basic question on definition of function
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on wikipedia it says that a function is a relation or process that associates each x of X an element y of Y. I can understand how a function is a relation defined by some equation but can't really understand the interpretation as a process.
Is it necessary to understand this and if so could someone make this clearer for me, thanks
soft-question
asked Sep 2 at 10:38
James Anthony
776
add a comment |Â
up vote
2
down vote
favorite
on wikipedia it says that a function is a relation or process that associates each x of X an element y of Y. I can understand how a function is a relation defined by some equation but can't really understand the interpretation as a process.
Is it necessary to understand this and if so could someone make this clearer for me, thanks
soft-question
asked Sep 2 at 10:38
James Anthony
776
In France there is two related definitions, functions and applications: your definition corresponds to application because the domain of the function is ALL the set X while, for example the real function $f(x)=dfrac 1x$ is not defined in $x=0$. In other words, many times you are concerned with an expression for which you have to define the domain of it while for your function the domain is given.
– Piquito
Sep 2 at 10:56
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
on wikipedia it says that a function is a relation or process that associates each x of X an element y of Y. I can understand how a function is a relation defined by some equation but can't really understand the interpretation as a process.
Is it necessary to understand this and if so could someone make this clearer for me, thanks
soft-question
asked Sep 2 at 10:38
James Anthony
776
on wikipedia it says that a function is a relation or process that associates each x of X an element y of Y. I can understand how a function is a relation defined by some equation but can't really understand the interpretation as a process.
Is it necessary to understand this and if so could someone make this clearer for me, thanks
soft-question
soft-question
asked Sep 2 at 10:38
James Anthony
776
asked Sep 2 at 10:38
James Anthony
776
asked Sep 2 at 10:38
James Anthony
776
asked Sep 2 at 10:38
James Anthony
776
asked Sep 2 at 10:38
James Anthony
776
776
In France there is two related definitions, functions and applications: your definition corresponds to application because the domain of the function is ALL the set X while, for example the real function $f(x)=dfrac 1x$ is not defined in $x=0$. In other words, many times you are concerned with an expression for which you have to define the domain of it while for your function the domain is given.
– Piquito
Sep 2 at 10:56
add a comment |Â
In France there is two related definitions, functions and applications: your definition corresponds to application because the domain of the function is ALL the set X while, for example the real function $f(x)=dfrac 1x$ is not defined in $x=0$. In other words, many times you are concerned with an expression for which you have to define the domain of it while for your function the domain is given.
– Piquito
Sep 2 at 10:56
In France there is two related definitions, functions and applications: your definition corresponds to application because the domain of the function is ALL the set X while, for example the real function $f(x)=dfrac 1x$ is not defined in $x=0$. In other words, many times you are concerned with an expression for which you have to define the domain of it while for your function the domain is given.
– Piquito
Sep 2 at 10:56
In France there is two related definitions, functions and applications: your definition corresponds to application because the domain of the function is ALL the set X while, for example the real function $f(x)=dfrac 1x$ is not defined in $x=0$. In other words, many times you are concerned with an expression for which you have to define the domain of it while for your function the domain is given.
– Piquito
Sep 2 at 10:56
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
More precisely by definition given $2$ sets $X$ and $Y$ a function $f:Xto Y$ is a "law" which associates to any value $xin X$ one and only one value $yin Y$.
Note that we don't need that $f$ is defined by an explicit formula or expression the definiton indeed works in a more general context.
For example we can consider the function $f$ which associates to any person the mother.
Or in a more mathematical context, we can consider the function $f:mathbbNtomathbbN$ which associates to any natural number $n$ the corresponding $n^th$ prime number.
Refer also to the related
What exactly is a function?
Why is $sin : mathbbR to [-5,5] $ different from $sin : mathbbR to mathbbR$?
answered Sep 2 at 10:42
gimusi
72.2k73888
thanks, if a function is a law then would that not imply that say f(x)=x+1 is the function?
– James Anthony
Sep 2 at 10:50
@JamesAnthony We always need to specify the in which sets we are considering the law, so $f:mathbbRto mathbbR$ such that $f(x)=x+1$ is a function.
– gimusi
Sep 2 at 10:52
So is function not a product of the law
– James Anthony
Sep 2 at 10:53
1
@JamesAnthony I've added some link where the topic is discussed in great detail.
– gimusi
Sep 2 at 10:54
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
More precisely by definition given $2$ sets $X$ and $Y$ a function $f:Xto Y$ is a "law" which associates to any value $xin X$ one and only one value $yin Y$.
Note that we don't need that $f$ is defined by an explicit formula or expression the definiton indeed works in a more general context.
For example we can consider the function $f$ which associates to any person the mother.
Or in a more mathematical context, we can consider the function $f:mathbbNtomathbbN$ which associates to any natural number $n$ the corresponding $n^th$ prime number.
Refer also to the related
What exactly is a function?
Why is $sin : mathbbR to [-5,5] $ different from $sin : mathbbR to mathbbR$?
answered Sep 2 at 10:42
gimusi
72.2k73888
thanks, if a function is a law then would that not imply that say f(x)=x+1 is the function?
– James Anthony
Sep 2 at 10:50
@JamesAnthony We always need to specify the in which sets we are considering the law, so $f:mathbbRto mathbbR$ such that $f(x)=x+1$ is a function.
– gimusi
Sep 2 at 10:52
So is function not a product of the law
– James Anthony
Sep 2 at 10:53
1
@JamesAnthony I've added some link where the topic is discussed in great detail.
