Clarification of Answers given in Spivak's Calculus, Chapter 1-24
$begingroup$
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
asked Dec 10 '18 at 15:15
JoshuaJoshua
613
$endgroup$
add a comment |
$begingroup$
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
asked Dec 10 '18 at 15:15
JoshuaJoshua
613
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 10 '18 at 15:15
JoshuaJoshua
613
$endgroup$
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
asked Dec 10 '18 at 15:15
JoshuaJoshua
613
asked Dec 10 '18 at 15:15
JoshuaJoshua
613
asked Dec 10 '18 at 15:15
JoshuaJoshua
613
asked Dec 10 '18 at 15:15
JoshuaJoshua
613
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
add a comment |
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
add a comment |
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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