Proving $f(x)=|x|$ is onto The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Determining if Function is 1:1 or OntoProving a bijection(injection and surjection) over a functionRigorous proof that surjectivity implies injectivity for finite setsProving continuity on Sobolev space with weak topologyProving or Disproving a function that is onto itself is one to one.How to prove $f(n)=lceilfracn2rceil$ is one-to-one and onto?Onto functions from Power set of Naturals$f(y)geq f(x)rightarrow (y-x)geq0$ in $mathbbR^2_+$ if weakly increasing?Is $f(x) = 3 -frac2x$ injective or surjective?Prove a function is onto if its domain is a Cartesian product
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Proving $f(x)=|x|$ is onto
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Determining if Function is 1:1 or OntoProving a bijection(injection and surjection) over a functionRigorous proof that surjectivity implies injectivity for finite setsProving continuity on Sobolev space with weak topologyProving or Disproving a function that is onto itself is one to one.How to prove $f(n)=lceilfracn2rceil$ is one-to-one and onto?Onto functions from Power set of Naturals$f(y)geq f(x)rightarrow (y-x)geq0$ in $mathbbR^2_+$ if weakly increasing?Is $f(x) = 3 -frac2x$ injective or surjective?Prove a function is onto if its domain is a Cartesian product
$begingroup$
I've been working on proving that this is a onto function:
$f$ : $mathbb R$ $to$ $mathbb R^geq0$ is defined by $f(x)=|x|$
My proof so far: Let $yinmathbb R$.
Rough work: $|x|=y Rightarrow sqrt x^2=y Rightarrow n^2=y^2 Rightarrow pm x=pm y$
Suppose $f(pm y)=|pm y|=y$.
I know that this function is definitely onto given the co-domain of $mathbb R^geq0$, but I feel like my proof is flawed. Am I supposed to individually account for the $-x$ and the $+x$ from $pm x=pm y$ when trying to solve $f(x) = y$?
Thanks!
proof-verification elementary-set-theory
$endgroup$
add a comment |
$begingroup$
I've been working on proving that this is a onto function:
$f$ : $mathbb R$ $to$ $mathbb R^geq0$ is defined by $f(x)=|x|$
My proof so far: Let $yinmathbb R$.
Rough work: $|x|=y Rightarrow sqrt x^2=y Rightarrow n^2=y^2 Rightarrow pm x=pm y$
Suppose $f(pm y)=|pm y|=y$.
I know that this function is definitely onto given the co-domain of $mathbb R^geq0$, but I feel like my proof is flawed. Am I supposed to individually account for the $-x$ and the $+x$ from $pm x=pm y$ when trying to solve $f(x) = y$?
Thanks!
proof-verification elementary-set-theory
$endgroup$
1
$begingroup$
Let $ain mathbb R ^geq 0$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
$endgroup$
– Arrow
Mar 24 at 15:23
6
$begingroup$
You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
$endgroup$
– symplectomorphic
Mar 24 at 15:24
add a comment |
$begingroup$
I've been working on proving that this is a onto function:
$f$ : $mathbb R$ $to$ $mathbb R^geq0$ is defined by $f(x)=|x|$
My proof so far: Let $yinmathbb R$.
Rough work: $|x|=y Rightarrow sqrt x^2=y Rightarrow n^2=y^2 Rightarrow pm x=pm y$
Suppose $f(pm y)=|pm y|=y$.
I know that this function is definitely onto given the co-domain of $mathbb R^geq0$, but I feel like my proof is flawed. Am I supposed to individually account for the $-x$ and the $+x$ from $pm x=pm y$ when trying to solve $f(x) = y$?
Thanks!
proof-verification elementary-set-theory
$endgroup$
I've been working on proving that this is a onto function:
$f$ : $mathbb R$ $to$ $mathbb R^geq0$ is defined by $f(x)=|x|$
My proof so far: Let $yinmathbb R$.
Rough work: $|x|=y Rightarrow sqrt x^2=y Rightarrow n^2=y^2 Rightarrow pm x=pm y$
Suppose $f(pm y)=|pm y|=y$.
I know that this function is definitely onto given the co-domain of $mathbb R^geq0$, but I feel like my proof is flawed. Am I supposed to individually account for the $-x$ and the $+x$ from $pm x=pm y$ when trying to solve $f(x) = y$?
