Mixing
Is $n^{ncdot n}$ equal to $(n^n)^n$?
0
$begingroup$
I have a simple question, is $n^{ncdot n}=(n^n)^n$?
I believe it does, because, for example $(n^2)^2 = n^{2cdot 2}$. Also $1^1 = 1cdot1, 2^2 = 2cdot2, 3^3 = 3cdot3, 4^4 = 4cdot 4$, so I suppose $n^n = ncdot n$, am i right?
exponential-function
edited Jan 20 at 4:53
Larry
2,41331129
asked Jan 20 at 4:14
we_morwe_mor
31
$endgroup$
add a comment |
0
$begingroup$
I have a simple question, is $n^{ncdot n}=(n^n)^n$?
I believe it does, because, for example $(n^2)^2 = n^{2cdot 2}$. Also $1^1 = 1cdot1, 2^2 = 2cdot2, 3^3 = 3cdot3, 4^4 = 4cdot 4$, so I suppose $n^n = ncdot n$, am i right?
exponential-function
edited Jan 20 at 4:53
Larry
2,41331129
asked Jan 20 at 4:14
we_morwe_mor
31
$endgroup$
1
$begingroup$
yes............................
$endgroup$
– Will Jagy
Jan 20 at 4:17
$begingroup$
But $3*3=9not=27=3^3$.
$endgroup$
– Jens Schwaiger
Jan 20 at 4:19
$begingroup$
@Jens Schwaiger oh you are right, thank you
$endgroup$
– we_mor
Jan 20 at 4:22
add a comment |
0
0
0
$begingroup$
I have a simple question, is $n^{ncdot n}=(n^n)^n$?
I believe it does, because, for example $(n^2)^2 = n^{2cdot 2}$. Also $1^1 = 1cdot1, 2^2 = 2cdot2, 3^3 = 3cdot3, 4^4 = 4cdot 4$, so I suppose $n^n = ncdot n$, am i right?
exponential-function
edited Jan 20 at 4:53
Larry
2,41331129
asked Jan 20 at 4:14
we_morwe_mor
31
$endgroup$
I have a simple question, is $n^{ncdot n}=(n^n)^n$?
I believe it does, because, for example $(n^2)^2 = n^{2cdot 2}$. Also $1^1 = 1cdot1, 2^2 = 2cdot2, 3^3 = 3cdot3, 4^4 = 4cdot 4$, so I suppose $n^n = ncdot n$, am i right?
exponential-function
exponential-function
edited Jan 20 at 4:53
Larry
2,41331129
asked Jan 20 at 4:14
we_morwe_mor
31
edited Jan 20 at 4:53
Larry
2,41331129
asked Jan 20 at 4:14
we_morwe_mor
31
edited Jan 20 at 4:53
Larry
2,41331129
edited Jan 20 at 4:53
Larry
2,41331129
edited Jan 20 at 4:53
Larry
2,41331129
2,41331129
asked Jan 20 at 4:14
we_morwe_mor
31
asked Jan 20 at 4:14
we_morwe_mor
31
asked Jan 20 at 4:14
we_morwe_mor
31
31
1
$begingroup$
yes............................
$endgroup$
– Will Jagy
Jan 20 at 4:17
$begingroup$
But $3*3=9not=27=3^3$.
$endgroup$
– Jens Schwaiger
Jan 20 at 4:19
$begingroup$
@Jens Schwaiger oh you are right, thank you
$endgroup$
– we_mor
Jan 20 at 4:22
add a comment |
1
$begingroup$
yes............................
$endgroup$
– Will Jagy
Jan 20 at 4:17
$begingroup$
But $3*3=9not=27=3^3$.
$endgroup$
– Jens Schwaiger
Jan 20 at 4:19
$begingroup$
@Jens Schwaiger oh you are right, thank you
$endgroup$
– we_mor
Jan 20 at 4:22
1
1
$begingroup$
yes............................
$endgroup$
– Will Jagy
Jan 20 at 4:17
$begingroup$
yes............................
$endgroup$
– Will Jagy
Jan 20 at 4:17
$begingroup$
But $3*3=9not=27=3^3$.
