Mixing
Can someone prove why $sqrt{ab}=sqrt{a}sqrt{b}$ is only valid when a and b are positive?
up vote
9
down vote
favorite
I have seen many people say that a and b can't be positive for example in this false proof :
$$1=sqrt{1}=sqrt{(-1)(-1)}=sqrt{-1} sqrt{-1} = i^2 = -1$$
Trust me, I understand that $1neq -1$ and also by seeing this, I believe and accept that a and b should be positive or greater than 0 (for $sqrt{ab}=sqrt{a}sqrt{b}$)
But I'm interested to know why is that, which is something I don't know? What is the proof ?
Thanks a lot.
complex-numbers arithmetic
asked Jan 8 '15 at 20:22
M.S.E
644836
|
show 9 more comments
up vote
9
down vote
favorite
I have seen many people say that a and b can't be positive for example in this false proof :
$$1=sqrt{1}=sqrt{(-1)(-1)}=sqrt{-1} sqrt{-1} = i^2 = -1$$
Trust me, I understand that $1neq -1$ and also by seeing this, I believe and accept that a and b should be positive or greater than 0 (for $sqrt{ab}=sqrt{a}sqrt{b}$)
But I'm interested to know why is that, which is something I don't know? What is the proof ?
Thanks a lot.
complex-numbers arithmetic
asked Jan 8 '15 at 20:22
M.S.E
644836
In general, $(ab)^x=a^xb^x$ for any $age 0$ and $b,xinmathbb C$. So no, you don't need both to be positive.
– user2345215
Jan 8 '15 at 20:27
1
If you work in $mathbb R$ then the square root of a negative number is not defined. If you work in $mathbb C$ then the square root is a "multi-valued" function, so $sqrt{-1}$ could be $i$ or $-i$.
– Martin R
Jan 8 '15 at 20:33
The real underlying reason is that "the square root" has to be carefully defined when you try to generalize it. Notice that there are two square roots of any number. Because of that, we have to be careful in our definition for the square root to be well-defined.
– Max
Jan 8 '15 at 20:34
@MartinR really? I was taught that $sqrt{-1}$ is $i$
– M.S.E
Jan 8 '15 at 20:35
1
@user2345215: $a^b$ is so far from being well-defined that it usually has infinitely many possible values. The best one can do is to define it to be $e^{b log(a)}$, but the complex valued logarithm function has infinitely many branches.
– Lee Mosher
Jan 8 '15 at 20:37
|
show 9 more comments
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I have seen many people say that a and b can't be positive for example in this false proof :
$$1=sqrt{1}=sqrt{(-1)(-1)}=sqrt{-1} sqrt{-1} = i^2 = -1$$
Trust me, I understand that $1neq -1$ and also by seeing this, I believe and accept that a and b should be positive or greater than 0 (for $sqrt{ab}=sqrt{a}sqrt{b}$)
But I'm interested to know why is that, which is something I don't know? What is the proof ?
Thanks a lot.
complex-numbers arithmetic
asked Jan 8 '15 at 20:22
M.S.E
644836
I have seen many people say that a and b can't be positive for example in this false proof :
$$1=sqrt{1}=sqrt{(-1)(-1)}=sqrt{-1} sqrt{-1} = i^2 = -1$$
Trust me, I understand that $1neq -1$ and also by seeing this, I believe and accept that a and b should be positive or greater than 0 (for $sqrt{ab}=sqrt{a}sqrt{b}$)
But I'm interested to know why is that, which is something I don't know? What is the proof ?
Thanks a lot.
complex-numbers arithmetic
complex-numbers arithmetic
asked Jan 8 '15 at 20:22
M.S.E
644836
asked Jan 8 '15 at 20:22
M.S.E
644836
asked Jan 8 '15 at 20:22
M.S.E
644836
asked Jan 8 '15 at 20:22
M.S.E
644836
asked Jan 8 '15 at 20:22
M.S.E
644836
644836
In general, $(ab)^x=a^xb^x$ for any $age 0$ and $b,xinmathbb C$. So no, you don't need both to be positive.
