Mixing
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What is wrong with the reasoning in $(-1)^ frac24 = sqrt[4](-1)^2 = sqrt[4]1 = 1$ and $(-1)^ frac24 = (-1)^ frac12 = i$?
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up vote
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$$(-1)^ frac24 = sqrt[4](-1)^2 = sqrt[4]1 = 1$$
$$(-1)^ frac24 = (-1)^ frac12 = i$$
Came across an interesting Y11 question that made pose this one to my self. I can't for the life of me, find the flaw in the initial reasoning.
complex-numbers exponentiation radicals
edited Jul 23 at 8:39
user21820
35.8k440137
asked Jul 23 at 7:20
allquiet1984
507
add a comment |Â
up vote
1
down vote
favorite
$$(-1)^ frac24 = sqrt[4](-1)^2 = sqrt[4]1 = 1$$
$$(-1)^ frac24 = (-1)^ frac12 = i$$
Came across an interesting Y11 question that made pose this one to my self. I can't for the life of me, find the flaw in the initial reasoning.
complex-numbers exponentiation radicals
edited Jul 23 at 8:39
user21820
35.8k440137
asked Jul 23 at 7:20
allquiet1984
507
1
math.stackexchange.com/a/317546/96384, math.stackexchange.com/a/608050/96384
– Torsten Schoeneberg
Jul 23 at 7:26
2
Upshot: For the life of you, don't expect any well-behaved exponentiation / power rules when the base is a negative number.
– Torsten Schoeneberg
Jul 23 at 7:40
1
Related post, and note the comment by Lubin.
– user21820
Jul 23 at 8:36
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$$(-1)^ frac24 = sqrt[4](-1)^2 = sqrt[4]1 = 1$$
$$(-1)^ frac24 = (-1)^ frac12 = i$$
Came across an interesting Y11 question that made pose this one to my self. I can't for the life of me, find the flaw in the initial reasoning.
complex-numbers exponentiation radicals
edited Jul 23 at 8:39
user21820
35.8k440137
asked Jul 23 at 7:20
allquiet1984
507
$$(-1)^ frac24 = sqrt[4](-1)^2 = sqrt[4]1 = 1$$
$$(-1)^ frac24 = (-1)^ frac12 = i$$
Came across an interesting Y11 question that made pose this one to my self. I can't for the life of me, find the flaw in the initial reasoning.
complex-numbers exponentiation radicals
edited Jul 23 at 8:39
user21820
35.8k440137
asked Jul 23 at 7:20
allquiet1984
507
edited Jul 23 at 8:39
user21820
35.8k440137
edited Jul 23 at 8:39
user21820
35.8k440137
edited Jul 23 at 8:39
user21820
35.8k440137
35.8k440137
asked Jul 23 at 7:20
allquiet1984
507
asked Jul 23 at 7:20
allquiet1984
507
asked Jul 23 at 7:20
allquiet1984
507
507
1
math.stackexchange.com/a/317546/96384, math.stackexchange.com/a/608050/96384
– Torsten Schoeneberg
Jul 23 at 7:26
2
Upshot: For the life of you, don't expect any well-behaved exponentiation / power rules when the base is a negative number.
– Torsten Schoeneberg
Jul 23 at 7:40
1
Related post, and note the comment by Lubin.
– user21820
Jul 23 at 8:36
add a comment |Â
1
math.stackexchange.com/a/317546/96384, math.stackexchange.com/a/608050/96384
– Torsten Schoeneberg
Jul 23 at 7:26
2
Upshot: For the life of you, don't expect any well-behaved exponentiation / power rules when the base is a negative number.
– Torsten Schoeneberg
Jul 23 at 7:40
1
Related post, and note the comment by Lubin.
– user21820
Jul 23 at 8:36
1
1
math.stackexchange.com/a/317546/96384, math.stackexchange.com/a/608050/96384
– Torsten Schoeneberg
Jul 23 at 7:26
math.stackexchange.com/a/317546/96384, math.stackexchange.com/a/608050/96384
– Torsten Schoeneberg
Jul 23 at 7:26
2
2
Upshot: For the life of you, don't expect any well-behaved exponentiation / power rules when the base is a negative number.
– Torsten Schoeneberg
Jul 23 at 7:40
Upshot: For the life of you, don't expect any well-behaved exponentiation / power rules when the base is a negative number.
– Torsten Schoeneberg
Jul 23 at 7:40
1
1
Related post, and note the comment by Lubin.