$endgroup$
– Jens Schwaiger
Jan 20 at 4:19
$begingroup$
But $3*3=9not=27=3^3$.
$endgroup$
– Jens Schwaiger
Jan 20 at 4:19
$begingroup$
@Jens Schwaiger oh you are right, thank you
$endgroup$
– we_mor
Jan 20 at 4:22
$begingroup$
@Jens Schwaiger oh you are right, thank you
$endgroup$
– we_mor
Jan 20 at 4:22
add a comment |
1 Answer
1
active
oldest
votes
1
$begingroup$
A law of exponents says that
$$
(a^b)^c = a^{bc}.
$$
In other words, iterated exponents multiply. Now let $a=b=c=n$ to get $(n^n)^n=n^{n^2}$. Your final claim of $n^n = n^2$ is wrong (let $n=3$) and doesn't follow from the previous (correct) fact.
answered Jan 20 at 4:23
RandallRandall
10.2k11230
$endgroup$
$begingroup$
Do you think it's required to state that the objects under exponentiation must adhere to associativity and commutativity for this result to be upheld? The OP has not specified what time of numbers is being used here... although it's a very safe assumption to say it's the set of Natural Numbers under addition/multiplication.
$endgroup$
– DavidG
Jan 20 at 5:25
$begingroup$
I think for OP it’s safe to assume something reasonable like natural numbers, yes. Anything more abstract could be more confusing.
$endgroup$
– Randall
Jan 20 at 5:34
$begingroup$
Yes, agreed. It's interesting that these properties are taught as being 'Natural' and universal and then with Linear Algebra they are challenged and brought back to the beginning.
$endgroup$
– DavidG
Jan 20 at 5:38
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
A law of exponents says that
$$
(a^b)^c = a^{bc}.
$$
In other words, iterated exponents multiply. Now let $a=b=c=n$ to get $(n^n)^n=n^{n^2}$. Your final claim of $n^n = n^2$ is wrong (let $n=3$) and doesn't follow from the previous (correct) fact.
answered Jan 20 at 4:23
RandallRandall
10.2k11230
$endgroup$
$begingroup$
Do you think it's required to state that the objects under exponentiation must adhere to associativity and commutativity for this result to be upheld? The OP has not specified what time of numbers is being used here... although it's a very safe assumption to say it's the set of Natural Numbers under addition/multiplication.
$endgroup$
– DavidG
Jan 20 at 5:25
$begingroup$
I think for OP it’s safe to assume something reasonable like natural numbers, yes. Anything more abstract could be more confusing.
$endgroup$
– Randall
Jan 20 at 5:34
$begingroup$
Yes, agreed. It's interesting that these properties are taught as being 'Natural' and universal and then with Linear Algebra they are challenged and brought back to the beginning.
$endgroup$
– DavidG
Jan 20 at 5:38
add a comment |
1
$begingroup$
A law of exponents says that
$$
(a^b)^c = a^{bc}.
$$
In other words, iterated exponents multiply. Now let $a=b=c=n$ to get $(n^n)^n=n^{n^2}$. Your final claim of $n^n = n^2$ is wrong (let $n=3$) and doesn't follow from the previous (correct) fact.
answered Jan 20 at 4:23
RandallRandall
10.2k11230
$endgroup$
$begingroup$
Do you think it's required to state that the objects under exponentiation must adhere to associativity and commutativity for this result to be upheld? The OP has not specified what time of numbers is being used here... although it's a very safe assumption to say it's the set of Natural Numbers under addition/multiplication.
$endgroup$
– DavidG
Jan 20 at 5:25
$begingroup$
I think for OP it’s safe to assume something reasonable like natural numbers, yes. Anything more abstract could be more confusing.
$endgroup$
– Randall
Jan 20 at 5:34
$begingroup$
Yes, agreed. It's interesting that these properties are taught as being 'Natural' and universal and then with Linear Algebra they are challenged and brought back to the beginning.