Thanks!
proof-verification elementary-set-theory
proof-verification elementary-set-theory
edited Mar 25 at 8:00
YuiTo Cheng
2,40641037
2,40641037
asked Mar 24 at 15:21
Nick SabiaNick Sabia
356
356
1
$begingroup$
Let $ain mathbb R ^geq 0$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
$endgroup$
– Arrow
Mar 24 at 15:23
6
$begingroup$
You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
$endgroup$
– symplectomorphic
Mar 24 at 15:24
add a comment |
1
$begingroup$
Let $ain mathbb R ^geq 0$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
$endgroup$
– Arrow
Mar 24 at 15:23
6
$begingroup$
You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
$endgroup$
– symplectomorphic
Mar 24 at 15:24
1
1
$begingroup$
Let $ain mathbb R ^geq 0$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
$endgroup$
– Arrow
Mar 24 at 15:23
$begingroup$
Let $ain mathbb R ^geq 0$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
$endgroup$
– Arrow
Mar 24 at 15:23
6
6
$begingroup$
You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
$endgroup$
– symplectomorphic
Mar 24 at 15:24
$begingroup$
You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
$endgroup$
– symplectomorphic
Mar 24 at 15:24
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
In order to show that the function is onto (surjective) it is enough to argue that for each $y$ in the codomain there is at least one $x$ in the domain that maps to it.
You seem to be trying to find all of the $x$ such that $f(x)=y$, which is more work than you need to do and creates a rather large detour.
You could just say:
Let $yinmathbb R^ge 0$ be given. Then $f(y)=|y|=y$, so by setting $x=y$ we find an $xinmathbb R$ such that $f(x)=y$. Since $y$ was arbitrary this proves that $f$ is surjective.
(Even this is more verbose than it really needs to be, but in exercises at this elementary level it is good to be explicit about the details, to show that you know what you're doing).
$endgroup$
2
$begingroup$
(+1) Because this is a vey pedagogical answer.
$endgroup$
– José Carlos Santos
Mar 24 at 15:33
add a comment |
$begingroup$
Am I supposed to individually account for the -x and the +x from ±x=±y when trying to solve f(x)=y?
No.
You are simply supposed to show that for any general arbitrary $y in mathbb R^ge 0$ that there is, at least (you don't have to find them all), one $xin mathbb R$ so that $|x| = y$.
As $|y| = y$ this is very easy. And you are done.
The proof is two lines:
1) Let $y in mathbb R^ge 0$.
2) $f(y) = |y| = y$.
$endgroup$
add a comment |
$begingroup$
You should not start with $yinmathbb R$ but rather with $yinmathbbR^geqslant0$. Then $y=f(y)$. Since this happens for every $yinmathbbR^geqslant 0$, $f$ is onto.
$endgroup$
add a comment |
$begingroup$
Well, you want to show that $f$ is onto. So take an arbitrary element $yinBbb R_geq 0$ from the image set and find a preimage. Here a preimage is $y$ itself or even $-y$, since $|-y|=y$.
$endgroup$
add a comment |
$begingroup$
More generally, If $B subseteq A$ and $f: A rightarrow B$ is a continuous mapping with $f(b) = b$ for all $b in B$ then $f$ is called a retraction of $A$ onto $B$. It is trivial that any such $f$ is onto since $f(b) = b$ for all $b in B$ immediately implies that all such $b$ are in the range. Note that continuity really plays no role in this. Thinking of $|x|$ as a retraction is perhaps overkill, but shows that the proof that it is onto the nonnegative reals is a special case of a more general (and equally easy to prove) result.
$endgroup$
add a comment |
Your Answer
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In order to show that the function is onto (surjective) it is enough to argue that for each $y$ in the codomain there is at least one $x$ in the domain that maps to it.
You seem to be trying to find all of the $x$ such that $f(x)=y$, which is more work than you need to do and creates a rather large detour.
You could just say:
Let $yinmathbb R^ge 0$ be given. Then $f(y)=|y|=y$, so by setting $x=y$ we find an $xinmathbb R$ such that $f(x)=y$. Since $y$ was arbitrary this proves that $f$ is surjective.
(Even this is more verbose than it really needs to be, but in exercises at this elementary level it is good to be explicit about the details, to show that you know what you're doing).
$endgroup$
2
$begingroup$
(+1) Because this is a vey pedagogical answer.
$endgroup$
– José Carlos Santos
Mar 24 at 15:33
add a comment |
$begingroup$
In order to show that the function is onto (surjective) it is enough to argue that for each $y$ in the codomain there is at least one $x$ in the domain that maps to it.
You seem to be trying to find all of the $x$ such that $f(x)=y$, which is more work than you need to do and creates a rather large detour.
You could just say:
Let $yinmathbb R^ge 0$ be given. Then $f(y)=|y|=y$, so by setting $x=y$ we find an $xinmathbb R$ such that $f(x)=y$. Since $y$ was arbitrary this proves that $f$ is surjective.
