Mixing
Given a finite collection of numbers, the products obtained by multiplying them in any order are all equal.
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How do I prove the following by induction?
Given a finite collection of numbers, the sums/products obtained by adding/multiplying them in any order are all equal.
induction
asked Oct 21 '18 at 12:04
yh05yh05
137
$endgroup$
add a comment |
$begingroup$
How do I prove the following by induction?
Given a finite collection of numbers, the sums/products obtained by adding/multiplying them in any order are all equal.
induction
asked Oct 21 '18 at 12:04
yh05yh05
137
$endgroup$
add a comment |
$begingroup$
How do I prove the following by induction?
Given a finite collection of numbers, the sums/products obtained by adding/multiplying them in any order are all equal.
induction
asked Oct 21 '18 at 12:04
yh05yh05
137
$endgroup$
How do I prove the following by induction?
Given a finite collection of numbers, the sums/products obtained by adding/multiplying them in any order are all equal.
induction
induction
asked Oct 21 '18 at 12:04
yh05yh05
137
asked Oct 21 '18 at 12:04
yh05yh05
137
edited Jan 11 at 9:54
yh05
asked Oct 21 '18 at 12:04
yh05yh05
137
asked Oct 21 '18 at 12:04
yh05yh05
137
asked Oct 21 '18 at 12:04
yh05yh05
137
137
add a comment |
add a comment |
2 Answers
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The proof for multiplication is identical to that for addition, so I'll only look at addition; it works for any commutative associative operation.
Inductive hypothesis: any sum of $n$ numbers is equal to the sum of those same $n$ numbers in sorted order, expressed as $a_1 + (a_2 + (a_3 + dots))$ where $a_1 leq a_2 leq dots$.
Then take a collection of $n+1$ numbers, ${a_1, a_2, dots, a_{n+1}}$. Their sum is expressed as $x + y$, where $x$ and $y$ are some nonempty collections of the $a_i$. There are two cases: $x$ is formally the sum of just one $a_i$, or it's the sum of more than one.
- If $x = a_i$, say, then by the inductive hypothesis, the sum of the $a_i$ comprising $y$ is equal to the same sum in ascending order and associating to the right. So we just need to combine $a_i$ with the sorted $a_1 + (a_2 + dots)$ excluding $a_i$. You can do this inductively: $$x + (a_1 + (a_2 + dots)) = (x + a_1) + (a_2 + dots) = (a_1 + x) + (a_2 + dots) = a_1 + (x + (a_2 + dots))$$
and proceed until $x = a_i$ is in the right place. (This case also can be used if $y$ is just one $a_i$, by using commutativity.) - If $x$ is a sum of more than one $a_i$, then by inductive hypothesis it is $a_{i_1} + (a_{i_2} + dots)$; and $y$ is $a_{j_1} + (a_{j_2} + dots)$. Use commutativity if necessary to ensure that $a_{i_1} leq a_{j_1}$; then use associativity to separate the sum into $a_{i_1} + ((a_{i_2} + dots) + (a_{j_1} + (a_{j_2} + dots)))$. Now use the inductive hypothesis on the non-$a_{i_1}$ term.
answered Oct 21 '18 at 17:59
Patrick StevensPatrick Stevens
28.7k52874
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Okay. So, we proceed by way of induction. You need to show the base case is true which in this case is $n=3$. So, suppose we have $3$ arbitrary numbers $m_1,m_2,m_3$. We know that for any two numbers $a,b$ that $ab=ba$ and that multiplication is associative. So
$$m_1m_2m_3=m_1(m_2m_3)=m_1(m_3m_2)=m_1m_3m_2=(m_1m_3)m_2=(m_3m_1)m_2=m_3m_1m_2$$
By associativity and commutativity (for the case $n=2$). You can continue in this fashion to get all $6$ rearrangements of the product easily and show the base case holds.
Now, we need to show that the $n$th case implies the $n+1$st case. So, suppose that multiplication commutes for any $n$ numbers and consider the collection $m_1,m_2, dots m_{n+1}$ of $n+1$ numbers. We want to show that any arbitrary arrangement of the product is equal to $m_1m_2dots m_nm_{n+1}$ (why is this good enough?). So, say we have some arbitrary arrangement
$$m_im_{i+1} dots m_{i+(n+1)}$$
Then we can group this (by associativity) however we like. So, if $m_1$ is in the $i+j$ position, then we can group all the numbers to its left and consider it as one big number (by our induction hypothesis) since there will be clearly be at most $n$ of them (why?). We can commute them and so we have
$$m_1(m_i dots m_{i+j-1} m_{i+j+1} dots m_{i+(n+1)})$$
Continuing in this fashion we recover
$$m_1m_2 dots m_{n+1}$$
Since this can be done for any arbitrary arrangement the result follows. Hence, multiplication commutes for any finite collection of numbers.
