Help proving there is a sequence of rational numbers
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I'm trying to prove the following:
Let $Bbb Q$ be the countable set of rational numbers and ${x_n}_{n=1}^infty$ be a sequence such that for every q $in$ $Bbb Q$ there is a $n in Bbb N$ with $x_n = q$. Prove that there is such a sequence ${x_n}$.
My initial thoughts for an attempt at a solution:
We know that $|Bbb Q| = |Bbb N|$ by the Cantor Theorem. Further, I can show that there is a bijection between $Bbb Q$ and $Bbb N$.
If I define ${x_n}_{n=1}^infty$ as some relation between $(p,q) in Bbb Z times Bbb Z$, will ths properly prove that such a sequence exists?
real-analysis proof-writing rational-numbers natural-numbers
$endgroup$
add a comment |
$begingroup$
I'm trying to prove the following:
Let $Bbb Q$ be the countable set of rational numbers and ${x_n}_{n=1}^infty$ be a sequence such that for every q $in$ $Bbb Q$ there is a $n in Bbb N$ with $x_n = q$. Prove that there is such a sequence ${x_n}$.
My initial thoughts for an attempt at a solution:
We know that $|Bbb Q| = |Bbb N|$ by the Cantor Theorem. Further, I can show that there is a bijection between $Bbb Q$ and $Bbb N$.
If I define ${x_n}_{n=1}^infty$ as some relation between $(p,q) in Bbb Z times Bbb Z$, will ths properly prove that such a sequence exists?
real-analysis proof-writing rational-numbers natural-numbers
$endgroup$
$begingroup$
For an explicit bijection, see -math.stackexchange.com/questions/7643/…
$endgroup$
– Thomas Shelby
Dec 9 '18 at 17:15
1
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Thanks Thomas! This is wonderful to see!
$endgroup$
– user624612
Dec 9 '18 at 17:20
$begingroup$
What is the definition of an infinite sequence ${x_n}_{nin Bbb N} $? It's a function $ f$ with domain $Bbb N,$ except we write $x_n$ for $f(n)$..... And $|Bbb Q|=|Bbb N$| is an abbreviation for, or an equivalent to, the statement that there exists a bijection $f:Bbb Nto Bbb Q.$ And that's all there is to it.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:19
$begingroup$
You may enjoy the small book Stories About Sets, by Vilenkin.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:22
add a comment |
$begingroup$
I'm trying to prove the following:
Let $Bbb Q$ be the countable set of rational numbers and ${x_n}_{n=1}^infty$ be a sequence such that for every q $in$ $Bbb Q$ there is a $n in Bbb N$ with $x_n = q$. Prove that there is such a sequence ${x_n}$.
My initial thoughts for an attempt at a solution:
We know that $|Bbb Q| = |Bbb N|$ by the Cantor Theorem. Further, I can show that there is a bijection between $Bbb Q$ and $Bbb N$.
If I define ${x_n}_{n=1}^infty$ as some relation between $(p,q) in Bbb Z times Bbb Z$, will ths properly prove that such a sequence exists?
real-analysis proof-writing rational-numbers natural-numbers
$endgroup$
I'm trying to prove the following:
Let $Bbb Q$ be the countable set of rational numbers and ${x_n}_{n=1}^infty$ be a sequence such that for every q $in$ $Bbb Q$ there is a $n in Bbb N$ with $x_n = q$. Prove that there is such a sequence ${x_n}$.
My initial thoughts for an attempt at a solution:
We know that $|Bbb Q| = |Bbb N|$ by the Cantor Theorem. Further, I can show that there is a bijection between $Bbb Q$ and $Bbb N$.
If I define ${x_n}_{n=1}^infty$ as some relation between $(p,q) in Bbb Z times Bbb Z$, will ths properly prove that such a sequence exists?
real-analysis proof-writing rational-numbers natural-numbers
real-analysis proof-writing rational-numbers natural-numbers
asked Dec 9 '18 at 17:01
user624612
$begingroup$
For an explicit bijection, see -math.stackexchange.com/questions/7643/…
$endgroup$
– Thomas Shelby
Dec 9 '18 at 17:15
1
$begingroup$
Thanks Thomas! This is wonderful to see!
$endgroup$
– user624612
Dec 9 '18 at 17:20
$begingroup$
What is the definition of an infinite sequence ${x_n}_{nin Bbb N} $? It's a function $ f$ with domain $Bbb N,$ except we write $x_n$ for $f(n)$..... And $|Bbb Q|=|Bbb N$| is an abbreviation for, or an equivalent to, the statement that there exists a bijection $f:Bbb Nto Bbb Q.$ And that's all there is to it.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:19
$begingroup$
You may enjoy the small book Stories About Sets, by Vilenkin.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:22
add a comment |
$begingroup$
For an explicit bijection, see -math.stackexchange.com/questions/7643/…
$endgroup$
– Thomas Shelby
Dec 9 '18 at 17:15
1
$begingroup$
Thanks Thomas! This is wonderful to see!
