function notation question
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So if a function is a process that associates each $xin X$, known as the domain of the function, a unique element $yin Y$ known as the the co-domain of the function.
So to outline a few things I think are correct (Please let me know if I have a wrong understanding):
A function is usually denoted by a single letter like $f$, $g$ or $h$.
$f(x)$ is the value of the function $f$ at $x$. You can say $f$ is a function of the variable $x$.
It seems as though, from the comments, that $f$ can be defined by saying what it does to a general element of its domain by stating an identity and then a domain for which it is true. for example:
- $f$ is defined by $f(x)=x+1$ for $x>0$
or, $g(x)= x + 1$ for all $x>2$ would define $g$
I think what I have done is write $2$ identities which means that both sides of the relation define the same function of $x$ on the domain stated.
Question 1: So $f$ is the function, $f(x)$ is the value of the function at $x$ and $x+1$ is the value of $f$ at $x$. Is it right that $f(x)$ tells you that $f$ is a function of $x $ and $x+1$ is a function of $x$. I find this hard to understand as I have just said that $x + 1$ is the value of $f$ at $x$.
elementary-set-theory notation
edited Aug 27 at 15:58
Arnaud D.
14.8k52142
asked Aug 26 at 19:52
James Anthony
776
 |Â
show 3 more comments
up vote
0
down vote
favorite
So if a function is a process that associates each $xin X$, known as the domain of the function, a unique element $yin Y$ known as the the co-domain of the function.
So to outline a few things I think are correct (Please let me know if I have a wrong understanding):
A function is usually denoted by a single letter like $f$, $g$ or $h$.
$f(x)$ is the value of the function $f$ at $x$. You can say $f$ is a function of the variable $x$.
It seems as though, from the comments, that $f$ can be defined by saying what it does to a general element of its domain by stating an identity and then a domain for which it is true. for example:
- $f$ is defined by $f(x)=x+1$ for $x>0$
or, $g(x)= x + 1$ for all $x>2$ would define $g$
I think what I have done is write $2$ identities which means that both sides of the relation define the same function of $x$ on the domain stated.
Question 1: So $f$ is the function, $f(x)$ is the value of the function at $x$ and $x+1$ is the value of $f$ at $x$. Is it right that $f(x)$ tells you that $f$ is a function of $x $ and $x+1$ is a function of $x$. I find this hard to understand as I have just said that $x + 1$ is the value of $f$ at $x$.
elementary-set-theory notation
edited Aug 27 at 15:58
Arnaud D.
14.8k52142
asked Aug 26 at 19:52
James Anthony
776
$f(x)= y= x+1.$ So $f(x) = y$ is in $Y$, which is in the codomain. $x$, as you said, comes from the domain of $X$. So given $x=2 in mathbb Z$, $f(x)= y= 2+1=3 in Y$. Indeed, $f(x) = x+1$ is a function expressed in terms of $x$. The only thing not specified is the set $X$. That is, $f: Xto Y$ describes a function such that for every $x in X$, there exists a unique value $f(x)$ given by $xmapsto x+1$. The codomain, or rather, the image of $f(x)$ will depend in part on the domain of $X$.
– amWhy
Aug 26 at 20:03
Your question $2$: $f$ has been implicitly defined by merely $f: Xto f[X]$, where $f[X]$ is a set, and is the image of $f(x)$, and consists of all elements $y$ such that such that $y=x+1$.
– amWhy
Aug 26 at 20:08
In both cases, $f(x)$ is a function of $x$.
– amWhy
Aug 26 at 20:09
2
Why the downvotes? This is clearly a question from someone who is trying to understand functions.
– John Douma
Aug 26 at 20:16
2
Not from me, @JohnDouma. I take the question as trying to understand the definition of a function, and whether or not a "function" without a codomain designated explicitly, is defined. I think it's a valid question, and it is clear that James has put some thought into expressing his question as best he can.
– amWhy
Aug 26 at 20:16
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So if a function is a process that associates each $xin X$, known as the domain of the function, a unique element $yin Y$ known as the the co-domain of the function.
So to outline a few things I think are correct (Please let me know if I have a wrong understanding):
A function is usually denoted by a single letter like $f$, $g$ or $h$.
$f(x)$ is the value of the function $f$ at $x$. You can say $f$ is a function of the variable $x$.
It seems as though, from the comments, that $f$ can be defined by saying what it does to a general element of its domain by stating an identity and then a domain for which it is true. for example:
- $f$ is defined by $f(x)=x+1$ for $x>0$
or, $g(x)= x + 1$ for all $x>2$ would define $g$
I think what I have done is write $2$ identities which means that both sides of the relation define the same function of $x$ on the domain stated.