Addition is done in a similar manner.
answered Oct 21 '18 at 13:18
RhythmInkRhythmInk
1,683423
$endgroup$
$begingroup$
My argument still works fine. By associativity of multiplication $m_2m_3(m_1m_4)=m_2m_3m_1m_4$
$endgroup$
– RhythmInk
Oct 21 '18 at 16:25
$begingroup$
The point is associativity of multiplication allows me to ignore paraentheticals or move them however I like.
$endgroup$
– RhythmInk
Oct 21 '18 at 16:51
$begingroup$
Clarify, are you asking us to prove commutativity or associativity by induction? I took any order to mean commutative.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:09
$begingroup$
Yes you are. That's either associativity or commutativity extended to an arbitrary finite collection of elements. The axioms give us the base case. I don't think you quite understand what the question is asking.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:13
$begingroup$
I didn't... please take another look at the problem and the proof and reclarify if need be. Also, one does not and cannot prove axioms. They are axioms. We are extending them to the arbitrary finite case.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:18
|
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2 Answers
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$begingroup$
The proof for multiplication is identical to that for addition, so I'll only look at addition; it works for any commutative associative operation.
Inductive hypothesis: any sum of $n$ numbers is equal to the sum of those same $n$ numbers in sorted order, expressed as $a_1 + (a_2 + (a_3 + dots))$ where $a_1 leq a_2 leq dots$.
Then take a collection of $n+1$ numbers, ${a_1, a_2, dots, a_{n+1}}$. Their sum is expressed as $x + y$, where $x$ and $y$ are some nonempty collections of the $a_i$. There are two cases: $x$ is formally the sum of just one $a_i$, or it's the sum of more than one.
- If $x = a_i$, say, then by the inductive hypothesis, the sum of the $a_i$ comprising $y$ is equal to the same sum in ascending order and associating to the right. So we just need to combine $a_i$ with the sorted $a_1 + (a_2 + dots)$ excluding $a_i$. You can do this inductively: $$x + (a_1 + (a_2 + dots)) = (x + a_1) + (a_2 + dots) = (a_1 + x) + (a_2 + dots) = a_1 + (x + (a_2 + dots))$$
and proceed until $x = a_i$ is in the right place. (This case also can be used if $y$ is just one $a_i$, by using commutativity.) - If $x$ is a sum of more than one $a_i$, then by inductive hypothesis it is $a_{i_1} + (a_{i_2} + dots)$; and $y$ is $a_{j_1} + (a_{j_2} + dots)$. Use commutativity if necessary to ensure that $a_{i_1} leq a_{j_1}$; then use associativity to separate the sum into $a_{i_1} + ((a_{i_2} + dots) + (a_{j_1} + (a_{j_2} + dots)))$. Now use the inductive hypothesis on the non-$a_{i_1}$ term.
answered Oct 21 '18 at 17:59
Patrick StevensPatrick Stevens
28.7k52874
$endgroup$
add a comment |
$begingroup$
The proof for multiplication is identical to that for addition, so I'll only look at addition; it works for any commutative associative operation.
Inductive hypothesis: any sum of $n$ numbers is equal to the sum of those same $n$ numbers in sorted order, expressed as $a_1 + (a_2 + (a_3 + dots))$ where $a_1 leq a_2 leq dots$.
Then take a collection of $n+1$ numbers, ${a_1, a_2, dots, a_{n+1}}$. Their sum is expressed as $x + y$, where $x$ and $y$ are some nonempty collections of the $a_i$. There are two cases: $x$ is formally the sum of just one $a_i$, or it's the sum of more than one.
