Left and right inverses of functions.
$begingroup$
Let $X$ be the set of real transcendental numbers. Define the realtion $sim$ on $X$ by $xsim y$ iff $x-y in mathbb{Q}$ is an equivalence relation.
Let $Y$ denote the set of equivalence classes generated by $sim$ defined above.
Define the function $P:X rightarrow Y$ by $f(x) = [x]$ where $[x]$ denotes the equivalence class of $x$. Does the function P have:
- a left inverse?
- a right inverse?
My work so far:
Since $P$ is surjective by definition of equivalence classes, $P$ has a right inverse.
Now if $P$ is injective then it will have a left inverse. I think it's not injective as two real transcendental numbers can end up in the same equivalence class, so $P$ doesn't have a left inverse, but I am not sure how to show this rigorously.
functions inverse equivalence-relations
asked Dec 30 '18 at 13:02
xAlyxAly
404
$endgroup$
add a comment |
$begingroup$
Let $X$ be the set of real transcendental numbers. Define the realtion $sim$ on $X$ by $xsim y$ iff $x-y in mathbb{Q}$ is an equivalence relation.
Let $Y$ denote the set of equivalence classes generated by $sim$ defined above.
Define the function $P:X rightarrow Y$ by $f(x) = [x]$ where $[x]$ denotes the equivalence class of $x$. Does the function P have:
- a left inverse?
- a right inverse?
My work so far:
Since $P$ is surjective by definition of equivalence classes, $P$ has a right inverse.
Now if $P$ is injective then it will have a left inverse. I think it's not injective as two real transcendental numbers can end up in the same equivalence class, so $P$ doesn't have a left inverse, but I am not sure how to show this rigorously.
functions inverse equivalence-relations
asked Dec 30 '18 at 13:02
xAlyxAly
404
$endgroup$
add a comment |
$begingroup$
Let $X$ be the set of real transcendental numbers. Define the realtion $sim$ on $X$ by $xsim y$ iff $x-y in mathbb{Q}$ is an equivalence relation.
Let $Y$ denote the set of equivalence classes generated by $sim$ defined above.
Define the function $P:X rightarrow Y$ by $f(x) = [x]$ where $[x]$ denotes the equivalence class of $x$. Does the function P have:
- a left inverse?
- a right inverse?
My work so far:
Since $P$ is surjective by definition of equivalence classes, $P$ has a right inverse.
Now if $P$ is injective then it will have a left inverse. I think it's not injective as two real transcendental numbers can end up in the same equivalence class, so $P$ doesn't have a left inverse, but I am not sure how to show this rigorously.
functions inverse equivalence-relations
asked Dec 30 '18 at 13:02
xAlyxAly
404
$endgroup$
Let $X$ be the set of real transcendental numbers. Define the realtion $sim$ on $X$ by $xsim y$ iff $x-y in mathbb{Q}$ is an equivalence relation.
Let $Y$ denote the set of equivalence classes generated by $sim$ defined above.
Define the function $P:X rightarrow Y$ by $f(x) = [x]$ where $[x]$ denotes the equivalence class of $x$. Does the function P have:
- a left inverse?
- a right inverse?
My work so far:
Since $P$ is surjective by definition of equivalence classes, $P$ has a right inverse.
Now if $P$ is injective then it will have a left inverse. I think it's not injective as two real transcendental numbers can end up in the same equivalence class, so $P$ doesn't have a left inverse, but I am not sure how to show this rigorously.
functions inverse equivalence-relations
functions inverse equivalence-relations
asked Dec 30 '18 at 13:02
xAlyxAly
404
asked Dec 30 '18 at 13:02
xAlyxAly
404
asked Dec 30 '18 at 13:02
xAlyxAly
404
asked Dec 30 '18 at 13:02
xAlyxAly
404
asked Dec 30 '18 at 13:02
xAlyxAly
404
404
add a comment |
add a comment |
1 Answer
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oldest
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$begingroup$
Hint 1: prove that if $xin X$, then also $x+1in X$. Since $x+1sim x$, the map $P$ is not injective
Hint 2: the sum of algebraic numbers is algebraic.
answered Dec 30 '18 at 13:49
egregegreg
185k1486206
$endgroup$
$begingroup$
I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
$endgroup$
– xAly
Dec 30 '18 at 19:22
$begingroup$
@xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
$endgroup$
– egreg
Dec 30 '18 at 20:10
$begingroup$
But $X$ is the set of real transcendental numbers, not algebraic numbers?
$endgroup$
– xAly
Dec 30 '18 at 21:16
$begingroup$
@xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
$endgroup$
– egreg
Dec 30 '18 at 21:52
$begingroup$
Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
$endgroup$
– xAly
Dec 30 '18 at 22:02
|
show 1 more comment
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint 1: prove that if $xin X$, then also $x+1in X$. Since $x+1sim x$, the map $P$ is not injective
Hint 2: the sum of algebraic numbers is algebraic.
answered Dec 30 '18 at 13:49
egregegreg
185k1486206
$endgroup$
$begingroup$
I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
$endgroup$
– xAly
Dec 30 '18 at 19:22
$begingroup$
@xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
$endgroup$
– egreg
Dec 30 '18 at 20:10
$begingroup$
But $X$ is the set of real transcendental numbers, not algebraic numbers?
