Mixing
Why is it that $frac abtimesfrac1c=frac abc$?
Clash Royale CLAN TAG#URR8PPP
up vote
13
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favorite
I know it may sound stupid but I'm just genuinely wondering about it....
$$frac abtimesfrac1c=frac abc$$ where $b,cne0$.
How can we multiply numerators by numerators and denominators by denominators?
Is it just a rule? Or can it be proved?
arithmetic fractions
edited 2 days ago
Blue
43.6k868141
asked 2 days ago
JUNG WON CHO
793
 |Â
show 3 more comments
up vote
13
down vote
favorite
I know it may sound stupid but I'm just genuinely wondering about it....
$$frac abtimesfrac1c=frac abc$$ where $b,cne0$.
How can we multiply numerators by numerators and denominators by denominators?
Is it just a rule? Or can it be proved?
arithmetic fractions
edited 2 days ago
Blue
43.6k868141
asked 2 days ago
JUNG WON CHO
793
6
I don't know why this was downvoted or voted off topic, I think this is a great question. It's important to understand where the rules of arithmetic come from, and that they aren't just things we have to memorize and believe because our teachers said so.
– littleO
2 days ago
This question says what I said in my answer. I didn't see any comment from OP which says I want this. All of people say their thoughts without what the OP really wanted.
– user 108128
2 days ago
I did downvote it because it is a simple axiom. In fact people voted their thoughts and they like such these posts because they like speak about that not about what the OP says!
– user 108128
2 days ago
2
@littleO What you said is so funny . . . Oh my God, this is very good question because I want that and like to speak about.
– user 108128
2 days ago
1
@user108128 I feel you leave the wrong impression to the reader. You say it's simply an axiom and that's all as if to say axioms can't be questioned, that's how it is, get used to it etc. Yes, axioms are something we regard to be true, therefore they don't require proof, but we can certainly question whether some notions or operations are well-defined. I think that is the question to be asked in this topic even if the OP doesn't realise it. I mean no offense.
– Alvin Lepik
2 days ago
 |Â
show 3 more comments
up vote
13
down vote
favorite
up vote
13
down vote
favorite
I know it may sound stupid but I'm just genuinely wondering about it....
$$frac abtimesfrac1c=frac abc$$ where $b,cne0$.
How can we multiply numerators by numerators and denominators by denominators?
Is it just a rule? Or can it be proved?
arithmetic fractions
edited 2 days ago
Blue
43.6k868141
asked 2 days ago
JUNG WON CHO
793
I know it may sound stupid but I'm just genuinely wondering about it....
$$frac abtimesfrac1c=frac abc$$ where $b,cne0$.
How can we multiply numerators by numerators and denominators by denominators?
Is it just a rule? Or can it be proved?
arithmetic fractions
edited 2 days ago
Blue
43.6k868141
asked 2 days ago
JUNG WON CHO
793
edited 2 days ago
Blue
43.6k868141
edited 2 days ago
Blue
43.6k868141
edited 2 days ago
Blue
43.6k868141
43.6k868141
asked 2 days ago
JUNG WON CHO
793
asked 2 days ago
JUNG WON CHO
793
asked 2 days ago
JUNG WON CHO
793
793
6
I don't know why this was downvoted or voted off topic, I think this is a great question. It's important to understand where the rules of arithmetic come from, and that they aren't just things we have to memorize and believe because our teachers said so.
– littleO
2 days ago
This question says what I said in my answer. I didn't see any comment from OP which says I want this. All of people say their thoughts without what the OP really wanted.
– user 108128
2 days ago
I did downvote it because it is a simple axiom. In fact people voted their thoughts and they like such these posts because they like speak about that not about what the OP says!
– user 108128
2 days ago
2
@littleO What you said is so funny . . . Oh my God, this is very good question because I want that and like to speak about.
– user 108128
2 days ago
1
@user108128 I feel you leave the wrong impression to the reader. You say it's simply an axiom and that's all as if to say axioms can't be questioned, that's how it is, get used to it etc. Yes, axioms are something we regard to be true, therefore they don't require proof, but we can certainly question whether some notions or operations are well-defined. I think that is the question to be asked in this topic even if the OP doesn't realise it. I mean no offense.