– user2345215
Jan 8 '15 at 20:27
1
If you work in $mathbb R$ then the square root of a negative number is not defined. If you work in $mathbb C$ then the square root is a "multi-valued" function, so $sqrt{-1}$ could be $i$ or $-i$.
– Martin R
Jan 8 '15 at 20:33
The real underlying reason is that "the square root" has to be carefully defined when you try to generalize it. Notice that there are two square roots of any number. Because of that, we have to be careful in our definition for the square root to be well-defined.
– Max
Jan 8 '15 at 20:34
@MartinR really? I was taught that $sqrt{-1}$ is $i$
– M.S.E
Jan 8 '15 at 20:35
1
@user2345215: $a^b$ is so far from being well-defined that it usually has infinitely many possible values. The best one can do is to define it to be $e^{b log(a)}$, but the complex valued logarithm function has infinitely many branches.
– Lee Mosher
Jan 8 '15 at 20:37
|
show 9 more comments
In general, $(ab)^x=a^xb^x$ for any $age 0$ and $b,xinmathbb C$. So no, you don't need both to be positive.
– user2345215
Jan 8 '15 at 20:27
1
If you work in $mathbb R$ then the square root of a negative number is not defined. If you work in $mathbb C$ then the square root is a "multi-valued" function, so $sqrt{-1}$ could be $i$ or $-i$.
– Martin R
Jan 8 '15 at 20:33
The real underlying reason is that "the square root" has to be carefully defined when you try to generalize it. Notice that there are two square roots of any number. Because of that, we have to be careful in our definition for the square root to be well-defined.
– Max
Jan 8 '15 at 20:34
@MartinR really? I was taught that $sqrt{-1}$ is $i$
– M.S.E
Jan 8 '15 at 20:35
1
@user2345215: $a^b$ is so far from being well-defined that it usually has infinitely many possible values. The best one can do is to define it to be $e^{b log(a)}$, but the complex valued logarithm function has infinitely many branches.
– Lee Mosher
Jan 8 '15 at 20:37
In general, $(ab)^x=a^xb^x$ for any $age 0$ and $b,xinmathbb C$. So no, you don't need both to be positive.
– user2345215
Jan 8 '15 at 20:27
In general, $(ab)^x=a^xb^x$ for any $age 0$ and $b,xinmathbb C$. So no, you don't need both to be positive.
– user2345215
Jan 8 '15 at 20:27
1
1
If you work in $mathbb R$ then the square root of a negative number is not defined. If you work in $mathbb C$ then the square root is a "multi-valued" function, so $sqrt{-1}$ could be $i$ or $-i$.
– Martin R
Jan 8 '15 at 20:33
If you work in $mathbb R$ then the square root of a negative number is not defined. If you work in $mathbb C$ then the square root is a "multi-valued" function, so $sqrt{-1}$ could be $i$ or $-i$.
– Martin R
Jan 8 '15 at 20:33
The real underlying reason is that "the square root" has to be carefully defined when you try to generalize it. Notice that there are two square roots of any number. Because of that, we have to be careful in our definition for the square root to be well-defined.
– Max
Jan 8 '15 at 20:34
The real underlying reason is that "the square root" has to be carefully defined when you try to generalize it. Notice that there are two square roots of any number. Because of that, we have to be careful in our definition for the square root to be well-defined.
– Max
Jan 8 '15 at 20:34
@MartinR really? I was taught that $sqrt{-1}$ is $i$
– M.S.E
Jan 8 '15 at 20:35
@MartinR really? I was taught that $sqrt{-1}$ is $i$
– M.S.E
Jan 8 '15 at 20:35
1
1
@user2345215: $a^b$ is so far from being well-defined that it usually has infinitely many possible values. The best one can do is to define it to be $e^{b log(a)}$, but the complex valued logarithm function has infinitely many branches.
