Mixing
Example of a bijection from the set of real numbers to a subset of irrationals
$begingroup$
I need an example of a bijection from the set of real numbers to a subset of the irrationals.
I tried something like
$f(x)=x+sqrt{2}$,
but where should I map $-sqrt{2}$?
functions
edited Jul 8 '14 at 18:10
Mathmo123
17.9k33166
asked Jul 8 '14 at 17:16
user 160756user 160756
551
$endgroup$
add a comment |
$begingroup$
I need an example of a bijection from the set of real numbers to a subset of the irrationals.
I tried something like
$f(x)=x+sqrt{2}$,
but where should I map $-sqrt{2}$?
functions
edited Jul 8 '14 at 18:10
Mathmo123
17.9k33166
asked Jul 8 '14 at 17:16
user 160756user 160756
551
$endgroup$
$begingroup$
There are problems with all numbers of the form $-sqrt 2 +r$, with $rin mathbb Q$.
$endgroup$
– Git Gud
Jul 8 '14 at 17:18
$begingroup$
Worse than that -- your function maps $mathbb R$ onto itself
$endgroup$
– MPW
Jul 8 '14 at 17:20
9
$begingroup$
Map the rationals to the odd non-zero multiples of $sqrt2$. Map the non-zero multiples of $sqrt 2$ to the even multiples of $sqrt 2$. Take the identity elsewhere.
$endgroup$
– David Mitra
Jul 8 '14 at 17:23
add a comment |
$begingroup$
I need an example of a bijection from the set of real numbers to a subset of the irrationals.
I tried something like
$f(x)=x+sqrt{2}$,
but where should I map $-sqrt{2}$?
functions
edited Jul 8 '14 at 18:10
Mathmo123
17.9k33166
asked Jul 8 '14 at 17:16
user 160756user 160756
551
$endgroup$
I need an example of a bijection from the set of real numbers to a subset of the irrationals.
I tried something like
$f(x)=x+sqrt{2}$,
but where should I map $-sqrt{2}$?
functions
functions
edited Jul 8 '14 at 18:10
Mathmo123
17.9k33166
asked Jul 8 '14 at 17:16
user 160756user 160756
551
edited Jul 8 '14 at 18:10
Mathmo123
17.9k33166
asked Jul 8 '14 at 17:16
user 160756user 160756
551
edited Jul 8 '14 at 18:10
Mathmo123
17.9k33166
edited Jul 8 '14 at 18:10
Mathmo123
17.9k33166
edited Jul 8 '14 at 18:10
Mathmo123
17.9k33166
17.9k33166
asked Jul 8 '14 at 17:16
user 160756user 160756
551
asked Jul 8 '14 at 17:16
user 160756user 160756
551
asked Jul 8 '14 at 17:16
user 160756user 160756
551
551
$begingroup$
There are problems with all numbers of the form $-sqrt 2 +r$, with $rin mathbb Q$.
$endgroup$
– Git Gud
Jul 8 '14 at 17:18
$begingroup$
Worse than that -- your function maps $mathbb R$ onto itself
$endgroup$
– MPW
Jul 8 '14 at 17:20
9
$begingroup$
Map the rationals to the odd non-zero multiples of $sqrt2$. Map the non-zero multiples of $sqrt 2$ to the even multiples of $sqrt 2$. Take the identity elsewhere.
$endgroup$
– David Mitra
Jul 8 '14 at 17:23
add a comment |
$begingroup$
There are problems with all numbers of the form $-sqrt 2 +r$, with $rin mathbb Q$.
$endgroup$
– Git Gud
Jul 8 '14 at 17:18
$begingroup$
Worse than that -- your function maps $mathbb R$ onto itself
$endgroup$
– MPW
Jul 8 '14 at 17:20
9
$begingroup$
Map the rationals to the odd non-zero multiples of $sqrt2$. Map the non-zero multiples of $sqrt 2$ to the even multiples of $sqrt 2$. Take the identity elsewhere.
$endgroup$
– David Mitra
Jul 8 '14 at 17:23
$begingroup$
There are problems with all numbers of the form $-sqrt 2 +r$, with $rin mathbb Q$.