- If $x = a_i$, say, then by the inductive hypothesis, the sum of the $a_i$ comprising $y$ is equal to the same sum in ascending order and associating to the right. So we just need to combine $a_i$ with the sorted $a_1 + (a_2 + dots)$ excluding $a_i$. You can do this inductively: $$x + (a_1 + (a_2 + dots)) = (x + a_1) + (a_2 + dots) = (a_1 + x) + (a_2 + dots) = a_1 + (x + (a_2 + dots))$$
and proceed until $x = a_i$ is in the right place. (This case also can be used if $y$ is just one $a_i$, by using commutativity.) - If $x$ is a sum of more than one $a_i$, then by inductive hypothesis it is $a_{i_1} + (a_{i_2} + dots)$; and $y$ is $a_{j_1} + (a_{j_2} + dots)$. Use commutativity if necessary to ensure that $a_{i_1} leq a_{j_1}$; then use associativity to separate the sum into $a_{i_1} + ((a_{i_2} + dots) + (a_{j_1} + (a_{j_2} + dots)))$. Now use the inductive hypothesis on the non-$a_{i_1}$ term.
answered Oct 21 '18 at 17:59
Patrick StevensPatrick Stevens
28.7k52874
$endgroup$
add a comment |
$begingroup$
The proof for multiplication is identical to that for addition, so I'll only look at addition; it works for any commutative associative operation.
Inductive hypothesis: any sum of $n$ numbers is equal to the sum of those same $n$ numbers in sorted order, expressed as $a_1 + (a_2 + (a_3 + dots))$ where $a_1 leq a_2 leq dots$.
Then take a collection of $n+1$ numbers, ${a_1, a_2, dots, a_{n+1}}$. Their sum is expressed as $x + y$, where $x$ and $y$ are some nonempty collections of the $a_i$. There are two cases: $x$ is formally the sum of just one $a_i$, or it's the sum of more than one.
- If $x = a_i$, say, then by the inductive hypothesis, the sum of the $a_i$ comprising $y$ is equal to the same sum in ascending order and associating to the right. So we just need to combine $a_i$ with the sorted $a_1 + (a_2 + dots)$ excluding $a_i$. You can do this inductively: $$x + (a_1 + (a_2 + dots)) = (x + a_1) + (a_2 + dots) = (a_1 + x) + (a_2 + dots) = a_1 + (x + (a_2 + dots))$$
and proceed until $x = a_i$ is in the right place. (This case also can be used if $y$ is just one $a_i$, by using commutativity.) - If $x$ is a sum of more than one $a_i$, then by inductive hypothesis it is $a_{i_1} + (a_{i_2} + dots)$; and $y$ is $a_{j_1} + (a_{j_2} + dots)$. Use commutativity if necessary to ensure that $a_{i_1} leq a_{j_1}$; then use associativity to separate the sum into $a_{i_1} + ((a_{i_2} + dots) + (a_{j_1} + (a_{j_2} + dots)))$. Now use the inductive hypothesis on the non-$a_{i_1}$ term.
answered Oct 21 '18 at 17:59
Patrick StevensPatrick Stevens
28.7k52874
$endgroup$
The proof for multiplication is identical to that for addition, so I'll only look at addition; it works for any commutative associative operation.
Inductive hypothesis: any sum of $n$ numbers is equal to the sum of those same $n$ numbers in sorted order, expressed as $a_1 + (a_2 + (a_3 + dots))$ where $a_1 leq a_2 leq dots$.
Then take a collection of $n+1$ numbers, ${a_1, a_2, dots, a_{n+1}}$. Their sum is expressed as $x + y$, where $x$ and $y$ are some nonempty collections of the $a_i$. There are two cases: $x$ is formally the sum of just one $a_i$, or it's the sum of more than one.
- If $x = a_i$, say, then by the inductive hypothesis, the sum of the $a_i$ comprising $y$ is equal to the same sum in ascending order and associating to the right. So we just need to combine $a_i$ with the sorted $a_1 + (a_2 + dots)$ excluding $a_i$. You can do this inductively: $$x + (a_1 + (a_2 + dots)) = (x + a_1) + (a_2 + dots) = (a_1 + x) + (a_2 + dots) = a_1 + (x + (a_2 + dots))$$
and proceed until $x = a_i$ is in the right place. (This case also can be used if $y$ is just one $a_i$, by using commutativity.) - If $x$ is a sum of more than one $a_i$, then by inductive hypothesis it is $a_{i_1} + (a_{i_2} + dots)$; and $y$ is $a_{j_1} + (a_{j_2} + dots)$. Use commutativity if necessary to ensure that $a_{i_1} leq a_{j_1}$; then use associativity to separate the sum into $a_{i_1} + ((a_{i_2} + dots) + (a_{j_1} + (a_{j_2} + dots)))$. Now use the inductive hypothesis on the non-$a_{i_1}$ term.