$endgroup$
– DavidG
Jan 20 at 5:38
add a comment |
1
1
1
$begingroup$
A law of exponents says that
$$
(a^b)^c = a^{bc}.
$$
In other words, iterated exponents multiply. Now let $a=b=c=n$ to get $(n^n)^n=n^{n^2}$. Your final claim of $n^n = n^2$ is wrong (let $n=3$) and doesn't follow from the previous (correct) fact.
answered Jan 20 at 4:23
RandallRandall
10.2k11230
$endgroup$
A law of exponents says that
$$
(a^b)^c = a^{bc}.
$$
In other words, iterated exponents multiply. Now let $a=b=c=n$ to get $(n^n)^n=n^{n^2}$. Your final claim of $n^n = n^2$ is wrong (let $n=3$) and doesn't follow from the previous (correct) fact.
answered Jan 20 at 4:23
RandallRandall
10.2k11230
edited Jan 20 at 4:54
answered Jan 20 at 4:23
RandallRandall
10.2k11230
answered Jan 20 at 4:23
RandallRandall
10.2k11230
answered Jan 20 at 4:23
RandallRandall
10.2k11230
10.2k11230
$begingroup$
Do you think it's required to state that the objects under exponentiation must adhere to associativity and commutativity for this result to be upheld? The OP has not specified what time of numbers is being used here... although it's a very safe assumption to say it's the set of Natural Numbers under addition/multiplication.
$endgroup$
– DavidG
Jan 20 at 5:25
$begingroup$
I think for OP it’s safe to assume something reasonable like natural numbers, yes. Anything more abstract could be more confusing.
$endgroup$
– Randall
Jan 20 at 5:34
$begingroup$
Yes, agreed. It's interesting that these properties are taught as being 'Natural' and universal and then with Linear Algebra they are challenged and brought back to the beginning.
$endgroup$
– DavidG
Jan 20 at 5:38
add a comment |
$begingroup$
Do you think it's required to state that the objects under exponentiation must adhere to associativity and commutativity for this result to be upheld? The OP has not specified what time of numbers is being used here... although it's a very safe assumption to say it's the set of Natural Numbers under addition/multiplication.
$endgroup$
– DavidG
Jan 20 at 5:25
$begingroup$
I think for OP it’s safe to assume something reasonable like natural numbers, yes. Anything more abstract could be more confusing.
$endgroup$
– Randall
Jan 20 at 5:34
$begingroup$
Yes, agreed. It's interesting that these properties are taught as being 'Natural' and universal and then with Linear Algebra they are challenged and brought back to the beginning.
$endgroup$
– DavidG
Jan 20 at 5:38
$begingroup$
Do you think it's required to state that the objects under exponentiation must adhere to associativity and commutativity for this result to be upheld? The OP has not specified what time of numbers is being used here... although it's a very safe assumption to say it's the set of Natural Numbers under addition/multiplication.
$endgroup$
– DavidG
Jan 20 at 5:25
$begingroup$
Do you think it's required to state that the objects under exponentiation must adhere to associativity and commutativity for this result to be upheld? The OP has not specified what time of numbers is being used here... although it's a very safe assumption to say it's the set of Natural Numbers under addition/multiplication.
$endgroup$
– DavidG
Jan 20 at 5:25
$begingroup$
I think for OP it’s safe to assume something reasonable like natural numbers, yes. Anything more abstract could be more confusing.
$endgroup$
– Randall
Jan 20 at 5:34
$begingroup$
I think for OP it’s safe to assume something reasonable like natural numbers, yes. Anything more abstract could be more confusing.
$endgroup$
– Randall
Jan 20 at 5:34
$begingroup$
Yes, agreed. It's interesting that these properties are taught as being 'Natural' and universal and then with Linear Algebra they are challenged and brought back to the beginning.
$endgroup$
– DavidG
Jan 20 at 5:38
$begingroup$
Yes, agreed. It's interesting that these properties are taught as being 'Natural' and universal and then with Linear Algebra they are challenged and brought back to the beginning.
$endgroup$
– DavidG
Jan 20 at 5:38
add a comment |
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1
$begingroup$
yes............................
$endgroup$
– Will Jagy
Jan 20 at 4:17
$begingroup$
But $3*3=9not=27=3^3$.
$endgroup$
– Jens Schwaiger
Jan 20 at 4:19
$begingroup$
@Jens Schwaiger oh you are right, thank you
$endgroup$
– we_mor
Jan 20 at 4:22