– Alvin Lepik
2 days ago
 |Â
show 3 more comments
6
I don't know why this was downvoted or voted off topic, I think this is a great question. It's important to understand where the rules of arithmetic come from, and that they aren't just things we have to memorize and believe because our teachers said so.
– littleO
2 days ago
This question says what I said in my answer. I didn't see any comment from OP which says I want this. All of people say their thoughts without what the OP really wanted.
– user 108128
2 days ago
I did downvote it because it is a simple axiom. In fact people voted their thoughts and they like such these posts because they like speak about that not about what the OP says!
– user 108128
2 days ago
2
@littleO What you said is so funny . . . Oh my God, this is very good question because I want that and like to speak about.
– user 108128
2 days ago
1
@user108128 I feel you leave the wrong impression to the reader. You say it's simply an axiom and that's all as if to say axioms can't be questioned, that's how it is, get used to it etc. Yes, axioms are something we regard to be true, therefore they don't require proof, but we can certainly question whether some notions or operations are well-defined. I think that is the question to be asked in this topic even if the OP doesn't realise it. I mean no offense.
– Alvin Lepik
2 days ago
6
6
I don't know why this was downvoted or voted off topic, I think this is a great question. It's important to understand where the rules of arithmetic come from, and that they aren't just things we have to memorize and believe because our teachers said so.
– littleO
2 days ago
I don't know why this was downvoted or voted off topic, I think this is a great question. It's important to understand where the rules of arithmetic come from, and that they aren't just things we have to memorize and believe because our teachers said so.
– littleO
2 days ago
This question says what I said in my answer. I didn't see any comment from OP which says I want this. All of people say their thoughts without what the OP really wanted.
– user 108128
2 days ago
This question says what I said in my answer. I didn't see any comment from OP which says I want this. All of people say their thoughts without what the OP really wanted.
– user 108128
2 days ago
I did downvote it because it is a simple axiom. In fact people voted their thoughts and they like such these posts because they like speak about that not about what the OP says!
– user 108128
2 days ago
I did downvote it because it is a simple axiom. In fact people voted their thoughts and they like such these posts because they like speak about that not about what the OP says!
– user 108128
2 days ago
2
2
@littleO What you said is so funny . . . Oh my God, this is very good question because I want that and like to speak about.
– user 108128
2 days ago
@littleO What you said is so funny . . . Oh my God, this is very good question because I want that and like to speak about.
– user 108128
2 days ago
1
1
@user108128 I feel you leave the wrong impression to the reader. You say it's simply an axiom and that's all as if to say axioms can't be questioned, that's how it is, get used to it etc. Yes, axioms are something we regard to be true, therefore they don't require proof, but we can certainly question whether some notions or operations are well-defined. I think that is the question to be asked in this topic even if the OP doesn't realise it. I mean no offense.
– Alvin Lepik
2 days ago
@user108128 I feel you leave the wrong impression to the reader. You say it's simply an axiom and that's all as if to say axioms can't be questioned, that's how it is, get used to it etc. Yes, axioms are something we regard to be true, therefore they don't require proof, but we can certainly question whether some notions or operations are well-defined. I think that is the question to be asked in this topic even if the OP doesn't realise it. I mean no offense.
– Alvin Lepik
2 days ago
 |Â
show 3 more comments
5 Answers
5
active
oldest
votes
up vote
9
down vote
You can think of multiplication as meaning "of". So what is $2/5$ of $3/7$ (for example)?
Draw a picture of a cake (a rectangular cake) sliced into 7 equal vertical slices, with $3$ of those slices having red frosting. That's $3/7$ of the cake.
Take that 3/7 of the cake and slice it horizontally into 5 equal pieces, and pour sprinkles on 2 of those 5 pieces. (When you're doing the horizontal slicing, slice the entire cake horizontally while you're at it.)
