Probability of a number being rationalProof that the irrational numbers are uncountableProbability and measure theoryHelp with Riemann integrationExample of $f,g: [0,1]to[0,1]$ and Riemann-integrable, but $gcirc f$ is not?Riemann Integrals and Periodic FunctionsLebesgue-integrability of piecewise function with random variableHow to show the following function is Riemann IntegrableExamine if a piecewise-defined function is Riemann integrableProving piecewise function is not Riemann integrableWhy this function is not Riemann integrable on $[0, 1]$Computation of Riemann integral
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Probability of a number being rational
Proof that the irrational numbers are uncountableProbability and measure theoryHelp with Riemann integrationExample of $f,g: [0,1]to[0,1]$ and Riemann-integrable, but $gcirc f$ is not?Riemann Integrals and Periodic FunctionsLebesgue-integrability of piecewise function with random variableHow to show the following function is Riemann IntegrableExamine if a piecewise-defined function is Riemann integrableProving piecewise function is not Riemann integrableWhy this function is not Riemann integrable on $[0, 1]$Computation of Riemann integral
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$begingroup$
If $x in [0, 1]$, what is $textP(xin mathbb Q)$?
In other words, what is the probability that $x$ is rational?
This is what I tried:
$$beginarrayrcltextP(x in mathbb Q) &=& displaystyle int^1_0 f(x),dx
endarray$$
where
$$f(x) = begincases1, & x in mathbb Q\0, & xnotinmathbb Qendcases$$
However, the function is not Riemann-integrable.
I want to try comparing the cardinalities of the rational set and the irrational set by using a one-to-one mapping between the two sets. But I don't have idea how I can do it. Can anyone give a hint?
probability integration elementary-set-theory rational-numbers
edited Oct 13 at 13:12
Rodrigo de Azevedo
13.8k4 gold badges23 silver badges68 bronze badges
asked Oct 12 at 7:27
Max WongMax Wong
35811 bronze badges
$endgroup$
add a comment
|
$begingroup$
If $x in [0, 1]$, what is $textP(xin mathbb Q)$?
In other words, what is the probability that $x$ is rational?
This is what I tried:
$$beginarrayrcltextP(x in mathbb Q) &=& displaystyle int^1_0 f(x),dx
endarray$$
where
$$f(x) = begincases1, & x in mathbb Q\0, & xnotinmathbb Qendcases$$
However, the function is not Riemann-integrable.
I want to try comparing the cardinalities of the rational set and the irrational set by using a one-to-one mapping between the two sets. But I don't have idea how I can do it. Can anyone give a hint?
probability integration elementary-set-theory rational-numbers
edited Oct 13 at 13:12
Rodrigo de Azevedo
13.8k4 gold badges23 silver badges68 bronze badges
asked Oct 12 at 7:27
Max WongMax Wong
35811 bronze badges
$endgroup$
add a comment
|
$begingroup$
If $x in [0, 1]$, what is $textP(xin mathbb Q)$?
In other words, what is the probability that $x$ is rational?
This is what I tried:
$$beginarrayrcltextP(x in mathbb Q) &=& displaystyle int^1_0 f(x),dx
endarray$$
where
$$f(x) = begincases1, & x in mathbb Q\0, & xnotinmathbb Qendcases$$
However, the function is not Riemann-integrable.
I want to try comparing the cardinalities of the rational set and the irrational set by using a one-to-one mapping between the two sets. But I don't have idea how I can do it. Can anyone give a hint?
probability integration elementary-set-theory rational-numbers
edited Oct 13 at 13:12
Rodrigo de Azevedo
13.8k4 gold badges23 silver badges68 bronze badges
asked Oct 12 at 7:27
Max WongMax Wong
35811 bronze badges
$endgroup$
If $x in [0, 1]$, what is $textP(xin mathbb Q)$?
In other words, what is the probability that $x$ is rational?
