Can anyone help me understand the notation of binary operator in the definition
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The definition goes,
Let $S$ be a non-empty set. Then a function $f : S times S to S$ is
a binary operator.
If I understand correctly, then it means $(a,b) in S times S Rightarrow f(a,b)=atext*b in S$ then $text*$ is a binary operator. Alright fine. But that doesn't explain that why is $f$ a binary operator?
Let me explain my doubt more clearly, I'm having trouble with notations here,
If $f(a,b) = a text* b$ where $text*$ is a binary operator. And if I say that $f$ is the same binary operator as well, then $f= text*$.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is not the image of $a$ under $f$. The correct expression (according to me) should be $(a,b)f(atext*b)$
So, $f$ can't be $text*$. I don't understand how can a relation (function in this case) be equal to a binary operator
functions
add a comment |Â
up vote
2
down vote
favorite
The definition goes,
Let $S$ be a non-empty set. Then a function $f : S times S to S$ is
a binary operator.
If I understand correctly, then it means $(a,b) in S times S Rightarrow f(a,b)=atext*b in S$ then $text*$ is a binary operator. Alright fine. But that doesn't explain that why is $f$ a binary operator?
Let me explain my doubt more clearly, I'm having trouble with notations here,
If $f(a,b) = a text* b$ where $text*$ is a binary operator. And if I say that $f$ is the same binary operator as well, then $f= text*$.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is not the image of $a$ under $f$. The correct expression (according to me) should be $(a,b)f(atext*b)$
So, $f$ can't be $text*$. I don't understand how can a relation (function in this case) be equal to a binary operator
functions
I hope I'm making myself clear.
â William
Aug 23 at 10:37
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The definition goes,
Let $S$ be a non-empty set. Then a function $f : S times S to S$ is
a binary operator.
If I understand correctly, then it means $(a,b) in S times S Rightarrow f(a,b)=atext*b in S$ then $text*$ is a binary operator. Alright fine. But that doesn't explain that why is $f$ a binary operator?
Let me explain my doubt more clearly, I'm having trouble with notations here,
If $f(a,b) = a text* b$ where $text*$ is a binary operator. And if I say that $f$ is the same binary operator as well, then $f= text*$.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is not the image of $a$ under $f$. The correct expression (according to me) should be $(a,b)f(atext*b)$
So, $f$ can't be $text*$. I don't understand how can a relation (function in this case) be equal to a binary operator
functions
The definition goes,
Let $S$ be a non-empty set. Then a function $f : S times S to S$ is
a binary operator.
If I understand correctly, then it means $(a,b) in S times S Rightarrow f(a,b)=atext*b in S$ then $text*$ is a binary operator. Alright fine. But that doesn't explain that why is $f$ a binary operator?
Let me explain my doubt more clearly, I'm having trouble with notations here,
If $f(a,b) = a text* b$ where $text*$ is a binary operator. And if I say that $f$ is the same binary operator as well, then $f= text*$.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is not the image of $a$ under $f$. The correct expression (according to me) should be $(a,b)f(atext*b)$
So, $f$ can't be $text*$. I don't understand how can a relation (function in this case) be equal to a binary operator
functions
asked Aug 23 at 10:29
William
903314
903314
I hope I'm making myself clear.
â William
Aug 23 at 10:37
add a comment |Â
I hope I'm making myself clear.
â William
Aug 23 at 10:37
I hope I'm making myself clear.
â William
Aug 23 at 10:37
I hope I'm making myself clear.
â William
Aug 23 at 10:37
add a comment |Â
4 Answers
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up vote
2
down vote
$*$ and $f$ are just different notations for the same underlying function. So
$a*b=c$
$f(a,b)=c$
$(a,b) xrightarrowf c$
all have the same meaning. A binary operator simply maps each ordered pair of elements from $S$ (i.e. each member of $S times S$) to an element of $S$.
add a comment |Â
up vote
1
down vote
It is called binary, because $f$ has two arguments. It acts on $2$-tuples.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is
not the image of $a$ under $f$.
$a f b$ is called infix notation. ($f a b$ is called prefix notation and $a b f$ postfix notation). All mean $f(a, b)$.
As usual, the interpretation of how you parse a given term depends on your context.
add a comment |Â
up vote
1
down vote
Perhaps it is better to use the viewpoint that some definitions can be relaxed, in some sense, and that notation changes depending on how we want to use an object.
Say that $ast$ is a binary operator. Then we can take two elements, say $a, b in S$, and combine them using $ast$ to get $aast bin S$. So a binary operator is something that takes two elements and combines them to give one.
Now define $fcolon Stimes S to S,enspace f(x, y) = xast y$. Then $f$ takes two elements and combines them to give one, and hence $f$ is a binary operator by using the binary operation $ast$, but we will use the notation $f(x, y)$ instead of $xast y$ to emphasise that we're using some function $f$, rather than just the binary operation $ast$.
