bijection between $mathbb{Q}$ and $mathbb{N}$ that preserve the order.
$begingroup$
I know there is a bijection between $mathbb{Q}$ and $mathbb{N}$. But is there a bijection $mathbb{Q}xrightarrow{f}mathbb{N}$ that preserves the order? Intuitively I think this is not possible. What I would think of is enumerating the rationals in a form ${q_n}_{n in mathbb{N}}$ where $q_n < q_m$ if $n<m$ (I think this is not possible but why?) and then map to elements of $mathbb{N}$.
rational-numbers
asked Jan 24 at 16:32
roi_saumonroi_saumon
59438
$endgroup$
add a comment |
$begingroup$
I know there is a bijection between $mathbb{Q}$ and $mathbb{N}$. But is there a bijection $mathbb{Q}xrightarrow{f}mathbb{N}$ that preserves the order? Intuitively I think this is not possible. What I would think of is enumerating the rationals in a form ${q_n}_{n in mathbb{N}}$ where $q_n < q_m$ if $n<m$ (I think this is not possible but why?) and then map to elements of $mathbb{N}$.
rational-numbers
asked Jan 24 at 16:32
roi_saumonroi_saumon
59438
$endgroup$
$begingroup$
And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:36
1
$begingroup$
Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
$endgroup$
– Clive Newstead
Jan 24 at 16:37
$begingroup$
Compare also with this question here.
$endgroup$
– Dietrich Burde
Jan 24 at 16:38
$begingroup$
@CliveNewstead Thanks, that was actually easyer than I thought.
$endgroup$
– roi_saumon
Jan 24 at 16:46
add a comment |
$begingroup$
I know there is a bijection between $mathbb{Q}$ and $mathbb{N}$. But is there a bijection $mathbb{Q}xrightarrow{f}mathbb{N}$ that preserves the order? Intuitively I think this is not possible. What I would think of is enumerating the rationals in a form ${q_n}_{n in mathbb{N}}$ where $q_n < q_m$ if $n<m$ (I think this is not possible but why?) and then map to elements of $mathbb{N}$.
rational-numbers
asked Jan 24 at 16:32
roi_saumonroi_saumon
59438
$endgroup$
I know there is a bijection between $mathbb{Q}$ and $mathbb{N}$. But is there a bijection $mathbb{Q}xrightarrow{f}mathbb{N}$ that preserves the order? Intuitively I think this is not possible. What I would think of is enumerating the rationals in a form ${q_n}_{n in mathbb{N}}$ where $q_n < q_m$ if $n<m$ (I think this is not possible but why?) and then map to elements of $mathbb{N}$.
rational-numbers
rational-numbers
asked Jan 24 at 16:32
roi_saumonroi_saumon
59438
asked Jan 24 at 16:32
roi_saumonroi_saumon
59438
asked Jan 24 at 16:32
roi_saumonroi_saumon
59438
asked Jan 24 at 16:32
roi_saumonroi_saumon
59438
asked Jan 24 at 16:32
roi_saumonroi_saumon
59438
59438
$begingroup$
And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:36
1
$begingroup$
Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
$endgroup$
– Clive Newstead
Jan 24 at 16:37
$begingroup$
Compare also with this question here.
$endgroup$
– Dietrich Burde
Jan 24 at 16:38
$begingroup$
@CliveNewstead Thanks, that was actually easyer than I thought.
$endgroup$
– roi_saumon
Jan 24 at 16:46
add a comment |
$begingroup$
And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:36
1
$begingroup$
Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
$endgroup$
– Clive Newstead
Jan 24 at 16:37
$begingroup$
Compare also with this question here.
$endgroup$
– Dietrich Burde
Jan 24 at 16:38
$begingroup$
@CliveNewstead Thanks, that was actually easyer than I thought.
$endgroup$
– roi_saumon
Jan 24 at 16:46
$begingroup$
And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:36
$begingroup$
And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:36
1
1
$begingroup$
Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
$endgroup$
– Clive Newstead
Jan 24 at 16:37
$begingroup$
Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
$endgroup$
– Clive Newstead
Jan 24 at 16:37
$begingroup$
Compare also with this question here.
$endgroup$
– Dietrich Burde
Jan 24 at 16:38
$begingroup$
Compare also with this question here.
$endgroup$
– Dietrich Burde
Jan 24 at 16:38
$begingroup$
@CliveNewstead Thanks, that was actually easyer than I thought.
