Quaternions are Non-Mathematical

[talanum1]

We have the equation: i^2=j^2=k^2=ijk=-1. Now solve the first three equations and substitute the result into the fourth. We get:

ijk = (+/-sqrt(-1))(+/-sqrt(-1))(+/-sqrt(-1)) = +/-sqrt(-1) not = -1.

Now the story is that you cannot make this substitution which is contrary to Mathematical rules.

 

[talanum1]

We have the equation: i^2=j^2=k^2=ijk=-1. Now solve the first three equations and substitute the result into the fourth. We get:

ijk = (+/-sqrt(-1))(+/-sqrt(-1))(+/-sqrt(-1)) = +/-sqrt(-1) not = -1.

Now the story is that you cannot make this substitution which is contrary to Mathematical rules.

Lemma: |i| = |j| = |k| is consistent with the axiom: i^2 = j^2 = k^2. Then the i, j, k cannot be distinct except where the sign on any two may differ.

Proof:

Assume i^2 = j^2 = k^2

Do the operation: drawing the square root to both sides of the equation:

sqrt(i^2) = sqrt(j^2) and sqrt(j^2) = sqrt(k^2)

or |i| = |j| = |k|

now two of them cannot be distinct because one of the following sentences must be true:

i = sqrt(-1), j = sqrt(-1), k = sqrt(-1)
i = -sqrt(-1), j = sqrt(-1), k = sqrt(-1)
i = sqrt(-1), j = -sqrt(-1), k = sqrt(-1)
i = sqrt(-1), j = sqrt(-1), k = -sqrt(-1)
i = -sqrt(-1), j = -sqrt(-1), k = sqrt(-1)
i = -sqrt(-1), j = sqrt(-1), k = -sqrt(-1)
i = sqrt(-1), j = -sqrt(-1), k = -sqrt(-1)
i = -sqrt(-1), j = -sqrt(-1), k = -sqrt(-1)

QED.

 

[talanum1]

From the above we can conclude they are commutative, contradicting a consequence of the axioms.

So it doesn't matter that I can't substitute +/-sqrt(-1) for i, j, k.

 

[talanum1]

Those formulas for i, j, k aren't the only values that satisfies i^2 = j^2 = k^2 = -1, in fact any point on the unit sphere would. So we have: i can be = j can be = k which contradicts the proof that they are always distinct.

 

[luiza]

Quateronions are Non-Mathematical​

You are right .
And it has us all in tears .

 

[talanum1]

Anyway saying there exists more than one sqrt(-1) such that sqrt(-1) not = sqrt(-1) is stretching the truth, therefore false.

 

[Churchill]

We have the equation: i^2=j^2=k^2=ijk=-1. Now solve the first three equations and substitute the result into the fourth. We get:

ijk = (+/-sqrt(-1))(+/-sqrt(-1))(+/-sqrt(-1)) = +/-sqrt(-1) not = -1.

Now the story is that you cannot make this substitution which is contrary to Mathematical rules.

 

[talanum1]

If they are not complex numbers, one shouldn't be able to prove that they are.

 

[Churchill]

If they are not complex numbers, one shouldn't be able to prove that they are.

One can't, one can present a flawed proof of course, that's entirely different.

 

[talanum1]

Drawing the square root on both sides of i^2 = -1 is not a flawed proof. What makes it a flaw?

 

[Churchill]

Drawing the square root on both sides of i^2 = -1 is not a flawed proof. What makes it a flaw?

I'm not that familair with quaternions, but from what I read this not true:

ijk = (+/-sqrt(-1))(+/-sqrt(-1))(+/-sqrt(-1)) = +/-sqrt(-1)

The situation is that there's more than one square root of -1 in quaternions, jsut as the there's more than one square root of 4.

Raise it with Copilot, it will explain more thoroughly, e.g.

 

[talanum1]

I've substituted 3 different solutions of i^2 = j^2 = k^2 = -1 into i, j, k, and still didn't get ijk = -1.

 

[Churchill]

I've substituted 3 different solutions of i^2 = j^2 = k^2 = -1 into i, j, k, and still didn't get ijk = -1.

Ahh, well...

ijk = -1 is not a deduction, not a derivation, it is a defintion, a premise, that's where you err.

Using i to mean one thing in complex algebra and something else in quaternion algebra is the source of the confusion, it seems it happened because Hamilton began all his stuff by trying to extend complex numbers to 3D.

That led nowhere so he devised a new algebra and ended up using i for one of the "dimensions" but it doesn't represent the usual root of -1.

 

[Churchill]

 

[talanum1]

ijk = -1 is not a deduction, not a derivation, it is a defintion, a premise, that's where you err.

The model must satisfy the axioms.

Say: sqrt (i^2) = i, sqrt (j^2) = (i+j)/sqrt(2) and sqrt(k^2) = (i-j)/sqrt(2) is a representation of the model. Then:

ijk = i (-1+1)/sqrt(2)^2 = 0 not = -1.

 

[Churchill]

The model must satisfy the axioms.

Say: sqrt (i^2) = i, sqrt (j^2) = (i+j)/sqrt(2) and sqrt(k^2) = (i-j)/sqrt(2) is a representation of the model. Then:

ijk = i (-1+1)/sqrt(2)^2 = 0 not = -1.

Do you accept that ij ≠ ji ?

 

[talanum1]

Do you accept that ij ≠ ji ?

I can't seem to wrap my head around that.

For that to happen the numbers must be different if read left-right than right-left.

 

[Churchill]

I can't seem to wrap my head around that.

For that to happen the numbers must be different if read left-right than right-left.

This feature is called non-commutativity. It captures the fact that if we had a point in 3D space and we:

move it angle A relative to say the Y axis and then angle B relative to say the X axis,

then the final position of the point will be different compares to if we:

move it angle B relative to say the X axis and then angle A relative to say the Y axis.

So the final position is different depending on the order we apply the operations. The multiplication is designed to represent the successive application of the two operations.

 

[talanum1]

What can i, j and k be equal to?

 

[Churchill]

What can i, j and k be equal to?

They are fixed values, like i is in a complex number but is not the same as i in a complex number.