– user21820
Jul 23 at 8:36
Related post, and note the comment by Lubin.
– user21820
Jul 23 at 8:36
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
8
down vote
accepted
The question is essentially the same as asking what's wrong with:
$1=sqrt(1)^2=sqrt(-1)^2=-1$.
The error stems from being ambiguous about the meaning of $sqrtx$ or, more generally, of $sqrt[p]x$. A (real or) complex number $xneq 0$ has a set of $p$ complex $p^th$ roots. If $x$ is a positive real number, exactly one of those roots is going to be a positive real number, so in some contexts (mostly when one is considering only positive real numbers), one talks about that single positive real root as the $p^th$ root of $x$, ignoring the other roots.
But you have to be consistent. You can't say "look, $-2$ is a square root of $4$, but the square root of $4$ is $2$, so $-2=2$".
answered Jul 23 at 7:23
Anonymous
4,8033940
add a comment |Â
up vote
3
down vote
The existing answers are not addressing the real issue, and hence are in my opinion misleading. The problem has nothing to do with multiple roots.
When you write "$a^b$", this is a notation that indicates exponentiation. In modern mathematics, this is typically understood as the exponentiation function applied to the two inputs $a$ and $b$. If we use $pow$ to denote this function, then $a^b = pow(a,b)$.
Now the whole idea of exponentiation is repeated multiplication, and that is why we usually define $pow$ such that for every real $x$ and natural $n$ we have:
$pow(x,0) = 1$
$pow(x,n+1) = pow(x,n) · x$.
To extend this to rational and then real exponents is very non-trivial, and one has to prove the properties that the resulting $pow$ function has. One cannot just assume that the properties that were true for natural exponents remains true for real exponents! (See the linked post for a brief sketch of how it is done.)
In particular, after defining $x^a = pow(x,a)$ for reals $x,a$ such that $x>0$, it will still take a lot of hard work to prove the following nice properties:
(1) $pow(x,a+b) = pow(x,a) · pow(x,b)$ for any real $x>0$ and reals $a,b$.
(2) $pow(x,a·b) = pow(pow(x,a),b)$ for any real $x>0$ and reals $a,b$.
I have purposely written the properties in this manner to make it clear that it is unjustifiable to assume that they hold unless you have proven them. Specifically, you have attempted in your first line to use property (2) for negative $x$:
(Wrong) $colorred pow(-1,2·frac14) = pow(pow(-1,2),frac14) $.
And that is the real mistake; you have moved the factor of "$2$" from the second input of the $pow$ function to the first input, without any justification. It turns out that it is false, and that is why you get nonsense from the first line.
In fact, in this previous post I gave the exact same example to show why one cannot blindly apply previously known facts about some objects to other objects, expecting them to still hold.
answered Jul 24 at 7:24
user21820
35.8k440137
add a comment |Â
up vote
1
down vote
We don't define rational power for negative numbers (accept odd roots, as user21820 noticed). Now, you can use branch cut (as user28120 noticed) to find root of $-1$ but if you do it so, you need to say that $(-1)^1/2=e^ln(-1)cdot1/2$, and when you are doing it in this way, paradoxes won't appear.
answered Jul 23 at 7:38
îріù ïрþш
981513
1
You post is simply wrong. Some authors define the cube-root function to be the inverse of the cube function on the whole real line. Same for odd roots. Secondly, complex exponentiation is typically defined via a branch-cut and so there is nothing wrong with saying that the principal square root of $-1$ is $i$, where we take the branch of the natural logarithm corresponding to the argument $(-À,À]$.
– user21820
Jul 23 at 8:33
Yes, you are right about odd root and branch-cut. I will edit my post.
– Ã®Ñ€Ñ–ù ïрþш
Jul 23 at 9:55
Thanks for editing your post. Unfortunately, it still doesn't answer the question. Indeed taking the principal branch $(-1)^1/2 = e^ln(-1)·1/2 = i$. So that is not the source of the error. The other answer by Anonymous is equally misleading as well...
– user21820
Jul 24 at 6:59
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
The question is essentially the same as asking what's wrong with:
$1=sqrt(1)^2=sqrt(-1)^2=-1$.
The error stems from being ambiguous about the meaning of $sqrtx$ or, more generally, of $sqrt[p]x$. A (real or) complex number $xneq 0$ has a set of $p$ complex $p^th$ roots. If $x$ is a positive real number, exactly one of those roots is going to be a positive real number, so in some contexts (mostly when one is considering only positive real numbers), one talks about that single positive real root as the $p^th$ root of $x$, ignoring the other roots.
