Can numbers as well as having roots in the Imaginary-Real plane also be said to have roots as quaternions or higher complex forms? If so how?

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This is an argand Diagram to show 2d Imaginary-real plane solutions to a said root of a number jut there to illustrate the normal complex solutions.

roots

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asked Jul 15 at 20:11

Rosco

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  When you ask "can numbers have roots" what do you mean? Usually, equations have roots which are numbers.
  – herb steinberg
  Jul 15 at 20:46
  

  
  
  
  

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up vote
-2
down vote

favorite

1

enter image description here
This is an argand Diagram to show 2d Imaginary-real plane solutions to a said root of a number jut there to illustrate the normal complex solutions.

roots

share|cite|improve this question

asked Jul 15 at 20:11

Rosco

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  When you ask "can numbers have roots" what do you mean? Usually, equations have roots which are numbers.
  – herb steinberg
  Jul 15 at 20:46
  

  
  
  
  

add a comment | 

up vote
-2
down vote

favorite

1

up vote
-2
down vote

favorite

1

1

enter image description here
This is an argand Diagram to show 2d Imaginary-real plane solutions to a said root of a number jut there to illustrate the normal complex solutions.

roots

share|cite|improve this question

asked Jul 15 at 20:11

Rosco

1

enter image description here
This is an argand Diagram to show 2d Imaginary-real plane solutions to a said root of a number jut there to illustrate the normal complex solutions.

roots

share|cite|improve this question

asked Jul 15 at 20:11

Rosco

1

share|cite|improve this question

share|cite|improve this question

asked Jul 15 at 20:11

Rosco

1

asked Jul 15 at 20:11

Rosco

1

asked Jul 15 at 20:11

Rosco

1

1

• 
  
  
  

  
  

  
  
  
  
  
  

  
  When you ask "can numbers have roots" what do you mean? Usually, equations have roots which are numbers.
  – herb steinberg
  Jul 15 at 20:46
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  When you ask "can numbers have roots" what do you mean? Usually, equations have roots which are numbers.
  – herb steinberg
  Jul 15 at 20:46
  

  
  
  
  

When you ask "can numbers have roots" what do you mean? Usually, equations have roots which are numbers.
– herb steinberg
Jul 15 at 20:46

When you ask "can numbers have roots" what do you mean? Usually, equations have roots which are numbers.
– herb steinberg
Jul 15 at 20:46

add a comment | 

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Sure. The situation's no different from how a real number can have more complex roots than it has real roots. E.g. the fourth roots of $1$ in the reals are $1$ and $-1$, but in the complex numbers we get $i$ and $-i$ as well.

By embedding an algebraic structure $mathcalA$ inside a larger algebraic structure $mathcalB$ we can often get more variety at the cost of having fewer nice properties. In this case, our additional variety is "having more roots," and the nice property we lose is that the proof that an $n$th-degree polynomial over $mathbbC$ has at most $n$ complex roots breaks down when we try to replace $mathbbC$ with $mathbbH$ (= the quaternions), and so there's no reason to not expect more roots to pop up. (As an exercise you might try to find out how many square roots $-1$ has in $mathbbC$.)

share|cite|improve this answer

answered Jul 15 at 20:47

Noah Schweber

111k9140264

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the response I am just wondering to whether that implies the idea of stating that a numbers nth roots are complex in the from (a + bi) is arbitrary and instead a more pure way could be used to describe this, e.g. something along the lines of " That the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system(1),". Or does this suggestion go too far and go into the realms of mathematical lunacy which happens on forums far too often?
  – Rosco
  Jul 16 at 15:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Rosco I'm not sure what you mean exactly - what do " a numbers nth roots are complex in the from (a + bi)" mean? and "the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system" mean? - but that doesn't mean it's a meaningless question, just that work is needed to make it precise. For example, the notion of an element (like $i$) generating a number system can be made precise via abstract algebra; field extensions and the Cayley-Dickson process are each relevant, and the latter extends to the quaternions too.
  – Noah Schweber
  Jul 16 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the reply, your replies have been of a very high quality, I do appreciate them. I have looked up the Cayley-Dickson process and it has helped to clarify my view. I do agree that my questions was vague. What I meant by the question was when we take an nth root of an rational number we see that there is a mixture of real and/or complex solutions, for example the 6th root of 7 we see that there two purely real solutions and 4 which are complex. Does it matter that the solutions MUST be complex or can they be in any form in which the base of the number system is made from a root of -1?
  – Rosco
  Jul 16 at 20:21
  