The portion of the cake with sprinkles is 2/5 of 3/7. But if you draw the picture, you see that the cake has been chopped into 35 equal pieces (5 groups of 7), and 6 of those 35 pieces have sprinkles. So,
$$
frac25 text of frac37 = frac2 times 35 times 7.
$$
answered 2 days ago
littleO
25.8k54099
add a comment |Â
up vote
2
down vote
There are three steps to this process
- Define what a fraction actually is
- Define what multiplying two numbers actually does
- Prove that multiplying fractions is done by multiplying numerator with numerator and denominator by denominator
Steps 1 and 2 can be done many different ways, and for each combination, step 3 will be done differently.
answered 2 days ago
Arthur
97.9k792173
But defining an arbitrary fraction $adiv b$ is letting it be equal to $a(1div b) = (adiv 1)(1div b)$, but now we are back to the question again: why are we multiplying? I guess you might have to reverse steps $1$ and $2$.
– user477343
2 days ago
1
@user477343: That's no definition. I think Arthur had in mind something like in this answer, for example.
– Hans Lundmark
2 days ago
@HansLundmark Or like in the article "Teaching fractions according to the common core standards" a bit down on this site. But yes.
– Arthur
2 days ago
@HansLundmark well, I mean, that is how I define it, with some thorough explanation of course. But a key essential is definitely multiplication; i.e., $adiv b=acdot b^-1$. Nonetheless, the answer in the link serves a top-notch explanation :)
– user477343
2 days ago
add a comment |Â
up vote
1
down vote
Great question! The short answer to this: it works because we defined it like this. (I assume we are talking about multiplication of rational numbers)
We are worried, however, whether the operation is well-defined. It means that the result
$$fracab cdot fracxy = fracaxby $$
must not depend on the choice of fractions. It can't be that this equality holds for some fractions, but not for some other fractions. That would make the operation ill-defined.
The process of verification is quite an involved one, however, especially for multiplication.
On a quick search I did find this which covers all one needs.
To give a slightly different spin on this problem. Intuition paves the way for how we want to define certain operations. Other answers give an intuitive explanation why multiplying two fractions produces a certain fraction. We used these intuitions to define how multiplication of two fractions behaves. But to be absolutely sure we didn't make a mistake, we must also verify the operation is well-defined and that is beyond reach for intuition.
This idea of well-definedness is very important in mathematics not just as a failsafe for addition and multiplication of (rational) numbers to be bulletproof.
answered 2 days ago
Alvin Lepik
2,025718
add a comment |Â
up vote
1
down vote
As @Arthur points out, understanding why fractions multiply as they do depends on understanding what a fraction is. That's a subtle question.
There are ways to answer your particular question if you choose to think of fractions as what you get when you cut up pies, but I think the best way starts with
defining (thinking about) $1/x$ as the number $?$ that solves the equation
$$
? times x = 1 .
$$
Then you can use the ordinary rules of arithmetic to show that the left side of your equation is a solution to the equation
$$
? times bc = a
$$
and so must equal $a/(bc)$.
Related
How to make sense of fractions?
answered 2 days ago
Ethan Bolker
35.6k54199
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up vote
0
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This is just how we've (in most cases, at any rate) chosen to define multiplication of rational numbers $mathbf Q.$ And it can be proven to work (by this I mean that it is a well-defined binary operation on $mathbf Q$). Of course, this definition has some intuition behind it about how rational numbers should behave under multiplication. However, all this can be seen quite neatly by a development of the system $left(mathbf Q, times right)$ from the natural numbers $mathbf N:=0,1,2,3,ldots$ and the operation $times$ defined on them in the usual recursive manner which eventually boils down to the primitive successor function. There may be other ways to effect this development, though, but I think this accords most with intuition.
answered 2 days ago
Allawonder
1,254412
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
You can think of multiplication as meaning "of". So what is $2/5$ of $3/7$ (for example)?
Draw a picture of a cake (a rectangular cake) sliced into 7 equal vertical slices, with $3$ of those slices having red frosting. That's $3/7$ of the cake.