– Lee Mosher
Jan 8 '15 at 20:37
@user2345215: $a^b$ is so far from being well-defined that it usually has infinitely many possible values. The best one can do is to define it to be $e^{b log(a)}$, but the complex valued logarithm function has infinitely many branches.
– Lee Mosher
Jan 8 '15 at 20:37
|
show 9 more comments
2 Answers
2
active
oldest
votes
up vote
8
down vote
accepted
If $a$ is a complex number and you write $sqrt{a}$ to denote a specific number, you have introduced a problem, as there are two complex numbeers whose square is $a$.
The problem shows up when you need to choose one of the two alternatives for all complex numbers at the same time, and this is unavoidable in some situations. For example, notice that for a statement such as
for all complex numbers $a$ and $b$ we have $sqrt asqrt b=sqrt{ab}$
to mean anything, we need to give sense to $sqrt{a}$ for all $a$s.
A good choice for $sqrt{-1}$ is certainly $i$. A good choice for $sqrt{1}$ is $1$, of course. Now $sqrt{-1}cdotsqrt{-1}$ is, according to these choices, equal to $icdot i$ which is $-1$. On the other hand $sqrt{(-1)(-1)}$ is $sqrt{1}$ and we chose that to be $1$; we have a problem.
Ok. Maybe we should have chosen $sqrt{1}$ to be $-1$? Let's see what happens. Now $sqrt{1}cdotsqrt{1}$ is $(-1)cdot(-1)$, which is $1$, and $sqrt{1cdot1}=sqrt{1}=-1$: oh no!
And you can go on like this...
Indeed, theere are four choices in all for what $sqrt{1}$ and $sqrt{-1}$ can mean:
| sqrt{1} sqrt{-1}
| 1 i
| 1 -i
| -1 i
| -1 -i
In each of these four options you can reach a problem.
edited Jan 8 '15 at 22:21
MJD
46.6k28205387
answered Jan 8 '15 at 20:40
Mariano Suárez-Álvarez
110k7154280
I don't think this is fully satisfactory. In the positive reals we also have the problem of $2$ roots, yet people define $sqrt x$ without problems. In $mathbb C$, we can still define $sqrt x$ (or any power) using the principal branch of the complex logarithm. While the formula $sqrt{a}sqrt{b}=sqrt{ab}$ doesn't hold for arbitrary $a,binmathbb C$, it holds if we require just one of them to be nonnegative real. I think this is still useful.
– user2345215
Jan 8 '15 at 20:48
2
When we only care about defininf square roots of positive real numbers, this does not arise: we can choose $sqrt{a}$ for $ageq0$ to denote the nonnegative real square root, and then we can actually prove our choices work out. But choosing square roots for a larger set than that is more demanding, because more equalities have to hold.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
@user2345215, yes , the formula has validity in several special situations. That is irrelevant, really.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
See also wikipedia en.wikipedia.org/wiki/…
– Lehs
Jan 8 '15 at 22:38
add a comment |
up vote
0
down vote
$sqrt{x}$ is defined in $mathbb{R}$ as long as $x geq 0$. Thus in the first place for $sqrt{ab}$ to be defined and make sense in $mathbb{R}$ we would need $ab geq 0.$ When $ab=0,$ the case is trivial. So let's consider the case when $ab>0$ which yeilds either $a>0,b>0$ or $a<0,b<0$. In the first case, both $sqrt{a}$ and $sqrt{b}$ are well defined and hence $sqrt{ab}=sqrt{a}sqrt{b}$ is true. However, in the second case, as both $a$ and $b$ are negative, neither of $sqrt{a}$ and $sqrt{b}$ are defined. Thus we cannot write $sqrt{ab}=sqrt{a}sqrt{b}$ when $a$ and $b$ are negative.