$endgroup$
– roi_saumon
Jan 24 at 16:46
$begingroup$
@CliveNewstead Thanks, that was actually easyer than I thought.
$endgroup$
– roi_saumon
Jan 24 at 16:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let $n in mathbb{N}$ and $q_n,q_{n+1} in mathbb{Q}$ where $q_n < q_{n+1}$.
Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $mathbb{Q}subset mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.
answered Jan 24 at 16:44
BeyBey
1464
$endgroup$
add a comment |
$begingroup$
The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.
If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $qneq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$
answered Jan 24 at 16:36
Kamil KiełczewskiKamil Kiełczewski
1267
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
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oldest
votes
$begingroup$
Let $n in mathbb{N}$ and $q_n,q_{n+1} in mathbb{Q}$ where $q_n < q_{n+1}$.
Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $mathbb{Q}subset mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.
answered Jan 24 at 16:44
BeyBey
1464
$endgroup$
add a comment |
$begingroup$
Let $n in mathbb{N}$ and $q_n,q_{n+1} in mathbb{Q}$ where $q_n < q_{n+1}$.
Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $mathbb{Q}subset mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.
answered Jan 24 at 16:44
BeyBey
1464
$endgroup$
add a comment |
$begingroup$
Let $n in mathbb{N}$ and $q_n,q_{n+1} in mathbb{Q}$ where $q_n < q_{n+1}$.
Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $mathbb{Q}subset mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.
answered Jan 24 at 16:44
BeyBey
1464
$endgroup$
Let $n in mathbb{N}$ and $q_n,q_{n+1} in mathbb{Q}$ where $q_n < q_{n+1}$.
Assume there does not exist another rational number $q_k$ where $q_n<q_k<q_{n+1}$ (i.e., we can order the rationals). Since $mathbb{Q}subset mathbb{R}$, this implies that the rationals are not dense in the reals, which is false. Therefore, our assumption is false.
answered Jan 24 at 16:44
BeyBey
1464
answered Jan 24 at 16:44
BeyBey
1464
answered Jan 24 at 16:44
BeyBey
1464
answered Jan 24 at 16:44
BeyBey
1464
1464
add a comment |
add a comment |
$begingroup$
The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.
If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $qneq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$
answered Jan 24 at 16:36
Kamil KiełczewskiKamil Kiełczewski
1267
$endgroup$
add a comment |
$begingroup$
The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.
If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $qneq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$
answered Jan 24 at 16:36
Kamil KiełczewskiKamil Kiełczewski
1267
$endgroup$
add a comment |
$begingroup$
The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.
If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $qneq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$
answered Jan 24 at 16:36
Kamil KiełczewskiKamil Kiełczewski
1267
$endgroup$
The reson is that if you have natural number n then you can find the NEXT number (=n+1), but if you have Rational number q then you cannot find NEXT number - because between any two different rational numbers is infinite number of other rational numbers.
If such bijection $f$ exists then when you find $q=f(n)$ then you shoud be able to find next rational number $p=f(n+1)$ but because $qneq p$ then between $q$ and $p$ exist infinite number of other rational numbers e.g. $r=(p+q)/2$
answered Jan 24 at 16:36
Kamil KiełczewskiKamil Kiełczewski
1267
edited Jan 24 at 16:49
answered Jan 24 at 16:36
Kamil KiełczewskiKamil Kiełczewski
1267
answered Jan 24 at 16:36
Kamil KiełczewskiKamil Kiełczewski
1267
answered Jan 24 at 16:36
Kamil KiełczewskiKamil Kiełczewski
1267
1267
add a comment |
add a comment |
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$begingroup$
And what of $q_0$ ( or $q_1$)? That is, there is a least natural number, but not a least rational.
$endgroup$
– астон вілла олоф мэллбэрг
Jan 24 at 16:36
1
$begingroup$
Your question is equivalent to this one, since defining an order-preserving bijection $mathbb{Q} to mathbb{N}$ is equivalent to listing the rational numbers in increasing order.
$endgroup$
– Clive Newstead
Jan 24 at 16:37
$begingroup$
Compare also with this question here.
$endgroup$
– Dietrich Burde
Jan 24 at 16:38
$begingroup$
@CliveNewstead Thanks, that was actually easyer than I thought.
$endgroup$
– roi_saumon
Jan 24 at 16:46