How does division “distribute”?
$begingroup$
What would be the expansion of
$$a/(btimes c)$$
I can figure this out by writing it out as a fraction or by using examples.
I am more interested in how this plays out just using the divison symbol.
e.g. for just multiplication, $atimes(btimes c)=atimes btimes c$
algebra-precalculus
asked Dec 11 '18 at 13:07
quadfirequadfire
83
$endgroup$
add a comment |
$begingroup$
What would be the expansion of
$$a/(btimes c)$$
I can figure this out by writing it out as a fraction or by using examples.
I am more interested in how this plays out just using the divison symbol.
e.g. for just multiplication, $atimes(btimes c)=atimes btimes c$
algebra-precalculus
asked Dec 11 '18 at 13:07
quadfirequadfire
83
$endgroup$
1
$begingroup$
$a/( b times c) = a times dfrac {1}{( b times c)}$.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:12
$begingroup$
$a/(b times c) = (a/b)/c$
$endgroup$
– Matti P.
Dec 11 '18 at 13:13
$begingroup$
See Multiplying fractions.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:15
add a comment |
$begingroup$
What would be the expansion of
$$a/(btimes c)$$
I can figure this out by writing it out as a fraction or by using examples.
I am more interested in how this plays out just using the divison symbol.
e.g. for just multiplication, $atimes(btimes c)=atimes btimes c$
algebra-precalculus
asked Dec 11 '18 at 13:07
quadfirequadfire
83
$endgroup$
What would be the expansion of
$$a/(btimes c)$$
I can figure this out by writing it out as a fraction or by using examples.
I am more interested in how this plays out just using the divison symbol.
e.g. for just multiplication, $atimes(btimes c)=atimes btimes c$
algebra-precalculus
algebra-precalculus
asked Dec 11 '18 at 13:07
quadfirequadfire
83
asked Dec 11 '18 at 13:07
quadfirequadfire
83
asked Dec 11 '18 at 13:07
quadfirequadfire
83
asked Dec 11 '18 at 13:07
quadfirequadfire
83
asked Dec 11 '18 at 13:07
quadfirequadfire
83
83
1
$begingroup$
$a/( b times c) = a times dfrac {1}{( b times c)}$.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:12
$begingroup$
$a/(b times c) = (a/b)/c$
$endgroup$
– Matti P.
Dec 11 '18 at 13:13
$begingroup$
See Multiplying fractions.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:15
add a comment |
1
$begingroup$
$a/( b times c) = a times dfrac {1}{( b times c)}$.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:12
$begingroup$
$a/(b times c) = (a/b)/c$
$endgroup$
– Matti P.
Dec 11 '18 at 13:13
$begingroup$
See Multiplying fractions.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:15
1
1
$begingroup$
$a/( b times c) = a times dfrac {1}{( b times c)}$.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:12
$begingroup$
$a/( b times c) = a times dfrac {1}{( b times c)}$.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:12
$begingroup$
$a/(b times c) = (a/b)/c$
$endgroup$
– Matti P.
Dec 11 '18 at 13:13
$begingroup$
$a/(b times c) = (a/b)/c$
$endgroup$
– Matti P.
Dec 11 '18 at 13:13
$begingroup$
See Multiplying fractions.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:15
$begingroup$
See Multiplying fractions.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:15
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If you only have multiplication and division, it works exactly the same way as if you had had only addition and subtraction:
$$
a-(b+c) = a-b-c iff a/(btimes c) = (a/b)/c
$$
(I like to use the brackets for repeated division, because division isn't as universally left-to-right as subtraction is). We can reason about it as follows: Whatever $a/(btimes c)$ is, we'll call it $x$. Then we have
$$
x = a/(btimes c)\
(btimes c)x = a\
(btimes c)x/b = a/b\
cx = a/b\
cx/c = (a/b)/c\
x = (a/b)/c
$$
And we see that $a/(btimes c)$ and $(a/b)/c$ are the same thing.
answered Dec 11 '18 at 13:13
ArthurArthur
115k7116198
$endgroup$
$begingroup$
Makes sense. Still, what is there to stop me writing the expression as $a/b/c$ and then evaluating $b/c$ first? It seems the parantheses are vital or is there some convention when doing divison I might be missing? Thanks!