This is what I tried:
$$beginarrayrcltextP(x in mathbb Q) &=& displaystyle int^1_0 f(x),dx
endarray$$
where
$$f(x) = begincases1, & x in mathbb Q\0, & xnotinmathbb Qendcases$$
However, the function is not Riemann-integrable.
I want to try comparing the cardinalities of the rational set and the irrational set by using a one-to-one mapping between the two sets. But I don't have idea how I can do it. Can anyone give a hint?
probability integration elementary-set-theory rational-numbers
probability integration elementary-set-theory rational-numbers
edited Oct 13 at 13:12
Rodrigo de Azevedo
13.8k4 gold badges23 silver badges68 bronze badges
asked Oct 12 at 7:27
Max WongMax Wong
35811 bronze badges
edited Oct 13 at 13:12
Rodrigo de Azevedo
13.8k4 gold badges23 silver badges68 bronze badges
asked Oct 12 at 7:27
Max WongMax Wong
35811 bronze badges
edited Oct 13 at 13:12
Rodrigo de Azevedo
13.8k4 gold badges23 silver badges68 bronze badges
edited Oct 13 at 13:12
Rodrigo de Azevedo
13.8k4 gold badges23 silver badges68 bronze badges
edited Oct 13 at 13:12
Rodrigo de Azevedo
13.8k4 gold badges23 silver badges68 bronze badges
13.8k4 gold badges23 silver badges68 bronze badges
asked Oct 12 at 7:27
Max WongMax Wong
35811 bronze badges
asked Oct 12 at 7:27
Max WongMax Wong
35811 bronze badges
asked Oct 12 at 7:27
Max WongMax Wong
35811 bronze badges
35811 bronze badges
add a comment
|
add a comment
|
3 Answers
3
active
oldest
votes
$begingroup$
The function $f$ is indeed not Riemann-integrable, but it is Lebesgue-integrable. And its integral is $0$. Therefore, the answer is $0$. That is natural, since $mathbb Qcap[0,1]$ is countable, whereas $[0,1]$ is uncountable. It follows that there is no bijection between $mathbb Qcap[0,1]$ and $[0,1]setminusmathbb Q$.
edited Oct 12 at 7:33
Thomas Shelby
6,8193 gold badges12 silver badges33 bronze badges
answered Oct 12 at 7:31
José Carlos SantosJosé Carlos Santos
222k27 gold badges172 silver badges298 bronze badges
$endgroup$
$begingroup$
What is the relation between a probability function and integration?
$endgroup$
– Cantor
Oct 12 at 17:23
4
$begingroup$
Probability theory is very much connected to measure theory. An we can compute the measure of a set by computing the integral of its characteristic function.
$endgroup$
– José Carlos Santos
Oct 12 at 17:29
$begingroup$
Thank you very much Dr. Santos. I'll inquire further on the subject. Can you recommend an article or book on the matter?
$endgroup$
– Cantor
Oct 12 at 17:31
2
$begingroup$
I think that this question and its answers are a good starting point.
$endgroup$
– José Carlos Santos
Oct 12 at 17:35
add a comment
|
$begingroup$
This answer might miss the real goal, but is inspired by the tag "probability".
In your question it is not explicitly mentioned how probability $P$ is defined.
So actually the question: "what is the probability that $x$ is rational?" makes no sense in this context.
Your try indicates that you are thinking of uniform distribution on interval $[0,1]$ where: $$P(B):=lambda(Bcap[0,1])$$ for every Borel set and $lambda$ denotes the Lebesgue measure on the $sigma$-algebra of Borel sets $Bsubseteqmathbb R$.
It is well known that $lambda(x)=0$ for any $xinmathbb R$ and consequently we have $lambda(S)=sum_xin Slambda(x)=0$ for every countable set, and of course $mathbb Q$ is countable.
So under uniform distribution on $[0,1]$ we have $P(mathbb Q)=0$.
In probability theory Riemann-integrability is (as far as I know) not practicized.