Take that 3/7 of the cake and slice it horizontally into 5 equal pieces, and pour sprinkles on 2 of those 5 pieces. (When you're doing the horizontal slicing, slice the entire cake horizontally while you're at it.)
The portion of the cake with sprinkles is 2/5 of 3/7. But if you draw the picture, you see that the cake has been chopped into 35 equal pieces (5 groups of 7), and 6 of those 35 pieces have sprinkles. So,
$$
frac25 text of frac37 = frac2 times 35 times 7.
$$
answered 2 days ago
littleO
25.8k54099
add a comment |Â
up vote
9
down vote
You can think of multiplication as meaning "of". So what is $2/5$ of $3/7$ (for example)?
Draw a picture of a cake (a rectangular cake) sliced into 7 equal vertical slices, with $3$ of those slices having red frosting. That's $3/7$ of the cake.
Take that 3/7 of the cake and slice it horizontally into 5 equal pieces, and pour sprinkles on 2 of those 5 pieces. (When you're doing the horizontal slicing, slice the entire cake horizontally while you're at it.)
The portion of the cake with sprinkles is 2/5 of 3/7. But if you draw the picture, you see that the cake has been chopped into 35 equal pieces (5 groups of 7), and 6 of those 35 pieces have sprinkles. So,
$$
frac25 text of frac37 = frac2 times 35 times 7.
$$
answered 2 days ago
littleO
25.8k54099
add a comment |Â
up vote
9
down vote
up vote
9
down vote
You can think of multiplication as meaning "of". So what is $2/5$ of $3/7$ (for example)?
Draw a picture of a cake (a rectangular cake) sliced into 7 equal vertical slices, with $3$ of those slices having red frosting. That's $3/7$ of the cake.
Take that 3/7 of the cake and slice it horizontally into 5 equal pieces, and pour sprinkles on 2 of those 5 pieces. (When you're doing the horizontal slicing, slice the entire cake horizontally while you're at it.)
The portion of the cake with sprinkles is 2/5 of 3/7. But if you draw the picture, you see that the cake has been chopped into 35 equal pieces (5 groups of 7), and 6 of those 35 pieces have sprinkles. So,
$$
frac25 text of frac37 = frac2 times 35 times 7.
$$
answered 2 days ago
littleO
25.8k54099
You can think of multiplication as meaning "of". So what is $2/5$ of $3/7$ (for example)?
Draw a picture of a cake (a rectangular cake) sliced into 7 equal vertical slices, with $3$ of those slices having red frosting. That's $3/7$ of the cake.
Take that 3/7 of the cake and slice it horizontally into 5 equal pieces, and pour sprinkles on 2 of those 5 pieces. (When you're doing the horizontal slicing, slice the entire cake horizontally while you're at it.)
The portion of the cake with sprinkles is 2/5 of 3/7. But if you draw the picture, you see that the cake has been chopped into 35 equal pieces (5 groups of 7), and 6 of those 35 pieces have sprinkles. So,
$$
frac25 text of frac37 = frac2 times 35 times 7.
$$
answered 2 days ago
littleO
25.8k54099
edited 2 days ago
answered 2 days ago
littleO
25.8k54099
answered 2 days ago
littleO
25.8k54099
answered 2 days ago
littleO
25.8k54099
25.8k54099
add a comment |Â
add a comment |Â
up vote
2
down vote
There are three steps to this process
- Define what a fraction actually is
- Define what multiplying two numbers actually does
- Prove that multiplying fractions is done by multiplying numerator with numerator and denominator by denominator
Steps 1 and 2 can be done many different ways, and for each combination, step 3 will be done differently.
answered 2 days ago
Arthur
97.9k792173
But defining an arbitrary fraction $adiv b$ is letting it be equal to $a(1div b) = (adiv 1)(1div b)$, but now we are back to the question again: why are we multiplying? I guess you might have to reverse steps $1$ and $2$.
– user477343
2 days ago
1
@user477343: That's no definition. I think Arthur had in mind something like in this answer, for example.
– Hans Lundmark
2 days ago
@HansLundmark Or like in the article "Teaching fractions according to the common core standards" a bit down on this site. But yes.