$endgroup$
– user624612
Dec 9 '18 at 17:20
$begingroup$
What is the definition of an infinite sequence ${x_n}_{nin Bbb N} $? It's a function $ f$ with domain $Bbb N,$ except we write $x_n$ for $f(n)$..... And $|Bbb Q|=|Bbb N$| is an abbreviation for, or an equivalent to, the statement that there exists a bijection $f:Bbb Nto Bbb Q.$ And that's all there is to it.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:19
$begingroup$
You may enjoy the small book Stories About Sets, by Vilenkin.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:22
$begingroup$
For an explicit bijection, see -math.stackexchange.com/questions/7643/…
$endgroup$
– Thomas Shelby
Dec 9 '18 at 17:15
$begingroup$
For an explicit bijection, see -math.stackexchange.com/questions/7643/…
$endgroup$
– Thomas Shelby
Dec 9 '18 at 17:15
1
1
$begingroup$
Thanks Thomas! This is wonderful to see!
$endgroup$
– user624612
Dec 9 '18 at 17:20
$begingroup$
Thanks Thomas! This is wonderful to see!
$endgroup$
– user624612
Dec 9 '18 at 17:20
$begingroup$
What is the definition of an infinite sequence ${x_n}_{nin Bbb N} $? It's a function $ f$ with domain $Bbb N,$ except we write $x_n$ for $f(n)$..... And $|Bbb Q|=|Bbb N$| is an abbreviation for, or an equivalent to, the statement that there exists a bijection $f:Bbb Nto Bbb Q.$ And that's all there is to it.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:19
$begingroup$
What is the definition of an infinite sequence ${x_n}_{nin Bbb N} $? It's a function $ f$ with domain $Bbb N,$ except we write $x_n$ for $f(n)$..... And $|Bbb Q|=|Bbb N$| is an abbreviation for, or an equivalent to, the statement that there exists a bijection $f:Bbb Nto Bbb Q.$ And that's all there is to it.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:19
$begingroup$
You may enjoy the small book Stories About Sets, by Vilenkin.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:22
$begingroup$
You may enjoy the small book Stories About Sets, by Vilenkin.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:22
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It is actually quit easy once you know that $|mathbb{Q}|=|mathbb{N}|$ as this implies that there exists a bijection $f:mathbb{N}rightarrow mathbb{Q}$. Now simply let $x_n = f(n)$.
answered Dec 9 '18 at 17:11
YankoYanko
7,0081629
$endgroup$
$begingroup$
Thank you! This makes sense! I didn't know if much more was needed other than defining the bijective mapping, but it seems that this is about it.
$endgroup$
– user624612
Dec 9 '18 at 17:23
add a comment |
$begingroup$
The answer to this question is going to depend upon what you have available as definitions. Where I come from, a sequence of elements from a set $S$ is just a function $f:mathbb{N}rightarrow S$; if that is the case for you, since you already know there is a bijection from $mathbb{N}$ to $mathbb{Q}$ you are done!
If you wanted to explicitly write down such a sequence, then yes, you will probably want to start with a surjection from $mathbb{N}$ to $mathbb{Z}times mathbb{Z}$, and then a surjection from $mathbb{Z}times mathbb{Z}$ to $mathbb{Q}$. The composition of these functions will be a sequence that hits every rational number.
How to build such surjections? You could define $f:mathbb{N}rightarrow mathbb{Z}times mathbb{Z}$ in pieces, say by sending numbers of the form $2^i3^j$ to $(i,j)$, numbers of the form $5^i7^j$ to $(-i,-j)$, numbers of the form $11^i13^j$ to $(i,-j)$, and numbers of the form $17^i19^j$ to $(-i,j)$, with everything else sent to $(0,0)$. You could define $g:mathbb{Z}times mathbb{Z} rightarrow mathbb{Q}$ by sending $(0,q)$ and $(p,0)$ to $0$, and all other $(p,q)$ to $p/q$.
answered Dec 9 '18 at 17:12
GenericMathematicianGenericMathematician
863
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is actually quit easy once you know that $|mathbb{Q}|=|mathbb{N}|$ as this implies that there exists a bijection $f:mathbb{N}rightarrow mathbb{Q}$. Now simply let $x_n = f(n)$.
answered Dec 9 '18 at 17:11
YankoYanko
7,0081629
$endgroup$
$begingroup$
Thank you! This makes sense! I didn't know if much more was needed other than defining the bijective mapping, but it seems that this is about it.
