Mixing
Square roots of $j$ and $ε$
$begingroup$
I know how to find the square root of the imaginary unit $i$, but I'm still learning about split-complex and dual numbers. I can't find any info anywhere about the square roots of $j$ and $ε$, if they have them.
radicals split-complex-numbers
edited Jan 31 at 1:50
Eevee Trainer
10k31740
asked Jan 31 at 1:49
A.J.A.J.
162
$endgroup$
add a comment |
$begingroup$
I know how to find the square root of the imaginary unit $i$, but I'm still learning about split-complex and dual numbers. I can't find any info anywhere about the square roots of $j$ and $ε$, if they have them.
radicals split-complex-numbers
edited Jan 31 at 1:50
Eevee Trainer
10k31740
asked Jan 31 at 1:49
A.J.A.J.
162
$endgroup$
3
$begingroup$
$j$ and $varepsilon$?
$endgroup$
– Chase Ryan Taylor
Jan 31 at 1:52
1
$begingroup$
@ChaseRyanTaylor I hadn't heard of them either, but dual numbers and split-complex numbers are a thing.
$endgroup$
– Theo Bendit
Jan 31 at 1:54
1
$begingroup$
He means the classes of $x$ in the quotients $Bbb R[x]/(x^2-1)$ and $Bbb R[x]/(x^2)$, respectively.
$endgroup$
– Ivo Terek
Jan 31 at 1:55
add a comment |
$begingroup$
I know how to find the square root of the imaginary unit $i$, but I'm still learning about split-complex and dual numbers. I can't find any info anywhere about the square roots of $j$ and $ε$, if they have them.
radicals split-complex-numbers
edited Jan 31 at 1:50
Eevee Trainer
10k31740
asked Jan 31 at 1:49
A.J.A.J.
162
$endgroup$
I know how to find the square root of the imaginary unit $i$, but I'm still learning about split-complex and dual numbers. I can't find any info anywhere about the square roots of $j$ and $ε$, if they have them.
radicals split-complex-numbers
radicals split-complex-numbers
edited Jan 31 at 1:50
Eevee Trainer
10k31740
asked Jan 31 at 1:49
A.J.A.J.
162
edited Jan 31 at 1:50
Eevee Trainer
10k31740
asked Jan 31 at 1:49
A.J.A.J.
162
edited Jan 31 at 1:50
Eevee Trainer
10k31740
edited Jan 31 at 1:50
Eevee Trainer
10k31740
edited Jan 31 at 1:50
Eevee Trainer
10k31740
10k31740
asked Jan 31 at 1:49
A.J.A.J.
162
asked Jan 31 at 1:49
A.J.A.J.
162
asked Jan 31 at 1:49
A.J.A.J.
162
162
3
$begingroup$
$j$ and $varepsilon$?
$endgroup$
– Chase Ryan Taylor
Jan 31 at 1:52
1
$begingroup$
@ChaseRyanTaylor I hadn't heard of them either, but dual numbers and split-complex numbers are a thing.
$endgroup$
– Theo Bendit
Jan 31 at 1:54
1
$begingroup$
He means the classes of $x$ in the quotients $Bbb R[x]/(x^2-1)$ and $Bbb R[x]/(x^2)$, respectively.
$endgroup$
– Ivo Terek
Jan 31 at 1:55
add a comment |
3
$begingroup$
$j$ and $varepsilon$?
$endgroup$
– Chase Ryan Taylor
Jan 31 at 1:52
1
$begingroup$
@ChaseRyanTaylor I hadn't heard of them either, but dual numbers and split-complex numbers are a thing.
$endgroup$
– Theo Bendit
Jan 31 at 1:54
1
$begingroup$
He means the classes of $x$ in the quotients $Bbb R[x]/(x^2-1)$ and $Bbb R[x]/(x^2)$, respectively.
$endgroup$
– Ivo Terek
Jan 31 at 1:55
3
3
$begingroup$
$j$ and $varepsilon$?
$endgroup$
– Chase Ryan Taylor
Jan 31 at 1:52
$begingroup$
$j$ and $varepsilon$?
$endgroup$
– Chase Ryan Taylor
Jan 31 at 1:52
1
1
$begingroup$
@ChaseRyanTaylor I hadn't heard of them either, but dual numbers and split-complex numbers are a thing.
$endgroup$
– Theo Bendit
Jan 31 at 1:54
$begingroup$
@ChaseRyanTaylor I hadn't heard of them either, but dual numbers and split-complex numbers are a thing.
$endgroup$
– Theo Bendit
Jan 31 at 1:54
1
1
$begingroup$
He means the classes of $x$ in the quotients $Bbb R[x]/(x^2-1)$ and $Bbb R[x]/(x^2)$, respectively.