$endgroup$
– Git Gud
Jul 8 '14 at 17:18
$begingroup$
There are problems with all numbers of the form $-sqrt 2 +r$, with $rin mathbb Q$.
$endgroup$
– Git Gud
Jul 8 '14 at 17:18
$begingroup$
Worse than that -- your function maps $mathbb R$ onto itself
$endgroup$
– MPW
Jul 8 '14 at 17:20
$begingroup$
Worse than that -- your function maps $mathbb R$ onto itself
$endgroup$
– MPW
Jul 8 '14 at 17:20
9
9
$begingroup$
Map the rationals to the odd non-zero multiples of $sqrt2$. Map the non-zero multiples of $sqrt 2$ to the even multiples of $sqrt 2$. Take the identity elsewhere.
$endgroup$
– David Mitra
Jul 8 '14 at 17:23
$begingroup$
Map the rationals to the odd non-zero multiples of $sqrt2$. Map the non-zero multiples of $sqrt 2$ to the even multiples of $sqrt 2$. Take the identity elsewhere.
$endgroup$
– David Mitra
Jul 8 '14 at 17:23
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let $f(x) = dfrac{arctan x}{pi}$, so $f^{-1}(x) = tan pi x$. $f$ maps $mathbb{R}$ to $(-dfrac12,dfrac12)$.
$$g(x) = begin{cases}
x in mathbb{Q} & x + sqrt{5}\
x notin mathbb{Q} & x\
end{cases}$$
$$g^{-1}(x) = begin{cases}
x > 1 & x - sqrt{5}\
x le 1 & x\
end{cases}$$
So $g circ f$ maps from $mathbb{R}$ to a certain subset of the irrationals between $-dfrac12$ and $dfrac12 + sqrt{5}$, and $f^{-1} circ g^{-1}$ maps the inverse.
answered Jul 8 '14 at 17:39
NovaDenizenNovaDenizen
3,437718
$endgroup$
$begingroup$
$gof$ is not onto!
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 17:48
$begingroup$
You are correct, but it doesn't have to be onto. The question asked for a bijective map from $mathbb{R}$ to an irrational subset of the reals.
$endgroup$
– NovaDenizen
Jul 8 '14 at 17:50
$begingroup$
Yeah! In this case, it is not interesting! :)
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 18:04
add a comment |
$begingroup$
Let $A_i = pi^i {mathbb Q}$ and $ A =cup_{ige0} A_i$. Then the map $h:{mathbb R}backslash{mathbb Q}rightarrow {mathbb R}$
$$
h(x) = left{
begin{array}{ll}
pi x & xin A \
x & x in B={mathbb R}backslash A\
end{array}
right.
$$
is a bijection.
answered Jul 8 '14 at 17:41
Mohammad KhosraviMohammad Khosravi
2,014716
$endgroup$
add a comment |
$begingroup$
Enumerate $Bbb Q$ as $q_n$ for $ninBbb N$. Now define,
$$f(x)=begin{cases}e^{2n+1} & x=q_n\e^{2n} & x=e^n, ninBbb N\x&text{otherwise}end{cases}$$
Key point here is that $e$ is transcendental, so $e^nnotinBbb Q$ for any $n>0$. You can effectively replace $e$ by any other number with this property. For example $pi,sqrt2^sqrt2$, Liouville number, and so on and so forth.
answered Jul 8 '14 at 17:47
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
$begingroup$
@David: $e$ is transcendental. So yes.
$endgroup$
– Asaf Karagila♦
Jul 8 '14 at 18:00
add a comment |
$begingroup$
Consider $A$ = { $pi + n : n in Bbb N$}.
Clearly $A$ is countable. ( Consider the bijection $n rightarrow pi + n$ , $forall n in Bbb N$).
So elements of $A$ can be expressed as, $A = {a_k : k in Bbb N}$.
And we also can have $Bbb Q = {b_k : k in Bbb N}$
Now consider the bijection
$phi : Bbb R rightarrow Bbb R - Bbb Q$ by,
$phi(x) := a_{2k+1}$ if $ x = b_k , x in Bbb Q$.
