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Why is the reciprocal used in fraction division?
How to make sense of fractions?How do I rewrite -100+1/2 as the mixed number -99 1/2?Fraction exponents in divisionfraction division understandingFraction and Decimal: Reciprocal of x's non-integerWhy was I taught to convert “improper fractions” into mixed numbers?The division of a fraction - Whole or Part?division by fraction proofWhen dividing by a fraction, why can you not take the reciprocal of term involving addition/subtraction?How to tell when a fraction does not end?Basic division problem: dividing a fraction by a fraction
$begingroup$
I don't know if this is a basic question or whatever, but I can't seem to find an answer.
As far as I understand the reciprocal of a number the inverse of that number, that still doesn't clarify why it is needed.
For many years I've only ever done math like if I were a robot. I just did it and never understood what I was doing. So when I went and divided fractions I just used the reciprocal, because "that was the way to do it". I want to understand math at a deeper level, especially subjects like probability, statistics, calculus, and linear algebra. To do that I have to understand the fundamentals however.
Any response is appreciated.
fractions
$endgroup$
add a comment |
$begingroup$
I don't know if this is a basic question or whatever, but I can't seem to find an answer.
As far as I understand the reciprocal of a number the inverse of that number, that still doesn't clarify why it is needed.
For many years I've only ever done math like if I were a robot. I just did it and never understood what I was doing. So when I went and divided fractions I just used the reciprocal, because "that was the way to do it". I want to understand math at a deeper level, especially subjects like probability, statistics, calculus, and linear algebra. To do that I have to understand the fundamentals however.
Any response is appreciated.
fractions
$endgroup$
$begingroup$
This may be helpful. math.stackexchange.com/questions/1127483/…
$endgroup$
– Ethan Bolker
2 hours ago
$begingroup$
If you are asking this question, it probably means that you do not have enough experience with algebra
$endgroup$
– rash
2 hours ago
$begingroup$
Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
$endgroup$
– Henning Makholm
2 hours ago
add a comment |
$begingroup$
I don't know if this is a basic question or whatever, but I can't seem to find an answer.
As far as I understand the reciprocal of a number the inverse of that number, that still doesn't clarify why it is needed.
For many years I've only ever done math like if I were a robot. I just did it and never understood what I was doing. So when I went and divided fractions I just used the reciprocal, because "that was the way to do it". I want to understand math at a deeper level, especially subjects like probability, statistics, calculus, and linear algebra. To do that I have to understand the fundamentals however.
Any response is appreciated.
fractions
$endgroup$
I don't know if this is a basic question or whatever, but I can't seem to find an answer.
As far as I understand the reciprocal of a number the inverse of that number, that still doesn't clarify why it is needed.
For many years I've only ever done math like if I were a robot. I just did it and never understood what I was doing. So when I went and divided fractions I just used the reciprocal, because "that was the way to do it". I want to understand math at a deeper level, especially subjects like probability, statistics, calculus, and linear algebra. To do that I have to understand the fundamentals however.
Any response is appreciated.
fractions
fractions
asked 2 hours ago
ArgusArgus
20418
20418
$begingroup$
This may be helpful. math.stackexchange.com/questions/1127483/…
$endgroup$
– Ethan Bolker
2 hours ago
$begingroup$
If you are asking this question, it probably means that you do not have enough experience with algebra
$endgroup$
– rash
2 hours ago
$begingroup$
Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
$endgroup$
– Henning Makholm
2 hours ago
add a comment |
$begingroup$
This may be helpful. math.stackexchange.com/questions/1127483/…
$endgroup$
– Ethan Bolker
2 hours ago
$begingroup$
If you are asking this question, it probably means that you do not have enough experience with algebra
$endgroup$
– rash
2 hours ago
$begingroup$
Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
$endgroup$
– Henning Makholm
2 hours ago
$begingroup$
This may be helpful. math.stackexchange.com/questions/1127483/…
$endgroup$
– Ethan Bolker
2 hours ago
$begingroup$
This may be helpful. math.stackexchange.com/questions/1127483/…
$endgroup$
– Ethan Bolker
2 hours ago
$begingroup$
If you are asking this question, it probably means that you do not have enough experience with algebra
$endgroup$
– rash
2 hours ago
$begingroup$
If you are asking this question, it probably means that you do not have enough experience with algebra
$endgroup$
– rash
2 hours ago
$begingroup$
Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
$endgroup$
– Henning Makholm
2 hours ago
$begingroup$
Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
$endgroup$
– Henning Makholm
2 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I think you're asking why the rule for division of fractions,
$$fracpq div fracrs = fracpq cdot fracsr,$$
works.