But you have to be consistent. You can't say "look, $-2$ is a square root of $4$, but the square root of $4$ is $2$, so $-2=2$".
answered Jul 23 at 7:23
Anonymous
4,8033940
add a comment |Â
up vote
8
down vote
accepted
The question is essentially the same as asking what's wrong with:
$1=sqrt(1)^2=sqrt(-1)^2=-1$.
The error stems from being ambiguous about the meaning of $sqrtx$ or, more generally, of $sqrt[p]x$. A (real or) complex number $xneq 0$ has a set of $p$ complex $p^th$ roots. If $x$ is a positive real number, exactly one of those roots is going to be a positive real number, so in some contexts (mostly when one is considering only positive real numbers), one talks about that single positive real root as the $p^th$ root of $x$, ignoring the other roots.
But you have to be consistent. You can't say "look, $-2$ is a square root of $4$, but the square root of $4$ is $2$, so $-2=2$".
answered Jul 23 at 7:23
Anonymous
4,8033940
add a comment |Â
up vote
8
down vote
accepted
up vote
8
down vote
accepted
The question is essentially the same as asking what's wrong with:
$1=sqrt(1)^2=sqrt(-1)^2=-1$.
The error stems from being ambiguous about the meaning of $sqrtx$ or, more generally, of $sqrt[p]x$. A (real or) complex number $xneq 0$ has a set of $p$ complex $p^th$ roots. If $x$ is a positive real number, exactly one of those roots is going to be a positive real number, so in some contexts (mostly when one is considering only positive real numbers), one talks about that single positive real root as the $p^th$ root of $x$, ignoring the other roots.
But you have to be consistent. You can't say "look, $-2$ is a square root of $4$, but the square root of $4$ is $2$, so $-2=2$".
answered Jul 23 at 7:23
Anonymous
4,8033940
The question is essentially the same as asking what's wrong with:
$1=sqrt(1)^2=sqrt(-1)^2=-1$.
The error stems from being ambiguous about the meaning of $sqrtx$ or, more generally, of $sqrt[p]x$. A (real or) complex number $xneq 0$ has a set of $p$ complex $p^th$ roots. If $x$ is a positive real number, exactly one of those roots is going to be a positive real number, so in some contexts (mostly when one is considering only positive real numbers), one talks about that single positive real root as the $p^th$ root of $x$, ignoring the other roots.
But you have to be consistent. You can't say "look, $-2$ is a square root of $4$, but the square root of $4$ is $2$, so $-2=2$".
answered Jul 23 at 7:23
Anonymous
4,8033940
edited Jul 23 at 7:50
answered Jul 23 at 7:23
Anonymous
4,8033940
answered Jul 23 at 7:23
Anonymous
4,8033940
answered Jul 23 at 7:23
Anonymous
4,8033940
4,8033940
add a comment |Â
add a comment |Â
up vote
3
down vote
The existing answers are not addressing the real issue, and hence are in my opinion misleading. The problem has nothing to do with multiple roots.
When you write "$a^b$", this is a notation that indicates exponentiation. In modern mathematics, this is typically understood as the exponentiation function applied to the two inputs $a$ and $b$. If we use $pow$ to denote this function, then $a^b = pow(a,b)$.
Now the whole idea of exponentiation is repeated multiplication, and that is why we usually define $pow$ such that for every real $x$ and natural $n$ we have:
$pow(x,0) = 1$
$pow(x,n+1) = pow(x,n) · x$.
To extend this to rational and then real exponents is very non-trivial, and one has to prove the properties that the resulting $pow$ function has. One cannot just assume that the properties that were true for natural exponents remains true for real exponents! (See the linked post for a brief sketch of how it is done.)
In particular, after defining $x^a = pow(x,a)$ for reals $x,a$ such that $x>0$, it will still take a lot of hard work to prove the following nice properties:
(1) $pow(x,a+b) = pow(x,a) · pow(x,b)$ for any real $x>0$ and reals $a,b$.
(2) $pow(x,a·b) = pow(pow(x,a),b)$ for any real $x>0$ and reals $a,b$.
I have purposely written the properties in this manner to make it clear that it is unjustifiable to assume that they hold unless you have proven them. Specifically, you have attempted in your first line to use property (2) for negative $x$:
(Wrong) $colorred pow(-1,2·frac14) = pow(pow(-1,2),frac14) $.
