Square roots of quaternions












13












$begingroup$


In class we saw that $-1$ has infinitely many square roots in the ring of quaternions. Is it possible to compute the square roots of a given nonreal quaternion? What do they look like?






abstract-algebra quaternions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Set up the equations for $q^2 = a$, see what you can get out of them.
    $endgroup$
    – vonbrand
    May 5 '13 at 19:06
















13












$begingroup$


In class we saw that $-1$ has infinitely many square roots in the ring of quaternions. Is it possible to compute the square roots of a given nonreal quaternion? What do they look like?






abstract-algebra quaternions







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Set up the equations for $q^2 = a$, see what you can get out of them.
    $endgroup$
    – vonbrand
    May 5 '13 at 19:06














13












13








13


2



$begingroup$


In class we saw that $-1$ has infinitely many square roots in the ring of quaternions. Is it possible to compute the square roots of a given nonreal quaternion? What do they look like?






abstract-algebra quaternions







share|cite|improve this question











$endgroup$




In class we saw that $-1$ has infinitely many square roots in the ring of quaternions. Is it possible to compute the square roots of a given nonreal quaternion? What do they look like?






abstract-algebra quaternions




abstract-algebra quaternions






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 14:47









Jyrki Lahtonen

110k13171379




110k13171379










asked May 5 '13 at 19:01









RubyBlueRubyBlue

684




684








  • 1




    $begingroup$
    Set up the equations for $q^2 = a$, see what you can get out of them.
    $endgroup$
    – vonbrand
    May 5 '13 at 19:06














  • 1




    $begingroup$
    Set up the equations for $q^2 = a$, see what you can get out of them.
    $endgroup$
    – vonbrand
    May 5 '13 at 19:06








1




1




$begingroup$
Set up the equations for $q^2 = a$, see what you can get out of them.
$endgroup$
– vonbrand
May 5 '13 at 19:06




$begingroup$
Set up the equations for $q^2 = a$, see what you can get out of them.
$endgroup$
– vonbrand
May 5 '13 at 19:06










1 Answer
1






active

oldest

votes


















10












$begingroup$

With non-real quaternions we can always find two square roots.



You can write any quaternion in the form
$$
q=a+bvec{u},
$$
where $a$ and $b$ are real, and $vec{u}$ is a unit vector. You probably know that as a quaternion $vec{u}^2=-1$. Therefore we can treat $vec{u}$ as if it were the usual imaginary unit $i$ of the complex numbers. So we can use the usual techniques of finding square roots of complex numbers.



Note that with non-real quaternions we only get two square roots. This is because the square of the above quaternion is
$$
q^2=(a^2-b^2)+2abvec{u}.
$$
For this to be non-real, we need both $a$ and $b$ to be non-zero. So whenever
$q_1^2=q_2$ for some quaternions $q_1,q_2$ where $q_2notinmathbb{R}$, they must both
lie in the same plane, i.e. they must be linear combinations of $1$ and the same unit vector $vec{u}$. This means that $q_1$ and $q_2$ must belong to the same copy of $mathbb{C}=mathbb{R}oplusmathbb{R}vec{u}$. As any complex number has only two square roots, the same applies to non-real quaternions because of this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Interesting. In the quaternion field $mathbb{H}$, only negative real numbers have infinitely many square roots, while all other quaternions have only two square roots. Is this true and does this generalize to $n$-th roots?
    $endgroup$
    – Zhuoran He
    Oct 30 '17 at 5:20










  • $begingroup$
    @ZhuoranHe I think that the above argument applies to $n$th roots of non-real quaternions as well. But if $n>2$ there are infinitely many $n$th roots of unity in $Bbb{H}$: to each unit vector $vec{u}$ we have $n-1$ non-real $n$th roots of unity in $Bbb{R}oplusBbb{R}vec{u}$. Therefore all reals have infinitely many $n$th roots in $Bbb{H}$.
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '17 at 7:26