$endgroup$
– Ivo Terek
Jan 31 at 1:55
$begingroup$
He means the classes of $x$ in the quotients $Bbb R[x]/(x^2-1)$ and $Bbb R[x]/(x^2)$, respectively.
$endgroup$
– Ivo Terek
Jan 31 at 1:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
There are no square roots for $j$. Say that $(a+bj)^2 = j$. Then $$a^2+2abj+b^2=j$$leads to $a^2+b^2 = 0$ and $2ab=1$, which has no solution. Similarly, if $(a+bepsilon)^2=epsilon$, then $$a^2+2abepsilon = epsilon$$leads to $a^2 = 0$ and $2ab=1$, which has no solution.
This probably owes to the fact that the split-complex algebra and the dual numbers algebra are not fields. In general, if $alpha$ and $beta$ are real numbers, the behavior of the set $$Bbb C_{alpha,beta} = {a+bmathfrak{u} mid a,b in Bbb Rmbox{ and }mathfrak{u}^2=alpha+betamathfrak{u}},$$equipped with the obvious operations, can be controled by the discriminant $Delta = beta^2+4alpha$.
- If $Delta < 0$, then $Bbb C_{alpha,beta}$ is a field.
- If $Delta=0$, the zero divisors of $Bbb C_{alpha,beta}$ are precisely the elements $a+bmathfrak{u}$ with $a+beta b/2=0$, while the others have inverses.
- If $Delta>0$, the zero divisors are precisely the elements $a+bmathfrak{u}$ such that $$a+(beta+sqrt{Delta})b/2 = 0 quadmbox{or}quad a+(beta-sqrt{Delta})b/2=0,$$while the others have inverses.
Clearly the last two cases are, in reality, a single one, but I think it is easier to see what happens if you state it like this. Proof of these facts? Exercise!
answered Jan 31 at 1:54
Ivo TerekIvo Terek
46.7k954146
$endgroup$
add a comment |
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$begingroup$
There are no square roots for $j$. Say that $(a+bj)^2 = j$. Then $$a^2+2abj+b^2=j$$leads to $a^2+b^2 = 0$ and $2ab=1$, which has no solution. Similarly, if $(a+bepsilon)^2=epsilon$, then $$a^2+2abepsilon = epsilon$$leads to $a^2 = 0$ and $2ab=1$, which has no solution.
This probably owes to the fact that the split-complex algebra and the dual numbers algebra are not fields. In general, if $alpha$ and $beta$ are real numbers, the behavior of the set $$Bbb C_{alpha,beta} = {a+bmathfrak{u} mid a,b in Bbb Rmbox{ and }mathfrak{u}^2=alpha+betamathfrak{u}},$$equipped with the obvious operations, can be controled by the discriminant $Delta = beta^2+4alpha$.
- If $Delta < 0$, then $Bbb C_{alpha,beta}$ is a field.
- If $Delta=0$, the zero divisors of $Bbb C_{alpha,beta}$ are precisely the elements $a+bmathfrak{u}$ with $a+beta b/2=0$, while the others have inverses.
- If $Delta>0$, the zero divisors are precisely the elements $a+bmathfrak{u}$ such that $$a+(beta+sqrt{Delta})b/2 = 0 quadmbox{or}quad a+(beta-sqrt{Delta})b/2=0,$$while the others have inverses.
Clearly the last two cases are, in reality, a single one, but I think it is easier to see what happens if you state it like this. Proof of these facts? Exercise!
answered Jan 31 at 1:54
Ivo TerekIvo Terek
46.7k954146
$endgroup$
add a comment |
$begingroup$
There are no square roots for $j$. Say that $(a+bj)^2 = j$. Then $$a^2+2abj+b^2=j$$leads to $a^2+b^2 = 0$ and $2ab=1$, which has no solution. Similarly, if $(a+bepsilon)^2=epsilon$, then $$a^2+2abepsilon = epsilon$$leads to $a^2 = 0$ and $2ab=1$, which has no solution.
This probably owes to the fact that the split-complex algebra and the dual numbers algebra are not fields. In general, if $alpha$ and $beta$ are real numbers, the behavior of the set $$Bbb C_{alpha,beta} = {a+bmathfrak{u} mid a,b in Bbb Rmbox{ and }mathfrak{u}^2=alpha+betamathfrak{u}},$$equipped with the obvious operations, can be controled by the discriminant $Delta = beta^2+4alpha$.
- If $Delta < 0$, then $Bbb C_{alpha,beta}$ is a field.
- If $Delta=0$, the zero divisors of $Bbb C_{alpha,beta}$ are precisely the elements $a+bmathfrak{u}$ with $a+beta b/2=0$, while the others have inverses.