(Even this is more verbose than it really needs to be, but in exercises at this elementary level it is good to be explicit about the details, to show that you know what you're doing).
$endgroup$
2
$begingroup$
(+1) Because this is a vey pedagogical answer.
$endgroup$
– José Carlos Santos
Mar 24 at 15:33
add a comment |
$begingroup$
In order to show that the function is onto (surjective) it is enough to argue that for each $y$ in the codomain there is at least one $x$ in the domain that maps to it.
You seem to be trying to find all of the $x$ such that $f(x)=y$, which is more work than you need to do and creates a rather large detour.
You could just say:
Let $yinmathbb R^ge 0$ be given. Then $f(y)=|y|=y$, so by setting $x=y$ we find an $xinmathbb R$ such that $f(x)=y$. Since $y$ was arbitrary this proves that $f$ is surjective.
(Even this is more verbose than it really needs to be, but in exercises at this elementary level it is good to be explicit about the details, to show that you know what you're doing).
$endgroup$
In order to show that the function is onto (surjective) it is enough to argue that for each $y$ in the codomain there is at least one $x$ in the domain that maps to it.
You seem to be trying to find all of the $x$ such that $f(x)=y$, which is more work than you need to do and creates a rather large detour.
You could just say:
Let $yinmathbb R^ge 0$ be given. Then $f(y)=|y|=y$, so by setting $x=y$ we find an $xinmathbb R$ such that $f(x)=y$. Since $y$ was arbitrary this proves that $f$ is surjective.
(Even this is more verbose than it really needs to be, but in exercises at this elementary level it is good to be explicit about the details, to show that you know what you're doing).
answered Mar 24 at 15:25
Henning MakholmHenning Makholm
243k17311555
243k17311555
2
$begingroup$
(+1) Because this is a vey pedagogical answer.
$endgroup$
– José Carlos Santos
Mar 24 at 15:33
add a comment |
2
$begingroup$
(+1) Because this is a vey pedagogical answer.
$endgroup$
– José Carlos Santos
Mar 24 at 15:33
2
2
$begingroup$
(+1) Because this is a vey pedagogical answer.
$endgroup$
– José Carlos Santos
Mar 24 at 15:33
$begingroup$
(+1) Because this is a vey pedagogical answer.
$endgroup$
– José Carlos Santos
Mar 24 at 15:33
add a comment |
$begingroup$
Am I supposed to individually account for the -x and the +x from ±x=±y when trying to solve f(x)=y?
No.
You are simply supposed to show that for any general arbitrary $y in mathbb R^ge 0$ that there is, at least (you don't have to find them all), one $xin mathbb R$ so that $|x| = y$.
As $|y| = y$ this is very easy. And you are done.
The proof is two lines:
1) Let $y in mathbb R^ge 0$.
2) $f(y) = |y| = y$.
$endgroup$
add a comment |
$begingroup$
Am I supposed to individually account for the -x and the +x from ±x=±y when trying to solve f(x)=y?
No.
You are simply supposed to show that for any general arbitrary $y in mathbb R^ge 0$ that there is, at least (you don't have to find them all), one $xin mathbb R$ so that $|x| = y$.
As $|y| = y$ this is very easy. And you are done.
The proof is two lines:
1) Let $y in mathbb R^ge 0$.
2) $f(y) = |y| = y$.
$endgroup$
add a comment |
$begingroup$
Am I supposed to individually account for the -x and the +x from ±x=±y when trying to solve f(x)=y?
No.
You are simply supposed to show that for any general arbitrary $y in mathbb R^ge 0$ that there is, at least (you don't have to find them all), one $xin mathbb R$ so that $|x| = y$.
As $|y| = y$ this is very easy. And you are done.
The proof is two lines:
1) Let $y in mathbb R^ge 0$.
2) $f(y) = |y| = y$.
$endgroup$
Am I supposed to individually account for the -x and the +x from ±x=±y when trying to solve f(x)=y?
No.
You are simply supposed to show that for any general arbitrary $y in mathbb R^ge 0$ that there is, at least (you don't have to find them all), one $xin mathbb R$ so that $|x| = y$.
As $|y| = y$ this is very easy. And you are done.
The proof is two lines:
1) Let $y in mathbb R^ge 0$.
2) $f(y) = |y| = y$.
answered Mar 24 at 16:04
fleabloodfleablood
1
1
add a comment |
add a comment |
$begingroup$
You should not start with $yinmathbb R$ but rather with $yinmathbbR^geqslant0$. Then $y=f(y)$. Since this happens for every $yinmathbbR^geqslant 0$, $f$ is onto.
$endgroup$
add a comment |
$begingroup$
You should not start with $yinmathbb R$ but rather with $yinmathbbR^geqslant0$. Then $y=f(y)$. Since this happens for every $yinmathbbR^geqslant 0$, $f$ is onto.