Notation $P(xin mathbb Q)$ only makes sense if $x$ denotes a random variable here.
So your problem setting: "if $xin[0,1]$ then what is $P(xinmathbb Q)$?" is not okay.
answered Oct 12 at 7:57
drhabdrhab
116k6 gold badges51 silver badges146 bronze badges
$endgroup$
add a comment
|
$begingroup$
I want to try comparing the cardinalities of the rational set and the irrational set by using a one-to-one mapping between the two sets
You can't. The set of rationals is countable. The set of irrationals is uncountable (see Proof that the irrational numbers are uncountable). Cantor proved by diagonalization that such that mapping cannot exist.
answered Oct 13 at 3:00
Laurence R. UgaldeLaurence R. Ugalde
1165 bronze badges
$endgroup$
add a comment
|
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The function $f$ is indeed not Riemann-integrable, but it is Lebesgue-integrable. And its integral is $0$. Therefore, the answer is $0$. That is natural, since $mathbb Qcap[0,1]$ is countable, whereas $[0,1]$ is uncountable. It follows that there is no bijection between $mathbb Qcap[0,1]$ and $[0,1]setminusmathbb Q$.
edited Oct 12 at 7:33
Thomas Shelby
6,8193 gold badges12 silver badges33 bronze badges
answered Oct 12 at 7:31
José Carlos SantosJosé Carlos Santos
222k27 gold badges172 silver badges298 bronze badges
$endgroup$
$begingroup$
What is the relation between a probability function and integration?
$endgroup$
– Cantor
Oct 12 at 17:23
4
$begingroup$
Probability theory is very much connected to measure theory. An we can compute the measure of a set by computing the integral of its characteristic function.
$endgroup$
– José Carlos Santos
Oct 12 at 17:29
$begingroup$
Thank you very much Dr. Santos. I'll inquire further on the subject. Can you recommend an article or book on the matter?
$endgroup$
– Cantor
Oct 12 at 17:31
2
$begingroup$
I think that this question and its answers are a good starting point.
$endgroup$
– José Carlos Santos
Oct 12 at 17:35
add a comment
|
$begingroup$
The function $f$ is indeed not Riemann-integrable, but it is Lebesgue-integrable. And its integral is $0$. Therefore, the answer is $0$. That is natural, since $mathbb Qcap[0,1]$ is countable, whereas $[0,1]$ is uncountable. It follows that there is no bijection between $mathbb Qcap[0,1]$ and $[0,1]setminusmathbb Q$.
edited Oct 12 at 7:33
Thomas Shelby
6,8193 gold badges12 silver badges33 bronze badges
answered Oct 12 at 7:31
José Carlos SantosJosé Carlos Santos
222k27 gold badges172 silver badges298 bronze badges
$endgroup$
$begingroup$
What is the relation between a probability function and integration?
$endgroup$
– Cantor
Oct 12 at 17:23
4
$begingroup$
Probability theory is very much connected to measure theory. An we can compute the measure of a set by computing the integral of its characteristic function.
$endgroup$
– José Carlos Santos
Oct 12 at 17:29
$begingroup$
Thank you very much Dr. Santos. I'll inquire further on the subject. Can you recommend an article or book on the matter?
$endgroup$
– Cantor
Oct 12 at 17:31
2
$begingroup$
I think that this question and its answers are a good starting point.