$endgroup$
– quadfire
Dec 11 '18 at 13:30
$begingroup$
@quadfire You will find many people here (and elsewhere) who adhere to the convention that divisions (nd multiplications) are evaluated left-to-right (just like addition and subtraction: for $a-b-c$ people would think you crazy if you first evaluated $b-c$ and then subtracted the result from $a$). I personally do not follow that convention and instead use brackets to make it clear what I mean. So when I see $a/b/c$ I consider that ambiguous exactly for the reason that you mention.
$endgroup$
– Arthur
Dec 11 '18 at 13:33
add a comment |
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
If you only have multiplication and division, it works exactly the same way as if you had had only addition and subtraction:
$$
a-(b+c) = a-b-c iff a/(btimes c) = (a/b)/c
$$
(I like to use the brackets for repeated division, because division isn't as universally left-to-right as subtraction is). We can reason about it as follows: Whatever $a/(btimes c)$ is, we'll call it $x$. Then we have
$$
x = a/(btimes c)\
(btimes c)x = a\
(btimes c)x/b = a/b\
cx = a/b\
cx/c = (a/b)/c\
x = (a/b)/c
$$
And we see that $a/(btimes c)$ and $(a/b)/c$ are the same thing.
answered Dec 11 '18 at 13:13
ArthurArthur
115k7116198
$endgroup$
$begingroup$
Makes sense. Still, what is there to stop me writing the expression as $a/b/c$ and then evaluating $b/c$ first? It seems the parantheses are vital or is there some convention when doing divison I might be missing? Thanks!
$endgroup$
– quadfire
Dec 11 '18 at 13:30
$begingroup$
@quadfire You will find many people here (and elsewhere) who adhere to the convention that divisions (nd multiplications) are evaluated left-to-right (just like addition and subtraction: for $a-b-c$ people would think you crazy if you first evaluated $b-c$ and then subtracted the result from $a$). I personally do not follow that convention and instead use brackets to make it clear what I mean. So when I see $a/b/c$ I consider that ambiguous exactly for the reason that you mention.
$endgroup$
– Arthur
Dec 11 '18 at 13:33
add a comment |
$begingroup$
If you only have multiplication and division, it works exactly the same way as if you had had only addition and subtraction:
$$
a-(b+c) = a-b-c iff a/(btimes c) = (a/b)/c
$$
(I like to use the brackets for repeated division, because division isn't as universally left-to-right as subtraction is). We can reason about it as follows: Whatever $a/(btimes c)$ is, we'll call it $x$. Then we have
$$
x = a/(btimes c)\
(btimes c)x = a\
(btimes c)x/b = a/b\
cx = a/b\
cx/c = (a/b)/c\
x = (a/b)/c
$$
And we see that $a/(btimes c)$ and $(a/b)/c$ are the same thing.
answered Dec 11 '18 at 13:13
ArthurArthur
115k7116198
$endgroup$
$begingroup$
Makes sense. Still, what is there to stop me writing the expression as $a/b/c$ and then evaluating $b/c$ first? It seems the parantheses are vital or is there some convention when doing divison I might be missing? Thanks!
$endgroup$
– quadfire
Dec 11 '18 at 13:30
$begingroup$
@quadfire You will find many people here (and elsewhere) who adhere to the convention that divisions (nd multiplications) are evaluated left-to-right (just like addition and subtraction: for $a-b-c$ people would think you crazy if you first evaluated $b-c$ and then subtracted the result from $a$). I personally do not follow that convention and instead use brackets to make it clear what I mean. So when I see $a/b/c$ I consider that ambiguous exactly for the reason that you mention.
$endgroup$
– Arthur
Dec 11 '18 at 13:33
add a comment |
$begingroup$
If you only have multiplication and division, it works exactly the same way as if you had had only addition and subtraction:
$$
a-(b+c) = a-b-c iff a/(btimes c) = (a/b)/c
$$
(I like to use the brackets for repeated division, because division isn't as universally left-to-right as subtraction is). We can reason about it as follows: Whatever $a/(btimes c)$ is, we'll call it $x$. Then we have
$$
x = a/(btimes c)\
(btimes c)x = a\
(btimes c)x/b = a/b\
cx = a/b\
cx/c = (a/b)/c\
x = (a/b)/c
$$
And we see that $a/(btimes c)$ and $(a/b)/c$ are the same thing.