$endgroup$
add a comment |
$begingroup$
You should not start with $yinmathbb R$ but rather with $yinmathbbR^geqslant0$. Then $y=f(y)$. Since this happens for every $yinmathbbR^geqslant 0$, $f$ is onto.
$endgroup$
You should not start with $yinmathbb R$ but rather with $yinmathbbR^geqslant0$. Then $y=f(y)$. Since this happens for every $yinmathbbR^geqslant 0$, $f$ is onto.
edited Mar 24 at 15:41
J. W. Tanner
4,7721420
4,7721420
answered Mar 24 at 15:26
José Carlos SantosJosé Carlos Santos
174k23134243
174k23134243
add a comment |
add a comment |
$begingroup$
Well, you want to show that $f$ is onto. So take an arbitrary element $yinBbb R_geq 0$ from the image set and find a preimage. Here a preimage is $y$ itself or even $-y$, since $|-y|=y$.
$endgroup$
add a comment |
$begingroup$
Well, you want to show that $f$ is onto. So take an arbitrary element $yinBbb R_geq 0$ from the image set and find a preimage. Here a preimage is $y$ itself or even $-y$, since $|-y|=y$.
$endgroup$
add a comment |
$begingroup$
Well, you want to show that $f$ is onto. So take an arbitrary element $yinBbb R_geq 0$ from the image set and find a preimage. Here a preimage is $y$ itself or even $-y$, since $|-y|=y$.
$endgroup$
Well, you want to show that $f$ is onto. So take an arbitrary element $yinBbb R_geq 0$ from the image set and find a preimage. Here a preimage is $y$ itself or even $-y$, since $|-y|=y$.
answered Mar 24 at 15:24
WuestenfuxWuestenfux
5,5131513
5,5131513
add a comment |
add a comment |
$begingroup$
More generally, If $B subseteq A$ and $f: A rightarrow B$ is a continuous mapping with $f(b) = b$ for all $b in B$ then $f$ is called a retraction of $A$ onto $B$. It is trivial that any such $f$ is onto since $f(b) = b$ for all $b in B$ immediately implies that all such $b$ are in the range. Note that continuity really plays no role in this. Thinking of $|x|$ as a retraction is perhaps overkill, but shows that the proof that it is onto the nonnegative reals is a special case of a more general (and equally easy to prove) result.
$endgroup$
add a comment |
$begingroup$
More generally, If $B subseteq A$ and $f: A rightarrow B$ is a continuous mapping with $f(b) = b$ for all $b in B$ then $f$ is called a retraction of $A$ onto $B$. It is trivial that any such $f$ is onto since $f(b) = b$ for all $b in B$ immediately implies that all such $b$ are in the range. Note that continuity really plays no role in this. Thinking of $|x|$ as a retraction is perhaps overkill, but shows that the proof that it is onto the nonnegative reals is a special case of a more general (and equally easy to prove) result.
$endgroup$
add a comment |
$begingroup$
More generally, If $B subseteq A$ and $f: A rightarrow B$ is a continuous mapping with $f(b) = b$ for all $b in B$ then $f$ is called a retraction of $A$ onto $B$. It is trivial that any such $f$ is onto since $f(b) = b$ for all $b in B$ immediately implies that all such $b$ are in the range. Note that continuity really plays no role in this. Thinking of $|x|$ as a retraction is perhaps overkill, but shows that the proof that it is onto the nonnegative reals is a special case of a more general (and equally easy to prove) result.
$endgroup$
More generally, If $B subseteq A$ and $f: A rightarrow B$ is a continuous mapping with $f(b) = b$ for all $b in B$ then $f$ is called a retraction of $A$ onto $B$. It is trivial that any such $f$ is onto since $f(b) = b$ for all $b in B$ immediately implies that all such $b$ are in the range. Note that continuity really plays no role in this. Thinking of $|x|$ as a retraction is perhaps overkill, but shows that the proof that it is onto the nonnegative reals is a special case of a more general (and equally easy to prove) result.
answered Mar 24 at 20:31
John ColemanJohn Coleman
4,00311224
4,00311224
add a comment |
add a comment |
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$begingroup$
Let $ain mathbb R ^geq 0$. Then by definition, $|a|=a$. Therefore, $a$ is in the image of $f:xmapsto |x|$.
$endgroup$
– Arrow
Mar 24 at 15:23
6
$begingroup$
You proof is wrong from the beginning. You need to assume y is in the nonnegative reals and show you can find a real x that maps to it.
$endgroup$
– symplectomorphic
Mar 24 at 15:24