$endgroup$
– user624612
Dec 9 '18 at 17:23
add a comment |
$begingroup$
It is actually quit easy once you know that $|mathbb{Q}|=|mathbb{N}|$ as this implies that there exists a bijection $f:mathbb{N}rightarrow mathbb{Q}$. Now simply let $x_n = f(n)$.
answered Dec 9 '18 at 17:11
YankoYanko
7,0081629
$endgroup$
$begingroup$
Thank you! This makes sense! I didn't know if much more was needed other than defining the bijective mapping, but it seems that this is about it.
$endgroup$
– user624612
Dec 9 '18 at 17:23
add a comment |
$begingroup$
It is actually quit easy once you know that $|mathbb{Q}|=|mathbb{N}|$ as this implies that there exists a bijection $f:mathbb{N}rightarrow mathbb{Q}$. Now simply let $x_n = f(n)$.
answered Dec 9 '18 at 17:11
YankoYanko
7,0081629
$endgroup$
It is actually quit easy once you know that $|mathbb{Q}|=|mathbb{N}|$ as this implies that there exists a bijection $f:mathbb{N}rightarrow mathbb{Q}$. Now simply let $x_n = f(n)$.
answered Dec 9 '18 at 17:11
YankoYanko
7,0081629
answered Dec 9 '18 at 17:11
YankoYanko
7,0081629
answered Dec 9 '18 at 17:11
YankoYanko
7,0081629
answered Dec 9 '18 at 17:11
YankoYanko
7,0081629
7,0081629
$begingroup$
Thank you! This makes sense! I didn't know if much more was needed other than defining the bijective mapping, but it seems that this is about it.
$endgroup$
– user624612
Dec 9 '18 at 17:23
add a comment |
$begingroup$
Thank you! This makes sense! I didn't know if much more was needed other than defining the bijective mapping, but it seems that this is about it.
$endgroup$
– user624612
Dec 9 '18 at 17:23
$begingroup$
Thank you! This makes sense! I didn't know if much more was needed other than defining the bijective mapping, but it seems that this is about it.
$endgroup$
– user624612
Dec 9 '18 at 17:23
$begingroup$
Thank you! This makes sense! I didn't know if much more was needed other than defining the bijective mapping, but it seems that this is about it.
$endgroup$
– user624612
Dec 9 '18 at 17:23
add a comment |
$begingroup$
The answer to this question is going to depend upon what you have available as definitions. Where I come from, a sequence of elements from a set $S$ is just a function $f:mathbb{N}rightarrow S$; if that is the case for you, since you already know there is a bijection from $mathbb{N}$ to $mathbb{Q}$ you are done!
If you wanted to explicitly write down such a sequence, then yes, you will probably want to start with a surjection from $mathbb{N}$ to $mathbb{Z}times mathbb{Z}$, and then a surjection from $mathbb{Z}times mathbb{Z}$ to $mathbb{Q}$. The composition of these functions will be a sequence that hits every rational number.
How to build such surjections? You could define $f:mathbb{N}rightarrow mathbb{Z}times mathbb{Z}$ in pieces, say by sending numbers of the form $2^i3^j$ to $(i,j)$, numbers of the form $5^i7^j$ to $(-i,-j)$, numbers of the form $11^i13^j$ to $(i,-j)$, and numbers of the form $17^i19^j$ to $(-i,j)$, with everything else sent to $(0,0)$. You could define $g:mathbb{Z}times mathbb{Z} rightarrow mathbb{Q}$ by sending $(0,q)$ and $(p,0)$ to $0$, and all other $(p,q)$ to $p/q$.
answered Dec 9 '18 at 17:12
GenericMathematicianGenericMathematician
863
$endgroup$
add a comment |
$begingroup$
The answer to this question is going to depend upon what you have available as definitions. Where I come from, a sequence of elements from a set $S$ is just a function $f:mathbb{N}rightarrow S$; if that is the case for you, since you already know there is a bijection from $mathbb{N}$ to $mathbb{Q}$ you are done!
If you wanted to explicitly write down such a sequence, then yes, you will probably want to start with a surjection from $mathbb{N}$ to $mathbb{Z}times mathbb{Z}$, and then a surjection from $mathbb{Z}times mathbb{Z}$ to $mathbb{Q}$. The composition of these functions will be a sequence that hits every rational number.
