Mixing
Euclidean division? ( $16=5cdot 3+1$ vs $16=3cdot 5+1$)
$begingroup$
Is the equality
$16=5cdot 3+1$
the euclidean division of $16$ by $3$ or not ?
This question is a point of discord between teachers where some them state that the divisor must be written in the first position (in this example, one has to write
$16=3cdot 5+1$ i.e must write the divisor first then the quotient).
What do you think ?
notation arithmetic euclidean-algorithm
edited Jan 17 at 16:00
gt6989b
34.3k22455
asked Jan 17 at 15:27
ProfahkProfahk
92
$endgroup$
add a comment |
$begingroup$
Is the equality
$16=5cdot 3+1$
the euclidean division of $16$ by $3$ or not ?
This question is a point of discord between teachers where some them state that the divisor must be written in the first position (in this example, one has to write
$16=3cdot 5+1$ i.e must write the divisor first then the quotient).
What do you think ?
notation arithmetic euclidean-algorithm
edited Jan 17 at 16:00
gt6989b
34.3k22455
asked Jan 17 at 15:27
ProfahkProfahk
92
$endgroup$
$begingroup$
is that a question on notation specifically? if so, add notation tag
$endgroup$
– gt6989b
Jan 17 at 15:32
$begingroup$
@gt6988b yes but I can't edit my post.
$endgroup$
– Profahk
Jan 17 at 15:38
add a comment |
$begingroup$
Is the equality
$16=5cdot 3+1$
the euclidean division of $16$ by $3$ or not ?
This question is a point of discord between teachers where some them state that the divisor must be written in the first position (in this example, one has to write
$16=3cdot 5+1$ i.e must write the divisor first then the quotient).
What do you think ?
notation arithmetic euclidean-algorithm
edited Jan 17 at 16:00
gt6989b
34.3k22455
asked Jan 17 at 15:27
ProfahkProfahk
92
$endgroup$
Is the equality
$16=5cdot 3+1$
the euclidean division of $16$ by $3$ or not ?
This question is a point of discord between teachers where some them state that the divisor must be written in the first position (in this example, one has to write
$16=3cdot 5+1$ i.e must write the divisor first then the quotient).
What do you think ?
notation arithmetic euclidean-algorithm
notation arithmetic euclidean-algorithm
edited Jan 17 at 16:00
gt6989b
34.3k22455
asked Jan 17 at 15:27
ProfahkProfahk
92
edited Jan 17 at 16:00
gt6989b
34.3k22455
asked Jan 17 at 15:27
ProfahkProfahk
92
edited Jan 17 at 16:00
gt6989b
34.3k22455
edited Jan 17 at 16:00
gt6989b
34.3k22455
edited Jan 17 at 16:00
gt6989b
34.3k22455
34.3k22455
asked Jan 17 at 15:27
ProfahkProfahk
92
asked Jan 17 at 15:27
ProfahkProfahk
92
asked Jan 17 at 15:27
ProfahkProfahk
92
92
$begingroup$
is that a question on notation specifically? if so, add notation tag
$endgroup$
– gt6989b
Jan 17 at 15:32
$begingroup$
@gt6988b yes but I can't edit my post.
$endgroup$
– Profahk
Jan 17 at 15:38
add a comment |
$begingroup$
is that a question on notation specifically? if so, add notation tag
$endgroup$
– gt6989b
Jan 17 at 15:32
$begingroup$
@gt6988b yes but I can't edit my post.
$endgroup$
– Profahk
Jan 17 at 15:38
$begingroup$
is that a question on notation specifically? if so, add notation tag
$endgroup$
– gt6989b
Jan 17 at 15:32
$begingroup$
is that a question on notation specifically? if so, add notation tag
$endgroup$
– gt6989b
Jan 17 at 15:32
$begingroup$
@gt6988b yes but I can't edit my post.
$endgroup$
– Profahk
Jan 17 at 15:38
$begingroup$
@gt6988b yes but I can't edit my post.
$endgroup$
– Profahk
Jan 17 at 15:38
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$ 16 = 5times 3 + 1$ could arise from dividing $16$ by $5$ or by $3$. Without any further context there is no way to determine if $5$ or $3$ is the intended divisor.
Though - as you mention - one could impose syntactic conventions that imply which is the intended divisor, this is usually not a good idea, since it violates referential transparency, i.e. replacing an expression with an equivalent expression should not change its meaning. But if we replace $3^2$ by $3times 3$ in $,37 = 3^2times 4 + 1$ then it changes the divisor from $9$ to $3$ (using your convention that the first factor is the divisor).