– Arthur
2 days ago
@HansLundmark well, I mean, that is how I define it, with some thorough explanation of course. But a key essential is definitely multiplication; i.e., $adiv b=acdot b^-1$. Nonetheless, the answer in the link serves a top-notch explanation :)
– user477343
2 days ago
add a comment |Â
up vote
2
down vote
There are three steps to this process
- Define what a fraction actually is
- Define what multiplying two numbers actually does
- Prove that multiplying fractions is done by multiplying numerator with numerator and denominator by denominator
Steps 1 and 2 can be done many different ways, and for each combination, step 3 will be done differently.
answered 2 days ago
Arthur
97.9k792173
But defining an arbitrary fraction $adiv b$ is letting it be equal to $a(1div b) = (adiv 1)(1div b)$, but now we are back to the question again: why are we multiplying? I guess you might have to reverse steps $1$ and $2$.
– user477343
2 days ago
1
@user477343: That's no definition. I think Arthur had in mind something like in this answer, for example.
– Hans Lundmark
2 days ago
@HansLundmark Or like in the article "Teaching fractions according to the common core standards" a bit down on this site. But yes.
– Arthur
2 days ago
@HansLundmark well, I mean, that is how I define it, with some thorough explanation of course. But a key essential is definitely multiplication; i.e., $adiv b=acdot b^-1$. Nonetheless, the answer in the link serves a top-notch explanation :)
– user477343
2 days ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
There are three steps to this process
- Define what a fraction actually is
- Define what multiplying two numbers actually does
- Prove that multiplying fractions is done by multiplying numerator with numerator and denominator by denominator
Steps 1 and 2 can be done many different ways, and for each combination, step 3 will be done differently.
answered 2 days ago
Arthur
97.9k792173
There are three steps to this process
- Define what a fraction actually is
- Define what multiplying two numbers actually does
- Prove that multiplying fractions is done by multiplying numerator with numerator and denominator by denominator
Steps 1 and 2 can be done many different ways, and for each combination, step 3 will be done differently.
answered 2 days ago
Arthur
97.9k792173
answered 2 days ago
Arthur
97.9k792173
answered 2 days ago
Arthur
97.9k792173
answered 2 days ago
Arthur
97.9k792173
97.9k792173
But defining an arbitrary fraction $adiv b$ is letting it be equal to $a(1div b) = (adiv 1)(1div b)$, but now we are back to the question again: why are we multiplying? I guess you might have to reverse steps $1$ and $2$.
– user477343
2 days ago
1
@user477343: That's no definition. I think Arthur had in mind something like in this answer, for example.
– Hans Lundmark
2 days ago
@HansLundmark Or like in the article "Teaching fractions according to the common core standards" a bit down on this site. But yes.
– Arthur
2 days ago
@HansLundmark well, I mean, that is how I define it, with some thorough explanation of course. But a key essential is definitely multiplication; i.e., $adiv b=acdot b^-1$. Nonetheless, the answer in the link serves a top-notch explanation :)
– user477343
2 days ago
add a comment |Â
But defining an arbitrary fraction $adiv b$ is letting it be equal to $a(1div b) = (adiv 1)(1div b)$, but now we are back to the question again: why are we multiplying? I guess you might have to reverse steps $1$ and $2$.
– user477343
2 days ago
1
@user477343: That's no definition. I think Arthur had in mind something like in this answer, for example.
– Hans Lundmark
2 days ago
@HansLundmark Or like in the article "Teaching fractions according to the common core standards" a bit down on this site. But yes.
– Arthur
2 days ago
@HansLundmark well, I mean, that is how I define it, with some thorough explanation of course. But a key essential is definitely multiplication; i.e., $adiv b=acdot b^-1$. Nonetheless, the answer in the link serves a top-notch explanation :)
– user477343
2 days ago
But defining an arbitrary fraction $adiv b$ is letting it be equal to $a(1div b) = (adiv 1)(1div b)$, but now we are back to the question again: why are we multiplying? I guess you might have to reverse steps $1$ and $2$.