And I'm assuming that you're already comfortable with how to multiply fractions.
We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $Adiv B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $Adiv B$ means the $X$ that solves the equation $$ Xcdot B = A $$
When our $A$ and $B$ are fraction, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $frac pq div frac rs$ we need to solve the equation
$$ X cdot frac rs = frac pq $$
And indeed setting $X=frac pqcdot frac sr = fracpsqr$ does this:
$$ fracpsqrcdotfrac rs = fracpscdot rqrcdot s = fracpcdot srqcdot sr = frac pq$$
like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).
This computation hopefully also gives some ides why it works, at least part way. In $fracpsqr$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.
Writing the solution $fracpsqr$ as $frac pqcdot frac sr$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.
$endgroup$
add a comment |
$begingroup$
Your question isn't completely clear but what I understood is that you don't get why $$fracfracabfraccd= fracab*fracdc$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^-1$$ with $f^-1$ the reciprocal of $f$, therefore $$fracfracabfraccd=fracab(fraccd)^-1$$ and since $$fraccd*fracdc=1$$ we have $$fracfracabfraccd= fracab*fracdc$$ our result
New contributor
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I think you're asking why the rule for division of fractions,
$$fracpq div fracrs = fracpq cdot fracsr,$$
works.
And I'm assuming that you're already comfortable with how to multiply fractions.
We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $Adiv B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $Adiv B$ means the $X$ that solves the equation $$ Xcdot B = A $$
When our $A$ and $B$ are fraction, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $frac pq div frac rs$ we need to solve the equation
$$ X cdot frac rs = frac pq $$
And indeed setting $X=frac pqcdot frac sr = fracpsqr$ does this:
$$ fracpsqrcdotfrac rs = fracpscdot rqrcdot s = fracpcdot srqcdot sr = frac pq$$
like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).
This computation hopefully also gives some ides why it works, at least part way. In $fracpsqr$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.
Writing the solution $fracpsqr$ as $frac pqcdot frac sr$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.
$endgroup$
add a comment |
$begingroup$
I think you're asking why the rule for division of fractions,
$$fracpq div fracrs = fracpq cdot fracsr,$$
works.
And I'm assuming that you're already comfortable with how to multiply fractions.
We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $Adiv B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $Adiv B$ means the $X$ that solves the equation $$ Xcdot B = A $$
When our $A$ and $B$ are fraction, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $frac pq div frac rs$ we need to solve the equation
$$ X cdot frac rs = frac pq $$
And indeed setting $X=frac pqcdot frac sr = fracpsqr$ does this:
$$ fracpsqrcdotfrac rs = fracpscdot rqrcdot s = fracpcdot srqcdot sr = frac pq$$
like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).
This computation hopefully also gives some ides why it works, at least part way. In $fracpsqr$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.
Writing the solution $fracpsqr$ as $frac pqcdot frac sr$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.
$endgroup$
add a comment |
$begingroup$
I think you're asking why the rule for division of fractions,
$$fracpq div fracrs = fracpq cdot fracsr,$$
works.
And I'm assuming that you're already comfortable with how to multiply fractions.
We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $Adiv B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $Adiv B$ means the $X$ that solves the equation $$ Xcdot B = A $$
When our $A$ and $B$ are fraction, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $frac pq div frac rs$ we need to solve the equation
$$ X cdot frac rs = frac pq $$
And indeed setting $X=frac pqcdot frac sr = fracpsqr$ does this:
$$ fracpsqrcdotfrac rs = fracpscdot rqrcdot s = fracpcdot srqcdot sr = frac pq$$
like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).
This computation hopefully also gives some ides why it works, at least part way. In $fracpsqr$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.
Writing the solution $fracpsqr$ as $frac pqcdot frac sr$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.