Further, conventions that force one to use specific commutations / associations of products may lead to increased complexity, e.g. instead of writing $, f(x) = (x^2+x+1)^n g(x) + 1$ we would be forced to instead write $,f(x) = (x^2+x+1)(x^2+x+1)^{n-1}g(x) + 1$ to denote that $,x^2+x+1,$ is the intended divisor.
As such, it is generally better to avoid such conventions and instead explicitly state the divisor.
answered Jan 17 at 16:18
Bill DubuqueBill Dubuque
211k29193647
$endgroup$
$begingroup$
Thank You.Do you have any reference (book, website,...) that I can convince my colleagues ?
$endgroup$
– Profahk
Jan 17 at 17:43
add a comment |
$begingroup$
The answer is neither yes nor no. “Euclidean division” is a concept, not an arithmetic expression. The quotient of the Euclidean division of $16$ by $5$ is $3$ and the remainder is $1$. You can prove it by any of the equalities $16=3times5+1$ or $16=5times3+1$ (they are equivalent, of course), but, again, none of them is the Euclidean division.
edited Jan 17 at 16:03
J. W. Tanner
2,4581117
answered Jan 17 at 15:32
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
$begingroup$
But we can write an equality to express an Euclidean division ?
$endgroup$
– Profahk
Jan 17 at 15:41
$begingroup$
Both equalities $16=3times5+1$ and $16=5times3+1$ explain why is it that the quotient is $5$ and the remainder is $1$. But none of them is the Euclidean division.
$endgroup$
– José Carlos Santos
Jan 17 at 15:43
$begingroup$
I agree that both of them explain why the quotient is 5 and the remainder 1.
$endgroup$
– Profahk
Jan 17 at 15:46
$begingroup$
I think I'm misunderstood.
$endgroup$
– Profahk
Jan 17 at 15:47
add a comment |
Your Answer
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
$ 16 = 5times 3 + 1$ could arise from dividing $16$ by $5$ or by $3$. Without any further context there is no way to determine if $5$ or $3$ is the intended divisor.
Though - as you mention - one could impose syntactic conventions that imply which is the intended divisor, this is usually not a good idea, since it violates referential transparency, i.e. replacing an expression with an equivalent expression should not change its meaning. But if we replace $3^2$ by $3times 3$ in $,37 = 3^2times 4 + 1$ then it changes the divisor from $9$ to $3$ (using your convention that the first factor is the divisor).
Further, conventions that force one to use specific commutations / associations of products may lead to increased complexity, e.g. instead of writing $, f(x) = (x^2+x+1)^n g(x) + 1$ we would be forced to instead write $,f(x) = (x^2+x+1)(x^2+x+1)^{n-1}g(x) + 1$ to denote that $,x^2+x+1,$ is the intended divisor.
As such, it is generally better to avoid such conventions and instead explicitly state the divisor.
answered Jan 17 at 16:18
Bill DubuqueBill Dubuque
211k29193647
$endgroup$
$begingroup$
Thank You.Do you have any reference (book, website,...) that I can convince my colleagues ?
$endgroup$
– Profahk
Jan 17 at 17:43
add a comment |
$begingroup$
$ 16 = 5times 3 + 1$ could arise from dividing $16$ by $5$ or by $3$. Without any further context there is no way to determine if $5$ or $3$ is the intended divisor.
Though - as you mention - one could impose syntactic conventions that imply which is the intended divisor, this is usually not a good idea, since it violates referential transparency, i.e. replacing an expression with an equivalent expression should not change its meaning. But if we replace $3^2$ by $3times 3$ in $,37 = 3^2times 4 + 1$ then it changes the divisor from $9$ to $3$ (using your convention that the first factor is the divisor).
Further, conventions that force one to use specific commutations / associations of products may lead to increased complexity, e.g. instead of writing $, f(x) = (x^2+x+1)^n g(x) + 1$ we would be forced to instead write $,f(x) = (x^2+x+1)(x^2+x+1)^{n-1}g(x) + 1$ to denote that $,x^2+x+1,$ is the intended divisor.
As such, it is generally better to avoid such conventions and instead explicitly state the divisor.
answered Jan 17 at 16:18
Bill DubuqueBill Dubuque
211k29193647
$endgroup$
$begingroup$
Thank You.Do you have any reference (book, website,...) that I can convince my colleagues ?
$endgroup$
– Profahk
Jan 17 at 17:43
add a comment |
$begingroup$
$ 16 = 5times 3 + 1$ could arise from dividing $16$ by $5$ or by $3$. Without any further context there is no way to determine if $5$ or $3$ is the intended divisor.
Though - as you mention - one could impose syntactic conventions that imply which is the intended divisor, this is usually not a good idea, since it violates referential transparency, i.e. replacing an expression with an equivalent expression should not change its meaning. But if we replace $3^2$ by $3times 3$ in $,37 = 3^2times 4 + 1$ then it changes the divisor from $9$ to $3$ (using your convention that the first factor is the divisor).