  
  
  
  

add a comment | 

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1 Answer
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1 Answer
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up vote
1
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Sure. The situation's no different from how a real number can have more complex roots than it has real roots. E.g. the fourth roots of $1$ in the reals are $1$ and $-1$, but in the complex numbers we get $i$ and $-i$ as well.

By embedding an algebraic structure $mathcalA$ inside a larger algebraic structure $mathcalB$ we can often get more variety at the cost of having fewer nice properties. In this case, our additional variety is "having more roots," and the nice property we lose is that the proof that an $n$th-degree polynomial over $mathbbC$ has at most $n$ complex roots breaks down when we try to replace $mathbbC$ with $mathbbH$ (= the quaternions), and so there's no reason to not expect more roots to pop up. (As an exercise you might try to find out how many square roots $-1$ has in $mathbbC$.)

share|cite|improve this answer

answered Jul 15 at 20:47

Noah Schweber

111k9140264

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the response I am just wondering to whether that implies the idea of stating that a numbers nth roots are complex in the from (a + bi) is arbitrary and instead a more pure way could be used to describe this, e.g. something along the lines of " That the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system(1),". Or does this suggestion go too far and go into the realms of mathematical lunacy which happens on forums far too often?
  – Rosco
  Jul 16 at 15:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Rosco I'm not sure what you mean exactly - what do " a numbers nth roots are complex in the from (a + bi)" mean? and "the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system" mean? - but that doesn't mean it's a meaningless question, just that work is needed to make it precise. For example, the notion of an element (like $i$) generating a number system can be made precise via abstract algebra; field extensions and the Cayley-Dickson process are each relevant, and the latter extends to the quaternions too.
  – Noah Schweber
  Jul 16 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the reply, your replies have been of a very high quality, I do appreciate them. I have looked up the Cayley-Dickson process and it has helped to clarify my view. I do agree that my questions was vague. What I meant by the question was when we take an nth root of an rational number we see that there is a mixture of real and/or complex solutions, for example the 6th root of 7 we see that there two purely real solutions and 4 which are complex. Does it matter that the solutions MUST be complex or can they be in any form in which the base of the number system is made from a root of -1?
  – Rosco
  Jul 16 at 20:21
  

  
  
  
  

add a comment | 

up vote
1
down vote

Sure. The situation's no different from how a real number can have more complex roots than it has real roots. E.g. the fourth roots of $1$ in the reals are $1$ and $-1$, but in the complex numbers we get $i$ and $-i$ as well.

By embedding an algebraic structure $mathcalA$ inside a larger algebraic structure $mathcalB$ we can often get more variety at the cost of having fewer nice properties. In this case, our additional variety is "having more roots," and the nice property we lose is that the proof that an $n$th-degree polynomial over $mathbbC$ has at most $n$ complex roots breaks down when we try to replace $mathbbC$ with $mathbbH$ (= the quaternions), and so there's no reason to not expect more roots to pop up. (As an exercise you might try to find out how many square roots $-1$ has in $mathbbC$.)