- If $Delta>0$, the zero divisors are precisely the elements $a+bmathfrak{u}$ such that $$a+(beta+sqrt{Delta})b/2 = 0 quadmbox{or}quad a+(beta-sqrt{Delta})b/2=0,$$while the others have inverses.
Clearly the last two cases are, in reality, a single one, but I think it is easier to see what happens if you state it like this. Proof of these facts? Exercise!
answered Jan 31 at 1:54
Ivo TerekIvo Terek
46.7k954146
$endgroup$
add a comment |
$begingroup$
There are no square roots for $j$. Say that $(a+bj)^2 = j$. Then $$a^2+2abj+b^2=j$$leads to $a^2+b^2 = 0$ and $2ab=1$, which has no solution. Similarly, if $(a+bepsilon)^2=epsilon$, then $$a^2+2abepsilon = epsilon$$leads to $a^2 = 0$ and $2ab=1$, which has no solution.
This probably owes to the fact that the split-complex algebra and the dual numbers algebra are not fields. In general, if $alpha$ and $beta$ are real numbers, the behavior of the set $$Bbb C_{alpha,beta} = {a+bmathfrak{u} mid a,b in Bbb Rmbox{ and }mathfrak{u}^2=alpha+betamathfrak{u}},$$equipped with the obvious operations, can be controled by the discriminant $Delta = beta^2+4alpha$.
- If $Delta < 0$, then $Bbb C_{alpha,beta}$ is a field.
- If $Delta=0$, the zero divisors of $Bbb C_{alpha,beta}$ are precisely the elements $a+bmathfrak{u}$ with $a+beta b/2=0$, while the others have inverses.
- If $Delta>0$, the zero divisors are precisely the elements $a+bmathfrak{u}$ such that $$a+(beta+sqrt{Delta})b/2 = 0 quadmbox{or}quad a+(beta-sqrt{Delta})b/2=0,$$while the others have inverses.
Clearly the last two cases are, in reality, a single one, but I think it is easier to see what happens if you state it like this. Proof of these facts? Exercise!
answered Jan 31 at 1:54
Ivo TerekIvo Terek
46.7k954146
$endgroup$
There are no square roots for $j$. Say that $(a+bj)^2 = j$. Then $$a^2+2abj+b^2=j$$leads to $a^2+b^2 = 0$ and $2ab=1$, which has no solution. Similarly, if $(a+bepsilon)^2=epsilon$, then $$a^2+2abepsilon = epsilon$$leads to $a^2 = 0$ and $2ab=1$, which has no solution.
This probably owes to the fact that the split-complex algebra and the dual numbers algebra are not fields. In general, if $alpha$ and $beta$ are real numbers, the behavior of the set $$Bbb C_{alpha,beta} = {a+bmathfrak{u} mid a,b in Bbb Rmbox{ and }mathfrak{u}^2=alpha+betamathfrak{u}},$$equipped with the obvious operations, can be controled by the discriminant $Delta = beta^2+4alpha$.
- If $Delta < 0$, then $Bbb C_{alpha,beta}$ is a field.
- If $Delta=0$, the zero divisors of $Bbb C_{alpha,beta}$ are precisely the elements $a+bmathfrak{u}$ with $a+beta b/2=0$, while the others have inverses.
- If $Delta>0$, the zero divisors are precisely the elements $a+bmathfrak{u}$ such that $$a+(beta+sqrt{Delta})b/2 = 0 quadmbox{or}quad a+(beta-sqrt{Delta})b/2=0,$$while the others have inverses.
Clearly the last two cases are, in reality, a single one, but I think it is easier to see what happens if you state it like this. Proof of these facts? Exercise!
answered Jan 31 at 1:54
Ivo TerekIvo Terek
46.7k954146
edited Jan 31 at 2:03
answered Jan 31 at 1:54
Ivo TerekIvo Terek
46.7k954146
answered Jan 31 at 1:54
Ivo TerekIvo Terek
46.7k954146
answered Jan 31 at 1:54
Ivo TerekIvo Terek
46.7k954146
46.7k954146
add a comment |
add a comment |
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3
$begingroup$
$j$ and $varepsilon$?
$endgroup$
– Chase Ryan Taylor
Jan 31 at 1:52
1
$begingroup$
@ChaseRyanTaylor I hadn't heard of them either, but dual numbers and split-complex numbers are a thing.
$endgroup$
– Theo Bendit
Jan 31 at 1:54
1
$begingroup$
He means the classes of $x$ in the quotients $Bbb R[x]/(x^2-1)$ and $Bbb R[x]/(x^2)$, respectively.
$endgroup$
– Ivo Terek
Jan 31 at 1:55