Question 1: So $f$ is the function, $f(x)$ is the value of the function at $x$ and $x+1$ is the value of $f$ at $x$. Is it right that $f(x)$ tells you that $f$ is a function of $x $ and $x+1$ is a function of $x$. I find this hard to understand as I have just said that $x + 1$ is the value of $f$ at $x$.
elementary-set-theory notation
edited Aug 27 at 15:58
Arnaud D.
14.8k52142
asked Aug 26 at 19:52
James Anthony
776
So if a function is a process that associates each $xin X$, known as the domain of the function, a unique element $yin Y$ known as the the co-domain of the function.
So to outline a few things I think are correct (Please let me know if I have a wrong understanding):
A function is usually denoted by a single letter like $f$, $g$ or $h$.
$f(x)$ is the value of the function $f$ at $x$. You can say $f$ is a function of the variable $x$.
It seems as though, from the comments, that $f$ can be defined by saying what it does to a general element of its domain by stating an identity and then a domain for which it is true. for example:
- $f$ is defined by $f(x)=x+1$ for $x>0$
or, $g(x)= x + 1$ for all $x>2$ would define $g$
I think what I have done is write $2$ identities which means that both sides of the relation define the same function of $x$ on the domain stated.
Question 1: So $f$ is the function, $f(x)$ is the value of the function at $x$ and $x+1$ is the value of $f$ at $x$. Is it right that $f(x)$ tells you that $f$ is a function of $x $ and $x+1$ is a function of $x$. I find this hard to understand as I have just said that $x + 1$ is the value of $f$ at $x$.
elementary-set-theory notation
edited Aug 27 at 15:58
Arnaud D.
14.8k52142
asked Aug 26 at 19:52
James Anthony
776
edited Aug 27 at 15:58
Arnaud D.
14.8k52142
edited Aug 27 at 15:58
Arnaud D.
14.8k52142
edited Aug 27 at 15:58
Arnaud D.
14.8k52142
14.8k52142
asked Aug 26 at 19:52
James Anthony
776
asked Aug 26 at 19:52
James Anthony
776
asked Aug 26 at 19:52
James Anthony
776
776
$f(x)= y= x+1.$ So $f(x) = y$ is in $Y$, which is in the codomain. $x$, as you said, comes from the domain of $X$. So given $x=2 in mathbb Z$, $f(x)= y= 2+1=3 in Y$. Indeed, $f(x) = x+1$ is a function expressed in terms of $x$. The only thing not specified is the set $X$. That is, $f: Xto Y$ describes a function such that for every $x in X$, there exists a unique value $f(x)$ given by $xmapsto x+1$. The codomain, or rather, the image of $f(x)$ will depend in part on the domain of $X$.
– amWhy
Aug 26 at 20:03
Your question $2$: $f$ has been implicitly defined by merely $f: Xto f[X]$, where $f[X]$ is a set, and is the image of $f(x)$, and consists of all elements $y$ such that such that $y=x+1$.
– amWhy
Aug 26 at 20:08
In both cases, $f(x)$ is a function of $x$.
– amWhy
Aug 26 at 20:09
2
Why the downvotes? This is clearly a question from someone who is trying to understand functions.
– John Douma
Aug 26 at 20:16
2
Not from me, @JohnDouma. I take the question as trying to understand the definition of a function, and whether or not a "function" without a codomain designated explicitly, is defined. I think it's a valid question, and it is clear that James has put some thought into expressing his question as best he can.
– amWhy
Aug 26 at 20:16
 |Â
show 3 more comments
$f(x)= y= x+1.$ So $f(x) = y$ is in $Y$, which is in the codomain. $x$, as you said, comes from the domain of $X$. So given $x=2 in mathbb Z$, $f(x)= y= 2+1=3 in Y$. Indeed, $f(x) = x+1$ is a function expressed in terms of $x$. The only thing not specified is the set $X$. That is, $f: Xto Y$ describes a function such that for every $x in X$, there exists a unique value $f(x)$ given by $xmapsto x+1$. The codomain, or rather, the image of $f(x)$ will depend in part on the domain of $X$.
– amWhy
Aug 26 at 20:03
Your question $2$: $f$ has been implicitly defined by merely $f: Xto f[X]$, where $f[X]$ is a set, and is the image of $f(x)$, and consists of all elements $y$ such that such that $y=x+1$.
– amWhy
Aug 26 at 20:08
In both cases, $f(x)$ is a function of $x$.
