How is rearranging 56 x 100 ÷ 8 into 56 ÷8 x100 allowed by the commutative property?
$begingroup$
So according to the commutative property for multiplication:
$a * b = b * a$
However this does not hold for division
$a ÷ b$ $!= b ÷a$
Why is it that in the following case:
$56 * 100 ÷ 8 = 56 ÷ 8* 100$
It seems like division is breaking the rule. There is something I am misunderstanding here.
Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?
If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?
arithmetic
asked 51 mins ago
SphygmomanometerSphygmomanometer
717
$endgroup$
add a comment |
$begingroup$
So according to the commutative property for multiplication:
$a * b = b * a$
However this does not hold for division
$a ÷ b$ $!= b ÷a$
Why is it that in the following case:
$56 * 100 ÷ 8 = 56 ÷ 8* 100$
It seems like division is breaking the rule. There is something I am misunderstanding here.
Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?
If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?
arithmetic
asked 51 mins ago
SphygmomanometerSphygmomanometer
717
$endgroup$
add a comment |
$begingroup$
So according to the commutative property for multiplication:
$a * b = b * a$
However this does not hold for division
$a ÷ b$ $!= b ÷a$
Why is it that in the following case:
$56 * 100 ÷ 8 = 56 ÷ 8* 100$
It seems like division is breaking the rule. There is something I am misunderstanding here.
Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?
If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?
arithmetic
asked 51 mins ago
SphygmomanometerSphygmomanometer
717
$endgroup$
So according to the commutative property for multiplication:
$a * b = b * a$
However this does not hold for division
$a ÷ b$ $!= b ÷a$
Why is it that in the following case:
$56 * 100 ÷ 8 = 56 ÷ 8* 100$
It seems like division is breaking the rule. There is something I am misunderstanding here.
Is it because $a*b÷c=a÷c*b$ is allowed since $b÷c$ are not being rearranged so that $c÷b$?
If this is the case are you allowed to rearrange values in equations so long as no values have the form $a ÷ b$ $= b ÷a$ and $a - b$ $= b -a$ ?
arithmetic
arithmetic
asked 51 mins ago
SphygmomanometerSphygmomanometer
717
asked 51 mins ago
SphygmomanometerSphygmomanometer
717
edited 49 mins ago
Sphygmomanometer
asked 51 mins ago
SphygmomanometerSphygmomanometer
717
asked 51 mins ago
SphygmomanometerSphygmomanometer
717
asked 51 mins ago
SphygmomanometerSphygmomanometer
717
717
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$
answered 42 mins ago
AndreiAndrei
12.1k21127
$endgroup$
add a comment |
$begingroup$
The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.
So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$
answered 42 mins ago
Angela RichardsonAngela Richardson
5,28911733
$endgroup$
add a comment |
$begingroup$
Hope this makes sense.
$$ atimes b ÷ c $$
$$=atimesdfrac{b}{c}$$
$$=atimes btimesdfrac{1}{c}$$
$$=(atimesdfrac{1}{c})times b$$
$$=dfrac{a}{c}times b$$
$$=a÷ctimes b$$
answered 37 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$
So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.
answered 37 mins ago
Rhys HughesRhys Hughes
6,7101530
$endgroup$
add a comment |
$begingroup$
Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.
So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.
Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.
So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.
answered 5 mins ago
Mark BennetMark Bennet
81.3k983180
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$
answered 42 mins ago
AndreiAndrei
12.1k21127
$endgroup$
add a comment |
$begingroup$
Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$
answered 42 mins ago
AndreiAndrei
12.1k21127
$endgroup$
add a comment |
$begingroup$
Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$
answered 42 mins ago
AndreiAndrei
12.1k21127
$endgroup$
Notice that you have always $div 8$, no matter the order of the other terms. You don't divide by a different number. It might help to think $div c$ as a multiplication with $d=1/c$. Then everything would look easier:
$$atimes bdiv c=atimes btimes d=atimes dtimes b=adiv ctimes b$$
answered 42 mins ago
AndreiAndrei
12.1k21127
answered 42 mins ago
AndreiAndrei
12.1k21127
answered 42 mins ago
AndreiAndrei
12.1k21127
answered 42 mins ago
AndreiAndrei
12.1k21127
12.1k21127
add a comment |
add a comment |
$begingroup$
The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.
So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$
answered 42 mins ago
Angela RichardsonAngela Richardson
5,28911733
$endgroup$
add a comment |
$begingroup$
The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.
So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$
answered 42 mins ago
Angela RichardsonAngela Richardson
5,28911733
$endgroup$
add a comment |
$begingroup$
The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.