As far as complex numbers are concerned, your question does not apply since complex numbers are not ordered in the usual sense. I mean, you cannot talk of a complex number being positive or negative. For example, if $i$ were positive, then $i^2 geq 0$ which is not true as $-1 ngeq 0$ and also if $i$ were negative, then again we get $i^2 geq 0$ which is not true as $-1 ngeq 0$.
answered Oct 10 '16 at 0:37
Ralph Lazarus
11
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
If $a$ is a complex number and you write $sqrt{a}$ to denote a specific number, you have introduced a problem, as there are two complex numbeers whose square is $a$.
The problem shows up when you need to choose one of the two alternatives for all complex numbers at the same time, and this is unavoidable in some situations. For example, notice that for a statement such as
for all complex numbers $a$ and $b$ we have $sqrt asqrt b=sqrt{ab}$
to mean anything, we need to give sense to $sqrt{a}$ for all $a$s.
A good choice for $sqrt{-1}$ is certainly $i$. A good choice for $sqrt{1}$ is $1$, of course. Now $sqrt{-1}cdotsqrt{-1}$ is, according to these choices, equal to $icdot i$ which is $-1$. On the other hand $sqrt{(-1)(-1)}$ is $sqrt{1}$ and we chose that to be $1$; we have a problem.
Ok. Maybe we should have chosen $sqrt{1}$ to be $-1$? Let's see what happens. Now $sqrt{1}cdotsqrt{1}$ is $(-1)cdot(-1)$, which is $1$, and $sqrt{1cdot1}=sqrt{1}=-1$: oh no!
And you can go on like this...
Indeed, theere are four choices in all for what $sqrt{1}$ and $sqrt{-1}$ can mean:
| sqrt{1} sqrt{-1}
| 1 i
| 1 -i
| -1 i
| -1 -i
In each of these four options you can reach a problem.
edited Jan 8 '15 at 22:21
MJD
46.6k28205387
answered Jan 8 '15 at 20:40
Mariano Suárez-Álvarez
110k7154280
I don't think this is fully satisfactory. In the positive reals we also have the problem of $2$ roots, yet people define $sqrt x$ without problems. In $mathbb C$, we can still define $sqrt x$ (or any power) using the principal branch of the complex logarithm. While the formula $sqrt{a}sqrt{b}=sqrt{ab}$ doesn't hold for arbitrary $a,binmathbb C$, it holds if we require just one of them to be nonnegative real. I think this is still useful.
– user2345215
Jan 8 '15 at 20:48
2
When we only care about defininf square roots of positive real numbers, this does not arise: we can choose $sqrt{a}$ for $ageq0$ to denote the nonnegative real square root, and then we can actually prove our choices work out. But choosing square roots for a larger set than that is more demanding, because more equalities have to hold.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
@user2345215, yes , the formula has validity in several special situations. That is irrelevant, really.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
See also wikipedia en.wikipedia.org/wiki/…
– Lehs
Jan 8 '15 at 22:38
add a comment |
up vote
8
down vote
accepted
If $a$ is a complex number and you write $sqrt{a}$ to denote a specific number, you have introduced a problem, as there are two complex numbeers whose square is $a$.
The problem shows up when you need to choose one of the two alternatives for all complex numbers at the same time, and this is unavoidable in some situations. For example, notice that for a statement such as
for all complex numbers $a$ and $b$ we have $sqrt asqrt b=sqrt{ab}$
to mean anything, we need to give sense to $sqrt{a}$ for all $a$s.
A good choice for $sqrt{-1}$ is certainly $i$. A good choice for $sqrt{1}$ is $1$, of course. Now $sqrt{-1}cdotsqrt{-1}$ is, according to these choices, equal to $icdot i$ which is $-1$. On the other hand $sqrt{(-1)(-1)}$ is $sqrt{1}$ and we chose that to be $1$; we have a problem.