How to build such surjections? You could define $f:mathbb{N}rightarrow mathbb{Z}times mathbb{Z}$ in pieces, say by sending numbers of the form $2^i3^j$ to $(i,j)$, numbers of the form $5^i7^j$ to $(-i,-j)$, numbers of the form $11^i13^j$ to $(i,-j)$, and numbers of the form $17^i19^j$ to $(-i,j)$, with everything else sent to $(0,0)$. You could define $g:mathbb{Z}times mathbb{Z} rightarrow mathbb{Q}$ by sending $(0,q)$ and $(p,0)$ to $0$, and all other $(p,q)$ to $p/q$.
answered Dec 9 '18 at 17:12
GenericMathematicianGenericMathematician
863
$endgroup$
add a comment |
$begingroup$
The answer to this question is going to depend upon what you have available as definitions. Where I come from, a sequence of elements from a set $S$ is just a function $f:mathbb{N}rightarrow S$; if that is the case for you, since you already know there is a bijection from $mathbb{N}$ to $mathbb{Q}$ you are done!
If you wanted to explicitly write down such a sequence, then yes, you will probably want to start with a surjection from $mathbb{N}$ to $mathbb{Z}times mathbb{Z}$, and then a surjection from $mathbb{Z}times mathbb{Z}$ to $mathbb{Q}$. The composition of these functions will be a sequence that hits every rational number.
How to build such surjections? You could define $f:mathbb{N}rightarrow mathbb{Z}times mathbb{Z}$ in pieces, say by sending numbers of the form $2^i3^j$ to $(i,j)$, numbers of the form $5^i7^j$ to $(-i,-j)$, numbers of the form $11^i13^j$ to $(i,-j)$, and numbers of the form $17^i19^j$ to $(-i,j)$, with everything else sent to $(0,0)$. You could define $g:mathbb{Z}times mathbb{Z} rightarrow mathbb{Q}$ by sending $(0,q)$ and $(p,0)$ to $0$, and all other $(p,q)$ to $p/q$.
answered Dec 9 '18 at 17:12
GenericMathematicianGenericMathematician
863
$endgroup$
The answer to this question is going to depend upon what you have available as definitions. Where I come from, a sequence of elements from a set $S$ is just a function $f:mathbb{N}rightarrow S$; if that is the case for you, since you already know there is a bijection from $mathbb{N}$ to $mathbb{Q}$ you are done!
If you wanted to explicitly write down such a sequence, then yes, you will probably want to start with a surjection from $mathbb{N}$ to $mathbb{Z}times mathbb{Z}$, and then a surjection from $mathbb{Z}times mathbb{Z}$ to $mathbb{Q}$. The composition of these functions will be a sequence that hits every rational number.
How to build such surjections? You could define $f:mathbb{N}rightarrow mathbb{Z}times mathbb{Z}$ in pieces, say by sending numbers of the form $2^i3^j$ to $(i,j)$, numbers of the form $5^i7^j$ to $(-i,-j)$, numbers of the form $11^i13^j$ to $(i,-j)$, and numbers of the form $17^i19^j$ to $(-i,j)$, with everything else sent to $(0,0)$. You could define $g:mathbb{Z}times mathbb{Z} rightarrow mathbb{Q}$ by sending $(0,q)$ and $(p,0)$ to $0$, and all other $(p,q)$ to $p/q$.
answered Dec 9 '18 at 17:12
GenericMathematicianGenericMathematician
863
edited Dec 9 '18 at 17:19
answered Dec 9 '18 at 17:12
GenericMathematicianGenericMathematician
863
answered Dec 9 '18 at 17:12
GenericMathematicianGenericMathematician
863
answered Dec 9 '18 at 17:12
GenericMathematicianGenericMathematician
863
863
add a comment |
add a comment |
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$begingroup$
For an explicit bijection, see -math.stackexchange.com/questions/7643/…
$endgroup$
– Thomas Shelby
Dec 9 '18 at 17:15
1
$begingroup$
Thanks Thomas! This is wonderful to see!
$endgroup$
– user624612
Dec 9 '18 at 17:20
$begingroup$
What is the definition of an infinite sequence ${x_n}_{nin Bbb N} $? It's a function $ f$ with domain $Bbb N,$ except we write $x_n$ for $f(n)$..... And $|Bbb Q|=|Bbb N$| is an abbreviation for, or an equivalent to, the statement that there exists a bijection $f:Bbb Nto Bbb Q.$ And that's all there is to it.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:19
$begingroup$
You may enjoy the small book Stories About Sets, by Vilenkin.
$endgroup$
– DanielWainfleet
Dec 9 '18 at 20:22