answered Oct 21 '18 at 17:59
Patrick StevensPatrick Stevens
28.7k52874
answered Oct 21 '18 at 17:59
Patrick StevensPatrick Stevens
28.7k52874
answered Oct 21 '18 at 17:59
Patrick StevensPatrick Stevens
28.7k52874
answered Oct 21 '18 at 17:59
Patrick StevensPatrick Stevens
28.7k52874
28.7k52874
add a comment |
add a comment |
$begingroup$
Okay. So, we proceed by way of induction. You need to show the base case is true which in this case is $n=3$. So, suppose we have $3$ arbitrary numbers $m_1,m_2,m_3$. We know that for any two numbers $a,b$ that $ab=ba$ and that multiplication is associative. So
$$m_1m_2m_3=m_1(m_2m_3)=m_1(m_3m_2)=m_1m_3m_2=(m_1m_3)m_2=(m_3m_1)m_2=m_3m_1m_2$$
By associativity and commutativity (for the case $n=2$). You can continue in this fashion to get all $6$ rearrangements of the product easily and show the base case holds.
Now, we need to show that the $n$th case implies the $n+1$st case. So, suppose that multiplication commutes for any $n$ numbers and consider the collection $m_1,m_2, dots m_{n+1}$ of $n+1$ numbers. We want to show that any arbitrary arrangement of the product is equal to $m_1m_2dots m_nm_{n+1}$ (why is this good enough?). So, say we have some arbitrary arrangement
$$m_im_{i+1} dots m_{i+(n+1)}$$
Then we can group this (by associativity) however we like. So, if $m_1$ is in the $i+j$ position, then we can group all the numbers to its left and consider it as one big number (by our induction hypothesis) since there will be clearly be at most $n$ of them (why?). We can commute them and so we have
$$m_1(m_i dots m_{i+j-1} m_{i+j+1} dots m_{i+(n+1)})$$
Continuing in this fashion we recover
$$m_1m_2 dots m_{n+1}$$
Since this can be done for any arbitrary arrangement the result follows. Hence, multiplication commutes for any finite collection of numbers.
Addition is done in a similar manner.
answered Oct 21 '18 at 13:18
RhythmInkRhythmInk
1,683423
$endgroup$
$begingroup$
My argument still works fine. By associativity of multiplication $m_2m_3(m_1m_4)=m_2m_3m_1m_4$
$endgroup$
– RhythmInk
Oct 21 '18 at 16:25
$begingroup$
The point is associativity of multiplication allows me to ignore paraentheticals or move them however I like.
$endgroup$
– RhythmInk
Oct 21 '18 at 16:51
$begingroup$
Clarify, are you asking us to prove commutativity or associativity by induction? I took any order to mean commutative.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:09
$begingroup$
Yes you are. That's either associativity or commutativity extended to an arbitrary finite collection of elements. The axioms give us the base case. I don't think you quite understand what the question is asking.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:13
$begingroup$
I didn't... please take another look at the problem and the proof and reclarify if need be. Also, one does not and cannot prove axioms. They are axioms. We are extending them to the arbitrary finite case.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:18
|
show 2 more comments
$begingroup$
Okay. So, we proceed by way of induction. You need to show the base case is true which in this case is $n=3$. So, suppose we have $3$ arbitrary numbers $m_1,m_2,m_3$. We know that for any two numbers $a,b$ that $ab=ba$ and that multiplication is associative. So
$$m_1m_2m_3=m_1(m_2m_3)=m_1(m_3m_2)=m_1m_3m_2=(m_1m_3)m_2=(m_3m_1)m_2=m_3m_1m_2$$
By associativity and commutativity (for the case $n=2$). You can continue in this fashion to get all $6$ rearrangements of the product easily and show the base case holds.
Now, we need to show that the $n$th case implies the $n+1$st case. So, suppose that multiplication commutes for any $n$ numbers and consider the collection $m_1,m_2, dots m_{n+1}$ of $n+1$ numbers. We want to show that any arbitrary arrangement of the product is equal to $m_1m_2dots m_nm_{n+1}$ (why is this good enough?). So, say we have some arbitrary arrangement
$$m_im_{i+1} dots m_{i+(n+1)}$$
Then we can group this (by associativity) however we like. So, if $m_1$ is in the $i+j$ position, then we can group all the numbers to its left and consider it as one big number (by our induction hypothesis) since there will be clearly be at most $n$ of them (why?). We can commute them and so we have
$$m_1(m_i dots m_{i+j-1} m_{i+j+1} dots m_{i+(n+1)})$$
Continuing in this fashion we recover
$$m_1m_2 dots m_{n+1}$$
Since this can be done for any arbitrary arrangement the result follows. Hence, multiplication commutes for any finite collection of numbers.