Ok. Maybe we should have chosen $sqrt{1}$ to be $-1$? Let's see what happens. Now $sqrt{1}cdotsqrt{1}$ is $(-1)cdot(-1)$, which is $1$, and $sqrt{1cdot1}=sqrt{1}=-1$: oh no!
And you can go on like this...
Indeed, theere are four choices in all for what $sqrt{1}$ and $sqrt{-1}$ can mean:
| sqrt{1} sqrt{-1}
| 1 i
| 1 -i
| -1 i
| -1 -i
In each of these four options you can reach a problem.
edited Jan 8 '15 at 22:21
MJD
46.6k28205387
answered Jan 8 '15 at 20:40
Mariano Suárez-Álvarez
110k7154280
I don't think this is fully satisfactory. In the positive reals we also have the problem of $2$ roots, yet people define $sqrt x$ without problems. In $mathbb C$, we can still define $sqrt x$ (or any power) using the principal branch of the complex logarithm. While the formula $sqrt{a}sqrt{b}=sqrt{ab}$ doesn't hold for arbitrary $a,binmathbb C$, it holds if we require just one of them to be nonnegative real. I think this is still useful.
– user2345215
Jan 8 '15 at 20:48
2
When we only care about defininf square roots of positive real numbers, this does not arise: we can choose $sqrt{a}$ for $ageq0$ to denote the nonnegative real square root, and then we can actually prove our choices work out. But choosing square roots for a larger set than that is more demanding, because more equalities have to hold.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
@user2345215, yes , the formula has validity in several special situations. That is irrelevant, really.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
See also wikipedia en.wikipedia.org/wiki/…
– Lehs
Jan 8 '15 at 22:38
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
If $a$ is a complex number and you write $sqrt{a}$ to denote a specific number, you have introduced a problem, as there are two complex numbeers whose square is $a$.
The problem shows up when you need to choose one of the two alternatives for all complex numbers at the same time, and this is unavoidable in some situations. For example, notice that for a statement such as
for all complex numbers $a$ and $b$ we have $sqrt asqrt b=sqrt{ab}$
to mean anything, we need to give sense to $sqrt{a}$ for all $a$s.
A good choice for $sqrt{-1}$ is certainly $i$. A good choice for $sqrt{1}$ is $1$, of course. Now $sqrt{-1}cdotsqrt{-1}$ is, according to these choices, equal to $icdot i$ which is $-1$. On the other hand $sqrt{(-1)(-1)}$ is $sqrt{1}$ and we chose that to be $1$; we have a problem.
Ok. Maybe we should have chosen $sqrt{1}$ to be $-1$? Let's see what happens. Now $sqrt{1}cdotsqrt{1}$ is $(-1)cdot(-1)$, which is $1$, and $sqrt{1cdot1}=sqrt{1}=-1$: oh no!
And you can go on like this...
Indeed, theere are four choices in all for what $sqrt{1}$ and $sqrt{-1}$ can mean:
| sqrt{1} sqrt{-1}
| 1 i
| 1 -i
| -1 i
| -1 -i
In each of these four options you can reach a problem.
edited Jan 8 '15 at 22:21
MJD
46.6k28205387
answered Jan 8 '15 at 20:40
Mariano Suárez-Álvarez
110k7154280
If $a$ is a complex number and you write $sqrt{a}$ to denote a specific number, you have introduced a problem, as there are two complex numbeers whose square is $a$.
The problem shows up when you need to choose one of the two alternatives for all complex numbers at the same time, and this is unavoidable in some situations. For example, notice that for a statement such as
for all complex numbers $a$ and $b$ we have $sqrt asqrt b=sqrt{ab}$
to mean anything, we need to give sense to $sqrt{a}$ for all $a$s.