This is like being able to write $f(a) = b$ and $a stackrelfmapsto b$ interchangeably. Sometimes it's better to write the former, and sometimes it's better to write the latter, but they both mean the same thing.
add a comment |Â
up vote
1
down vote
When you write $afb$ then you seem to confuse the infix notation of a relation (of which a function is a special kind) and the infix notation of a binary operator (like $+$).
What the definition tries to say is that a binary operator on a set $S$ is a function $S times S to S.$ It doesn't however say anything about notation. But as you know, we seldom write $+(3,5) = 8,$ but instead $3+5 = 8.$ That does not mean that $+$ relate $3$ and $5$ like $3 < 5$ does.
If you want to use the infix notation of the function $+$ as a relation, then the correct way to write it is $(3,5)+8.$ But that's a really confusing way to write the identity that is normally written $3+5=8.$
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
$*$ and $f$ are just different notations for the same underlying function. So
$a*b=c$
$f(a,b)=c$
$(a,b) xrightarrowf c$
all have the same meaning. A binary operator simply maps each ordered pair of elements from $S$ (i.e. each member of $S times S$) to an element of $S$.
add a comment |Â
up vote
2
down vote
$*$ and $f$ are just different notations for the same underlying function. So
$a*b=c$
$f(a,b)=c$
$(a,b) xrightarrowf c$
all have the same meaning. A binary operator simply maps each ordered pair of elements from $S$ (i.e. each member of $S times S$) to an element of $S$.
add a comment |Â
up vote
2
down vote
up vote
2
down vote
$*$ and $f$ are just different notations for the same underlying function. So
$a*b=c$
$f(a,b)=c$
$(a,b) xrightarrowf c$
all have the same meaning. A binary operator simply maps each ordered pair of elements from $S$ (i.e. each member of $S times S$) to an element of $S$.
$*$ and $f$ are just different notations for the same underlying function. So
$a*b=c$
$f(a,b)=c$
$(a,b) xrightarrowf c$
all have the same meaning. A binary operator simply maps each ordered pair of elements from $S$ (i.e. each member of $S times S$) to an element of $S$.
answered Aug 23 at 10:45
gandalf61
5,966522
5,966522
add a comment |Â
add a comment |Â
up vote
1
down vote
It is called binary, because $f$ has two arguments. It acts on $2$-tuples.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is
not the image of $a$ under $f$.
$a f b$ is called infix notation. ($f a b$ is called prefix notation and $a b f$ postfix notation). All mean $f(a, b)$.
As usual, the interpretation of how you parse a given term depends on your context.
add a comment |Â
up vote
1
down vote
It is called binary, because $f$ has two arguments. It acts on $2$-tuples.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is
not the image of $a$ under $f$.
$a f b$ is called infix notation. ($f a b$ is called prefix notation and $a b f$ postfix notation). All mean $f(a, b)$.
As usual, the interpretation of how you parse a given term depends on your context.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
It is called binary, because $f$ has two arguments. It acts on $2$-tuples.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is
not the image of $a$ under $f$.
$a f b$ is called infix notation. ($f a b$ is called prefix notation and $a b f$ postfix notation). All mean $f(a, b)$.
As usual, the interpretation of how you parse a given term depends on your context.
It is called binary, because $f$ has two arguments. It acts on $2$-tuples.
So I can write $f(a,b)= afb$ which seems to be absurd because $b$ is
not the image of $a$ under $f$.
$a f b$ is called infix notation. ($f a b$ is called prefix notation and $a b f$ postfix notation). All mean $f(a, b)$.
As usual, the interpretation of how you parse a given term depends on your context.
edited Aug 23 at 10:40
answered Aug 23 at 10:35
mvw
30.8k22252
30.8k22252
add a comment |Â
add a comment |Â
up vote
1
down vote
Perhaps it is better to use the viewpoint that some definitions can be relaxed, in some sense, and that notation changes depending on how we want to use an object.
Say that $ast$ is a binary operator. Then we can take two elements, say $a, b in S$, and combine them using $ast$ to get $aast bin S$. So a binary operator is something that takes two elements and combines them to give one.
Now define $fcolon Stimes S to S,enspace f(x, y) = xast y$. Then $f$ takes two elements and combines them to give one, and hence $f$ is a binary operator by using the binary operation $ast$, but we will use the notation $f(x, y)$ instead of $xast y$ to emphasise that we're using some function $f$, rather than just the binary operation $ast$.
This is like being able to write $f(a) = b$ and $a stackrelfmapsto b$ interchangeably. Sometimes it's better to write the former, and sometimes it's better to write the latter, but they both mean the same thing.
add a comment |Â
up vote
1
down vote
Perhaps it is better to use the viewpoint that some definitions can be relaxed, in some sense, and that notation changes depending on how we want to use an object.
Say that $ast$ is a binary operator. Then we can take two elements, say $a, b in S$, and combine them using $ast$ to get $aast bin S$. So a binary operator is something that takes two elements and combines them to give one.