= $a_{2k}$ if $x = a_k , x in A$
= $x$ if $x in Bbb R - (A cup Bbb Q)$
The construction in my example can be seen as a very general bijection from a general uncountable set to a proper uncountable subset of it.
answered May 6 '17 at 15:41
CoherentCoherent
1,147623
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $f(x) = dfrac{arctan x}{pi}$, so $f^{-1}(x) = tan pi x$. $f$ maps $mathbb{R}$ to $(-dfrac12,dfrac12)$.
$$g(x) = begin{cases}
x in mathbb{Q} & x + sqrt{5}\
x notin mathbb{Q} & x\
end{cases}$$
$$g^{-1}(x) = begin{cases}
x > 1 & x - sqrt{5}\
x le 1 & x\
end{cases}$$
So $g circ f$ maps from $mathbb{R}$ to a certain subset of the irrationals between $-dfrac12$ and $dfrac12 + sqrt{5}$, and $f^{-1} circ g^{-1}$ maps the inverse.
answered Jul 8 '14 at 17:39
NovaDenizenNovaDenizen
3,437718
$endgroup$
$begingroup$
$gof$ is not onto!
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 17:48
$begingroup$
You are correct, but it doesn't have to be onto. The question asked for a bijective map from $mathbb{R}$ to an irrational subset of the reals.
$endgroup$
– NovaDenizen
Jul 8 '14 at 17:50
$begingroup$
Yeah! In this case, it is not interesting! :)
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 18:04
add a comment |
$begingroup$
Let $f(x) = dfrac{arctan x}{pi}$, so $f^{-1}(x) = tan pi x$. $f$ maps $mathbb{R}$ to $(-dfrac12,dfrac12)$.
$$g(x) = begin{cases}
x in mathbb{Q} & x + sqrt{5}\
x notin mathbb{Q} & x\
end{cases}$$
$$g^{-1}(x) = begin{cases}
x > 1 & x - sqrt{5}\
x le 1 & x\
end{cases}$$
So $g circ f$ maps from $mathbb{R}$ to a certain subset of the irrationals between $-dfrac12$ and $dfrac12 + sqrt{5}$, and $f^{-1} circ g^{-1}$ maps the inverse.
answered Jul 8 '14 at 17:39
NovaDenizenNovaDenizen
3,437718
$endgroup$
$begingroup$
$gof$ is not onto!
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 17:48
$begingroup$
You are correct, but it doesn't have to be onto. The question asked for a bijective map from $mathbb{R}$ to an irrational subset of the reals.
$endgroup$
– NovaDenizen
Jul 8 '14 at 17:50
$begingroup$
Yeah! In this case, it is not interesting! :)
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 18:04
add a comment |
$begingroup$
Let $f(x) = dfrac{arctan x}{pi}$, so $f^{-1}(x) = tan pi x$. $f$ maps $mathbb{R}$ to $(-dfrac12,dfrac12)$.
$$g(x) = begin{cases}
x in mathbb{Q} & x + sqrt{5}\
x notin mathbb{Q} & x\
end{cases}$$
$$g^{-1}(x) = begin{cases}
x > 1 & x - sqrt{5}\
x le 1 & x\
end{cases}$$
So $g circ f$ maps from $mathbb{R}$ to a certain subset of the irrationals between $-dfrac12$ and $dfrac12 + sqrt{5}$, and $f^{-1} circ g^{-1}$ maps the inverse.
answered Jul 8 '14 at 17:39
NovaDenizenNovaDenizen
3,437718
$endgroup$
Let $f(x) = dfrac{arctan x}{pi}$, so $f^{-1}(x) = tan pi x$. $f$ maps $mathbb{R}$ to $(-dfrac12,dfrac12)$.
$$g(x) = begin{cases}
x in mathbb{Q} & x + sqrt{5}\
x notin mathbb{Q} & x\
end{cases}$$
$$g^{-1}(x) = begin{cases}
x > 1 & x - sqrt{5}\
x le 1 & x\
end{cases}$$
So $g circ f$ maps from $mathbb{R}$ to a certain subset of the irrationals between $-dfrac12$ and $dfrac12 + sqrt{5}$, and $f^{-1} circ g^{-1}$ maps the inverse.
answered Jul 8 '14 at 17:39
NovaDenizenNovaDenizen
3,437718
edited Jul 8 '14 at 17:46
answered Jul 8 '14 at 17:39
NovaDenizenNovaDenizen
3,437718
answered Jul 8 '14 at 17:39
NovaDenizenNovaDenizen
3,437718
answered Jul 8 '14 at 17:39
NovaDenizenNovaDenizen
3,437718
3,437718
$begingroup$
$gof$ is not onto!