$endgroup$
– José Carlos Santos
Oct 12 at 17:35
add a comment
|
$begingroup$
The function $f$ is indeed not Riemann-integrable, but it is Lebesgue-integrable. And its integral is $0$. Therefore, the answer is $0$. That is natural, since $mathbb Qcap[0,1]$ is countable, whereas $[0,1]$ is uncountable. It follows that there is no bijection between $mathbb Qcap[0,1]$ and $[0,1]setminusmathbb Q$.
edited Oct 12 at 7:33
Thomas Shelby
6,8193 gold badges12 silver badges33 bronze badges
answered Oct 12 at 7:31
José Carlos SantosJosé Carlos Santos
222k27 gold badges172 silver badges298 bronze badges
$endgroup$
The function $f$ is indeed not Riemann-integrable, but it is Lebesgue-integrable. And its integral is $0$. Therefore, the answer is $0$. That is natural, since $mathbb Qcap[0,1]$ is countable, whereas $[0,1]$ is uncountable. It follows that there is no bijection between $mathbb Qcap[0,1]$ and $[0,1]setminusmathbb Q$.
edited Oct 12 at 7:33
Thomas Shelby
6,8193 gold badges12 silver badges33 bronze badges
answered Oct 12 at 7:31
José Carlos SantosJosé Carlos Santos
222k27 gold badges172 silver badges298 bronze badges
edited Oct 12 at 7:33
Thomas Shelby
6,8193 gold badges12 silver badges33 bronze badges
edited Oct 12 at 7:33
Thomas Shelby
6,8193 gold badges12 silver badges33 bronze badges
edited Oct 12 at 7:33
Thomas Shelby
6,8193 gold badges12 silver badges33 bronze badges
6,8193 gold badges12 silver badges33 bronze badges
answered Oct 12 at 7:31
José Carlos SantosJosé Carlos Santos
222k27 gold badges172 silver badges298 bronze badges
answered Oct 12 at 7:31
José Carlos SantosJosé Carlos Santos
222k27 gold badges172 silver badges298 bronze badges
answered Oct 12 at 7:31
José Carlos SantosJosé Carlos Santos
222k27 gold badges172 silver badges298 bronze badges
222k27 gold badges172 silver badges298 bronze badges
$begingroup$
What is the relation between a probability function and integration?
$endgroup$
– Cantor
Oct 12 at 17:23
4
$begingroup$
Probability theory is very much connected to measure theory. An we can compute the measure of a set by computing the integral of its characteristic function.
$endgroup$
– José Carlos Santos
Oct 12 at 17:29
$begingroup$
Thank you very much Dr. Santos. I'll inquire further on the subject. Can you recommend an article or book on the matter?
$endgroup$
– Cantor
Oct 12 at 17:31
2
$begingroup$
I think that this question and its answers are a good starting point.
$endgroup$
– José Carlos Santos
Oct 12 at 17:35
add a comment
|
$begingroup$
What is the relation between a probability function and integration?
$endgroup$
– Cantor
Oct 12 at 17:23
4
$begingroup$
Probability theory is very much connected to measure theory. An we can compute the measure of a set by computing the integral of its characteristic function.
$endgroup$
– José Carlos Santos
Oct 12 at 17:29
$begingroup$
Thank you very much Dr. Santos. I'll inquire further on the subject. Can you recommend an article or book on the matter?
$endgroup$
– Cantor
Oct 12 at 17:31
2
$begingroup$
I think that this question and its answers are a good starting point.
$endgroup$
– José Carlos Santos
Oct 12 at 17:35
$begingroup$
What is the relation between a probability function and integration?
$endgroup$
– Cantor
Oct 12 at 17:23
$begingroup$
What is the relation between a probability function and integration?
$endgroup$
– Cantor
Oct 12 at 17:23
4
4
$begingroup$
Probability theory is very much connected to measure theory. An we can compute the measure of a set by computing the integral of its characteristic function.
$endgroup$
– José Carlos Santos
Oct 12 at 17:29
$begingroup$
Probability theory is very much connected to measure theory. An we can compute the measure of a set by computing the integral of its characteristic function.
$endgroup$
– José Carlos Santos
Oct 12 at 17:29
$begingroup$
Thank you very much Dr. Santos. I'll inquire further on the subject. Can you recommend an article or book on the matter?
$endgroup$
– Cantor
Oct 12 at 17:31
$begingroup$
Thank you very much Dr. Santos. I'll inquire further on the subject. Can you recommend an article or book on the matter?