  • $begingroup$
    I get it. Whenever there's symmetry breaking, e.g. $sqrt{-1}=pm i$, $1^{1/3}=1, e^{pm2pi i/3}$ in $mathbb{C}$, quaternions would say that the $pm i$ is in fact an infinite number of directions $ai+bj+ckinmathbb{H}$ with $a^2+b^2+c^2=1$.
    $endgroup$
    – Zhuoran He
    Oct 30 '17 at 17:47










  • $begingroup$
    I think there should be a "where" in between $q_2,q_2$ where it says "$q_1,q_2,q_2inmathbb{R}$." I can't make the edit due to the character minimum on edits.
    $endgroup$
    – j0equ1nn
    Nov 25 '17 at 4:02










  • $begingroup$
    Thanks @j0equ1nn. Such details are anything but clear to me.
    $endgroup$
    – Jyrki Lahtonen
    Nov 25 '17 at 5:47











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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active

oldest

votes









10












$begingroup$

With non-real quaternions we can always find two square roots.



You can write any quaternion in the form
$$
q=a+bvec{u},
$$
where $a$ and $b$ are real, and $vec{u}$ is a unit vector. You probably know that as a quaternion $vec{u}^2=-1$. Therefore we can treat $vec{u}$ as if it were the usual imaginary unit $i$ of the complex numbers. So we can use the usual techniques of finding square roots of complex numbers.



Note that with non-real quaternions we only get two square roots. This is because the square of the above quaternion is
$$
q^2=(a^2-b^2)+2abvec{u}.
$$
For this to be non-real, we need both $a$ and $b$ to be non-zero. So whenever
$q_1^2=q_2$ for some quaternions $q_1,q_2$ where $q_2notinmathbb{R}$, they must both
lie in the same plane, i.e. they must be linear combinations of $1$ and the same unit vector $vec{u}$. This means that $q_1$ and $q_2$ must belong to the same copy of $mathbb{C}=mathbb{R}oplusmathbb{R}vec{u}$. As any complex number has only two square roots, the same applies to non-real quaternions because of this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Interesting. In the quaternion field $mathbb{H}$, only negative real numbers have infinitely many square roots, while all other quaternions have only two square roots. Is this true and does this generalize to $n$-th roots?
    $endgroup$
    – Zhuoran He
    Oct 30 '17 at 5:20










  • $begingroup$
    @ZhuoranHe I think that the above argument applies to $n$th roots of non-real quaternions as well. But if $n>2$ there are infinitely many $n$th roots of unity in $Bbb{H}$: to each unit vector $vec{u}$ we have $n-1$ non-real $n$th roots of unity in $Bbb{R}oplusBbb{R}vec{u}$. Therefore all reals have infinitely many $n$th roots in $Bbb{H}$.
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '17 at 7:26












  • $begingroup$
    I get it. Whenever there's symmetry breaking, e.g. $sqrt{-1}=pm i$, $1^{1/3}=1, e^{pm2pi i/3}$ in $mathbb{C}$, quaternions would say that the $pm i$ is in fact an infinite number of directions $ai+bj+ckinmathbb{H}$ with $a^2+b^2+c^2=1$.
    $endgroup$
    – Zhuoran He
    Oct 30 '17 at 17:47










  • $begingroup$
    I think there should be a "where" in between $q_2,q_2$ where it says "$q_1,q_2,q_2inmathbb{R}$." I can't make the edit due to the character minimum on edits.
    $endgroup$
    – j0equ1nn
    Nov 25 '17 at 4:02










  • $begingroup$
    Thanks @j0equ1nn. Such details are anything but clear to me.
    $endgroup$
    – Jyrki Lahtonen
    Nov 25 '17 at 5:47
















10












$begingroup$

With non-real quaternions we can always find two square roots.