– user477343
2 days ago
But defining an arbitrary fraction $adiv b$ is letting it be equal to $a(1div b) = (adiv 1)(1div b)$, but now we are back to the question again: why are we multiplying? I guess you might have to reverse steps $1$ and $2$.
– user477343
2 days ago
1
1
@user477343: That's no definition. I think Arthur had in mind something like in this answer, for example.
– Hans Lundmark
2 days ago
@user477343: That's no definition. I think Arthur had in mind something like in this answer, for example.
– Hans Lundmark
2 days ago
@HansLundmark Or like in the article "Teaching fractions according to the common core standards" a bit down on this site. But yes.
– Arthur
2 days ago
@HansLundmark Or like in the article "Teaching fractions according to the common core standards" a bit down on this site. But yes.
– Arthur
2 days ago
@HansLundmark well, I mean, that is how I define it, with some thorough explanation of course. But a key essential is definitely multiplication; i.e., $adiv b=acdot b^-1$. Nonetheless, the answer in the link serves a top-notch explanation :)
– user477343
2 days ago
@HansLundmark well, I mean, that is how I define it, with some thorough explanation of course. But a key essential is definitely multiplication; i.e., $adiv b=acdot b^-1$. Nonetheless, the answer in the link serves a top-notch explanation :)
– user477343
2 days ago
add a comment |Â
up vote
1
down vote
Great question! The short answer to this: it works because we defined it like this. (I assume we are talking about multiplication of rational numbers)
We are worried, however, whether the operation is well-defined. It means that the result
$$fracab cdot fracxy = fracaxby $$
must not depend on the choice of fractions. It can't be that this equality holds for some fractions, but not for some other fractions. That would make the operation ill-defined.
The process of verification is quite an involved one, however, especially for multiplication.
On a quick search I did find this which covers all one needs.
To give a slightly different spin on this problem. Intuition paves the way for how we want to define certain operations. Other answers give an intuitive explanation why multiplying two fractions produces a certain fraction. We used these intuitions to define how multiplication of two fractions behaves. But to be absolutely sure we didn't make a mistake, we must also verify the operation is well-defined and that is beyond reach for intuition.
This idea of well-definedness is very important in mathematics not just as a failsafe for addition and multiplication of (rational) numbers to be bulletproof.
answered 2 days ago
Alvin Lepik
2,025718
add a comment |Â
up vote
1
down vote
Great question! The short answer to this: it works because we defined it like this. (I assume we are talking about multiplication of rational numbers)
We are worried, however, whether the operation is well-defined. It means that the result
$$fracab cdot fracxy = fracaxby $$
must not depend on the choice of fractions. It can't be that this equality holds for some fractions, but not for some other fractions. That would make the operation ill-defined.
The process of verification is quite an involved one, however, especially for multiplication.
On a quick search I did find this which covers all one needs.
To give a slightly different spin on this problem. Intuition paves the way for how we want to define certain operations. Other answers give an intuitive explanation why multiplying two fractions produces a certain fraction. We used these intuitions to define how multiplication of two fractions behaves. But to be absolutely sure we didn't make a mistake, we must also verify the operation is well-defined and that is beyond reach for intuition.
This idea of well-definedness is very important in mathematics not just as a failsafe for addition and multiplication of (rational) numbers to be bulletproof.
answered 2 days ago
Alvin Lepik
2,025718
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Great question! The short answer to this: it works because we defined it like this. (I assume we are talking about multiplication of rational numbers)
We are worried, however, whether the operation is well-defined. It means that the result
$$fracab cdot fracxy = fracaxby $$
must not depend on the choice of fractions. It can't be that this equality holds for some fractions, but not for some other fractions. That would make the operation ill-defined.
The process of verification is quite an involved one, however, especially for multiplication.
On a quick search I did find this which covers all one needs.
To give a slightly different spin on this problem. Intuition paves the way for how we want to define certain operations. Other answers give an intuitive explanation why multiplying two fractions produces a certain fraction. We used these intuitions to define how multiplication of two fractions behaves. But to be absolutely sure we didn't make a mistake, we must also verify the operation is well-defined and that is beyond reach for intuition.
