Mixing
Why isn't $i * i = 1$? [duplicate]
This question already has an answer here:
Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?
9 answers
When we studied complex numbers they told us that $i * i = -1$ because $i = sqrt -1$ and
$i * i = i^2$, so the square removes the root.
However we can say as well that $i * i = sqrt {-1} * sqrt {-1} = sqrt {-1 * -1} = sqrt 1 = 1$.
In any case both are valid math, right?
Why can't I follow the second reasoming?!
complex-numbers roots
edited Nov 20 '18 at 17:57
Monstrous Moonshiner
2,25011337
asked Nov 20 '18 at 17:41
Raafat Abualazm
1
marked as duplicate by Adrian Keister, Xander Henderson, Arthur, rschwieb, Key Flex Nov 20 '18 at 17:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?
9 answers
When we studied complex numbers they told us that $i * i = -1$ because $i = sqrt -1$ and
$i * i = i^2$, so the square removes the root.
However we can say as well that $i * i = sqrt {-1} * sqrt {-1} = sqrt {-1 * -1} = sqrt 1 = 1$.
In any case both are valid math, right?
Why can't I follow the second reasoming?!
complex-numbers roots
edited Nov 20 '18 at 17:57
Monstrous Moonshiner
2,25011337
asked Nov 20 '18 at 17:41
Raafat Abualazm
1
marked as duplicate by Adrian Keister, Xander Henderson, Arthur, rschwieb, Key Flex Nov 20 '18 at 17:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
No, this is not valid math. In the first place because there is no real number such that $r^2=-1$, so that nothing allows you to generalize the calculation rules for reals to this "beast".
– Yves Daoust
Nov 20 '18 at 17:45
add a comment |
This question already has an answer here:
Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?
9 answers
When we studied complex numbers they told us that $i * i = -1$ because $i = sqrt -1$ and
$i * i = i^2$, so the square removes the root.
However we can say as well that $i * i = sqrt {-1} * sqrt {-1} = sqrt {-1 * -1} = sqrt 1 = 1$.
In any case both are valid math, right?
Why can't I follow the second reasoming?!
complex-numbers roots
edited Nov 20 '18 at 17:57
Monstrous Moonshiner
2,25011337
asked Nov 20 '18 at 17:41
Raafat Abualazm
1
This question already has an answer here:
Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?
9 answers
When we studied complex numbers they told us that $i * i = -1$ because $i = sqrt -1$ and
$i * i = i^2$, so the square removes the root.
However we can say as well that $i * i = sqrt {-1} * sqrt {-1} = sqrt {-1 * -1} = sqrt 1 = 1$.
In any case both are valid math, right?
Why can't I follow the second reasoming?!
This question already has an answer here:
Why $sqrt{-1 times {-1}} neq sqrt{-1}^2$?
9 answers
complex-numbers roots
complex-numbers roots
edited Nov 20 '18 at 17:57
Monstrous Moonshiner
2,25011337
asked Nov 20 '18 at 17:41
Raafat Abualazm
1
edited Nov 20 '18 at 17:57
Monstrous Moonshiner
2,25011337
asked Nov 20 '18 at 17:41
Raafat Abualazm
1
edited Nov 20 '18 at 17:57
Monstrous Moonshiner
2,25011337
edited Nov 20 '18 at 17:57
Monstrous Moonshiner
2,25011337
edited Nov 20 '18 at 17:57
Monstrous Moonshiner
2,25011337
2,25011337
asked Nov 20 '18 at 17:41
Raafat Abualazm
1
asked Nov 20 '18 at 17:41
Raafat Abualazm
1
asked Nov 20 '18 at 17:41
Raafat Abualazm
1
1
marked as duplicate by Adrian Keister, Xander Henderson, Arthur, rschwieb, Key Flex Nov 20 '18 at 17:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Adrian Keister, Xander Henderson, Arthur, rschwieb, Key Flex Nov 20 '18 at 17:57
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
No, this is not valid math. In the first place because there is no real number such that $r^2=-1$, so that nothing allows you to generalize the calculation rules for reals to this "beast".
– Yves Daoust
Nov 20 '18 at 17:45
add a comment |
1
No, this is not valid math. In the first place because there is no real number such that $r^2=-1$, so that nothing allows you to generalize the calculation rules for reals to this "beast".
– Yves Daoust
Nov 20 '18 at 17:45
1
1
No, this is not valid math. In the first place because there is no real number such that $r^2=-1$, so that nothing allows you to generalize the calculation rules for reals to this "beast".
