What does ${mathbb R^n}$ mean?
$begingroup$
I know that $mathbb R$ refers to all real numbers. But what does $mathbb R^2$ mean?
For that matter, what does $mathbb R^n$ mean when $n$ equals any natural number?
real-numbers
$endgroup$
add a comment |
$begingroup$
I know that $mathbb R$ refers to all real numbers. But what does $mathbb R^2$ mean?
For that matter, what does $mathbb R^n$ mean when $n$ equals any natural number?
real-numbers
$endgroup$
2
$begingroup$
en.wikipedia.org/wiki/Real_coordinate_space
$endgroup$
– T. Bongers
Dec 2 '18 at 22:40
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Its an $n$ dimensional vector over the reals
$endgroup$
– Seth
Dec 2 '18 at 22:43
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$mathbb R^n$ is the set of all ordered $n$-tuples of real numbers.
$endgroup$
– littleO
Dec 2 '18 at 22:47
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These questions may also help: math.stackexchange.com/questions/2423055/… math.stackexchange.com/questions/855437/what-does-mathbbrj-mean
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– Jean-Claude Arbaut
Dec 2 '18 at 22:47
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Thank you all for these answers
$endgroup$
– Xavier Stanton
Dec 3 '18 at 2:07
add a comment |
$begingroup$
I know that $mathbb R$ refers to all real numbers. But what does $mathbb R^2$ mean?
For that matter, what does $mathbb R^n$ mean when $n$ equals any natural number?
real-numbers
$endgroup$
I know that $mathbb R$ refers to all real numbers. But what does $mathbb R^2$ mean?
For that matter, what does $mathbb R^n$ mean when $n$ equals any natural number?
real-numbers
real-numbers
edited Dec 3 '18 at 0:11
amWhy
1
1
asked Dec 2 '18 at 22:38
Xavier StantonXavier Stanton
311211
311211
2
$begingroup$
en.wikipedia.org/wiki/Real_coordinate_space
$endgroup$
– T. Bongers
Dec 2 '18 at 22:40
$begingroup$
Its an $n$ dimensional vector over the reals
$endgroup$
– Seth
Dec 2 '18 at 22:43
$begingroup$
$mathbb R^n$ is the set of all ordered $n$-tuples of real numbers.
$endgroup$
– littleO
Dec 2 '18 at 22:47
$begingroup$
These questions may also help: math.stackexchange.com/questions/2423055/… math.stackexchange.com/questions/855437/what-does-mathbbrj-mean
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 22:47
$begingroup$
Thank you all for these answers
$endgroup$
– Xavier Stanton
Dec 3 '18 at 2:07
add a comment |
2
$begingroup$
en.wikipedia.org/wiki/Real_coordinate_space
$endgroup$
– T. Bongers
Dec 2 '18 at 22:40
$begingroup$
Its an $n$ dimensional vector over the reals
$endgroup$
– Seth
Dec 2 '18 at 22:43
$begingroup$
$mathbb R^n$ is the set of all ordered $n$-tuples of real numbers.
$endgroup$
– littleO
Dec 2 '18 at 22:47
$begingroup$
These questions may also help: math.stackexchange.com/questions/2423055/… math.stackexchange.com/questions/855437/what-does-mathbbrj-mean
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 22:47
$begingroup$
Thank you all for these answers
$endgroup$
– Xavier Stanton
Dec 3 '18 at 2:07
2
2
$begingroup$
en.wikipedia.org/wiki/Real_coordinate_space
$endgroup$
– T. Bongers
Dec 2 '18 at 22:40
$begingroup$
en.wikipedia.org/wiki/Real_coordinate_space
$endgroup$
– T. Bongers
Dec 2 '18 at 22:40
$begingroup$
Its an $n$ dimensional vector over the reals
$endgroup$
– Seth
Dec 2 '18 at 22:43
$begingroup$
Its an $n$ dimensional vector over the reals
$endgroup$
– Seth
Dec 2 '18 at 22:43
$begingroup$
$mathbb R^n$ is the set of all ordered $n$-tuples of real numbers.
$endgroup$
– littleO
Dec 2 '18 at 22:47
$begingroup$
$mathbb R^n$ is the set of all ordered $n$-tuples of real numbers.
$endgroup$
– littleO
Dec 2 '18 at 22:47
$begingroup$
These questions may also help: math.stackexchange.com/questions/2423055/… math.stackexchange.com/questions/855437/what-does-mathbbrj-mean
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 22:47
$begingroup$
These questions may also help: math.stackexchange.com/questions/2423055/… math.stackexchange.com/questions/855437/what-does-mathbbrj-mean
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 22:47
$begingroup$
Thank you all for these answers
$endgroup$
– Xavier Stanton
Dec 3 '18 at 2:07
$begingroup$
Thank you all for these answers
$endgroup$
– Xavier Stanton
Dec 3 '18 at 2:07
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
In general a set $X$ will have a Cartesian product with itself sometimes called $X^2$. We can again take the Cartesian product of $X^2$ and $X$ to get $X^3$
and so on.