A good choice for $sqrt{-1}$ is certainly $i$. A good choice for $sqrt{1}$ is $1$, of course. Now $sqrt{-1}cdotsqrt{-1}$ is, according to these choices, equal to $icdot i$ which is $-1$. On the other hand $sqrt{(-1)(-1)}$ is $sqrt{1}$ and we chose that to be $1$; we have a problem.
Ok. Maybe we should have chosen $sqrt{1}$ to be $-1$? Let's see what happens. Now $sqrt{1}cdotsqrt{1}$ is $(-1)cdot(-1)$, which is $1$, and $sqrt{1cdot1}=sqrt{1}=-1$: oh no!
And you can go on like this...
Indeed, theere are four choices in all for what $sqrt{1}$ and $sqrt{-1}$ can mean:
| sqrt{1} sqrt{-1}
| 1 i
| 1 -i
| -1 i
| -1 -i
In each of these four options you can reach a problem.
edited Jan 8 '15 at 22:21
MJD
46.6k28205387
answered Jan 8 '15 at 20:40
Mariano Suárez-Álvarez
110k7154280
edited Jan 8 '15 at 22:21
MJD
46.6k28205387
edited Jan 8 '15 at 22:21
MJD
46.6k28205387
edited Jan 8 '15 at 22:21
MJD
46.6k28205387
46.6k28205387
answered Jan 8 '15 at 20:40
Mariano Suárez-Álvarez
110k7154280
answered Jan 8 '15 at 20:40
Mariano Suárez-Álvarez
110k7154280
answered Jan 8 '15 at 20:40
Mariano Suárez-Álvarez
110k7154280
110k7154280
I don't think this is fully satisfactory. In the positive reals we also have the problem of $2$ roots, yet people define $sqrt x$ without problems. In $mathbb C$, we can still define $sqrt x$ (or any power) using the principal branch of the complex logarithm. While the formula $sqrt{a}sqrt{b}=sqrt{ab}$ doesn't hold for arbitrary $a,binmathbb C$, it holds if we require just one of them to be nonnegative real. I think this is still useful.
– user2345215
Jan 8 '15 at 20:48
2
When we only care about defininf square roots of positive real numbers, this does not arise: we can choose $sqrt{a}$ for $ageq0$ to denote the nonnegative real square root, and then we can actually prove our choices work out. But choosing square roots for a larger set than that is more demanding, because more equalities have to hold.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
@user2345215, yes , the formula has validity in several special situations. That is irrelevant, really.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
See also wikipedia en.wikipedia.org/wiki/…
– Lehs
Jan 8 '15 at 22:38
add a comment |
I don't think this is fully satisfactory. In the positive reals we also have the problem of $2$ roots, yet people define $sqrt x$ without problems. In $mathbb C$, we can still define $sqrt x$ (or any power) using the principal branch of the complex logarithm. While the formula $sqrt{a}sqrt{b}=sqrt{ab}$ doesn't hold for arbitrary $a,binmathbb C$, it holds if we require just one of them to be nonnegative real. I think this is still useful.
– user2345215
Jan 8 '15 at 20:48
2
When we only care about defininf square roots of positive real numbers, this does not arise: we can choose $sqrt{a}$ for $ageq0$ to denote the nonnegative real square root, and then we can actually prove our choices work out. But choosing square roots for a larger set than that is more demanding, because more equalities have to hold.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
@user2345215, yes , the formula has validity in several special situations. That is irrelevant, really.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
See also wikipedia en.wikipedia.org/wiki/…
– Lehs
Jan 8 '15 at 22:38
I don't think this is fully satisfactory. In the positive reals we also have the problem of $2$ roots, yet people define $sqrt x$ without problems. In $mathbb C$, we can still define $sqrt x$ (or any power) using the principal branch of the complex logarithm. While the formula $sqrt{a}sqrt{b}=sqrt{ab}$ doesn't hold for arbitrary $a,binmathbb C$, it holds if we require just one of them to be nonnegative real. I think this is still useful.