You can write any quaternion in the form
$$
q=a+bvec{u},
$$
where $a$ and $b$ are real, and $vec{u}$ is a unit vector. You probably know that as a quaternion $vec{u}^2=-1$. Therefore we can treat $vec{u}$ as if it were the usual imaginary unit $i$ of the complex numbers. So we can use the usual techniques of finding square roots of complex numbers.



Note that with non-real quaternions we only get two square roots. This is because the square of the above quaternion is
$$
q^2=(a^2-b^2)+2abvec{u}.
$$
For this to be non-real, we need both $a$ and $b$ to be non-zero. So whenever
$q_1^2=q_2$ for some quaternions $q_1,q_2$ where $q_2notinmathbb{R}$, they must both
lie in the same plane, i.e. they must be linear combinations of $1$ and the same unit vector $vec{u}$. This means that $q_1$ and $q_2$ must belong to the same copy of $mathbb{C}=mathbb{R}oplusmathbb{R}vec{u}$. As any complex number has only two square roots, the same applies to non-real quaternions because of this.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Interesting. In the quaternion field $mathbb{H}$, only negative real numbers have infinitely many square roots, while all other quaternions have only two square roots. Is this true and does this generalize to $n$-th roots?
    $endgroup$
    – Zhuoran He
    Oct 30 '17 at 5:20










  • $begingroup$
    @ZhuoranHe I think that the above argument applies to $n$th roots of non-real quaternions as well. But if $n>2$ there are infinitely many $n$th roots of unity in $Bbb{H}$: to each unit vector $vec{u}$ we have $n-1$ non-real $n$th roots of unity in $Bbb{R}oplusBbb{R}vec{u}$. Therefore all reals have infinitely many $n$th roots in $Bbb{H}$.
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '17 at 7:26












  • $begingroup$
    I get it. Whenever there's symmetry breaking, e.g. $sqrt{-1}=pm i$, $1^{1/3}=1, e^{pm2pi i/3}$ in $mathbb{C}$, quaternions would say that the $pm i$ is in fact an infinite number of directions $ai+bj+ckinmathbb{H}$ with $a^2+b^2+c^2=1$.
    $endgroup$
    – Zhuoran He
    Oct 30 '17 at 17:47










  • $begingroup$
    I think there should be a "where" in between $q_2,q_2$ where it says "$q_1,q_2,q_2inmathbb{R}$." I can't make the edit due to the character minimum on edits.
    $endgroup$
    – j0equ1nn
    Nov 25 '17 at 4:02










  • $begingroup$
    Thanks @j0equ1nn. Such details are anything but clear to me.
    $endgroup$
    – Jyrki Lahtonen
    Nov 25 '17 at 5:47














10












10








10





$begingroup$

With non-real quaternions we can always find two square roots.



You can write any quaternion in the form
$$
q=a+bvec{u},
$$
where $a$ and $b$ are real, and $vec{u}$ is a unit vector. You probably know that as a quaternion $vec{u}^2=-1$. Therefore we can treat $vec{u}$ as if it were the usual imaginary unit $i$ of the complex numbers. So we can use the usual techniques of finding square roots of complex numbers.



Note that with non-real quaternions we only get two square roots. This is because the square of the above quaternion is
$$
q^2=(a^2-b^2)+2abvec{u}.
$$
For this to be non-real, we need both $a$ and $b$ to be non-zero. So whenever
$q_1^2=q_2$ for some quaternions $q_1,q_2$ where $q_2notinmathbb{R}$, they must both
lie in the same plane, i.e. they must be linear combinations of $1$ and the same unit vector $vec{u}$. This means that $q_1$ and $q_2$ must belong to the same copy of $mathbb{C}=mathbb{R}oplusmathbb{R}vec{u}$. As any complex number has only two square roots, the same applies to non-real quaternions because of this.






share|cite|improve this answer











$endgroup$



With non-real quaternions we can always find two square roots.