Another way to interpret this notation is by considering two sets $X$ and $Y$ then define $Y^X$ as the set of all function from $X$ to $Y$. Then we can interpret $X^2$ with $2$ meaning the set of $2$ elements rather than as the number $2$ itself. This can be done with $X^n$ being the set of function from a set of $n$ elements to $X$ since they are all unique up to isomorphism.
$endgroup$
add a comment |
$begingroup$
$mathbb R^n$ refers to the space of all $n$-dimensional vectors $(x_1, x_2, dots, x_n)$ where each coordinate $x_i$ is a real number i.e. $x_i in mathbb R$.
$endgroup$
add a comment |
$begingroup$
Without giving you a word definition, let's see if you understand this way:
$xinmathbb{R}$
$(x_1, x_2) in mathbb{R^2}$
$(x_1, x_2, x_3) in mathbb{R^3}$
Therefore,
$(x_1, x_2, x_3, dots , x_n) in mathbb{R^n}$
where each $x_i$ is a member of the real numbers $mathbb{R}$.
Thus, When we say something is a member of $mathbb{R}^n$, its saying that its a $n$ dimensional vector with elements of type $mathbb{R}$
$endgroup$
add a comment |
$begingroup$
$mathbb{R}^n$ is the set of all n-tuples with real elements. They are NOT a vector space by themselves, just a set. For a vector space, we would need an extra scalar field and 2 operations: addition between the vectors (elements of $mathbb{R}^n$) and multiplication between the scalars and vectors. But usually we just denote the vector space of $mathbb{R}^n$ over the $mathbb{R}$, with the usual product and sum as $mathbb{R}^n$ for simplicity reasons, but they are not the same.
$endgroup$
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
$begingroup$
In general a set $X$ will have a Cartesian product with itself sometimes called $X^2$. We can again take the Cartesian product of $X^2$ and $X$ to get $X^3$
and so on.
Another way to interpret this notation is by considering two sets $X$ and $Y$ then define $Y^X$ as the set of all function from $X$ to $Y$. Then we can interpret $X^2$ with $2$ meaning the set of $2$ elements rather than as the number $2$ itself. This can be done with $X^n$ being the set of function from a set of $n$ elements to $X$ since they are all unique up to isomorphism.
$endgroup$
add a comment |
$begingroup$
In general a set $X$ will have a Cartesian product with itself sometimes called $X^2$. We can again take the Cartesian product of $X^2$ and $X$ to get $X^3$
and so on.
Another way to interpret this notation is by considering two sets $X$ and $Y$ then define $Y^X$ as the set of all function from $X$ to $Y$. Then we can interpret $X^2$ with $2$ meaning the set of $2$ elements rather than as the number $2$ itself. This can be done with $X^n$ being the set of function from a set of $n$ elements to $X$ since they are all unique up to isomorphism.
$endgroup$
add a comment |
$begingroup$
In general a set $X$ will have a Cartesian product with itself sometimes called $X^2$. We can again take the Cartesian product of $X^2$ and $X$ to get $X^3$
and so on.
Another way to interpret this notation is by considering two sets $X$ and $Y$ then define $Y^X$ as the set of all function from $X$ to $Y$. Then we can interpret $X^2$ with $2$ meaning the set of $2$ elements rather than as the number $2$ itself. This can be done with $X^n$ being the set of function from a set of $n$ elements to $X$ since they are all unique up to isomorphism.
$endgroup$
In general a set $X$ will have a Cartesian product with itself sometimes called $X^2$. We can again take the Cartesian product of $X^2$ and $X$ to get $X^3$
and so on.
Another way to interpret this notation is by considering two sets $X$ and $Y$ then define $Y^X$ as the set of all function from $X$ to $Y$. Then we can interpret $X^2$ with $2$ meaning the set of $2$ elements rather than as the number $2$ itself. This can be done with $X^n$ being the set of function from a set of $n$ elements to $X$ since they are all unique up to isomorphism.
answered Dec 2 '18 at 22:52
CyclotomicFieldCyclotomicField
2,3481314
2,3481314
add a comment |
add a comment |
$begingroup$
$mathbb R^n$ refers to the space of all $n$-dimensional vectors $(x_1, x_2, dots, x_n)$ where each coordinate $x_i$ is a real number i.e. $x_i in mathbb R$.
$endgroup$
add a comment |
$begingroup$
$mathbb R^n$ refers to the space of all $n$-dimensional vectors $(x_1, x_2, dots, x_n)$ where each coordinate $x_i$ is a real number i.e. $x_i in mathbb R$.
$endgroup$
add a comment |
$begingroup$
$mathbb R^n$ refers to the space of all $n$-dimensional vectors $(x_1, x_2, dots, x_n)$ where each coordinate $x_i$ is a real number i.e. $x_i in mathbb R$.
$endgroup$
$mathbb R^n$ refers to the space of all $n$-dimensional vectors $(x_1, x_2, dots, x_n)$ where each coordinate $x_i$ is a real number i.e. $x_i in mathbb R$.
edited Dec 2 '18 at 22:49
answered Dec 2 '18 at 22:45
RebellosRebellos
14.5k31246
14.5k31246
add a comment |
add a comment |
$begingroup$
Without giving you a word definition, let's see if you understand this way:
$xinmathbb{R}$
$(x_1, x_2) in mathbb{R^2}$
$(x_1, x_2, x_3) in mathbb{R^3}$
Therefore,
$(x_1, x_2, x_3, dots , x_n) in mathbb{R^n}$
where each $x_i$ is a member of the real numbers $mathbb{R}$.