– Yves Daoust
Nov 20 '18 at 17:45
No, this is not valid math. In the first place because there is no real number such that $r^2=-1$, so that nothing allows you to generalize the calculation rules for reals to this "beast".
– Yves Daoust
Nov 20 '18 at 17:45
add a comment |
2 Answers
2
active
oldest
votes
You've learned the wrong definition of $i$. The true definition of $i$ is that $icdot i=-1$, and that's it. No $sqrt{phantom{-1}}$ in sight.
In fact, in my opinion, $sqrt{phantom{-1}}$ doesn't belong at all when dealing with complex numbers, and should be avoided whenever possible. For one thing because it isn't well-defined in any canonical way, and for another because even if you try to define it, you lose many of the properties of square roots that you usually take for granted, like $sqrt{ab}=sqrt asqrt b$.
answered Nov 20 '18 at 17:42
Arthur
111k7105186
So when they first thought of i it was a hack to solve x^2 + 1 = 0. Why my textbook then says that i = √-1
– Raafat Abualazm
Nov 21 '18 at 6:05
@RaafatAbualazam Mathematicians first came face to face with complex numbers during intermediate calculations when solving cubic equations. And they would appear as $sqrt{-15}$ and similar. So historically, $sqrt{-1}$ was first. But the current formal definition is closer to "a hack to solve x^2+1=0", yes. And your textbook (as well as every introductory textbook ever) says $i=sqrt{-1}$ because they think that's easier for students to understand. It looks somewhat more concrete than $i^2=-1$, but it leads directly to questions and misunderstandings like yours.
– Arthur
Nov 21 '18 at 6:49
I see what you are implying here. i is its own number and roots properties aren't the same.
– Raafat Abualazm
Nov 21 '18 at 19:12
add a comment |
To be precise,
you are right to state $icdot i=i^2=-1$, which is essentially a definition of $i$;
$i=sqrt{-1}$ is acceptable provided you acknowledge to use the "principal branch" of the complex square root (another possible choice is $i=-sqrt{-1}$);
$(sqrt{-1})^2=-1$ also works, by definition of the square root;
$sqrt{-1}sqrt{-1}=sqrt{(-1)(-1)}$ is wrong, because the principal square root does not enjoy the property $sqrt asqrt b=sqrt{ab}$;
- finally, $sqrt1=1$ remains true.
answered Nov 20 '18 at 17:54
Yves Daoust
124k671221
Why then √a √b != √ab. How come power can't be distributed over multiplied numbers. I have seen weird explanations online, I just can't understand them.
– Raafat Abualazm
Nov 21 '18 at 6:02
@RaafatAbualazam Why? Because your calculation showing $-1=1$ can't possibly be true, and $sqrt asqrt b=sqrt{ab}$ was the only non-definition assumption you used. So if we want all the definitions (like what multiplication means, what $-1$ means, what $sqrt{phantom{-1}}$ means and what $i$ means) to exist and make sense, then we cannot have $sqrt asqrt b=sqrt{ab}$.
– Arthur
Nov 21 '18 at 6:56
@RaafatAbualazm: a property is false until you have proven it true. For instance, regarding the product of matrices, $abne ba$. Don't take for granted that properties of the reals carry over to other objects.
– Yves Daoust
Nov 21 '18 at 9:09
@Arthur I saw that I was doing wonders with math, but I did not understand why.
– Raafat Abualazm
Nov 21 '18 at 19:14
Also one thing. Why can we take the -1 in or out of the root as we like but the identity is false. √-6 = √-1 √6, but √ab != √a √b, if not a,b >= 0. Must a and b be greater than zero or one being more than would suffice?
– Raafat Abualazm
Nov 21 '18 at 19:19
|
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
You've learned the wrong definition of $i$. The true definition of $i$ is that $icdot i=-1$, and that's it. No $sqrt{phantom{-1}}$ in sight.
In fact, in my opinion, $sqrt{phantom{-1}}$ doesn't belong at all when dealing with complex numbers, and should be avoided whenever possible. For one thing because it isn't well-defined in any canonical way, and for another because even if you try to define it, you lose many of the properties of square roots that you usually take for granted, like $sqrt{ab}=sqrt asqrt b$.
answered Nov 20 '18 at 17:42
Arthur
111k7105186
So when they first thought of i it was a hack to solve x^2 + 1 = 0. Why my textbook then says that i = √-1
– Raafat Abualazm
Nov 21 '18 at 6:05
@RaafatAbualazam Mathematicians first came face to face with complex numbers during intermediate calculations when solving cubic equations. And they would appear as $sqrt{-15}$ and similar. So historically, $sqrt{-1}$ was first. But the current formal definition is closer to "a hack to solve x^2+1=0", yes. And your textbook (as well as every introductory textbook ever) says $i=sqrt{-1}$ because they think that's easier for students to understand. It looks somewhat more concrete than $i^2=-1$, but it leads directly to questions and misunderstandings like yours.