This idea of well-definedness is very important in mathematics not just as a failsafe for addition and multiplication of (rational) numbers to be bulletproof.
answered 2 days ago
Alvin Lepik
2,025718
Great question! The short answer to this: it works because we defined it like this. (I assume we are talking about multiplication of rational numbers)
We are worried, however, whether the operation is well-defined. It means that the result
$$fracab cdot fracxy = fracaxby $$
must not depend on the choice of fractions. It can't be that this equality holds for some fractions, but not for some other fractions. That would make the operation ill-defined.
The process of verification is quite an involved one, however, especially for multiplication.
On a quick search I did find this which covers all one needs.
To give a slightly different spin on this problem. Intuition paves the way for how we want to define certain operations. Other answers give an intuitive explanation why multiplying two fractions produces a certain fraction. We used these intuitions to define how multiplication of two fractions behaves. But to be absolutely sure we didn't make a mistake, we must also verify the operation is well-defined and that is beyond reach for intuition.
This idea of well-definedness is very important in mathematics not just as a failsafe for addition and multiplication of (rational) numbers to be bulletproof.
answered 2 days ago
Alvin Lepik
2,025718
edited 2 days ago
answered 2 days ago
Alvin Lepik
2,025718
answered 2 days ago
Alvin Lepik
2,025718
answered 2 days ago
Alvin Lepik
2,025718
2,025718
add a comment |Â
add a comment |Â
up vote
1
down vote
As @Arthur points out, understanding why fractions multiply as they do depends on understanding what a fraction is. That's a subtle question.
There are ways to answer your particular question if you choose to think of fractions as what you get when you cut up pies, but I think the best way starts with
defining (thinking about) $1/x$ as the number $?$ that solves the equation
$$
? times x = 1 .
$$
Then you can use the ordinary rules of arithmetic to show that the left side of your equation is a solution to the equation
$$
? times bc = a
$$
and so must equal $a/(bc)$.
Related
How to make sense of fractions?
answered 2 days ago
Ethan Bolker
35.6k54199
add a comment |Â
up vote
1
down vote
As @Arthur points out, understanding why fractions multiply as they do depends on understanding what a fraction is. That's a subtle question.
There are ways to answer your particular question if you choose to think of fractions as what you get when you cut up pies, but I think the best way starts with
defining (thinking about) $1/x$ as the number $?$ that solves the equation
$$
? times x = 1 .
$$
Then you can use the ordinary rules of arithmetic to show that the left side of your equation is a solution to the equation
$$
? times bc = a
$$
and so must equal $a/(bc)$.
Related
How to make sense of fractions?
answered 2 days ago
Ethan Bolker
35.6k54199
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As @Arthur points out, understanding why fractions multiply as they do depends on understanding what a fraction is. That's a subtle question.
There are ways to answer your particular question if you choose to think of fractions as what you get when you cut up pies, but I think the best way starts with
defining (thinking about) $1/x$ as the number $?$ that solves the equation
$$
? times x = 1 .
$$
Then you can use the ordinary rules of arithmetic to show that the left side of your equation is a solution to the equation
$$
? times bc = a
$$
and so must equal $a/(bc)$.
Related
How to make sense of fractions?
answered 2 days ago
Ethan Bolker
35.6k54199
As @Arthur points out, understanding why fractions multiply as they do depends on understanding what a fraction is. That's a subtle question.
There are ways to answer your particular question if you choose to think of fractions as what you get when you cut up pies, but I think the best way starts with
defining (thinking about) $1/x$ as the number $?$ that solves the equation
$$
? times x = 1 .
$$
Then you can use the ordinary rules of arithmetic to show that the left side of your equation is a solution to the equation
$$
? times bc = a
$$
and so must equal $a/(bc)$.