$endgroup$
I think you're asking why the rule for division of fractions,
$$fracpq div fracrs = fracpq cdot fracsr,$$
works.
And I'm assuming that you're already comfortable with how to multiply fractions.
We need to go back to what division is supposed to achieve in the first place. When we look into that, the answer is that $Adiv B$ means something that gives $A$ when we multiply it by $B$ -- or, written in symbols, $Adiv B$ means the $X$ that solves the equation $$ Xcdot B = A $$
When our $A$ and $B$ are fraction, the "reciprocal" division rule can be regarded as a trick that happens to produce an $X$ that works. It's easy enough to see that it does work: If we're dividing $frac pq div frac rs$ we need to solve the equation
$$ X cdot frac rs = frac pq $$
And indeed setting $X=frac pqcdot frac sr = fracpsqr$ does this:
$$ fracpsqrcdotfrac rs = fracpscdot rqrcdot s = fracpcdot srqcdot sr = frac pq$$
like we want. (I'm also assuming that you're comfortable with cancelling the common factor $sr$ in the middle fraction).
This computation hopefully also gives some ides why it works, at least part way. In $fracpsqr$ the $p$ and $q$ are what we want to end up with, and the $s$ and $r$ are there to "neutralize" the $r$ and $s$ we have but want to discard. By making sure that the product has exactly one $r$ and one $s$ on each side of the fraction bar they make sure we can cancel them away.
Writing the solution $fracpsqr$ as $frac pqcdot frac sr$ might be best understood as just an easy way to remember what goes where. But this memory trick itself then also serves as motivation for considering the reciprocal to be an interesting operation in its own right in higher algebra.
edited 2 hours ago
answered 2 hours ago
Henning MakholmHenning Makholm
246k17316561
246k17316561
add a comment |
add a comment |
$begingroup$
Your question isn't completely clear but what I understood is that you don't get why $$fracfracabfraccd= fracab*fracdc$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^-1$$ with $f^-1$ the reciprocal of $f$, therefore $$fracfracabfraccd=fracab(fraccd)^-1$$ and since $$fraccd*fracdc=1$$ we have $$fracfracabfraccd= fracab*fracdc$$ our result
New contributor
$endgroup$
add a comment |
$begingroup$
Your question isn't completely clear but what I understood is that you don't get why $$fracfracabfraccd= fracab*fracdc$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^-1$$ with $f^-1$ the reciprocal of $f$, therefore $$fracfracabfraccd=fracab(fraccd)^-1$$ and since $$fraccd*fracdc=1$$ we have $$fracfracabfraccd= fracab*fracdc$$ our result
New contributor
$endgroup$
add a comment |
$begingroup$
Your question isn't completely clear but what I understood is that you don't get why $$fracfracabfraccd= fracab*fracdc$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^-1$$ with $f^-1$ the reciprocal of $f$, therefore $$fracfracabfraccd=fracab(fraccd)^-1$$ and since $$fraccd*fracdc=1$$ we have $$fracfracabfraccd= fracab*fracdc$$ our result
New contributor
$endgroup$
Your question isn't completely clear but what I understood is that you don't get why $$fracfracabfraccd= fracab*fracdc$$ the answer it's located in the axioms of the real numbers, a number $b$ it's the reciprocal of a number $d$ if $$ d*b=1$$ now, let's see the definition of fraction $$e/f=e*f^-1$$ with $f^-1$ the reciprocal of $f$, therefore $$fracfracabfraccd=fracab(fraccd)^-1$$ and since $$fraccd*fracdc=1$$ we have $$fracfracabfraccd= fracab*fracdc$$ our result
New contributor
New contributor
answered 2 hours ago
Mario AldeanMario Aldean
135
135
New contributor
New contributor
add a comment |
add a comment |
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$begingroup$
This may be helpful. math.stackexchange.com/questions/1127483/…
$endgroup$
– Ethan Bolker
2 hours ago
$begingroup$
If you are asking this question, it probably means that you do not have enough experience with algebra
$endgroup$
– rash
2 hours ago
$begingroup$
Also see How to explain the flipping of division by a fraction? on Mathematics Educators, showcasing many attempts at an intuitive and elementary explanation.
$endgroup$
– Henning Makholm
2 hours ago