Now define $fcolon Stimes S to S,enspace f(x, y) = xast y$. Then $f$ takes two elements and combines them to give one, and hence $f$ is a binary operator by using the binary operation $ast$, but we will use the notation $f(x, y)$ instead of $xast y$ to emphasise that we're using some function $f$, rather than just the binary operation $ast$.
This is like being able to write $f(a) = b$ and $a stackrelfmapsto b$ interchangeably. Sometimes it's better to write the former, and sometimes it's better to write the latter, but they both mean the same thing.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Perhaps it is better to use the viewpoint that some definitions can be relaxed, in some sense, and that notation changes depending on how we want to use an object.
Say that $ast$ is a binary operator. Then we can take two elements, say $a, b in S$, and combine them using $ast$ to get $aast bin S$. So a binary operator is something that takes two elements and combines them to give one.
Now define $fcolon Stimes S to S,enspace f(x, y) = xast y$. Then $f$ takes two elements and combines them to give one, and hence $f$ is a binary operator by using the binary operation $ast$, but we will use the notation $f(x, y)$ instead of $xast y$ to emphasise that we're using some function $f$, rather than just the binary operation $ast$.
This is like being able to write $f(a) = b$ and $a stackrelfmapsto b$ interchangeably. Sometimes it's better to write the former, and sometimes it's better to write the latter, but they both mean the same thing.
Perhaps it is better to use the viewpoint that some definitions can be relaxed, in some sense, and that notation changes depending on how we want to use an object.
Say that $ast$ is a binary operator. Then we can take two elements, say $a, b in S$, and combine them using $ast$ to get $aast bin S$. So a binary operator is something that takes two elements and combines them to give one.
Now define $fcolon Stimes S to S,enspace f(x, y) = xast y$. Then $f$ takes two elements and combines them to give one, and hence $f$ is a binary operator by using the binary operation $ast$, but we will use the notation $f(x, y)$ instead of $xast y$ to emphasise that we're using some function $f$, rather than just the binary operation $ast$.
This is like being able to write $f(a) = b$ and $a stackrelfmapsto b$ interchangeably. Sometimes it's better to write the former, and sometimes it's better to write the latter, but they both mean the same thing.
answered Aug 23 at 10:43
Bill Wallis
2,2361826
2,2361826
add a comment |Â
add a comment |Â
up vote
1
down vote
When you write $afb$ then you seem to confuse the infix notation of a relation (of which a function is a special kind) and the infix notation of a binary operator (like $+$).
What the definition tries to say is that a binary operator on a set $S$ is a function $S times S to S.$ It doesn't however say anything about notation. But as you know, we seldom write $+(3,5) = 8,$ but instead $3+5 = 8.$ That does not mean that $+$ relate $3$ and $5$ like $3 < 5$ does.
If you want to use the infix notation of the function $+$ as a relation, then the correct way to write it is $(3,5)+8.$ But that's a really confusing way to write the identity that is normally written $3+5=8.$
add a comment |Â
up vote
1
down vote
When you write $afb$ then you seem to confuse the infix notation of a relation (of which a function is a special kind) and the infix notation of a binary operator (like $+$).
What the definition tries to say is that a binary operator on a set $S$ is a function $S times S to S.$ It doesn't however say anything about notation. But as you know, we seldom write $+(3,5) = 8,$ but instead $3+5 = 8.$ That does not mean that $+$ relate $3$ and $5$ like $3 < 5$ does.
If you want to use the infix notation of the function $+$ as a relation, then the correct way to write it is $(3,5)+8.$ But that's a really confusing way to write the identity that is normally written $3+5=8.$
add a comment |Â
up vote
1
down vote
up vote
1
down vote
When you write $afb$ then you seem to confuse the infix notation of a relation (of which a function is a special kind) and the infix notation of a binary operator (like $+$).
What the definition tries to say is that a binary operator on a set $S$ is a function $S times S to S.$ It doesn't however say anything about notation. But as you know, we seldom write $+(3,5) = 8,$ but instead $3+5 = 8.$ That does not mean that $+$ relate $3$ and $5$ like $3 < 5$ does.
If you want to use the infix notation of the function $+$ as a relation, then the correct way to write it is $(3,5)+8.$ But that's a really confusing way to write the identity that is normally written $3+5=8.$
When you write $afb$ then you seem to confuse the infix notation of a relation (of which a function is a special kind) and the infix notation of a binary operator (like $+$).
What the definition tries to say is that a binary operator on a set $S$ is a function $S times S to S.$ It doesn't however say anything about notation. But as you know, we seldom write $+(3,5) = 8,$ but instead $3+5 = 8.$ That does not mean that $+$ relate $3$ and $5$ like $3 < 5$ does.
If you want to use the infix notation of the function $+$ as a relation, then the correct way to write it is $(3,5)+8.$ But that's a really confusing way to write the identity that is normally written $3+5=8.$
answered Aug 23 at 12:31
md2perpe
6,40811023
6,40811023
add a comment |Â
add a comment |Â
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I hope I'm making myself clear.
â William
Aug 23 at 10:37