And that is the real mistake; you have moved the factor of "$2$" from the second input of the $pow$ function to the first input, without any justification. It turns out that it is false, and that is why you get nonsense from the first line.
In fact, in this previous post I gave the exact same example to show why one cannot blindly apply previously known facts about some objects to other objects, expecting them to still hold.
answered Jul 24 at 7:24
user21820
35.8k440137
add a comment |Â
up vote
3
down vote
The existing answers are not addressing the real issue, and hence are in my opinion misleading. The problem has nothing to do with multiple roots.
When you write "$a^b$", this is a notation that indicates exponentiation. In modern mathematics, this is typically understood as the exponentiation function applied to the two inputs $a$ and $b$. If we use $pow$ to denote this function, then $a^b = pow(a,b)$.
Now the whole idea of exponentiation is repeated multiplication, and that is why we usually define $pow$ such that for every real $x$ and natural $n$ we have:
$pow(x,0) = 1$
$pow(x,n+1) = pow(x,n) · x$.
To extend this to rational and then real exponents is very non-trivial, and one has to prove the properties that the resulting $pow$ function has. One cannot just assume that the properties that were true for natural exponents remains true for real exponents! (See the linked post for a brief sketch of how it is done.)
In particular, after defining $x^a = pow(x,a)$ for reals $x,a$ such that $x>0$, it will still take a lot of hard work to prove the following nice properties:
(1) $pow(x,a+b) = pow(x,a) · pow(x,b)$ for any real $x>0$ and reals $a,b$.
(2) $pow(x,a·b) = pow(pow(x,a),b)$ for any real $x>0$ and reals $a,b$.
I have purposely written the properties in this manner to make it clear that it is unjustifiable to assume that they hold unless you have proven them. Specifically, you have attempted in your first line to use property (2) for negative $x$:
(Wrong) $colorred pow(-1,2·frac14) = pow(pow(-1,2),frac14) $.
And that is the real mistake; you have moved the factor of "$2$" from the second input of the $pow$ function to the first input, without any justification. It turns out that it is false, and that is why you get nonsense from the first line.
In fact, in this previous post I gave the exact same example to show why one cannot blindly apply previously known facts about some objects to other objects, expecting them to still hold.
answered Jul 24 at 7:24
user21820
35.8k440137
add a comment |Â
up vote
3
down vote
up vote
3
down vote
The existing answers are not addressing the real issue, and hence are in my opinion misleading. The problem has nothing to do with multiple roots.
When you write "$a^b$", this is a notation that indicates exponentiation. In modern mathematics, this is typically understood as the exponentiation function applied to the two inputs $a$ and $b$. If we use $pow$ to denote this function, then $a^b = pow(a,b)$.
Now the whole idea of exponentiation is repeated multiplication, and that is why we usually define $pow$ such that for every real $x$ and natural $n$ we have:
$pow(x,0) = 1$
$pow(x,n+1) = pow(x,n) · x$.
To extend this to rational and then real exponents is very non-trivial, and one has to prove the properties that the resulting $pow$ function has. One cannot just assume that the properties that were true for natural exponents remains true for real exponents! (See the linked post for a brief sketch of how it is done.)
In particular, after defining $x^a = pow(x,a)$ for reals $x,a$ such that $x>0$, it will still take a lot of hard work to prove the following nice properties:
(1) $pow(x,a+b) = pow(x,a) · pow(x,b)$ for any real $x>0$ and reals $a,b$.
(2) $pow(x,a·b) = pow(pow(x,a),b)$ for any real $x>0$ and reals $a,b$.
I have purposely written the properties in this manner to make it clear that it is unjustifiable to assume that they hold unless you have proven them. Specifically, you have attempted in your first line to use property (2) for negative $x$:
(Wrong) $colorred pow(-1,2·frac14) = pow(pow(-1,2),frac14) $.
And that is the real mistake; you have moved the factor of "$2$" from the second input of the $pow$ function to the first input, without any justification. It turns out that it is false, and that is why you get nonsense from the first line.
In fact, in this previous post I gave the exact same example to show why one cannot blindly apply previously known facts about some objects to other objects, expecting them to still hold.
answered Jul 24 at 7:24
user21820
35.8k440137
The existing answers are not addressing the real issue, and hence are in my opinion misleading. The problem has nothing to do with multiple roots.
When you write "$a^b$", this is a notation that indicates exponentiation. In modern mathematics, this is typically understood as the exponentiation function applied to the two inputs $a$ and $b$. If we use $pow$ to denote this function, then $a^b = pow(a,b)$.