share|cite|improve this answer

answered Jul 15 at 20:47

Noah Schweber

111k9140264

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the response I am just wondering to whether that implies the idea of stating that a numbers nth roots are complex in the from (a + bi) is arbitrary and instead a more pure way could be used to describe this, e.g. something along the lines of " That the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system(1),". Or does this suggestion go too far and go into the realms of mathematical lunacy which happens on forums far too often?
  – Rosco
  Jul 16 at 15:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Rosco I'm not sure what you mean exactly - what do " a numbers nth roots are complex in the from (a + bi)" mean? and "the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system" mean? - but that doesn't mean it's a meaningless question, just that work is needed to make it precise. For example, the notion of an element (like $i$) generating a number system can be made precise via abstract algebra; field extensions and the Cayley-Dickson process are each relevant, and the latter extends to the quaternions too.
  – Noah Schweber
  Jul 16 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the reply, your replies have been of a very high quality, I do appreciate them. I have looked up the Cayley-Dickson process and it has helped to clarify my view. I do agree that my questions was vague. What I meant by the question was when we take an nth root of an rational number we see that there is a mixture of real and/or complex solutions, for example the 6th root of 7 we see that there two purely real solutions and 4 which are complex. Does it matter that the solutions MUST be complex or can they be in any form in which the base of the number system is made from a root of -1?
  – Rosco
  Jul 16 at 20:21
  

  
  
  
  

add a comment | 

up vote
1
down vote

up vote
1
down vote

Sure. The situation's no different from how a real number can have more complex roots than it has real roots. E.g. the fourth roots of $1$ in the reals are $1$ and $-1$, but in the complex numbers we get $i$ and $-i$ as well.

By embedding an algebraic structure $mathcalA$ inside a larger algebraic structure $mathcalB$ we can often get more variety at the cost of having fewer nice properties. In this case, our additional variety is "having more roots," and the nice property we lose is that the proof that an $n$th-degree polynomial over $mathbbC$ has at most $n$ complex roots breaks down when we try to replace $mathbbC$ with $mathbbH$ (= the quaternions), and so there's no reason to not expect more roots to pop up. (As an exercise you might try to find out how many square roots $-1$ has in $mathbbC$.)

share|cite|improve this answer

answered Jul 15 at 20:47

Noah Schweber

111k9140264

Sure. The situation's no different from how a real number can have more complex roots than it has real roots. E.g. the fourth roots of $1$ in the reals are $1$ and $-1$, but in the complex numbers we get $i$ and $-i$ as well.

By embedding an algebraic structure $mathcalA$ inside a larger algebraic structure $mathcalB$ we can often get more variety at the cost of having fewer nice properties. In this case, our additional variety is "having more roots," and the nice property we lose is that the proof that an $n$th-degree polynomial over $mathbbC$ has at most $n$ complex roots breaks down when we try to replace $mathbbC$ with $mathbbH$ (= the quaternions), and so there's no reason to not expect more roots to pop up. (As an exercise you might try to find out how many square roots $-1$ has in $mathbbC$.)