– user2345215
Jan 8 '15 at 20:48
I don't think this is fully satisfactory. In the positive reals we also have the problem of $2$ roots, yet people define $sqrt x$ without problems. In $mathbb C$, we can still define $sqrt x$ (or any power) using the principal branch of the complex logarithm. While the formula $sqrt{a}sqrt{b}=sqrt{ab}$ doesn't hold for arbitrary $a,binmathbb C$, it holds if we require just one of them to be nonnegative real. I think this is still useful.
– user2345215
Jan 8 '15 at 20:48
2
2
When we only care about defininf square roots of positive real numbers, this does not arise: we can choose $sqrt{a}$ for $ageq0$ to denote the nonnegative real square root, and then we can actually prove our choices work out. But choosing square roots for a larger set than that is more demanding, because more equalities have to hold.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
When we only care about defininf square roots of positive real numbers, this does not arise: we can choose $sqrt{a}$ for $ageq0$ to denote the nonnegative real square root, and then we can actually prove our choices work out. But choosing square roots for a larger set than that is more demanding, because more equalities have to hold.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
@user2345215, yes , the formula has validity in several special situations. That is irrelevant, really.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
@user2345215, yes , the formula has validity in several special situations. That is irrelevant, really.
– Mariano Suárez-Álvarez
Jan 8 '15 at 20:49
See also wikipedia en.wikipedia.org/wiki/…
– Lehs
Jan 8 '15 at 22:38
See also wikipedia en.wikipedia.org/wiki/…
– Lehs
Jan 8 '15 at 22:38
add a comment |
up vote
0
down vote
$sqrt{x}$ is defined in $mathbb{R}$ as long as $x geq 0$. Thus in the first place for $sqrt{ab}$ to be defined and make sense in $mathbb{R}$ we would need $ab geq 0.$ When $ab=0,$ the case is trivial. So let's consider the case when $ab>0$ which yeilds either $a>0,b>0$ or $a<0,b<0$. In the first case, both $sqrt{a}$ and $sqrt{b}$ are well defined and hence $sqrt{ab}=sqrt{a}sqrt{b}$ is true. However, in the second case, as both $a$ and $b$ are negative, neither of $sqrt{a}$ and $sqrt{b}$ are defined. Thus we cannot write $sqrt{ab}=sqrt{a}sqrt{b}$ when $a$ and $b$ are negative.
As far as complex numbers are concerned, your question does not apply since complex numbers are not ordered in the usual sense. I mean, you cannot talk of a complex number being positive or negative. For example, if $i$ were positive, then $i^2 geq 0$ which is not true as $-1 ngeq 0$ and also if $i$ were negative, then again we get $i^2 geq 0$ which is not true as $-1 ngeq 0$.
answered Oct 10 '16 at 0:37
Ralph Lazarus
11
add a comment |
up vote
0
down vote
$sqrt{x}$ is defined in $mathbb{R}$ as long as $x geq 0$. Thus in the first place for $sqrt{ab}$ to be defined and make sense in $mathbb{R}$ we would need $ab geq 0.$ When $ab=0,$ the case is trivial. So let's consider the case when $ab>0$ which yeilds either $a>0,b>0$ or $a<0,b<0$. In the first case, both $sqrt{a}$ and $sqrt{b}$ are well defined and hence $sqrt{ab}=sqrt{a}sqrt{b}$ is true. However, in the second case, as both $a$ and $b$ are negative, neither of $sqrt{a}$ and $sqrt{b}$ are defined. Thus we cannot write $sqrt{ab}=sqrt{a}sqrt{b}$ when $a$ and $b$ are negative.