– Arthur
Nov 21 '18 at 6:49
I see what you are implying here. i is its own number and roots properties aren't the same.
– Raafat Abualazm
Nov 21 '18 at 19:12
add a comment |
You've learned the wrong definition of $i$. The true definition of $i$ is that $icdot i=-1$, and that's it. No $sqrt{phantom{-1}}$ in sight.
In fact, in my opinion, $sqrt{phantom{-1}}$ doesn't belong at all when dealing with complex numbers, and should be avoided whenever possible. For one thing because it isn't well-defined in any canonical way, and for another because even if you try to define it, you lose many of the properties of square roots that you usually take for granted, like $sqrt{ab}=sqrt asqrt b$.
answered Nov 20 '18 at 17:42
Arthur
111k7105186
So when they first thought of i it was a hack to solve x^2 + 1 = 0. Why my textbook then says that i = √-1
– Raafat Abualazm
Nov 21 '18 at 6:05
@RaafatAbualazam Mathematicians first came face to face with complex numbers during intermediate calculations when solving cubic equations. And they would appear as $sqrt{-15}$ and similar. So historically, $sqrt{-1}$ was first. But the current formal definition is closer to "a hack to solve x^2+1=0", yes. And your textbook (as well as every introductory textbook ever) says $i=sqrt{-1}$ because they think that's easier for students to understand. It looks somewhat more concrete than $i^2=-1$, but it leads directly to questions and misunderstandings like yours.
– Arthur
Nov 21 '18 at 6:49
I see what you are implying here. i is its own number and roots properties aren't the same.
– Raafat Abualazm
Nov 21 '18 at 19:12
add a comment |
You've learned the wrong definition of $i$. The true definition of $i$ is that $icdot i=-1$, and that's it. No $sqrt{phantom{-1}}$ in sight.
In fact, in my opinion, $sqrt{phantom{-1}}$ doesn't belong at all when dealing with complex numbers, and should be avoided whenever possible. For one thing because it isn't well-defined in any canonical way, and for another because even if you try to define it, you lose many of the properties of square roots that you usually take for granted, like $sqrt{ab}=sqrt asqrt b$.
answered Nov 20 '18 at 17:42
Arthur
111k7105186
You've learned the wrong definition of $i$. The true definition of $i$ is that $icdot i=-1$, and that's it. No $sqrt{phantom{-1}}$ in sight.
In fact, in my opinion, $sqrt{phantom{-1}}$ doesn't belong at all when dealing with complex numbers, and should be avoided whenever possible. For one thing because it isn't well-defined in any canonical way, and for another because even if you try to define it, you lose many of the properties of square roots that you usually take for granted, like $sqrt{ab}=sqrt asqrt b$.
answered Nov 20 '18 at 17:42
Arthur
111k7105186
edited Nov 20 '18 at 17:49
answered Nov 20 '18 at 17:42
Arthur
111k7105186
answered Nov 20 '18 at 17:42
Arthur
111k7105186
answered Nov 20 '18 at 17:42
Arthur
111k7105186
111k7105186
So when they first thought of i it was a hack to solve x^2 + 1 = 0. Why my textbook then says that i = √-1
– Raafat Abualazm
Nov 21 '18 at 6:05
@RaafatAbualazam Mathematicians first came face to face with complex numbers during intermediate calculations when solving cubic equations. And they would appear as $sqrt{-15}$ and similar. So historically, $sqrt{-1}$ was first. But the current formal definition is closer to "a hack to solve x^2+1=0", yes. And your textbook (as well as every introductory textbook ever) says $i=sqrt{-1}$ because they think that's easier for students to understand. It looks somewhat more concrete than $i^2=-1$, but it leads directly to questions and misunderstandings like yours.
– Arthur
Nov 21 '18 at 6:49
I see what you are implying here. i is its own number and roots properties aren't the same.