$endgroup$
– Cantor
Oct 12 at 17:31
2
2
$begingroup$
I think that this question and its answers are a good starting point.
$endgroup$
– José Carlos Santos
Oct 12 at 17:35
$begingroup$
I think that this question and its answers are a good starting point.
$endgroup$
– José Carlos Santos
Oct 12 at 17:35
add a comment
|
$begingroup$
This answer might miss the real goal, but is inspired by the tag "probability".
In your question it is not explicitly mentioned how probability $P$ is defined.
So actually the question: "what is the probability that $x$ is rational?" makes no sense in this context.
Your try indicates that you are thinking of uniform distribution on interval $[0,1]$ where: $$P(B):=lambda(Bcap[0,1])$$ for every Borel set and $lambda$ denotes the Lebesgue measure on the $sigma$-algebra of Borel sets $Bsubseteqmathbb R$.
It is well known that $lambda(x)=0$ for any $xinmathbb R$ and consequently we have $lambda(S)=sum_xin Slambda(x)=0$ for every countable set, and of course $mathbb Q$ is countable.
So under uniform distribution on $[0,1]$ we have $P(mathbb Q)=0$.
In probability theory Riemann-integrability is (as far as I know) not practicized.
Notation $P(xin mathbb Q)$ only makes sense if $x$ denotes a random variable here.
So your problem setting: "if $xin[0,1]$ then what is $P(xinmathbb Q)$?" is not okay.
answered Oct 12 at 7:57
drhabdrhab
116k6 gold badges51 silver badges146 bronze badges
$endgroup$
add a comment
|
$begingroup$
This answer might miss the real goal, but is inspired by the tag "probability".
In your question it is not explicitly mentioned how probability $P$ is defined.
So actually the question: "what is the probability that $x$ is rational?" makes no sense in this context.
Your try indicates that you are thinking of uniform distribution on interval $[0,1]$ where: $$P(B):=lambda(Bcap[0,1])$$ for every Borel set and $lambda$ denotes the Lebesgue measure on the $sigma$-algebra of Borel sets $Bsubseteqmathbb R$.
It is well known that $lambda(x)=0$ for any $xinmathbb R$ and consequently we have $lambda(S)=sum_xin Slambda(x)=0$ for every countable set, and of course $mathbb Q$ is countable.
So under uniform distribution on $[0,1]$ we have $P(mathbb Q)=0$.
In probability theory Riemann-integrability is (as far as I know) not practicized.
Notation $P(xin mathbb Q)$ only makes sense if $x$ denotes a random variable here.
So your problem setting: "if $xin[0,1]$ then what is $P(xinmathbb Q)$?" is not okay.
answered Oct 12 at 7:57
drhabdrhab
116k6 gold badges51 silver badges146 bronze badges
$endgroup$
add a comment
|
$begingroup$
This answer might miss the real goal, but is inspired by the tag "probability".
In your question it is not explicitly mentioned how probability $P$ is defined.
So actually the question: "what is the probability that $x$ is rational?" makes no sense in this context.
Your try indicates that you are thinking of uniform distribution on interval $[0,1]$ where: $$P(B):=lambda(Bcap[0,1])$$ for every Borel set and $lambda$ denotes the Lebesgue measure on the $sigma$-algebra of Borel sets $Bsubseteqmathbb R$.
It is well known that $lambda(x)=0$ for any $xinmathbb R$ and consequently we have $lambda(S)=sum_xin Slambda(x)=0$ for every countable set, and of course $mathbb Q$ is countable.
So under uniform distribution on $[0,1]$ we have $P(mathbb Q)=0$.
In probability theory Riemann-integrability is (as far as I know) not practicized.
Notation $P(xin mathbb Q)$ only makes sense if $x$ denotes a random variable here.