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 17:48
$begingroup$
You are correct, but it doesn't have to be onto. The question asked for a bijective map from $mathbb{R}$ to an irrational subset of the reals.
$endgroup$
– NovaDenizen
Jul 8 '14 at 17:50
$begingroup$
Yeah! In this case, it is not interesting! :)
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 18:04
add a comment |
$begingroup$
$gof$ is not onto!
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 17:48
$begingroup$
You are correct, but it doesn't have to be onto. The question asked for a bijective map from $mathbb{R}$ to an irrational subset of the reals.
$endgroup$
– NovaDenizen
Jul 8 '14 at 17:50
$begingroup$
Yeah! In this case, it is not interesting! :)
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 18:04
$begingroup$
$gof$ is not onto!
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 17:48
$begingroup$
$gof$ is not onto!
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 17:48
$begingroup$
You are correct, but it doesn't have to be onto. The question asked for a bijective map from $mathbb{R}$ to an irrational subset of the reals.
$endgroup$
– NovaDenizen
Jul 8 '14 at 17:50
$begingroup$
You are correct, but it doesn't have to be onto. The question asked for a bijective map from $mathbb{R}$ to an irrational subset of the reals.
$endgroup$
– NovaDenizen
Jul 8 '14 at 17:50
$begingroup$
Yeah! In this case, it is not interesting! :)
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 18:04
$begingroup$
Yeah! In this case, it is not interesting! :)
$endgroup$
– Mohammad Khosravi
Jul 8 '14 at 18:04
add a comment |
$begingroup$
Let $A_i = pi^i {mathbb Q}$ and $ A =cup_{ige0} A_i$. Then the map $h:{mathbb R}backslash{mathbb Q}rightarrow {mathbb R}$
$$
h(x) = left{
begin{array}{ll}
pi x & xin A \
x & x in B={mathbb R}backslash A\
end{array}
right.
$$
is a bijection.
answered Jul 8 '14 at 17:41
Mohammad KhosraviMohammad Khosravi
2,014716
$endgroup$
add a comment |
$begingroup$
Let $A_i = pi^i {mathbb Q}$ and $ A =cup_{ige0} A_i$. Then the map $h:{mathbb R}backslash{mathbb Q}rightarrow {mathbb R}$
$$
h(x) = left{
begin{array}{ll}
pi x & xin A \
x & x in B={mathbb R}backslash A\
end{array}
right.
$$
is a bijection.
answered Jul 8 '14 at 17:41
Mohammad KhosraviMohammad Khosravi
2,014716
$endgroup$
add a comment |
$begingroup$
Let $A_i = pi^i {mathbb Q}$ and $ A =cup_{ige0} A_i$. Then the map $h:{mathbb R}backslash{mathbb Q}rightarrow {mathbb R}$
$$
h(x) = left{
begin{array}{ll}
pi x & xin A \
x & x in B={mathbb R}backslash A\
end{array}
right.
$$
is a bijection.
answered Jul 8 '14 at 17:41
Mohammad KhosraviMohammad Khosravi
2,014716
$endgroup$
Let $A_i = pi^i {mathbb Q}$ and $ A =cup_{ige0} A_i$. Then the map $h:{mathbb R}backslash{mathbb Q}rightarrow {mathbb R}$
$$
h(x) = left{
begin{array}{ll}
pi x & xin A \
x & x in B={mathbb R}backslash A\
end{array}
right.
$$
is a bijection.