$endgroup$
– xAly
Dec 30 '18 at 21:16
$begingroup$
@xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
$endgroup$
– egreg
Dec 30 '18 at 21:52
$begingroup$
Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
$endgroup$
– xAly
Dec 30 '18 at 22:02
|
show 1 more comment
$begingroup$
Hint 1: prove that if $xin X$, then also $x+1in X$. Since $x+1sim x$, the map $P$ is not injective
Hint 2: the sum of algebraic numbers is algebraic.
answered Dec 30 '18 at 13:49
egregegreg
185k1486206
$endgroup$
$begingroup$
I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
$endgroup$
– xAly
Dec 30 '18 at 19:22
$begingroup$
@xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
$endgroup$
– egreg
Dec 30 '18 at 20:10
$begingroup$
But $X$ is the set of real transcendental numbers, not algebraic numbers?
$endgroup$
– xAly
Dec 30 '18 at 21:16
$begingroup$
@xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
$endgroup$
– egreg
Dec 30 '18 at 21:52
$begingroup$
Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
$endgroup$
– xAly
Dec 30 '18 at 22:02
|
show 1 more comment
$begingroup$
Hint 1: prove that if $xin X$, then also $x+1in X$. Since $x+1sim x$, the map $P$ is not injective
Hint 2: the sum of algebraic numbers is algebraic.
answered Dec 30 '18 at 13:49
egregegreg
185k1486206
$endgroup$
Hint 1: prove that if $xin X$, then also $x+1in X$. Since $x+1sim x$, the map $P$ is not injective
Hint 2: the sum of algebraic numbers is algebraic.
answered Dec 30 '18 at 13:49
egregegreg
185k1486206
answered Dec 30 '18 at 13:49
egregegreg
185k1486206
answered Dec 30 '18 at 13:49
egregegreg
185k1486206
answered Dec 30 '18 at 13:49
egregegreg
185k1486206
185k1486206
$begingroup$
I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
$endgroup$
– xAly
Dec 30 '18 at 19:22
$begingroup$
@xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
$endgroup$
– egreg
Dec 30 '18 at 20:10
$begingroup$
But $X$ is the set of real transcendental numbers, not algebraic numbers?
$endgroup$
– xAly
Dec 30 '18 at 21:16
$begingroup$
@xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
$endgroup$
– egreg
Dec 30 '18 at 21:52
$begingroup$
Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
$endgroup$
– xAly
Dec 30 '18 at 22:02
|
show 1 more comment
$begingroup$
I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
$endgroup$
– xAly
Dec 30 '18 at 19:22
$begingroup$
@xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
$endgroup$
– egreg
Dec 30 '18 at 20:10
$begingroup$
But $X$ is the set of real transcendental numbers, not algebraic numbers?
$endgroup$
– xAly
Dec 30 '18 at 21:16
$begingroup$
@xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
$endgroup$
– egreg
Dec 30 '18 at 21:52
$begingroup$
Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
$endgroup$
– xAly
Dec 30 '18 at 22:02
$begingroup$
I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
$endgroup$
– xAly
Dec 30 '18 at 19:22
$begingroup$
I am having trouble doing the first bit. I know the algebraic numbers form a field. So the sum of algebraic numbers is algebraic. So since $x$ is transcendental $x+1$ can't be algebraic?
$endgroup$
– xAly
Dec 30 '18 at 19:22
$begingroup$
@xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
$endgroup$
– egreg
Dec 30 '18 at 20:10
$begingroup$
@xAly $(x+1)-x=1inmathbb{Q}$, so $x+1sim x$. If $x+1$ is algebraic, then also $(x+1)-1=x$ is algebraic.
$endgroup$
– egreg
Dec 30 '18 at 20:10
$begingroup$
But $X$ is the set of real transcendental numbers, not algebraic numbers?
$endgroup$
– xAly
Dec 30 '18 at 21:16
$begingroup$
But $X$ is the set of real transcendental numbers, not algebraic numbers?
$endgroup$
– xAly
Dec 30 '18 at 21:16
$begingroup$
@xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
$endgroup$
– egreg
Dec 30 '18 at 21:52
$begingroup$
@xAly Yes. This proves that if $x$ is transcendental, then also $x+1$ is.
$endgroup$
– egreg
Dec 30 '18 at 21:52
$begingroup$
Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
$endgroup$
– xAly
Dec 30 '18 at 22:02
$begingroup$
Ah ok. If $x$ is transcendental and $x+1$ were algebraic, that would be impossible because then $x$ would have to be algebraic. That took me a bit...
$endgroup$
– xAly
Dec 30 '18 at 22:02
|
show 1 more comment
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