Now the whole idea of exponentiation is repeated multiplication, and that is why we usually define $pow$ such that for every real $x$ and natural $n$ we have:
$pow(x,0) = 1$
$pow(x,n+1) = pow(x,n) · x$.
To extend this to rational and then real exponents is very non-trivial, and one has to prove the properties that the resulting $pow$ function has. One cannot just assume that the properties that were true for natural exponents remains true for real exponents! (See the linked post for a brief sketch of how it is done.)
In particular, after defining $x^a = pow(x,a)$ for reals $x,a$ such that $x>0$, it will still take a lot of hard work to prove the following nice properties:
(1) $pow(x,a+b) = pow(x,a) · pow(x,b)$ for any real $x>0$ and reals $a,b$.
(2) $pow(x,a·b) = pow(pow(x,a),b)$ for any real $x>0$ and reals $a,b$.
I have purposely written the properties in this manner to make it clear that it is unjustifiable to assume that they hold unless you have proven them. Specifically, you have attempted in your first line to use property (2) for negative $x$:
(Wrong) $colorred pow(-1,2·frac14) = pow(pow(-1,2),frac14) $.
And that is the real mistake; you have moved the factor of "$2$" from the second input of the $pow$ function to the first input, without any justification. It turns out that it is false, and that is why you get nonsense from the first line.
In fact, in this previous post I gave the exact same example to show why one cannot blindly apply previously known facts about some objects to other objects, expecting them to still hold.
answered Jul 24 at 7:24
user21820
35.8k440137
answered Jul 24 at 7:24
user21820
35.8k440137
answered Jul 24 at 7:24
user21820
35.8k440137
answered Jul 24 at 7:24
user21820
35.8k440137
35.8k440137
add a comment |Â
add a comment |Â
up vote
1
down vote
We don't define rational power for negative numbers (accept odd roots, as user21820 noticed). Now, you can use branch cut (as user28120 noticed) to find root of $-1$ but if you do it so, you need to say that $(-1)^1/2=e^ln(-1)cdot1/2$, and when you are doing it in this way, paradoxes won't appear.
answered Jul 23 at 7:38
îріù ïрþш
981513
1
You post is simply wrong. Some authors define the cube-root function to be the inverse of the cube function on the whole real line. Same for odd roots. Secondly, complex exponentiation is typically defined via a branch-cut and so there is nothing wrong with saying that the principal square root of $-1$ is $i$, where we take the branch of the natural logarithm corresponding to the argument $(-À,À]$.
– user21820
Jul 23 at 8:33
Yes, you are right about odd root and branch-cut. I will edit my post.
– Ã®Ñ€Ñ–ù ïрþш
Jul 23 at 9:55
Thanks for editing your post. Unfortunately, it still doesn't answer the question. Indeed taking the principal branch $(-1)^1/2 = e^ln(-1)·1/2 = i$. So that is not the source of the error. The other answer by Anonymous is equally misleading as well...
– user21820
Jul 24 at 6:59
add a comment |Â
up vote
1
down vote
We don't define rational power for negative numbers (accept odd roots, as user21820 noticed). Now, you can use branch cut (as user28120 noticed) to find root of $-1$ but if you do it so, you need to say that $(-1)^1/2=e^ln(-1)cdot1/2$, and when you are doing it in this way, paradoxes won't appear.
answered Jul 23 at 7:38
îріù ïрþш
981513
1
You post is simply wrong. Some authors define the cube-root function to be the inverse of the cube function on the whole real line. Same for odd roots. Secondly, complex exponentiation is typically defined via a branch-cut and so there is nothing wrong with saying that the principal square root of $-1$ is $i$, where we take the branch of the natural logarithm corresponding to the argument $(-À,À]$.
– user21820
Jul 23 at 8:33
Yes, you are right about odd root and branch-cut. I will edit my post.
– Ã®Ñ€Ñ–ù ïрþш
Jul 23 at 9:55
Thanks for editing your post. Unfortunately, it still doesn't answer the question. Indeed taking the principal branch $(-1)^1/2 = e^ln(-1)·1/2 = i$. So that is not the source of the error. The other answer by Anonymous is equally misleading as well...