You can write any quaternion in the form
$$
q=a+bvec{u},
$$
where $a$ and $b$ are real, and $vec{u}$ is a unit vector. You probably know that as a quaternion $vec{u}^2=-1$. Therefore we can treat $vec{u}$ as if it were the usual imaginary unit $i$ of the complex numbers. So we can use the usual techniques of finding square roots of complex numbers.



Note that with non-real quaternions we only get two square roots. This is because the square of the above quaternion is
$$
q^2=(a^2-b^2)+2abvec{u}.
$$
For this to be non-real, we need both $a$ and $b$ to be non-zero. So whenever
$q_1^2=q_2$ for some quaternions $q_1,q_2$ where $q_2notinmathbb{R}$, they must both
lie in the same plane, i.e. they must be linear combinations of $1$ and the same unit vector $vec{u}$. This means that $q_1$ and $q_2$ must belong to the same copy of $mathbb{C}=mathbb{R}oplusmathbb{R}vec{u}$. As any complex number has only two square roots, the same applies to non-real quaternions because of this.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 25 '17 at 5:45

























answered May 5 '13 at 19:10









Jyrki LahtonenJyrki Lahtonen

110k13171379




110k13171379












  • $begingroup$
    Interesting. In the quaternion field $mathbb{H}$, only negative real numbers have infinitely many square roots, while all other quaternions have only two square roots. Is this true and does this generalize to $n$-th roots?
    $endgroup$
    – Zhuoran He
    Oct 30 '17 at 5:20










  • $begingroup$
    @ZhuoranHe I think that the above argument applies to $n$th roots of non-real quaternions as well. But if $n>2$ there are infinitely many $n$th roots of unity in $Bbb{H}$: to each unit vector $vec{u}$ we have $n-1$ non-real $n$th roots of unity in $Bbb{R}oplusBbb{R}vec{u}$. Therefore all reals have infinitely many $n$th roots in $Bbb{H}$.
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '17 at 7:26












  • $begingroup$
    I get it. Whenever there's symmetry breaking, e.g. $sqrt{-1}=pm i$, $1^{1/3}=1, e^{pm2pi i/3}$ in $mathbb{C}$, quaternions would say that the $pm i$ is in fact an infinite number of directions $ai+bj+ckinmathbb{H}$ with $a^2+b^2+c^2=1$.
    $endgroup$
    – Zhuoran He
    Oct 30 '17 at 17:47










  • $begingroup$
    I think there should be a "where" in between $q_2,q_2$ where it says "$q_1,q_2,q_2inmathbb{R}$." I can't make the edit due to the character minimum on edits.
    $endgroup$
    – j0equ1nn
    Nov 25 '17 at 4:02










  • $begingroup$
    Thanks @j0equ1nn. Such details are anything but clear to me.
    $endgroup$
    – Jyrki Lahtonen
    Nov 25 '17 at 5:47


















  • $begingroup$
    Interesting. In the quaternion field $mathbb{H}$, only negative real numbers have infinitely many square roots, while all other quaternions have only two square roots. Is this true and does this generalize to $n$-th roots?
    $endgroup$
    – Zhuoran He
    Oct 30 '17 at 5:20










  • $begingroup$
    @ZhuoranHe I think that the above argument applies to $n$th roots of non-real quaternions as well. But if $n>2$ there are infinitely many $n$th roots of unity in $Bbb{H}$: to each unit vector $vec{u}$ we have $n-1$ non-real $n$th roots of unity in $Bbb{R}oplusBbb{R}vec{u}$. Therefore all reals have infinitely many $n$th roots in $Bbb{H}$.
    $endgroup$
    – Jyrki Lahtonen
    Oct 30 '17 at 7:26












  • $begingroup$
    I get it. Whenever there's symmetry breaking, e.g. $sqrt{-1}=pm i$, $1^{1/3}=1, e^{pm2pi i/3}$ in $mathbb{C}$, quaternions would say that the $pm i$ is in fact an infinite number of directions $ai+bj+ckinmathbb{H}$ with $a^2+b^2+c^2=1$.
    $endgroup$
    – Zhuoran He
    Oct 30 '17 at 17:47