answered Dec 11 '18 at 13:13
ArthurArthur
115k7116198
$endgroup$
If you only have multiplication and division, it works exactly the same way as if you had had only addition and subtraction:
$$
a-(b+c) = a-b-c iff a/(btimes c) = (a/b)/c
$$
(I like to use the brackets for repeated division, because division isn't as universally left-to-right as subtraction is). We can reason about it as follows: Whatever $a/(btimes c)$ is, we'll call it $x$. Then we have
$$
x = a/(btimes c)\
(btimes c)x = a\
(btimes c)x/b = a/b\
cx = a/b\
cx/c = (a/b)/c\
x = (a/b)/c
$$
And we see that $a/(btimes c)$ and $(a/b)/c$ are the same thing.
answered Dec 11 '18 at 13:13
ArthurArthur
115k7116198
answered Dec 11 '18 at 13:13
ArthurArthur
115k7116198
answered Dec 11 '18 at 13:13
ArthurArthur
115k7116198
answered Dec 11 '18 at 13:13
ArthurArthur
115k7116198
115k7116198
$begingroup$
Makes sense. Still, what is there to stop me writing the expression as $a/b/c$ and then evaluating $b/c$ first? It seems the parantheses are vital or is there some convention when doing divison I might be missing? Thanks!
$endgroup$
– quadfire
Dec 11 '18 at 13:30
$begingroup$
@quadfire You will find many people here (and elsewhere) who adhere to the convention that divisions (nd multiplications) are evaluated left-to-right (just like addition and subtraction: for $a-b-c$ people would think you crazy if you first evaluated $b-c$ and then subtracted the result from $a$). I personally do not follow that convention and instead use brackets to make it clear what I mean. So when I see $a/b/c$ I consider that ambiguous exactly for the reason that you mention.
$endgroup$
– Arthur
Dec 11 '18 at 13:33
add a comment |
$begingroup$
Makes sense. Still, what is there to stop me writing the expression as $a/b/c$ and then evaluating $b/c$ first? It seems the parantheses are vital or is there some convention when doing divison I might be missing? Thanks!
$endgroup$
– quadfire
Dec 11 '18 at 13:30
$begingroup$
@quadfire You will find many people here (and elsewhere) who adhere to the convention that divisions (nd multiplications) are evaluated left-to-right (just like addition and subtraction: for $a-b-c$ people would think you crazy if you first evaluated $b-c$ and then subtracted the result from $a$). I personally do not follow that convention and instead use brackets to make it clear what I mean. So when I see $a/b/c$ I consider that ambiguous exactly for the reason that you mention.
$endgroup$
– Arthur
Dec 11 '18 at 13:33
$begingroup$
Makes sense. Still, what is there to stop me writing the expression as $a/b/c$ and then evaluating $b/c$ first? It seems the parantheses are vital or is there some convention when doing divison I might be missing? Thanks!
$endgroup$
– quadfire
Dec 11 '18 at 13:30
$begingroup$
Makes sense. Still, what is there to stop me writing the expression as $a/b/c$ and then evaluating $b/c$ first? It seems the parantheses are vital or is there some convention when doing divison I might be missing? Thanks!
$endgroup$
– quadfire
Dec 11 '18 at 13:30
$begingroup$
@quadfire You will find many people here (and elsewhere) who adhere to the convention that divisions (nd multiplications) are evaluated left-to-right (just like addition and subtraction: for $a-b-c$ people would think you crazy if you first evaluated $b-c$ and then subtracted the result from $a$). I personally do not follow that convention and instead use brackets to make it clear what I mean. So when I see $a/b/c$ I consider that ambiguous exactly for the reason that you mention.
$endgroup$
– Arthur
Dec 11 '18 at 13:33
$begingroup$
@quadfire You will find many people here (and elsewhere) who adhere to the convention that divisions (nd multiplications) are evaluated left-to-right (just like addition and subtraction: for $a-b-c$ people would think you crazy if you first evaluated $b-c$ and then subtracted the result from $a$). I personally do not follow that convention and instead use brackets to make it clear what I mean. So when I see $a/b/c$ I consider that ambiguous exactly for the reason that you mention.
$endgroup$
– Arthur
Dec 11 '18 at 13:33
add a comment |
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1
$begingroup$
$a/( b times c) = a times dfrac {1}{( b times c)}$.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:12
$begingroup$
$a/(b times c) = (a/b)/c$
$endgroup$
– Matti P.
Dec 11 '18 at 13:13
$begingroup$
See Multiplying fractions.
$endgroup$
– Mauro ALLEGRANZA
Dec 11 '18 at 13:15