answered Jul 8 '14 at 17:41
Mohammad KhosraviMohammad Khosravi
2,014716
edited Jul 8 '14 at 17:49
answered Jul 8 '14 at 17:41
Mohammad KhosraviMohammad Khosravi
2,014716
answered Jul 8 '14 at 17:41
Mohammad KhosraviMohammad Khosravi
2,014716
answered Jul 8 '14 at 17:41
Mohammad KhosraviMohammad Khosravi
2,014716
2,014716
add a comment |
add a comment |
$begingroup$
Enumerate $Bbb Q$ as $q_n$ for $ninBbb N$. Now define,
$$f(x)=begin{cases}e^{2n+1} & x=q_n\e^{2n} & x=e^n, ninBbb N\x&text{otherwise}end{cases}$$
Key point here is that $e$ is transcendental, so $e^nnotinBbb Q$ for any $n>0$. You can effectively replace $e$ by any other number with this property. For example $pi,sqrt2^sqrt2$, Liouville number, and so on and so forth.
answered Jul 8 '14 at 17:47
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
$begingroup$
@David: $e$ is transcendental. So yes.
$endgroup$
– Asaf Karagila♦
Jul 8 '14 at 18:00
add a comment |
$begingroup$
Enumerate $Bbb Q$ as $q_n$ for $ninBbb N$. Now define,
$$f(x)=begin{cases}e^{2n+1} & x=q_n\e^{2n} & x=e^n, ninBbb N\x&text{otherwise}end{cases}$$
Key point here is that $e$ is transcendental, so $e^nnotinBbb Q$ for any $n>0$. You can effectively replace $e$ by any other number with this property. For example $pi,sqrt2^sqrt2$, Liouville number, and so on and so forth.
answered Jul 8 '14 at 17:47
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
$begingroup$
@David: $e$ is transcendental. So yes.
$endgroup$
– Asaf Karagila♦
Jul 8 '14 at 18:00
add a comment |
$begingroup$
Enumerate $Bbb Q$ as $q_n$ for $ninBbb N$. Now define,
$$f(x)=begin{cases}e^{2n+1} & x=q_n\e^{2n} & x=e^n, ninBbb N\x&text{otherwise}end{cases}$$
Key point here is that $e$ is transcendental, so $e^nnotinBbb Q$ for any $n>0$. You can effectively replace $e$ by any other number with this property. For example $pi,sqrt2^sqrt2$, Liouville number, and so on and so forth.
answered Jul 8 '14 at 17:47
Asaf Karagila♦Asaf Karagila
306k33438769
$endgroup$
Enumerate $Bbb Q$ as $q_n$ for $ninBbb N$. Now define,
$$f(x)=begin{cases}e^{2n+1} & x=q_n\e^{2n} & x=e^n, ninBbb N\x&text{otherwise}end{cases}$$
Key point here is that $e$ is transcendental, so $e^nnotinBbb Q$ for any $n>0$. You can effectively replace $e$ by any other number with this property. For example $pi,sqrt2^sqrt2$, Liouville number, and so on and so forth.
answered Jul 8 '14 at 17:47
Asaf Karagila♦Asaf Karagila
306k33438769
edited Jul 8 '14 at 18:03
answered Jul 8 '14 at 17:47
Asaf Karagila♦Asaf Karagila
306k33438769
answered Jul 8 '14 at 17:47
Asaf Karagila♦Asaf Karagila
306k33438769
answered Jul 8 '14 at 17:47
Asaf Karagila♦Asaf Karagila
306k33438769
306k33438769
$begingroup$
@David: $e$ is transcendental. So yes.
$endgroup$
– Asaf Karagila♦
Jul 8 '14 at 18:00
add a comment |
$begingroup$
@David: $e$ is transcendental. So yes.
$endgroup$
– Asaf Karagila♦
Jul 8 '14 at 18:00
$begingroup$
@David: $e$ is transcendental. So yes.
$endgroup$
– Asaf Karagila♦
Jul 8 '14 at 18:00
$begingroup$
@David: $e$ is transcendental. So yes.
$endgroup$
– Asaf Karagila♦
Jul 8 '14 at 18:00
add a comment |
$begingroup$
Consider $A$ = { $pi + n : n in Bbb N$}.
Clearly $A$ is countable. ( Consider the bijection $n rightarrow pi + n$ , $forall n in Bbb N$).