Related
How to make sense of fractions?
answered 2 days ago
Ethan Bolker
35.6k54199
answered 2 days ago
Ethan Bolker
35.6k54199
answered 2 days ago
Ethan Bolker
35.6k54199
answered 2 days ago
Ethan Bolker
35.6k54199
35.6k54199
add a comment |Â
add a comment |Â
up vote
0
down vote
This is just how we've (in most cases, at any rate) chosen to define multiplication of rational numbers $mathbf Q.$ And it can be proven to work (by this I mean that it is a well-defined binary operation on $mathbf Q$). Of course, this definition has some intuition behind it about how rational numbers should behave under multiplication. However, all this can be seen quite neatly by a development of the system $left(mathbf Q, times right)$ from the natural numbers $mathbf N:=0,1,2,3,ldots$ and the operation $times$ defined on them in the usual recursive manner which eventually boils down to the primitive successor function. There may be other ways to effect this development, though, but I think this accords most with intuition.
answered 2 days ago
Allawonder
1,254412
add a comment |Â
up vote
0
down vote
This is just how we've (in most cases, at any rate) chosen to define multiplication of rational numbers $mathbf Q.$ And it can be proven to work (by this I mean that it is a well-defined binary operation on $mathbf Q$). Of course, this definition has some intuition behind it about how rational numbers should behave under multiplication. However, all this can be seen quite neatly by a development of the system $left(mathbf Q, times right)$ from the natural numbers $mathbf N:=0,1,2,3,ldots$ and the operation $times$ defined on them in the usual recursive manner which eventually boils down to the primitive successor function. There may be other ways to effect this development, though, but I think this accords most with intuition.
answered 2 days ago
Allawonder
1,254412
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This is just how we've (in most cases, at any rate) chosen to define multiplication of rational numbers $mathbf Q.$ And it can be proven to work (by this I mean that it is a well-defined binary operation on $mathbf Q$). Of course, this definition has some intuition behind it about how rational numbers should behave under multiplication. However, all this can be seen quite neatly by a development of the system $left(mathbf Q, times right)$ from the natural numbers $mathbf N:=0,1,2,3,ldots$ and the operation $times$ defined on them in the usual recursive manner which eventually boils down to the primitive successor function. There may be other ways to effect this development, though, but I think this accords most with intuition.
answered 2 days ago
Allawonder
1,254412
This is just how we've (in most cases, at any rate) chosen to define multiplication of rational numbers $mathbf Q.$ And it can be proven to work (by this I mean that it is a well-defined binary operation on $mathbf Q$). Of course, this definition has some intuition behind it about how rational numbers should behave under multiplication. However, all this can be seen quite neatly by a development of the system $left(mathbf Q, times right)$ from the natural numbers $mathbf N:=0,1,2,3,ldots$ and the operation $times$ defined on them in the usual recursive manner which eventually boils down to the primitive successor function. There may be other ways to effect this development, though, but I think this accords most with intuition.
answered 2 days ago
Allawonder
1,254412
answered 2 days ago
Allawonder
1,254412
answered 2 days ago
Allawonder
1,254412
answered 2 days ago
Allawonder
1,254412
1,254412
add a comment |Â
add a comment |Â
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6
I don't know why this was downvoted or voted off topic, I think this is a great question. It's important to understand where the rules of arithmetic come from, and that they aren't just things we have to memorize and believe because our teachers said so.
– littleO
2 days ago
This question says what I said in my answer. I didn't see any comment from OP which says I want this. All of people say their thoughts without what the OP really wanted.
– user 108128
2 days ago
I did downvote it because it is a simple axiom. In fact people voted their thoughts and they like such these posts because they like speak about that not about what the OP says!
– user 108128
2 days ago
2
@littleO What you said is so funny . . . Oh my God, this is very good question because I want that and like to speak about.
– user 108128
2 days ago
1
@user108128 I feel you leave the wrong impression to the reader. You say it's simply an axiom and that's all as if to say axioms can't be questioned, that's how it is, get used to it etc. Yes, axioms are something we regard to be true, therefore they don't require proof, but we can certainly question whether some notions or operations are well-defined. I think that is the question to be asked in this topic even if the OP doesn't realise it. I mean no offense.
– Alvin Lepik
2 days ago