#quaternion exponents are a bit off #shorts #math

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[theelk]

you can’t just do that exponent stuff willy nilly when you don’t have commutativity

[shanathered]

the exponent rule i^(j-1) ≠ i^j*i^-1 does not apply, as quaternions are noncommutative.

[OriginalSuschi]

On the left we applied inverse i on the right ok the left side while we applied it on the left on the right side of the equation. Since quaternions are generally not-commutative, we can‘t simply say that a*b=b*a for a and b being quaternionic numbers.

[ianmathwiz]

When you multiplied by i^(-1), you multiplied it on the left. But when you brought the i^(-1) back out, you had it on the right. Multiplication is not commutative in quaternions.

[deinauge]

you cannot add exponents with noncommutative numbers.
a^b * a^c != a^(b+c)
therefore i^(-1)*i^j = i^(j-1) is the first wrong step you took here

[fforstupid]

I'm a music student, that's when it went wrong

[Djaketooth]

It goes wrong in the second step. When you have i^j=i*j and you apply i^(-1) on the left you get i^(-1)*i^j=j. But exponentiating i is not distributive to addition so we don't have i^(-1)*i^j=i^(j-1). Even by converting i to e^ln(i) (taking the principal branch gives pi/2*i) we're still stuck: (e^(pi/2*i This is because we also don't have (e^x)^y=e^(x*y) for quaternions.

[chrisng]

Quaternions don't commute, so we need to keep track of left and right

[SlipperyTeeth]

Everyone seems to know that it's the commutativity, so I won't repeat that.
Just, naively, I thought it would work, because it's power associative. I've never realized that obviously that doesn't say anything about exponents that aren't natural numbers (my focus was on polynomials when I was learning about power associativity).

[linuxp]

You have to consider the anticommutativity, multiplying by i^(-1) on the left on both sides of the equation, kind like matrix multiplication (which quaternions are isomorphic to, anyways).

Done right, this easily shows:

I^J = I * J
I^(-1) * I ^ J = I ^(-1) * I * J
-I * (I ^ J) = J
-I * K = K * I = J

[kou---]

It goes wrong to multiply i^(-1) to different side in both side of the equation, since the multiply operation of quarterion is not commutative.

[zyklos]

Ah ok non-commutativity was broken when you divided by i from the left, but ended with i^-1 in the right, which switched the sign of the expression.
So something went wrong in the exponential, where addition seems to become non-commutative too (for a reason)

[sugarfrosted]

Quarnionic non integer powers are multivalued, so you're preserving the set of solutions with these operations, just not the specific chosen values. More or less what happens in complex analysis.

[dhzdhz]

i^j=k=ij leads to i^(-1)i^j=i^(-1)ij=j. i^(-1) is on the left of i^j. This order is important.

[yurigouveawagner]

if the product doesn't commute, you can't just sum the exponents. (i^(-1))×i×j = j, but i×j×(i^(-1)) = (-j)×i×(i^(-1)) = -j, so ij and i^(-1) anticommute. so if i^j = ij, i^j anticommutes w/ i^(-1) and you can't do i^(-1)×i^(j) = i^(j-1) since the sum commutes.

[Tehom]

In step one, we multiplied by i^-1 on the left, but we expressed the -1 in the exponent (ambiguous) and in step 3 we took the same thing out on the right. This is just wrong because quaternions don't commute, and in fact purely imaginary quaternions such as i^-1 anticommute. The switcheroo is what gives us the minus sign in the result.

[damianalex]

Are you going to bring a quaternion series?

I've recently started getting interest towards these numbers but I'm having a quite hard time finding material to study and exercise on them. Do you maybe have any suggestion on quaternions and quaternionic analysis?

[noumanegaou]

If we see a quaternion field as her matrix representation and definition of expent as exp of matrix we dont have nesserly exp(A+B) = exp(A) exp(B)
i^j =exp(jlog(i)) not equal

[PRIYANSH_SUTHAR]

I think the problem arise because in the second step, you are assuming that there is comuutativity and this causes the left and right sides to be different. i^(–1).i.j ≠ i.j.i^(–1).

[qschroed]

We never introduced how to handle exponents for quaternions outside of that one case here we had i^j, so multiplying by i^(-1) does not let us replace it by i^(-1+j)