Dec 10 '18 at 16:03












  • $begingroup$
    Thank you for the reply. I would then ask why does $a_1 + ldots + a_n$ have to be added from the back and not, say, from the front? Or should the logic be that is could be added in both ways?
    $endgroup$
    – Joshua
    Dec 11 '18 at 0:35










  • $begingroup$
    Ok I see it now, treating $b = a_{k+1} + a_{k+2}$, applying Spivak's definition, and finally expanding $b$ would result in the equality. Thanks!
    $endgroup$
    – Joshua
    Dec 11 '18 at 0:42














1












1








1





$begingroup$


I am self-studying Spivak's Calculus.



In Chapter 1, Question 24, $a_1 + ldots + a_n$ is defined as $a_1 + (a_2 + (a_3 + ldots + (a_{n-2} + (a_{n-1} + a_n))) ldots )$.



In part (a), we are asked to proof that $(a_1 + ldots + a_k) + a_{k+1} = a_1 + ldots + a_{k+1}$. The hint given was to use induction.



In the answers, it is written that:



begin{align} (a_1 + ldots + a_{k+1}) + a_{k+2} &= [(a_1 + ldots + a_{k}) + a_{k+1}] + a_{k+2} \ &= (a_1 + ldots + a_{k}) + (a_{k+1} + a_{k+2}) \ &= a_1 + ldots + a_{k} + (a_{k+1} + a_{k+2}) \ &= a_1 ldots + a_{k+2} end{align}



For the first and third equality, it is true because the equation holds for $k$. For the second equality, it is true because $a + (b + c) = (a + b) + c$. However, for the last equality, it is written that it is true by the definition of $a_1 + ldots + a_{k+2}$.



I do not understand how the last inequality is derived from the definition. Any help will be appreciated.






elementary-number-theory induction arithmetic







share|cite|improve this question









$endgroup$




I am self-studying Spivak's Calculus.



In Chapter 1, Question 24, $a_1 + ldots + a_n$ is defined as $a_1 + (a_2 + (a_3 + ldots + (a_{n-2} + (a_{n-1} + a_n))) ldots )$.



In part (a), we are asked to proof that $(a_1 + ldots + a_k) + a_{k+1} = a_1 + ldots + a_{k+1}$. The hint given was to use induction.



In the answers, it is written that:



begin{align} (a_1 + ldots + a_{k+1}) + a_{k+2} &= [(a_1 + ldots + a_{k}) + a_{k+1}] + a_{k+2} \ &= (a_1 + ldots + a_{k}) + (a_{k+1} + a_{k+2}) \ &= a_1 + ldots + a_{k} + (a_{k+1} + a_{k+2}) \ &= a_1 ldots + a_{k+2} end{align}



For the first and third equality, it is true because the equation holds for $k$. For the second equality, it is true because $a + (b + c) = (a + b) + c$. However, for the last equality, it is written that it is true by the definition of $a_1 + ldots + a_{k+2}$.



I do not understand how the last inequality is derived from the definition. Any help will be appreciated.






elementary-number-theory induction arithmetic




elementary-number-theory induction arithmetic






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 15:15









JoshuaJoshua

613




613








  • 1




    $begingroup$
    To make it more clear, let say for the moment that $b=a_{k+1}+a_{k+2}$. Then on your third line you have $a_1+cdots +a_k+b$ which is by definition is $a_1+left(a_2+cdots left(a_{k-1}+(a_k+b)right)right)$ which is the same as $a_1+left(a_2+cdots left(a_{k-1}+(a_k+(a_{k+1}+a_{k+2})right)right)$ but this is the very definition of $a_1+cdots +a_{k+2}$.
    $endgroup$
    – user9077
    Dec 10 '18 at 16:03












  • $begingroup$
    Thank you for the reply. I would then ask why does $a_1 + ldots + a_n$ have to be added from the back and not, say, from the front? Or should the logic be that is could be added in both ways?
    $endgroup$
    – Joshua
    Dec 11 '18 at 0:35










  • $begingroup$
    Ok I see it now, treating $b = a_{k+1} + a_{k+2}$, applying Spivak's definition, and finally expanding $b$ would result in the equality. Thanks!
    $endgroup$
    – Joshua
    Dec 11 '18 at 0:42














  • 1




    $begingroup$
    To make it more clear, let say for the moment that $b=a_{k+1}+a_{k+2}$. Then on your third line you have $a_1+cdots +a_k+b$ which is by definition is $a_1+left(a_2+cdots left(a_{k-1}+(a_k+b)right)right)$ which is the same as $a_1+left(a_2+cdots left(a_{k-1}+(a_k+(a_{k+1}+a_{k+2})right)right)$ but this is the very definition of $a_1+cdots +a_{k+2}$.
    $endgroup$
    – user9077
    Dec 10 '18 at 16:03












  • $begingroup$
    Thank you for the reply. I would then ask why does $a_1 + ldots + a_n$ have to be added from the back and not, say, from the front? Or should the logic be that is could be added in both ways?
    $endgroup$
    – Joshua
    Dec 11 '18 at 0:35










  • $begingroup$
    Ok I see it now, treating $b = a_{k+1} + a_{k+2}$, applying Spivak's definition, and finally expanding $b$ would result in the equality. Thanks!
    $endgroup$
    – Joshua
    Dec 11 '18 at 0:42








1




1




$begingroup$
To make it more clear, let say for the moment that $b=a_{k+1}+a_{k+2}$. Then on your third line you have $a_1+cdots +a_k+b$ which is by definition is $a_1+left(a_2+cdots left(a_{k-1}+(a_k+b)right)right)$ which is the same as $a_1+left(a_2+cdots left(a_{k-1}+(a_k+(a_{k+1}+a_{k+2})right)right)$ but this is the very definition of $a_1+cdots +a_{k+2}$.
$endgroup$
– user9077
Dec 10 '18 at 16:03






$begingroup$
To make it more clear, let say for the moment that $b=a_{k+1}+a_{k+2}$. Then on your third line you have $a_1+cdots +a_k+b$ which is by definition is $a_1+left(a_2+cdots left(a_{k-1}+(a_k+b)right)right)$ which is the same as $a_1+left(a_2+cdots left(a_{k-1}+(a_k+(a_{k+1}+a_{k+2})right)right)$ but this is the very definition of $a_1+cdots +a_{k+2}$.
$endgroup$
– user9077
Dec 10 '18 at 16:03














$begingroup$
Thank you for the reply. I would then ask why does $a_1 + ldots + a_n$ have to be added from the back and not, say, from the front? Or should the logic be that is could be added in both ways?
$endgroup$
– Joshua
Dec 11 '18 at 0:35




$begingroup$
Thank you for the reply. I would then ask why does $a_1 + ldots + a_n$ have to be added from the back and not, say, from the front? Or should the logic be that is could be added in both ways?
$endgroup$
– Joshua
Dec 11 '18 at 0:35












$begingroup$
Ok I see it now, treating $b = a_{k+1} + a_{k+2}$, applying Spivak's definition, and finally expanding $b$ would result in the equality. Thanks!
$endgroup$
– Joshua
Dec 11 '18 at 0:42




$begingroup$
Ok I see it now, treating $b = a_{k+1} + a_{k+2}$, applying Spivak's definition, and finally expanding $b$ would result in the equality. Thanks!
$endgroup$
– Joshua
Dec 11 '18 at 0:42










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