Thus, When we say something is a member of $mathbb{R}^n$, its saying that its a $n$ dimensional vector with elements of type $mathbb{R}$
$endgroup$
add a comment |
$begingroup$
Without giving you a word definition, let's see if you understand this way:
$xinmathbb{R}$
$(x_1, x_2) in mathbb{R^2}$
$(x_1, x_2, x_3) in mathbb{R^3}$
Therefore,
$(x_1, x_2, x_3, dots , x_n) in mathbb{R^n}$
where each $x_i$ is a member of the real numbers $mathbb{R}$.
Thus, When we say something is a member of $mathbb{R}^n$, its saying that its a $n$ dimensional vector with elements of type $mathbb{R}$
$endgroup$
add a comment |
$begingroup$
Without giving you a word definition, let's see if you understand this way:
$xinmathbb{R}$
$(x_1, x_2) in mathbb{R^2}$
$(x_1, x_2, x_3) in mathbb{R^3}$
Therefore,
$(x_1, x_2, x_3, dots , x_n) in mathbb{R^n}$
where each $x_i$ is a member of the real numbers $mathbb{R}$.
Thus, When we say something is a member of $mathbb{R}^n$, its saying that its a $n$ dimensional vector with elements of type $mathbb{R}$
$endgroup$
Without giving you a word definition, let's see if you understand this way:
$xinmathbb{R}$
$(x_1, x_2) in mathbb{R^2}$
$(x_1, x_2, x_3) in mathbb{R^3}$
Therefore,
$(x_1, x_2, x_3, dots , x_n) in mathbb{R^n}$
where each $x_i$ is a member of the real numbers $mathbb{R}$.
Thus, When we say something is a member of $mathbb{R}^n$, its saying that its a $n$ dimensional vector with elements of type $mathbb{R}$
answered Dec 2 '18 at 22:52
K Split XK Split X
4,20611131
4,20611131
add a comment |
add a comment |
$begingroup$
$mathbb{R}^n$ is the set of all n-tuples with real elements. They are NOT a vector space by themselves, just a set. For a vector space, we would need an extra scalar field and 2 operations: addition between the vectors (elements of $mathbb{R}^n$) and multiplication between the scalars and vectors. But usually we just denote the vector space of $mathbb{R}^n$ over the $mathbb{R}$, with the usual product and sum as $mathbb{R}^n$ for simplicity reasons, but they are not the same.
$endgroup$
add a comment |
$begingroup$
$mathbb{R}^n$ is the set of all n-tuples with real elements. They are NOT a vector space by themselves, just a set. For a vector space, we would need an extra scalar field and 2 operations: addition between the vectors (elements of $mathbb{R}^n$) and multiplication between the scalars and vectors. But usually we just denote the vector space of $mathbb{R}^n$ over the $mathbb{R}$, with the usual product and sum as $mathbb{R}^n$ for simplicity reasons, but they are not the same.
$endgroup$
add a comment |
$begingroup$
$mathbb{R}^n$ is the set of all n-tuples with real elements. They are NOT a vector space by themselves, just a set. For a vector space, we would need an extra scalar field and 2 operations: addition between the vectors (elements of $mathbb{R}^n$) and multiplication between the scalars and vectors. But usually we just denote the vector space of $mathbb{R}^n$ over the $mathbb{R}$, with the usual product and sum as $mathbb{R}^n$ for simplicity reasons, but they are not the same.
$endgroup$
$mathbb{R}^n$ is the set of all n-tuples with real elements. They are NOT a vector space by themselves, just a set. For a vector space, we would need an extra scalar field and 2 operations: addition between the vectors (elements of $mathbb{R}^n$) and multiplication between the scalars and vectors. But usually we just denote the vector space of $mathbb{R}^n$ over the $mathbb{R}$, with the usual product and sum as $mathbb{R}^n$ for simplicity reasons, but they are not the same.
answered Dec 2 '18 at 22:55
BotondBotond
5,6822732
5,6822732
add a comment |
add a comment |
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2
$begingroup$
en.wikipedia.org/wiki/Real_coordinate_space
$endgroup$
– T. Bongers
Dec 2 '18 at 22:40
$begingroup$
Its an $n$ dimensional vector over the reals
$endgroup$
– Seth
Dec 2 '18 at 22:43
$begingroup$
$mathbb R^n$ is the set of all ordered $n$-tuples of real numbers.
$endgroup$
– littleO
Dec 2 '18 at 22:47
$begingroup$
These questions may also help: math.stackexchange.com/questions/2423055/… math.stackexchange.com/questions/855437/what-does-mathbbrj-mean
$endgroup$
– Jean-Claude Arbaut
Dec 2 '18 at 22:47
$begingroup$
Thank you all for these answers
$endgroup$
– Xavier Stanton
Dec 3 '18 at 2:07