As far as complex numbers are concerned, your question does not apply since complex numbers are not ordered in the usual sense. I mean, you cannot talk of a complex number being positive or negative. For example, if $i$ were positive, then $i^2 geq 0$ which is not true as $-1 ngeq 0$ and also if $i$ were negative, then again we get $i^2 geq 0$ which is not true as $-1 ngeq 0$.
answered Oct 10 '16 at 0:37
Ralph Lazarus
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$sqrt{x}$ is defined in $mathbb{R}$ as long as $x geq 0$. Thus in the first place for $sqrt{ab}$ to be defined and make sense in $mathbb{R}$ we would need $ab geq 0.$ When $ab=0,$ the case is trivial. So let's consider the case when $ab>0$ which yeilds either $a>0,b>0$ or $a<0,b<0$. In the first case, both $sqrt{a}$ and $sqrt{b}$ are well defined and hence $sqrt{ab}=sqrt{a}sqrt{b}$ is true. However, in the second case, as both $a$ and $b$ are negative, neither of $sqrt{a}$ and $sqrt{b}$ are defined. Thus we cannot write $sqrt{ab}=sqrt{a}sqrt{b}$ when $a$ and $b$ are negative.
As far as complex numbers are concerned, your question does not apply since complex numbers are not ordered in the usual sense. I mean, you cannot talk of a complex number being positive or negative. For example, if $i$ were positive, then $i^2 geq 0$ which is not true as $-1 ngeq 0$ and also if $i$ were negative, then again we get $i^2 geq 0$ which is not true as $-1 ngeq 0$.
answered Oct 10 '16 at 0:37
Ralph Lazarus
11
$sqrt{x}$ is defined in $mathbb{R}$ as long as $x geq 0$. Thus in the first place for $sqrt{ab}$ to be defined and make sense in $mathbb{R}$ we would need $ab geq 0.$ When $ab=0,$ the case is trivial. So let's consider the case when $ab>0$ which yeilds either $a>0,b>0$ or $a<0,b<0$. In the first case, both $sqrt{a}$ and $sqrt{b}$ are well defined and hence $sqrt{ab}=sqrt{a}sqrt{b}$ is true. However, in the second case, as both $a$ and $b$ are negative, neither of $sqrt{a}$ and $sqrt{b}$ are defined. Thus we cannot write $sqrt{ab}=sqrt{a}sqrt{b}$ when $a$ and $b$ are negative.
As far as complex numbers are concerned, your question does not apply since complex numbers are not ordered in the usual sense. I mean, you cannot talk of a complex number being positive or negative. For example, if $i$ were positive, then $i^2 geq 0$ which is not true as $-1 ngeq 0$ and also if $i$ were negative, then again we get $i^2 geq 0$ which is not true as $-1 ngeq 0$.
answered Oct 10 '16 at 0:37
Ralph Lazarus
11
edited Oct 10 '16 at 1:27
answered Oct 10 '16 at 0:37
Ralph Lazarus
11
answered Oct 10 '16 at 0:37
Ralph Lazarus
11
answered Oct 10 '16 at 0:37
Ralph Lazarus
11
11
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In general, $(ab)^x=a^xb^x$ for any $age 0$ and $b,xinmathbb C$. So no, you don't need both to be positive.
– user2345215
Jan 8 '15 at 20:27
1
If you work in $mathbb R$ then the square root of a negative number is not defined. If you work in $mathbb C$ then the square root is a "multi-valued" function, so $sqrt{-1}$ could be $i$ or $-i$.
– Martin R
Jan 8 '15 at 20:33
The real underlying reason is that "the square root" has to be carefully defined when you try to generalize it. Notice that there are two square roots of any number. Because of that, we have to be careful in our definition for the square root to be well-defined.
– Max
Jan 8 '15 at 20:34
@MartinR really? I was taught that $sqrt{-1}$ is $i$
– M.S.E
Jan 8 '15 at 20:35
1
@user2345215: $a^b$ is so far from being well-defined that it usually has infinitely many possible values. The best one can do is to define it to be $e^{b log(a)}$, but the complex valued logarithm function has infinitely many branches.
– Lee Mosher
Jan 8 '15 at 20:37