Addition is done in a similar manner.
answered Oct 21 '18 at 13:18
RhythmInkRhythmInk
1,683423
$endgroup$
$begingroup$
My argument still works fine. By associativity of multiplication $m_2m_3(m_1m_4)=m_2m_3m_1m_4$
$endgroup$
– RhythmInk
Oct 21 '18 at 16:25
$begingroup$
The point is associativity of multiplication allows me to ignore paraentheticals or move them however I like.
$endgroup$
– RhythmInk
Oct 21 '18 at 16:51
$begingroup$
Clarify, are you asking us to prove commutativity or associativity by induction? I took any order to mean commutative.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:09
$begingroup$
Yes you are. That's either associativity or commutativity extended to an arbitrary finite collection of elements. The axioms give us the base case. I don't think you quite understand what the question is asking.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:13
$begingroup$
I didn't... please take another look at the problem and the proof and reclarify if need be. Also, one does not and cannot prove axioms. They are axioms. We are extending them to the arbitrary finite case.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:18
|
show 2 more comments
$begingroup$
Okay. So, we proceed by way of induction. You need to show the base case is true which in this case is $n=3$. So, suppose we have $3$ arbitrary numbers $m_1,m_2,m_3$. We know that for any two numbers $a,b$ that $ab=ba$ and that multiplication is associative. So
$$m_1m_2m_3=m_1(m_2m_3)=m_1(m_3m_2)=m_1m_3m_2=(m_1m_3)m_2=(m_3m_1)m_2=m_3m_1m_2$$
By associativity and commutativity (for the case $n=2$). You can continue in this fashion to get all $6$ rearrangements of the product easily and show the base case holds.
Now, we need to show that the $n$th case implies the $n+1$st case. So, suppose that multiplication commutes for any $n$ numbers and consider the collection $m_1,m_2, dots m_{n+1}$ of $n+1$ numbers. We want to show that any arbitrary arrangement of the product is equal to $m_1m_2dots m_nm_{n+1}$ (why is this good enough?). So, say we have some arbitrary arrangement
$$m_im_{i+1} dots m_{i+(n+1)}$$
Then we can group this (by associativity) however we like. So, if $m_1$ is in the $i+j$ position, then we can group all the numbers to its left and consider it as one big number (by our induction hypothesis) since there will be clearly be at most $n$ of them (why?). We can commute them and so we have
$$m_1(m_i dots m_{i+j-1} m_{i+j+1} dots m_{i+(n+1)})$$
Continuing in this fashion we recover
$$m_1m_2 dots m_{n+1}$$
Since this can be done for any arbitrary arrangement the result follows. Hence, multiplication commutes for any finite collection of numbers.
Addition is done in a similar manner.
answered Oct 21 '18 at 13:18
RhythmInkRhythmInk
1,683423
$endgroup$
Okay. So, we proceed by way of induction. You need to show the base case is true which in this case is $n=3$. So, suppose we have $3$ arbitrary numbers $m_1,m_2,m_3$. We know that for any two numbers $a,b$ that $ab=ba$ and that multiplication is associative. So
$$m_1m_2m_3=m_1(m_2m_3)=m_1(m_3m_2)=m_1m_3m_2=(m_1m_3)m_2=(m_3m_1)m_2=m_3m_1m_2$$
By associativity and commutativity (for the case $n=2$). You can continue in this fashion to get all $6$ rearrangements of the product easily and show the base case holds.
Now, we need to show that the $n$th case implies the $n+1$st case. So, suppose that multiplication commutes for any $n$ numbers and consider the collection $m_1,m_2, dots m_{n+1}$ of $n+1$ numbers. We want to show that any arbitrary arrangement of the product is equal to $m_1m_2dots m_nm_{n+1}$ (why is this good enough?). So, say we have some arbitrary arrangement
$$m_im_{i+1} dots m_{i+(n+1)}$$
Then we can group this (by associativity) however we like. So, if $m_1$ is in the $i+j$ position, then we can group all the numbers to its left and consider it as one big number (by our induction hypothesis) since there will be clearly be at most $n$ of them (why?). We can commute them and so we have
$$m_1(m_i dots m_{i+j-1} m_{i+j+1} dots m_{i+(n+1)})$$
Continuing in this fashion we recover
$$m_1m_2 dots m_{n+1}$$
Since this can be done for any arbitrary arrangement the result follows. Hence, multiplication commutes for any finite collection of numbers.