– gimusi
Sep 2 at 10:54
add a comment |Â
up vote
2
down vote
accepted
More precisely by definition given $2$ sets $X$ and $Y$ a function $f:Xto Y$ is a "law" which associates to any value $xin X$ one and only one value $yin Y$.
Note that we don't need that $f$ is defined by an explicit formula or expression the definiton indeed works in a more general context.
For example we can consider the function $f$ which associates to any person the mother.
Or in a more mathematical context, we can consider the function $f:mathbbNtomathbbN$ which associates to any natural number $n$ the corresponding $n^th$ prime number.
Refer also to the related
What exactly is a function?
Why is $sin : mathbbR to [-5,5] $ different from $sin : mathbbR to mathbbR$?
answered Sep 2 at 10:42
gimusi
72.2k73888
thanks, if a function is a law then would that not imply that say f(x)=x+1 is the function?
– James Anthony
Sep 2 at 10:50
@JamesAnthony We always need to specify the in which sets we are considering the law, so $f:mathbbRto mathbbR$ such that $f(x)=x+1$ is a function.
– gimusi
Sep 2 at 10:52
So is function not a product of the law
– James Anthony
Sep 2 at 10:53
1
@JamesAnthony I've added some link where the topic is discussed in great detail.
– gimusi
Sep 2 at 10:54
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
More precisely by definition given $2$ sets $X$ and $Y$ a function $f:Xto Y$ is a "law" which associates to any value $xin X$ one and only one value $yin Y$.
Note that we don't need that $f$ is defined by an explicit formula or expression the definiton indeed works in a more general context.
For example we can consider the function $f$ which associates to any person the mother.
Or in a more mathematical context, we can consider the function $f:mathbbNtomathbbN$ which associates to any natural number $n$ the corresponding $n^th$ prime number.
Refer also to the related
What exactly is a function?
Why is $sin : mathbbR to [-5,5] $ different from $sin : mathbbR to mathbbR$?
answered Sep 2 at 10:42
gimusi
72.2k73888
More precisely by definition given $2$ sets $X$ and $Y$ a function $f:Xto Y$ is a "law" which associates to any value $xin X$ one and only one value $yin Y$.
Note that we don't need that $f$ is defined by an explicit formula or expression the definiton indeed works in a more general context.
For example we can consider the function $f$ which associates to any person the mother.
Or in a more mathematical context, we can consider the function $f:mathbbNtomathbbN$ which associates to any natural number $n$ the corresponding $n^th$ prime number.
Refer also to the related
What exactly is a function?
Why is $sin : mathbbR to [-5,5] $ different from $sin : mathbbR to mathbbR$?
answered Sep 2 at 10:42
gimusi
72.2k73888
edited Sep 2 at 10:53
answered Sep 2 at 10:42
gimusi
72.2k73888
answered Sep 2 at 10:42
gimusi
72.2k73888
answered Sep 2 at 10:42
gimusi
72.2k73888
72.2k73888
thanks, if a function is a law then would that not imply that say f(x)=x+1 is the function?
– James Anthony
Sep 2 at 10:50
@JamesAnthony We always need to specify the in which sets we are considering the law, so $f:mathbbRto mathbbR$ such that $f(x)=x+1$ is a function.
– gimusi
Sep 2 at 10:52
So is function not a product of the law
– James Anthony
Sep 2 at 10:53
1
@JamesAnthony I've added some link where the topic is discussed in great detail.
– gimusi
Sep 2 at 10:54
add a comment |Â
thanks, if a function is a law then would that not imply that say f(x)=x+1 is the function?
– James Anthony
Sep 2 at 10:50
@JamesAnthony We always need to specify the in which sets we are considering the law, so $f:mathbbRto mathbbR$ such that $f(x)=x+1$ is a function.
– gimusi
Sep 2 at 10:52
So is function not a product of the law
– James Anthony
Sep 2 at 10:53
1
@JamesAnthony I've added some link where the topic is discussed in great detail.
– gimusi
Sep 2 at 10:54
thanks, if a function is a law then would that not imply that say f(x)=x+1 is the function?
– James Anthony
Sep 2 at 10:50
thanks, if a function is a law then would that not imply that say f(x)=x+1 is the function?
– James Anthony
Sep 2 at 10:50
@JamesAnthony We always need to specify the in which sets we are considering the law, so $f:mathbbRto mathbbR$ such that $f(x)=x+1$ is a function.
– gimusi
Sep 2 at 10:52
@JamesAnthony We always need to specify the in which sets we are considering the law, so $f:mathbbRto mathbbR$ such that $f(x)=x+1$ is a function.
– gimusi
Sep 2 at 10:52
So is function not a product of the law
– James Anthony
Sep 2 at 10:53
So is function not a product of the law
– James Anthony
Sep 2 at 10:53
1
1
@JamesAnthony I've added some link where the topic is discussed in great detail.
– gimusi
Sep 2 at 10:54
@JamesAnthony I've added some link where the topic is discussed in great detail.
– gimusi
Sep 2 at 10:54
add a comment |Â
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In France there is two related definitions, functions and applications: your definition corresponds to application because the domain of the function is ALL the set X while, for example the real function $f(x)=dfrac 1x$ is not defined in $x=0$. In other words, many times you are concerned with an expression for which you have to define the domain of it while for your function the domain is given.
– Piquito
Sep 2 at 10:56