– Raafat Abualazm
Nov 21 '18 at 19:12
add a comment |
So when they first thought of i it was a hack to solve x^2 + 1 = 0. Why my textbook then says that i = √-1
– Raafat Abualazm
Nov 21 '18 at 6:05
@RaafatAbualazam Mathematicians first came face to face with complex numbers during intermediate calculations when solving cubic equations. And they would appear as $sqrt{-15}$ and similar. So historically, $sqrt{-1}$ was first. But the current formal definition is closer to "a hack to solve x^2+1=0", yes. And your textbook (as well as every introductory textbook ever) says $i=sqrt{-1}$ because they think that's easier for students to understand. It looks somewhat more concrete than $i^2=-1$, but it leads directly to questions and misunderstandings like yours.
– Arthur
Nov 21 '18 at 6:49
I see what you are implying here. i is its own number and roots properties aren't the same.
– Raafat Abualazm
Nov 21 '18 at 19:12
So when they first thought of i it was a hack to solve x^2 + 1 = 0. Why my textbook then says that i = √-1
– Raafat Abualazm
Nov 21 '18 at 6:05
So when they first thought of i it was a hack to solve x^2 + 1 = 0. Why my textbook then says that i = √-1
– Raafat Abualazm
Nov 21 '18 at 6:05
@RaafatAbualazam Mathematicians first came face to face with complex numbers during intermediate calculations when solving cubic equations. And they would appear as $sqrt{-15}$ and similar. So historically, $sqrt{-1}$ was first. But the current formal definition is closer to "a hack to solve x^2+1=0", yes. And your textbook (as well as every introductory textbook ever) says $i=sqrt{-1}$ because they think that's easier for students to understand. It looks somewhat more concrete than $i^2=-1$, but it leads directly to questions and misunderstandings like yours.
– Arthur
Nov 21 '18 at 6:49
@RaafatAbualazam Mathematicians first came face to face with complex numbers during intermediate calculations when solving cubic equations. And they would appear as $sqrt{-15}$ and similar. So historically, $sqrt{-1}$ was first. But the current formal definition is closer to "a hack to solve x^2+1=0", yes. And your textbook (as well as every introductory textbook ever) says $i=sqrt{-1}$ because they think that's easier for students to understand. It looks somewhat more concrete than $i^2=-1$, but it leads directly to questions and misunderstandings like yours.
– Arthur
Nov 21 '18 at 6:49
I see what you are implying here. i is its own number and roots properties aren't the same.
– Raafat Abualazm
Nov 21 '18 at 19:12
I see what you are implying here. i is its own number and roots properties aren't the same.
– Raafat Abualazm
Nov 21 '18 at 19:12
add a comment |
To be precise,
you are right to state $icdot i=i^2=-1$, which is essentially a definition of $i$;
$i=sqrt{-1}$ is acceptable provided you acknowledge to use the "principal branch" of the complex square root (another possible choice is $i=-sqrt{-1}$);
$(sqrt{-1})^2=-1$ also works, by definition of the square root;
$sqrt{-1}sqrt{-1}=sqrt{(-1)(-1)}$ is wrong, because the principal square root does not enjoy the property $sqrt asqrt b=sqrt{ab}$;
- finally, $sqrt1=1$ remains true.
answered Nov 20 '18 at 17:54
Yves Daoust
124k671221
Why then √a √b != √ab. How come power can't be distributed over multiplied numbers. I have seen weird explanations online, I just can't understand them.
– Raafat Abualazm
Nov 21 '18 at 6:02
@RaafatAbualazam Why? Because your calculation showing $-1=1$ can't possibly be true, and $sqrt asqrt b=sqrt{ab}$ was the only non-definition assumption you used. So if we want all the definitions (like what multiplication means, what $-1$ means, what $sqrt{phantom{-1}}$ means and what $i$ means) to exist and make sense, then we cannot have $sqrt asqrt b=sqrt{ab}$.
– Arthur
Nov 21 '18 at 6:56
@RaafatAbualazm: a property is false until you have proven it true. For instance, regarding the product of matrices, $abne ba$. Don't take for granted that properties of the reals carry over to other objects.
– Yves Daoust
Nov 21 '18 at 9:09
@Arthur I saw that I was doing wonders with math, but I did not understand why.
– Raafat Abualazm
Nov 21 '18 at 19:14
Also one thing. Why can we take the -1 in or out of the root as we like but the identity is false. √-6 = √-1 √6, but √ab != √a √b, if not a,b >= 0. Must a and b be greater than zero or one being more than would suffice?