Further, conventions that force one to use specific commutations / associations of products may lead to increased complexity, e.g. instead of writing $, f(x) = (x^2+x+1)^n g(x) + 1$ we would be forced to instead write $,f(x) = (x^2+x+1)(x^2+x+1)^{n-1}g(x) + 1$ to denote that $,x^2+x+1,$ is the intended divisor.
As such, it is generally better to avoid such conventions and instead explicitly state the divisor.
answered Jan 17 at 16:18
Bill DubuqueBill Dubuque
211k29193647
$endgroup$
$ 16 = 5times 3 + 1$ could arise from dividing $16$ by $5$ or by $3$. Without any further context there is no way to determine if $5$ or $3$ is the intended divisor.
Though - as you mention - one could impose syntactic conventions that imply which is the intended divisor, this is usually not a good idea, since it violates referential transparency, i.e. replacing an expression with an equivalent expression should not change its meaning. But if we replace $3^2$ by $3times 3$ in $,37 = 3^2times 4 + 1$ then it changes the divisor from $9$ to $3$ (using your convention that the first factor is the divisor).
Further, conventions that force one to use specific commutations / associations of products may lead to increased complexity, e.g. instead of writing $, f(x) = (x^2+x+1)^n g(x) + 1$ we would be forced to instead write $,f(x) = (x^2+x+1)(x^2+x+1)^{n-1}g(x) + 1$ to denote that $,x^2+x+1,$ is the intended divisor.
As such, it is generally better to avoid such conventions and instead explicitly state the divisor.
answered Jan 17 at 16:18
Bill DubuqueBill Dubuque
211k29193647
edited Jan 17 at 16:45
answered Jan 17 at 16:18
Bill DubuqueBill Dubuque
211k29193647
answered Jan 17 at 16:18
Bill DubuqueBill Dubuque
211k29193647
answered Jan 17 at 16:18
Bill DubuqueBill Dubuque
211k29193647
211k29193647
$begingroup$
Thank You.Do you have any reference (book, website,...) that I can convince my colleagues ?
$endgroup$
– Profahk
Jan 17 at 17:43
add a comment |
$begingroup$
Thank You.Do you have any reference (book, website,...) that I can convince my colleagues ?
$endgroup$
– Profahk
Jan 17 at 17:43
$begingroup$
Thank You.Do you have any reference (book, website,...) that I can convince my colleagues ?
$endgroup$
– Profahk
Jan 17 at 17:43
$begingroup$
Thank You.Do you have any reference (book, website,...) that I can convince my colleagues ?
$endgroup$
– Profahk
Jan 17 at 17:43
add a comment |
$begingroup$
The answer is neither yes nor no. “Euclidean division” is a concept, not an arithmetic expression. The quotient of the Euclidean division of $16$ by $5$ is $3$ and the remainder is $1$. You can prove it by any of the equalities $16=3times5+1$ or $16=5times3+1$ (they are equivalent, of course), but, again, none of them is the Euclidean division.
edited Jan 17 at 16:03
J. W. Tanner
2,4581117
answered Jan 17 at 15:32
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
$begingroup$
But we can write an equality to express an Euclidean division ?
$endgroup$
– Profahk
Jan 17 at 15:41
$begingroup$
Both equalities $16=3times5+1$ and $16=5times3+1$ explain why is it that the quotient is $5$ and the remainder is $1$. But none of them is the Euclidean division.
$endgroup$
– José Carlos Santos
Jan 17 at 15:43
$begingroup$
I agree that both of them explain why the quotient is 5 and the remainder 1.
$endgroup$
– Profahk
Jan 17 at 15:46
$begingroup$
I think I'm misunderstood.
$endgroup$
– Profahk
Jan 17 at 15:47
add a comment |
$begingroup$
The answer is neither yes nor no. “Euclidean division” is a concept, not an arithmetic expression. The quotient of the Euclidean division of $16$ by $5$ is $3$ and the remainder is $1$. You can prove it by any of the equalities $16=3times5+1$ or $16=5times3+1$ (they are equivalent, of course), but, again, none of them is the Euclidean division.
edited Jan 17 at 16:03
J. W. Tanner
2,4581117
answered Jan 17 at 15:32
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
$begingroup$
But we can write an equality to express an Euclidean division ?
$endgroup$
– Profahk
Jan 17 at 15:41
$begingroup$
Both equalities $16=3times5+1$ and $16=5times3+1$ explain why is it that the quotient is $5$ and the remainder is $1$. But none of them is the Euclidean division.