– amWhy
Aug 26 at 20:09
2
Why the downvotes? This is clearly a question from someone who is trying to understand functions.
– John Douma
Aug 26 at 20:16
2
Not from me, @JohnDouma. I take the question as trying to understand the definition of a function, and whether or not a "function" without a codomain designated explicitly, is defined. I think it's a valid question, and it is clear that James has put some thought into expressing his question as best he can.
– amWhy
Aug 26 at 20:16
$f(x)= y= x+1.$ So $f(x) = y$ is in $Y$, which is in the codomain. $x$, as you said, comes from the domain of $X$. So given $x=2 in mathbb Z$, $f(x)= y= 2+1=3 in Y$. Indeed, $f(x) = x+1$ is a function expressed in terms of $x$. The only thing not specified is the set $X$. That is, $f: Xto Y$ describes a function such that for every $x in X$, there exists a unique value $f(x)$ given by $xmapsto x+1$. The codomain, or rather, the image of $f(x)$ will depend in part on the domain of $X$.
– amWhy
Aug 26 at 20:03
$f(x)= y= x+1.$ So $f(x) = y$ is in $Y$, which is in the codomain. $x$, as you said, comes from the domain of $X$. So given $x=2 in mathbb Z$, $f(x)= y= 2+1=3 in Y$. Indeed, $f(x) = x+1$ is a function expressed in terms of $x$. The only thing not specified is the set $X$. That is, $f: Xto Y$ describes a function such that for every $x in X$, there exists a unique value $f(x)$ given by $xmapsto x+1$. The codomain, or rather, the image of $f(x)$ will depend in part on the domain of $X$.
– amWhy
Aug 26 at 20:03
Your question $2$: $f$ has been implicitly defined by merely $f: Xto f[X]$, where $f[X]$ is a set, and is the image of $f(x)$, and consists of all elements $y$ such that such that $y=x+1$.
– amWhy
Aug 26 at 20:08
Your question $2$: $f$ has been implicitly defined by merely $f: Xto f[X]$, where $f[X]$ is a set, and is the image of $f(x)$, and consists of all elements $y$ such that such that $y=x+1$.
– amWhy
Aug 26 at 20:08
In both cases, $f(x)$ is a function of $x$.
– amWhy
Aug 26 at 20:09
In both cases, $f(x)$ is a function of $x$.
– amWhy
Aug 26 at 20:09
2
2
Why the downvotes? This is clearly a question from someone who is trying to understand functions.
– John Douma
Aug 26 at 20:16
Why the downvotes? This is clearly a question from someone who is trying to understand functions.
– John Douma
Aug 26 at 20:16
2
2
Not from me, @JohnDouma. I take the question as trying to understand the definition of a function, and whether or not a "function" without a codomain designated explicitly, is defined. I think it's a valid question, and it is clear that James has put some thought into expressing his question as best he can.
– amWhy
Aug 26 at 20:16
Not from me, @JohnDouma. I take the question as trying to understand the definition of a function, and whether or not a "function" without a codomain designated explicitly, is defined. I think it's a valid question, and it is clear that James has put some thought into expressing his question as best he can.
– amWhy
Aug 26 at 20:16
 |Â
show 3 more comments
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$f(x)= y= x+1.$ So $f(x) = y$ is in $Y$, which is in the codomain. $x$, as you said, comes from the domain of $X$. So given $x=2 in mathbb Z$, $f(x)= y= 2+1=3 in Y$. Indeed, $f(x) = x+1$ is a function expressed in terms of $x$. The only thing not specified is the set $X$. That is, $f: Xto Y$ describes a function such that for every $x in X$, there exists a unique value $f(x)$ given by $xmapsto x+1$. The codomain, or rather, the image of $f(x)$ will depend in part on the domain of $X$.
– amWhy
Aug 26 at 20:03
Your question $2$: $f$ has been implicitly defined by merely $f: Xto f[X]$, where $f[X]$ is a set, and is the image of $f(x)$, and consists of all elements $y$ such that such that $y=x+1$.
– amWhy
Aug 26 at 20:08
In both cases, $f(x)$ is a function of $x$.
– amWhy
Aug 26 at 20:09
2
Why the downvotes? This is clearly a question from someone who is trying to understand functions.
– John Douma
Aug 26 at 20:16
2
Not from me, @JohnDouma. I take the question as trying to understand the definition of a function, and whether or not a "function" without a codomain designated explicitly, is defined. I think it's a valid question, and it is clear that James has put some thought into expressing his question as best he can.
– amWhy
Aug 26 at 20:16