– Raafat Abualazm
Nov 21 '18 at 19:19
|
show 1 more comment
To be precise,
you are right to state $icdot i=i^2=-1$, which is essentially a definition of $i$;
$i=sqrt{-1}$ is acceptable provided you acknowledge to use the "principal branch" of the complex square root (another possible choice is $i=-sqrt{-1}$);
$(sqrt{-1})^2=-1$ also works, by definition of the square root;
$sqrt{-1}sqrt{-1}=sqrt{(-1)(-1)}$ is wrong, because the principal square root does not enjoy the property $sqrt asqrt b=sqrt{ab}$;
- finally, $sqrt1=1$ remains true.
answered Nov 20 '18 at 17:54
Yves Daoust
124k671221
Why then √a √b != √ab. How come power can't be distributed over multiplied numbers. I have seen weird explanations online, I just can't understand them.
– Raafat Abualazm
Nov 21 '18 at 6:02
@RaafatAbualazam Why? Because your calculation showing $-1=1$ can't possibly be true, and $sqrt asqrt b=sqrt{ab}$ was the only non-definition assumption you used. So if we want all the definitions (like what multiplication means, what $-1$ means, what $sqrt{phantom{-1}}$ means and what $i$ means) to exist and make sense, then we cannot have $sqrt asqrt b=sqrt{ab}$.
– Arthur
Nov 21 '18 at 6:56
@RaafatAbualazm: a property is false until you have proven it true. For instance, regarding the product of matrices, $abne ba$. Don't take for granted that properties of the reals carry over to other objects.
– Yves Daoust
Nov 21 '18 at 9:09
@Arthur I saw that I was doing wonders with math, but I did not understand why.
– Raafat Abualazm
Nov 21 '18 at 19:14
Also one thing. Why can we take the -1 in or out of the root as we like but the identity is false. √-6 = √-1 √6, but √ab != √a √b, if not a,b >= 0. Must a and b be greater than zero or one being more than would suffice?
– Raafat Abualazm
Nov 21 '18 at 19:19
|
show 1 more comment
To be precise,
you are right to state $icdot i=i^2=-1$, which is essentially a definition of $i$;
$i=sqrt{-1}$ is acceptable provided you acknowledge to use the "principal branch" of the complex square root (another possible choice is $i=-sqrt{-1}$);
$(sqrt{-1})^2=-1$ also works, by definition of the square root;
$sqrt{-1}sqrt{-1}=sqrt{(-1)(-1)}$ is wrong, because the principal square root does not enjoy the property $sqrt asqrt b=sqrt{ab}$;
- finally, $sqrt1=1$ remains true.
answered Nov 20 '18 at 17:54
Yves Daoust
124k671221
To be precise,
you are right to state $icdot i=i^2=-1$, which is essentially a definition of $i$;
$i=sqrt{-1}$ is acceptable provided you acknowledge to use the "principal branch" of the complex square root (another possible choice is $i=-sqrt{-1}$);
$(sqrt{-1})^2=-1$ also works, by definition of the square root;
$sqrt{-1}sqrt{-1}=sqrt{(-1)(-1)}$ is wrong, because the principal square root does not enjoy the property $sqrt asqrt b=sqrt{ab}$;
- finally, $sqrt1=1$ remains true.
answered Nov 20 '18 at 17:54
Yves Daoust
124k671221
answered Nov 20 '18 at 17:54
Yves Daoust
124k671221
answered Nov 20 '18 at 17:54
Yves Daoust
124k671221
answered Nov 20 '18 at 17:54
Yves Daoust
124k671221
124k671221
Why then √a √b != √ab. How come power can't be distributed over multiplied numbers. I have seen weird explanations online, I just can't understand them.
– Raafat Abualazm
Nov 21 '18 at 6:02
@RaafatAbualazam Why? Because your calculation showing $-1=1$ can't possibly be true, and $sqrt asqrt b=sqrt{ab}$ was the only non-definition assumption you used. So if we want all the definitions (like what multiplication means, what $-1$ means, what $sqrt{phantom{-1}}$ means and what $i$ means) to exist and make sense, then we cannot have $sqrt asqrt b=sqrt{ab}$.
– Arthur
Nov 21 '18 at 6:56
@RaafatAbualazm: a property is false until you have proven it true. For instance, regarding the product of matrices, $abne ba$. Don't take for granted that properties of the reals carry over to other objects.
– Yves Daoust
Nov 21 '18 at 9:09
@Arthur I saw that I was doing wonders with math, but I did not understand why.