So your problem setting: "if $xin[0,1]$ then what is $P(xinmathbb Q)$?" is not okay.
answered Oct 12 at 7:57
drhabdrhab
116k6 gold badges51 silver badges146 bronze badges
$endgroup$
This answer might miss the real goal, but is inspired by the tag "probability".
In your question it is not explicitly mentioned how probability $P$ is defined.
So actually the question: "what is the probability that $x$ is rational?" makes no sense in this context.
Your try indicates that you are thinking of uniform distribution on interval $[0,1]$ where: $$P(B):=lambda(Bcap[0,1])$$ for every Borel set and $lambda$ denotes the Lebesgue measure on the $sigma$-algebra of Borel sets $Bsubseteqmathbb R$.
It is well known that $lambda(x)=0$ for any $xinmathbb R$ and consequently we have $lambda(S)=sum_xin Slambda(x)=0$ for every countable set, and of course $mathbb Q$ is countable.
So under uniform distribution on $[0,1]$ we have $P(mathbb Q)=0$.
In probability theory Riemann-integrability is (as far as I know) not practicized.
Notation $P(xin mathbb Q)$ only makes sense if $x$ denotes a random variable here.
So your problem setting: "if $xin[0,1]$ then what is $P(xinmathbb Q)$?" is not okay.
answered Oct 12 at 7:57
drhabdrhab
116k6 gold badges51 silver badges146 bronze badges
answered Oct 12 at 7:57
drhabdrhab
116k6 gold badges51 silver badges146 bronze badges
answered Oct 12 at 7:57
drhabdrhab
116k6 gold badges51 silver badges146 bronze badges
answered Oct 12 at 7:57
drhabdrhab
116k6 gold badges51 silver badges146 bronze badges
116k6 gold badges51 silver badges146 bronze badges
add a comment
|
add a comment
|
$begingroup$
I want to try comparing the cardinalities of the rational set and the irrational set by using a one-to-one mapping between the two sets
You can't. The set of rationals is countable. The set of irrationals is uncountable (see Proof that the irrational numbers are uncountable). Cantor proved by diagonalization that such that mapping cannot exist.
answered Oct 13 at 3:00
Laurence R. UgaldeLaurence R. Ugalde
1165 bronze badges
$endgroup$
add a comment
|
$begingroup$
I want to try comparing the cardinalities of the rational set and the irrational set by using a one-to-one mapping between the two sets
You can't. The set of rationals is countable. The set of irrationals is uncountable (see Proof that the irrational numbers are uncountable). Cantor proved by diagonalization that such that mapping cannot exist.
answered Oct 13 at 3:00
Laurence R. UgaldeLaurence R. Ugalde
1165 bronze badges
$endgroup$
add a comment
|
$begingroup$
I want to try comparing the cardinalities of the rational set and the irrational set by using a one-to-one mapping between the two sets
You can't. The set of rationals is countable. The set of irrationals is uncountable (see Proof that the irrational numbers are uncountable). Cantor proved by diagonalization that such that mapping cannot exist.
answered Oct 13 at 3:00
Laurence R. UgaldeLaurence R. Ugalde
1165 bronze badges
$endgroup$
I want to try comparing the cardinalities of the rational set and the irrational set by using a one-to-one mapping between the two sets
You can't. The set of rationals is countable. The set of irrationals is uncountable (see Proof that the irrational numbers are uncountable). Cantor proved by diagonalization that such that mapping cannot exist.
answered Oct 13 at 3:00
Laurence R. UgaldeLaurence R. Ugalde
1165 bronze badges
edited Oct 13 at 3:08
answered Oct 13 at 3:00
Laurence R. UgaldeLaurence R. Ugalde
1165 bronze badges
answered Oct 13 at 3:00
Laurence R. UgaldeLaurence R. Ugalde
1165 bronze badges
answered Oct 13 at 3:00
Laurence R. UgaldeLaurence R. Ugalde
1165 bronze badges
1165 bronze badges
add a comment
|
add a comment
|
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