  • $begingroup$
    I think there should be a "where" in between $q_2,q_2$ where it says "$q_1,q_2,q_2inmathbb{R}$." I can't make the edit due to the character minimum on edits.
    $endgroup$
    – j0equ1nn
    Nov 25 '17 at 4:02










  • $begingroup$
    Thanks @j0equ1nn. Such details are anything but clear to me.
    $endgroup$
    – Jyrki Lahtonen
    Nov 25 '17 at 5:47
















$begingroup$
Interesting. In the quaternion field $mathbb{H}$, only negative real numbers have infinitely many square roots, while all other quaternions have only two square roots. Is this true and does this generalize to $n$-th roots?
$endgroup$
– Zhuoran He
Oct 30 '17 at 5:20




$begingroup$
Interesting. In the quaternion field $mathbb{H}$, only negative real numbers have infinitely many square roots, while all other quaternions have only two square roots. Is this true and does this generalize to $n$-th roots?
$endgroup$
– Zhuoran He
Oct 30 '17 at 5:20












$begingroup$
@ZhuoranHe I think that the above argument applies to $n$th roots of non-real quaternions as well. But if $n>2$ there are infinitely many $n$th roots of unity in $Bbb{H}$: to each unit vector $vec{u}$ we have $n-1$ non-real $n$th roots of unity in $Bbb{R}oplusBbb{R}vec{u}$. Therefore all reals have infinitely many $n$th roots in $Bbb{H}$.
$endgroup$
– Jyrki Lahtonen
Oct 30 '17 at 7:26






$begingroup$
@ZhuoranHe I think that the above argument applies to $n$th roots of non-real quaternions as well. But if $n>2$ there are infinitely many $n$th roots of unity in $Bbb{H}$: to each unit vector $vec{u}$ we have $n-1$ non-real $n$th roots of unity in $Bbb{R}oplusBbb{R}vec{u}$. Therefore all reals have infinitely many $n$th roots in $Bbb{H}$.
$endgroup$
– Jyrki Lahtonen
Oct 30 '17 at 7:26














$begingroup$
I get it. Whenever there's symmetry breaking, e.g. $sqrt{-1}=pm i$, $1^{1/3}=1, e^{pm2pi i/3}$ in $mathbb{C}$, quaternions would say that the $pm i$ is in fact an infinite number of directions $ai+bj+ckinmathbb{H}$ with $a^2+b^2+c^2=1$.
$endgroup$
– Zhuoran He
Oct 30 '17 at 17:47




$begingroup$
I get it. Whenever there's symmetry breaking, e.g. $sqrt{-1}=pm i$, $1^{1/3}=1, e^{pm2pi i/3}$ in $mathbb{C}$, quaternions would say that the $pm i$ is in fact an infinite number of directions $ai+bj+ckinmathbb{H}$ with $a^2+b^2+c^2=1$.
$endgroup$
– Zhuoran He
Oct 30 '17 at 17:47












$begingroup$
I think there should be a "where" in between $q_2,q_2$ where it says "$q_1,q_2,q_2inmathbb{R}$." I can't make the edit due to the character minimum on edits.
$endgroup$
– j0equ1nn
Nov 25 '17 at 4:02




$begingroup$
I think there should be a "where" in between $q_2,q_2$ where it says "$q_1,q_2,q_2inmathbb{R}$." I can't make the edit due to the character minimum on edits.
$endgroup$
– j0equ1nn
Nov 25 '17 at 4:02












$begingroup$
Thanks @j0equ1nn. Such details are anything but clear to me.
$endgroup$
– Jyrki Lahtonen
Nov 25 '17 at 5:47




$begingroup$
Thanks @j0equ1nn. Such details are anything but clear to me.
$endgroup$
– Jyrki Lahtonen
Nov 25 '17 at 5:47


















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