share|cite|improve this answer

answered Jul 15 at 20:47

Noah Schweber

111k9140264

share|cite|improve this answer

share|cite|improve this answer

answered Jul 15 at 20:47

Noah Schweber

111k9140264

answered Jul 15 at 20:47

Noah Schweber

111k9140264

answered Jul 15 at 20:47

Noah Schweber

111k9140264

111k9140264

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the response I am just wondering to whether that implies the idea of stating that a numbers nth roots are complex in the from (a + bi) is arbitrary and instead a more pure way could be used to describe this, e.g. something along the lines of " That the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system(1),". Or does this suggestion go too far and go into the realms of mathematical lunacy which happens on forums far too often?
  – Rosco
  Jul 16 at 15:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Rosco I'm not sure what you mean exactly - what do " a numbers nth roots are complex in the from (a + bi)" mean? and "the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system" mean? - but that doesn't mean it's a meaningless question, just that work is needed to make it precise. For example, the notion of an element (like $i$) generating a number system can be made precise via abstract algebra; field extensions and the Cayley-Dickson process are each relevant, and the latter extends to the quaternions too.
  – Noah Schweber
  Jul 16 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the reply, your replies have been of a very high quality, I do appreciate them. I have looked up the Cayley-Dickson process and it has helped to clarify my view. I do agree that my questions was vague. What I meant by the question was when we take an nth root of an rational number we see that there is a mixture of real and/or complex solutions, for example the 6th root of 7 we see that there two purely real solutions and 4 which are complex. Does it matter that the solutions MUST be complex or can they be in any form in which the base of the number system is made from a root of -1?
  – Rosco
  Jul 16 at 20:21
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thank you for the response I am just wondering to whether that implies the idea of stating that a numbers nth roots are complex in the from (a + bi) is arbitrary and instead a more pure way could be used to describe this, e.g. something along the lines of " That the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system(1),". Or does this suggestion go too far and go into the realms of mathematical lunacy which happens on forums far too often?
  – Rosco
  Jul 16 at 15:29
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Rosco I'm not sure what you mean exactly - what do " a numbers nth roots are complex in the from (a + bi)" mean? and "the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system" mean? - but that doesn't mean it's a meaningless question, just that work is needed to make it precise. For example, the notion of an element (like $i$) generating a number system can be made precise via abstract algebra; field extensions and the Cayley-Dickson process are each relevant, and the latter extends to the quaternions too.
  – Noah Schweber
  Jul 16 at 15:45
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks for the reply, your replies have been of a very high quality, I do appreciate them. I have looked up the Cayley-Dickson process and it has helped to clarify my view. I do agree that my questions was vague. What I meant by the question was when we take an nth root of an rational number we see that there is a mixture of real and/or complex solutions, for example the 6th root of 7 we see that there two purely real solutions and 4 which are complex. Does it matter that the solutions MUST be complex or can they be in any form in which the base of the number system is made from a root of -1?
  – Rosco
  Jul 16 at 20:21
  

  
  
  
  

Thank you for the response I am just wondering to whether that implies the idea of stating that a numbers nth roots are complex in the from (a + bi) is arbitrary and instead a more pure way could be used to describe this, e.g. something along the lines of " That the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system(1),". Or does this suggestion go too far and go into the realms of mathematical lunacy which happens on forums far too often?
– Rosco
Jul 16 at 15:29

Thank you for the response I am just wondering to whether that implies the idea of stating that a numbers nth roots are complex in the from (a + bi) is arbitrary and instead a more pure way could be used to describe this, e.g. something along the lines of " That the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system(1),". Or does this suggestion go too far and go into the realms of mathematical lunacy which happens on forums far too often?
– Rosco
Jul 16 at 15:29

@Rosco I'm not sure what you mean exactly - what do " a numbers nth roots are complex in the from (a + bi)" mean? and "the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system" mean? - but that doesn't mean it's a meaningless question, just that work is needed to make it precise. For example, the notion of an element (like $i$) generating a number system can be made precise via abstract algebra; field extensions and the Cayley-Dickson process are each relevant, and the latter extends to the quaternions too.
– Noah Schweber
Jul 16 at 15:45

@Rosco I'm not sure what you mean exactly - what do " a numbers nth roots are complex in the from (a + bi)" mean? and "the roots must have properties that exist in dimensions in which the square root of -1 is treated as being the base unit of that number system" mean? - but that doesn't mean it's a meaningless question, just that work is needed to make it precise. For example, the notion of an element (like $i$) generating a number system can be made precise via abstract algebra; field extensions and the Cayley-Dickson process are each relevant, and the latter extends to the quaternions too.
– Noah Schweber
Jul 16 at 15:45

Thanks for the reply, your replies have been of a very high quality, I do appreciate them. I have looked up the Cayley-Dickson process and it has helped to clarify my view. I do agree that my questions was vague. What I meant by the question was when we take an nth root of an rational number we see that there is a mixture of real and/or complex solutions, for example the 6th root of 7 we see that there two purely real solutions and 4 which are complex. Does it matter that the solutions MUST be complex or can they be in any form in which the base of the number system is made from a root of -1?
– Rosco
Jul 16 at 20:21

Thanks for the reply, your replies have been of a very high quality, I do appreciate them. I have looked up the Cayley-Dickson process and it has helped to clarify my view. I do agree that my questions was vague. What I meant by the question was when we take an nth root of an rational number we see that there is a mixture of real and/or complex solutions, for example the 6th root of 7 we see that there two purely real solutions and 4 which are complex. Does it matter that the solutions MUST be complex or can they be in any form in which the base of the number system is made from a root of -1?
– Rosco
Jul 16 at 20:21

add a comment | 

 

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