So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$
answered 42 mins ago
Angela RichardsonAngela Richardson
5,28911733
$endgroup$
The non-zero real numbers form an abelian group under multiplication. The notation $adiv b$ is shorthand for $ab^{-1}$.
So $ab^{-1}ne ba^{-1}$ but $56times100 times8^{-1}=56times 8^{-1}times 100$
answered 42 mins ago
Angela RichardsonAngela Richardson
5,28911733
answered 42 mins ago
Angela RichardsonAngela Richardson
5,28911733
answered 42 mins ago
Angela RichardsonAngela Richardson
5,28911733
answered 42 mins ago
Angela RichardsonAngela Richardson
5,28911733
5,28911733
add a comment |
add a comment |
$begingroup$
Hope this makes sense.
$$ atimes b ÷ c $$
$$=atimesdfrac{b}{c}$$
$$=atimes btimesdfrac{1}{c}$$
$$=(atimesdfrac{1}{c})times b$$
$$=dfrac{a}{c}times b$$
$$=a÷ctimes b$$
answered 37 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Hope this makes sense.
$$ atimes b ÷ c $$
$$=atimesdfrac{b}{c}$$
$$=atimes btimesdfrac{1}{c}$$
$$=(atimesdfrac{1}{c})times b$$
$$=dfrac{a}{c}times b$$
$$=a÷ctimes b$$
answered 37 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Hope this makes sense.
$$ atimes b ÷ c $$
$$=atimesdfrac{b}{c}$$
$$=atimes btimesdfrac{1}{c}$$
$$=(atimesdfrac{1}{c})times b$$
$$=dfrac{a}{c}times b$$
$$=a÷ctimes b$$
answered 37 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Hope this makes sense.
$$ atimes b ÷ c $$
$$=atimesdfrac{b}{c}$$
$$=atimes btimesdfrac{1}{c}$$
$$=(atimesdfrac{1}{c})times b$$
$$=dfrac{a}{c}times b$$
$$=a÷ctimes b$$
answered 37 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 37 mins ago
AdityaAditya
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 37 mins ago
AdityaAditya
112
answered 37 mins ago
AdityaAditya
112
112
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Aditya is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
$begingroup$
Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$
So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.
answered 37 mins ago
Rhys HughesRhys Hughes
6,7101530
$endgroup$
add a comment |
$begingroup$
Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$
So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.
answered 37 mins ago
Rhys HughesRhys Hughes
6,7101530
$endgroup$
add a comment |
$begingroup$
Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$
So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.
answered 37 mins ago
Rhys HughesRhys Hughes
6,7101530
$endgroup$
Because division is the inverse of multiplication, that is: $$X div Y =Xcdot frac 1Y$$
So you have: $$56cdot 100 cdot frac18 =56cdotfrac18cdot100$$
Which is obvious.
answered 37 mins ago
Rhys HughesRhys Hughes
6,7101530
answered 37 mins ago
Rhys HughesRhys Hughes
6,7101530
answered 37 mins ago
Rhys HughesRhys Hughes
6,7101530
answered 37 mins ago
Rhys HughesRhys Hughes
6,7101530
6,7101530
add a comment |
add a comment |
$begingroup$
Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.
So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.
Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.
So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.
answered 5 mins ago
Mark BennetMark Bennet
81.3k983180
$endgroup$
add a comment |
$begingroup$
Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.
So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.
Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.
So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.
answered 5 mins ago
Mark BennetMark Bennet
81.3k983180
$endgroup$
add a comment |
$begingroup$
Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.
So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.
Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.
So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.
answered 5 mins ago
Mark BennetMark Bennet
81.3k983180
$endgroup$
Others have answered the direct question, in that multiplication is commutative and that applies also to multiplication by a reciprocal (the equivalent of division). However there is an issue here with associativity and division which I think is worth mentioning. This has to do with the order in which operations are carried out.
So $adiv b times c$ is being interpreted from left to right as $(adiv b)times c=cfrac {ac}b$, but done from right to left $adiv (btimes c)=cfrac a{bc}$ and the two results are not the same.
Likewise with $adiv b div c$ we have $(adiv b)div c=cfrac a{bc} neq adiv (bdiv c)=cfrac {ac}b$.
So the conventional assumption that multiplication and division are operations of equal status and are carried out from left to right does make a difference in these cases and changes the result. This may be what is feeding your intuition that there is a potential problem with order.
answered 5 mins ago
Mark BennetMark Bennet
81.3k983180
answered 5 mins ago
Mark BennetMark Bennet
81.3k983180
answered 5 mins ago
Mark BennetMark Bennet
81.3k983180
answered 5 mins ago
Mark BennetMark Bennet
81.3k983180
81.3k983180
add a comment |
add a comment |
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