– Raafat Abualazm
Nov 21 '18 at 19:14
Also one thing. Why can we take the -1 in or out of the root as we like but the identity is false. √-6 = √-1 √6, but √ab != √a √b, if not a,b >= 0. Must a and b be greater than zero or one being more than would suffice?
– Raafat Abualazm
Nov 21 '18 at 19:19
|
show 1 more comment
Why then √a √b != √ab. How come power can't be distributed over multiplied numbers. I have seen weird explanations online, I just can't understand them.
– Raafat Abualazm
Nov 21 '18 at 6:02
@RaafatAbualazam Why? Because your calculation showing $-1=1$ can't possibly be true, and $sqrt asqrt b=sqrt{ab}$ was the only non-definition assumption you used. So if we want all the definitions (like what multiplication means, what $-1$ means, what $sqrt{phantom{-1}}$ means and what $i$ means) to exist and make sense, then we cannot have $sqrt asqrt b=sqrt{ab}$.
– Arthur
Nov 21 '18 at 6:56
@RaafatAbualazm: a property is false until you have proven it true. For instance, regarding the product of matrices, $abne ba$. Don't take for granted that properties of the reals carry over to other objects.
– Yves Daoust
Nov 21 '18 at 9:09
@Arthur I saw that I was doing wonders with math, but I did not understand why.
– Raafat Abualazm
Nov 21 '18 at 19:14
Also one thing. Why can we take the -1 in or out of the root as we like but the identity is false. √-6 = √-1 √6, but √ab != √a √b, if not a,b >= 0. Must a and b be greater than zero or one being more than would suffice?
– Raafat Abualazm
Nov 21 '18 at 19:19
Why then √a √b != √ab. How come power can't be distributed over multiplied numbers. I have seen weird explanations online, I just can't understand them.
– Raafat Abualazm
Nov 21 '18 at 6:02
Why then √a √b != √ab. How come power can't be distributed over multiplied numbers. I have seen weird explanations online, I just can't understand them.
– Raafat Abualazm
Nov 21 '18 at 6:02
@RaafatAbualazam Why? Because your calculation showing $-1=1$ can't possibly be true, and $sqrt asqrt b=sqrt{ab}$ was the only non-definition assumption you used. So if we want all the definitions (like what multiplication means, what $-1$ means, what $sqrt{phantom{-1}}$ means and what $i$ means) to exist and make sense, then we cannot have $sqrt asqrt b=sqrt{ab}$.
– Arthur
Nov 21 '18 at 6:56
@RaafatAbualazam Why? Because your calculation showing $-1=1$ can't possibly be true, and $sqrt asqrt b=sqrt{ab}$ was the only non-definition assumption you used. So if we want all the definitions (like what multiplication means, what $-1$ means, what $sqrt{phantom{-1}}$ means and what $i$ means) to exist and make sense, then we cannot have $sqrt asqrt b=sqrt{ab}$.
– Arthur
Nov 21 '18 at 6:56
@RaafatAbualazm: a property is false until you have proven it true. For instance, regarding the product of matrices, $abne ba$. Don't take for granted that properties of the reals carry over to other objects.
– Yves Daoust
Nov 21 '18 at 9:09
@RaafatAbualazm: a property is false until you have proven it true. For instance, regarding the product of matrices, $abne ba$. Don't take for granted that properties of the reals carry over to other objects.
– Yves Daoust
Nov 21 '18 at 9:09
@Arthur I saw that I was doing wonders with math, but I did not understand why.
– Raafat Abualazm
Nov 21 '18 at 19:14
@Arthur I saw that I was doing wonders with math, but I did not understand why.
– Raafat Abualazm
Nov 21 '18 at 19:14
Also one thing. Why can we take the -1 in or out of the root as we like but the identity is false. √-6 = √-1 √6, but √ab != √a √b, if not a,b >= 0. Must a and b be greater than zero or one being more than would suffice?
– Raafat Abualazm
Nov 21 '18 at 19:19
Also one thing. Why can we take the -1 in or out of the root as we like but the identity is false. √-6 = √-1 √6, but √ab != √a √b, if not a,b >= 0. Must a and b be greater than zero or one being more than would suffice?
– Raafat Abualazm
Nov 21 '18 at 19:19
|
show 1 more comment
1
No, this is not valid math. In the first place because there is no real number such that $r^2=-1$, so that nothing allows you to generalize the calculation rules for reals to this "beast".
– Yves Daoust
Nov 20 '18 at 17:45