– user21820
Jul 24 at 6:59
add a comment |Â
up vote
1
down vote
up vote
1
down vote
We don't define rational power for negative numbers (accept odd roots, as user21820 noticed). Now, you can use branch cut (as user28120 noticed) to find root of $-1$ but if you do it so, you need to say that $(-1)^1/2=e^ln(-1)cdot1/2$, and when you are doing it in this way, paradoxes won't appear.
answered Jul 23 at 7:38
îріù ïрþш
981513
We don't define rational power for negative numbers (accept odd roots, as user21820 noticed). Now, you can use branch cut (as user28120 noticed) to find root of $-1$ but if you do it so, you need to say that $(-1)^1/2=e^ln(-1)cdot1/2$, and when you are doing it in this way, paradoxes won't appear.
answered Jul 23 at 7:38
îріù ïрþш
981513
edited Jul 23 at 10:07
answered Jul 23 at 7:38
îріù ïрþш
981513
answered Jul 23 at 7:38
îріù ïрþш
981513
answered Jul 23 at 7:38
îріù ïрþш
981513
981513
1
You post is simply wrong. Some authors define the cube-root function to be the inverse of the cube function on the whole real line. Same for odd roots. Secondly, complex exponentiation is typically defined via a branch-cut and so there is nothing wrong with saying that the principal square root of $-1$ is $i$, where we take the branch of the natural logarithm corresponding to the argument $(-À,À]$.
– user21820
Jul 23 at 8:33
Yes, you are right about odd root and branch-cut. I will edit my post.
– Ã®Ñ€Ñ–ù ïрþш
Jul 23 at 9:55
Thanks for editing your post. Unfortunately, it still doesn't answer the question. Indeed taking the principal branch $(-1)^1/2 = e^ln(-1)·1/2 = i$. So that is not the source of the error. The other answer by Anonymous is equally misleading as well...
– user21820
Jul 24 at 6:59
add a comment |Â
1
You post is simply wrong. Some authors define the cube-root function to be the inverse of the cube function on the whole real line. Same for odd roots. Secondly, complex exponentiation is typically defined via a branch-cut and so there is nothing wrong with saying that the principal square root of $-1$ is $i$, where we take the branch of the natural logarithm corresponding to the argument $(-À,À]$.
– user21820
Jul 23 at 8:33
Yes, you are right about odd root and branch-cut. I will edit my post.
– Ã®Ñ€Ñ–ù ïрþш
Jul 23 at 9:55
Thanks for editing your post. Unfortunately, it still doesn't answer the question. Indeed taking the principal branch $(-1)^1/2 = e^ln(-1)·1/2 = i$. So that is not the source of the error. The other answer by Anonymous is equally misleading as well...
– user21820
Jul 24 at 6:59
1
1
You post is simply wrong. Some authors define the cube-root function to be the inverse of the cube function on the whole real line. Same for odd roots. Secondly, complex exponentiation is typically defined via a branch-cut and so there is nothing wrong with saying that the principal square root of $-1$ is $i$, where we take the branch of the natural logarithm corresponding to the argument $(-À,À]$.
– user21820
Jul 23 at 8:33
You post is simply wrong. Some authors define the cube-root function to be the inverse of the cube function on the whole real line. Same for odd roots. Secondly, complex exponentiation is typically defined via a branch-cut and so there is nothing wrong with saying that the principal square root of $-1$ is $i$, where we take the branch of the natural logarithm corresponding to the argument $(-À,À]$.
– user21820
Jul 23 at 8:33
Yes, you are right about odd root and branch-cut. I will edit my post.
– Ã®Ñ€Ñ–ù ïрþш
Jul 23 at 9:55
Yes, you are right about odd root and branch-cut. I will edit my post.
– Ã®Ñ€Ñ–ù ïрþш
Jul 23 at 9:55
Thanks for editing your post. Unfortunately, it still doesn't answer the question. Indeed taking the principal branch $(-1)^1/2 = e^ln(-1)·1/2 = i$. So that is not the source of the error. The other answer by Anonymous is equally misleading as well...
– user21820
Jul 24 at 6:59
Thanks for editing your post. Unfortunately, it still doesn't answer the question. Indeed taking the principal branch $(-1)^1/2 = e^ln(-1)·1/2 = i$. So that is not the source of the error. The other answer by Anonymous is equally misleading as well...
– user21820
Jul 24 at 6:59
add a comment |Â
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1
math.stackexchange.com/a/317546/96384, math.stackexchange.com/a/608050/96384
– Torsten Schoeneberg
Jul 23 at 7:26
2
Upshot: For the life of you, don't expect any well-behaved exponentiation / power rules when the base is a negative number.
– Torsten Schoeneberg
Jul 23 at 7:40
1
Related post, and note the comment by Lubin.
– user21820
Jul 23 at 8:36