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How to prove the Cardinal Number.
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Assume that $A$ and $B$ are finte sets and $card(A)=n$ and $card(B)=m$,
we will show that $card(A^B)=m^n$.
Pleas, hint me to prove this, i try to prove by induction(fiexd set$B$) but i can't use induction hypothesis, or can create a bijection $f:A^Bto ? $
Thank for give me .
linear-algebra
asked Aug 2 at 16:23
Chung wow
102
add a comment |Â
up vote
1
down vote
favorite
Assume that $A$ and $B$ are finte sets and $card(A)=n$ and $card(B)=m$,
we will show that $card(A^B)=m^n$.
Pleas, hint me to prove this, i try to prove by induction(fiexd set$B$) but i can't use induction hypothesis, or can create a bijection $f:A^Bto ? $
Thank for give me .
linear-algebra
asked Aug 2 at 16:23
Chung wow
102
It's not entirely clear what you're asking. Are $A$ and $B$ supposed to be finite sets? Do you mean to show that $card(A^B) = n^m$?
– Daniel Mroz
Aug 2 at 16:27
1
Hint: Try induction on the cardinality of $B$, with the induction hypothesis "for all finite sets $A$, $card(A^B) = card(A)^card(B)$. Use the fact that $n^m+1 = n*n^m$.
– Daniel Mroz
Aug 2 at 16:36
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Assume that $A$ and $B$ are finte sets and $card(A)=n$ and $card(B)=m$,
we will show that $card(A^B)=m^n$.
Pleas, hint me to prove this, i try to prove by induction(fiexd set$B$) but i can't use induction hypothesis, or can create a bijection $f:A^Bto ? $
Thank for give me .
linear-algebra
asked Aug 2 at 16:23
Chung wow
102
Assume that $A$ and $B$ are finte sets and $card(A)=n$ and $card(B)=m$,
we will show that $card(A^B)=m^n$.
Pleas, hint me to prove this, i try to prove by induction(fiexd set$B$) but i can't use induction hypothesis, or can create a bijection $f:A^Bto ? $
Thank for give me .
linear-algebra
asked Aug 2 at 16:23
Chung wow
102
edited Aug 2 at 16:29
asked Aug 2 at 16:23
Chung wow
102
asked Aug 2 at 16:23
Chung wow
102
asked Aug 2 at 16:23
Chung wow
102
102
It's not entirely clear what you're asking. Are $A$ and $B$ supposed to be finite sets? Do you mean to show that $card(A^B) = n^m$?
– Daniel Mroz
Aug 2 at 16:27
1
Hint: Try induction on the cardinality of $B$, with the induction hypothesis "for all finite sets $A$, $card(A^B) = card(A)^card(B)$. Use the fact that $n^m+1 = n*n^m$.
– Daniel Mroz
Aug 2 at 16:36
add a comment |Â
It's not entirely clear what you're asking. Are $A$ and $B$ supposed to be finite sets? Do you mean to show that $card(A^B) = n^m$?
– Daniel Mroz
Aug 2 at 16:27
1
Hint: Try induction on the cardinality of $B$, with the induction hypothesis "for all finite sets $A$, $card(A^B) = card(A)^card(B)$. Use the fact that $n^m+1 = n*n^m$.
– Daniel Mroz
Aug 2 at 16:36
It's not entirely clear what you're asking. Are $A$ and $B$ supposed to be finite sets? Do you mean to show that $card(A^B) = n^m$?
– Daniel Mroz
Aug 2 at 16:27
It's not entirely clear what you're asking. Are $A$ and $B$ supposed to be finite sets? Do you mean to show that $card(A^B) = n^m$?
– Daniel Mroz
Aug 2 at 16:27
1
1
Hint: Try induction on the cardinality of $B$, with the induction hypothesis "for all finite sets $A$, $card(A^B) = card(A)^card(B)$. Use the fact that $n^m+1 = n*n^m$.
– Daniel Mroz
Aug 2 at 16:36
Hint: Try induction on the cardinality of $B$, with the induction hypothesis "for all finite sets $A$, $card(A^B) = card(A)^card(B)$. Use the fact that $n^m+1 = n*n^m$.
– Daniel Mroz
Aug 2 at 16:36
add a comment |Â
1 Answer
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You can create the following bijection: first, order the elements of $B$ as, say, $B=b_1,dots,b_m$. Then any element of $A^B$, which is a map $p$ from $B$ to $A$, is determined by giving all the images of the elements of $B$.
Define
$f: A^B to A^m$
by $f(p)=(p(b_1),dots,p(b_m))$. It is a bijection.
Now, the cardinal of the product $A^m$ is $n^m$.
answered Aug 2 at 17:16
xarles
83548
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
You can create the following bijection: first, order the elements of $B$ as, say, $B=b_1,dots,b_m$. Then any element of $A^B$, which is a map $p$ from $B$ to $A$, is determined by giving all the images of the elements of $B$.
Define
$f: A^B to A^m$
by $f(p)=(p(b_1),dots,p(b_m))$. It is a bijection.
Now, the cardinal of the product $A^m$ is $n^m$.
answered Aug 2 at 17:16
xarles
83548
add a comment |Â
up vote
2
down vote
You can create the following bijection: first, order the elements of $B$ as, say, $B=b_1,dots,b_m$. Then any element of $A^B$, which is a map $p$ from $B$ to $A$, is determined by giving all the images of the elements of $B$.
Define
$f: A^B to A^m$
by $f(p)=(p(b_1),dots,p(b_m))$. It is a bijection.
Now, the cardinal of the product $A^m$ is $n^m$.
answered Aug 2 at 17:16
xarles
83548
add a comment |Â
up vote
2
down vote
up vote
2
down vote
You can create the following bijection: first, order the elements of $B$ as, say, $B=b_1,dots,b_m$. Then any element of $A^B$, which is a map $p$ from $B$ to $A$, is determined by giving all the images of the elements of $B$.
Define
$f: A^B to A^m$
by $f(p)=(p(b_1),dots,p(b_m))$. It is a bijection.
Now, the cardinal of the product $A^m$ is $n^m$.
answered Aug 2 at 17:16
xarles
83548
You can create the following bijection: first, order the elements of $B$ as, say, $B=b_1,dots,b_m$. Then any element of $A^B$, which is a map $p$ from $B$ to $A$, is determined by giving all the images of the elements of $B$.
Define
$f: A^B to A^m$
by $f(p)=(p(b_1),dots,p(b_m))$. It is a bijection.
Now, the cardinal of the product $A^m$ is $n^m$.
answered Aug 2 at 17:16
xarles
83548
answered Aug 2 at 17:16
xarles
83548
answered Aug 2 at 17:16
xarles
83548
answered Aug 2 at 17:16
xarles
83548
83548
add a comment |Â
add a comment |Â
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It's not entirely clear what you're asking. Are $A$ and $B$ supposed to be finite sets? Do you mean to show that $card(A^B) = n^m$?
– Daniel Mroz
Aug 2 at 16:27
1
Hint: Try induction on the cardinality of $B$, with the induction hypothesis "for all finite sets $A$, $card(A^B) = card(A)^card(B)$. Use the fact that $n^m+1 = n*n^m$.
– Daniel Mroz
Aug 2 at 16:36