So elements of $A$ can be expressed as, $A = {a_k : k in Bbb N}$.
And we also can have $Bbb Q = {b_k : k in Bbb N}$
Now consider the bijection
$phi : Bbb R rightarrow Bbb R - Bbb Q$ by,
$phi(x) := a_{2k+1}$ if $ x = b_k , x in Bbb Q$.
= $a_{2k}$ if $x = a_k , x in A$
= $x$ if $x in Bbb R - (A cup Bbb Q)$
The construction in my example can be seen as a very general bijection from a general uncountable set to a proper uncountable subset of it.
answered May 6 '17 at 15:41
CoherentCoherent
1,147623
$endgroup$
add a comment |
$begingroup$
Consider $A$ = { $pi + n : n in Bbb N$}.
Clearly $A$ is countable. ( Consider the bijection $n rightarrow pi + n$ , $forall n in Bbb N$).
So elements of $A$ can be expressed as, $A = {a_k : k in Bbb N}$.
And we also can have $Bbb Q = {b_k : k in Bbb N}$
Now consider the bijection
$phi : Bbb R rightarrow Bbb R - Bbb Q$ by,
$phi(x) := a_{2k+1}$ if $ x = b_k , x in Bbb Q$.
= $a_{2k}$ if $x = a_k , x in A$
= $x$ if $x in Bbb R - (A cup Bbb Q)$
The construction in my example can be seen as a very general bijection from a general uncountable set to a proper uncountable subset of it.
answered May 6 '17 at 15:41
CoherentCoherent
1,147623
$endgroup$
add a comment |
$begingroup$
Consider $A$ = { $pi + n : n in Bbb N$}.
Clearly $A$ is countable. ( Consider the bijection $n rightarrow pi + n$ , $forall n in Bbb N$).
So elements of $A$ can be expressed as, $A = {a_k : k in Bbb N}$.
And we also can have $Bbb Q = {b_k : k in Bbb N}$
Now consider the bijection
$phi : Bbb R rightarrow Bbb R - Bbb Q$ by,
$phi(x) := a_{2k+1}$ if $ x = b_k , x in Bbb Q$.
= $a_{2k}$ if $x = a_k , x in A$
= $x$ if $x in Bbb R - (A cup Bbb Q)$
The construction in my example can be seen as a very general bijection from a general uncountable set to a proper uncountable subset of it.
answered May 6 '17 at 15:41
CoherentCoherent
1,147623
$endgroup$
Consider $A$ = { $pi + n : n in Bbb N$}.
Clearly $A$ is countable. ( Consider the bijection $n rightarrow pi + n$ , $forall n in Bbb N$).
So elements of $A$ can be expressed as, $A = {a_k : k in Bbb N}$.
And we also can have $Bbb Q = {b_k : k in Bbb N}$
Now consider the bijection
$phi : Bbb R rightarrow Bbb R - Bbb Q$ by,
$phi(x) := a_{2k+1}$ if $ x = b_k , x in Bbb Q$.
= $a_{2k}$ if $x = a_k , x in A$
= $x$ if $x in Bbb R - (A cup Bbb Q)$
The construction in my example can be seen as a very general bijection from a general uncountable set to a proper uncountable subset of it.
answered May 6 '17 at 15:41
CoherentCoherent
1,147623
edited Jan 22 at 13:20
answered May 6 '17 at 15:41
CoherentCoherent
1,147623
answered May 6 '17 at 15:41
CoherentCoherent
1,147623
answered May 6 '17 at 15:41
CoherentCoherent
1,147623
1,147623
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$begingroup$
There are problems with all numbers of the form $-sqrt 2 +r$, with $rin mathbb Q$.
$endgroup$
– Git Gud
Jul 8 '14 at 17:18
$begingroup$
Worse than that -- your function maps $mathbb R$ onto itself
$endgroup$
– MPW
Jul 8 '14 at 17:20
9
$begingroup$
Map the rationals to the odd non-zero multiples of $sqrt2$. Map the non-zero multiples of $sqrt 2$ to the even multiples of $sqrt 2$. Take the identity elsewhere.
$endgroup$
– David Mitra
Jul 8 '14 at 17:23