$endgroup$
– José Carlos Santos
Jan 17 at 15:43
$begingroup$
I agree that both of them explain why the quotient is 5 and the remainder 1.
$endgroup$
– Profahk
Jan 17 at 15:46
$begingroup$
I think I'm misunderstood.
$endgroup$
– Profahk
Jan 17 at 15:47
add a comment |
$begingroup$
The answer is neither yes nor no. “Euclidean division” is a concept, not an arithmetic expression. The quotient of the Euclidean division of $16$ by $5$ is $3$ and the remainder is $1$. You can prove it by any of the equalities $16=3times5+1$ or $16=5times3+1$ (they are equivalent, of course), but, again, none of them is the Euclidean division.
edited Jan 17 at 16:03
J. W. Tanner
2,4581117
answered Jan 17 at 15:32
José Carlos SantosJosé Carlos Santos
163k22131234
$endgroup$
The answer is neither yes nor no. “Euclidean division” is a concept, not an arithmetic expression. The quotient of the Euclidean division of $16$ by $5$ is $3$ and the remainder is $1$. You can prove it by any of the equalities $16=3times5+1$ or $16=5times3+1$ (they are equivalent, of course), but, again, none of them is the Euclidean division.
edited Jan 17 at 16:03
J. W. Tanner
2,4581117
answered Jan 17 at 15:32
José Carlos SantosJosé Carlos Santos
163k22131234
edited Jan 17 at 16:03
J. W. Tanner
2,4581117
edited Jan 17 at 16:03
J. W. Tanner
2,4581117
edited Jan 17 at 16:03
J. W. Tanner
2,4581117
2,4581117
answered Jan 17 at 15:32
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 17 at 15:32
José Carlos SantosJosé Carlos Santos
163k22131234
answered Jan 17 at 15:32
José Carlos SantosJosé Carlos Santos
163k22131234
163k22131234
$begingroup$
But we can write an equality to express an Euclidean division ?
$endgroup$
– Profahk
Jan 17 at 15:41
$begingroup$
Both equalities $16=3times5+1$ and $16=5times3+1$ explain why is it that the quotient is $5$ and the remainder is $1$. But none of them is the Euclidean division.
$endgroup$
– José Carlos Santos
Jan 17 at 15:43
$begingroup$
I agree that both of them explain why the quotient is 5 and the remainder 1.
$endgroup$
– Profahk
Jan 17 at 15:46
$begingroup$
I think I'm misunderstood.
$endgroup$
– Profahk
Jan 17 at 15:47
add a comment |
$begingroup$
But we can write an equality to express an Euclidean division ?
$endgroup$
– Profahk
Jan 17 at 15:41
$begingroup$
Both equalities $16=3times5+1$ and $16=5times3+1$ explain why is it that the quotient is $5$ and the remainder is $1$. But none of them is the Euclidean division.
$endgroup$
– José Carlos Santos
Jan 17 at 15:43
$begingroup$
I agree that both of them explain why the quotient is 5 and the remainder 1.
$endgroup$
– Profahk
Jan 17 at 15:46
$begingroup$
I think I'm misunderstood.
$endgroup$
– Profahk
Jan 17 at 15:47
$begingroup$
But we can write an equality to express an Euclidean division ?
$endgroup$
– Profahk
Jan 17 at 15:41
$begingroup$
But we can write an equality to express an Euclidean division ?
$endgroup$
– Profahk
Jan 17 at 15:41
$begingroup$
Both equalities $16=3times5+1$ and $16=5times3+1$ explain why is it that the quotient is $5$ and the remainder is $1$. But none of them is the Euclidean division.
$endgroup$
– José Carlos Santos
Jan 17 at 15:43
$begingroup$
Both equalities $16=3times5+1$ and $16=5times3+1$ explain why is it that the quotient is $5$ and the remainder is $1$. But none of them is the Euclidean division.
$endgroup$
– José Carlos Santos
Jan 17 at 15:43
$begingroup$
I agree that both of them explain why the quotient is 5 and the remainder 1.
$endgroup$
– Profahk
Jan 17 at 15:46
$begingroup$
I agree that both of them explain why the quotient is 5 and the remainder 1.
$endgroup$
– Profahk
Jan 17 at 15:46
$begingroup$
I think I'm misunderstood.
$endgroup$
– Profahk
Jan 17 at 15:47
$begingroup$
I think I'm misunderstood.
$endgroup$
– Profahk
Jan 17 at 15:47
add a comment |
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$begingroup$
is that a question on notation specifically? if so, add notation tag
$endgroup$
– gt6989b
Jan 17 at 15:32
$begingroup$
@gt6988b yes but I can't edit my post.
$endgroup$
– Profahk
Jan 17 at 15:38