Addition is done in a similar manner.
answered Oct 21 '18 at 13:18
RhythmInkRhythmInk
1,683423
edited Oct 21 '18 at 17:32
answered Oct 21 '18 at 13:18
RhythmInkRhythmInk
1,683423
answered Oct 21 '18 at 13:18
RhythmInkRhythmInk
1,683423
answered Oct 21 '18 at 13:18
RhythmInkRhythmInk
1,683423
1,683423
$begingroup$
My argument still works fine. By associativity of multiplication $m_2m_3(m_1m_4)=m_2m_3m_1m_4$
$endgroup$
– RhythmInk
Oct 21 '18 at 16:25
$begingroup$
The point is associativity of multiplication allows me to ignore paraentheticals or move them however I like.
$endgroup$
– RhythmInk
Oct 21 '18 at 16:51
$begingroup$
Clarify, are you asking us to prove commutativity or associativity by induction? I took any order to mean commutative.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:09
$begingroup$
Yes you are. That's either associativity or commutativity extended to an arbitrary finite collection of elements. The axioms give us the base case. I don't think you quite understand what the question is asking.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:13
$begingroup$
I didn't... please take another look at the problem and the proof and reclarify if need be. Also, one does not and cannot prove axioms. They are axioms. We are extending them to the arbitrary finite case.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:18
|
show 2 more comments
$begingroup$
My argument still works fine. By associativity of multiplication $m_2m_3(m_1m_4)=m_2m_3m_1m_4$
$endgroup$
– RhythmInk
Oct 21 '18 at 16:25
$begingroup$
The point is associativity of multiplication allows me to ignore paraentheticals or move them however I like.
$endgroup$
– RhythmInk
Oct 21 '18 at 16:51
$begingroup$
Clarify, are you asking us to prove commutativity or associativity by induction? I took any order to mean commutative.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:09
$begingroup$
Yes you are. That's either associativity or commutativity extended to an arbitrary finite collection of elements. The axioms give us the base case. I don't think you quite understand what the question is asking.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:13
$begingroup$
I didn't... please take another look at the problem and the proof and reclarify if need be. Also, one does not and cannot prove axioms. They are axioms. We are extending them to the arbitrary finite case.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:18
$begingroup$
My argument still works fine. By associativity of multiplication $m_2m_3(m_1m_4)=m_2m_3m_1m_4$
$endgroup$
– RhythmInk
Oct 21 '18 at 16:25
$begingroup$
My argument still works fine. By associativity of multiplication $m_2m_3(m_1m_4)=m_2m_3m_1m_4$
$endgroup$
– RhythmInk
Oct 21 '18 at 16:25
$begingroup$
The point is associativity of multiplication allows me to ignore paraentheticals or move them however I like.
$endgroup$
– RhythmInk
Oct 21 '18 at 16:51
$begingroup$
The point is associativity of multiplication allows me to ignore paraentheticals or move them however I like.
$endgroup$
– RhythmInk
Oct 21 '18 at 16:51
$begingroup$
Clarify, are you asking us to prove commutativity or associativity by induction? I took any order to mean commutative.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:09
$begingroup$
Clarify, are you asking us to prove commutativity or associativity by induction? I took any order to mean commutative.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:09
$begingroup$
Yes you are. That's either associativity or commutativity extended to an arbitrary finite collection of elements. The axioms give us the base case. I don't think you quite understand what the question is asking.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:13
$begingroup$
Yes you are. That's either associativity or commutativity extended to an arbitrary finite collection of elements. The axioms give us the base case. I don't think you quite understand what the question is asking.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:13
$begingroup$
I didn't... please take another look at the problem and the proof and reclarify if need be. Also, one does not and cannot prove axioms. They are axioms. We are extending them to the arbitrary finite case.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:18
$begingroup$
I didn't... please take another look at the problem and the proof and reclarify if need be. Also, one does not and cannot prove axioms. They are axioms. We are extending them to the arbitrary finite case.
$endgroup$
– RhythmInk
Oct 21 '18 at 17:18
|
show 2 more comments
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