Debates - Potential Faulty Science

RedViking said:
AI is wrong most of the time with stuff like this. Best is to look at Academic Articles
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I wouldn't know, none of the AI responses posted by me (long ago now) or others in these threads were wrong. But otherwise agree, better to go to the sources.
 
You can plainly see the fault is in ijk=-1, I just used the mathematical rules.
 
talanum1 said:
You can plainly see the fault is in ijk=-1, I just used the mathematical rules.
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Problem is, you only get educated responses from the (very few) science boffins here sometimes. So who knows... RedViking is right, check with authoritative sources.

"Can plainly see" has let you down a lot before on this forum.
Not that there's anything bad about getting things wrong, but you do want the actual facts...

Anyway, enough from me, you do you :thumbsup:
 
Brian_G said:
Problem is, you only get educated responses from the (very few) science boffins here sometimes. So who knows... RedViking is right, check with authoritative sources.

"Can plainly see" has let you down a lot before on this forum.
Not that there's anything bad about getting things wrong, but you do want the actual facts...

Anyway, enough from me, you do you :thumbsup:
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I might not be an educated science boffin. But at least I know I am right.
 
OK, so you can't substitute sqrt(-1) for i, j, k. However there is another problem:

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 = -1

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) = sqrt(-1)

or |i| = |j| = |k| = sqrt(-1)

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.

From the formulas above we can conclude that their products are commutative, contradicting the result that they are not.
 
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Also i + j + k not only = i + j + k: also a contradiction.
 
How is your ChatGPT looking?

Short answer: the conclusion is wrong, and the reasoning is sloppy. Whether he’s being pretentious or just confused is up for debate—but mathematically, this doesn’t hold up.

Let’s break it cleanly.

🔴 The core mistake

The argument hinges on this step:

“take square roots: √(i²) = √(-1) ⇒ |i| = √(-1)”

That’s already a problem.

1. Misuse of square roots in complex numbers

In real numbers:

√(x²) = |x|

But in complex numbers, square roots are multi-valued. More importantly:

√(-1) is not a single number unless you define a branch (like choosing i specifically).

So writing:

|i| = √(-1)

is nonsense.

Correct facts:

|i| = 1
√(-1) = ±i (multi-valued)

So he’s mixing up magnitude (absolute value) with square roots. Those are completely different operations.

2. Treating i, j, k like ordinary numbers

He assumes:

i² = j² = k² = -1 ⇒ they must be ±√(-1)

That would be true in ℂ (complex numbers).

But i, j, k usually refer to quaternions, where:

i² = j² = k² = -1
BUT also:
ij = k
jk = i
ki = j
and importantly: ij ≠ ji (non-commutative)

So he's applying complex-number intuition to quaternions, which is invalid.

3. False conclusion about commutativity

He concludes:

“they must be ±√(-1), so they’re essentially the same ⇒ commutative”

That’s just wrong.

In quaternions:

i, j, k are distinct basis elements
They don’t commute:
ij = k
ji = -k

His “proof” never even touches multiplication between i, j, k—so it can’t possibly say anything about commutativity.

🧠 What’s really going on

He’s implicitly assuming:

“Anything whose square is -1 must just be ±i”

That’s only true in ℂ.

But in richer algebraic systems (like quaternions), you can have multiple distinct elements squaring to -1.

⚖️ Verdict
The argument is mathematically unsound
The key errors:
Misusing square roots
Confusing absolute value with roots
Applying complex-number rules to quaternions
Jumping to conclusions about commutativity without examining multiplication
💬 Straight talk

This isn’t deep math—it’s someone confidently misapplying basic concepts.

So yeah… the tone might be a bit “look at my lemma,” but the real issue is they don’t understand the structures they’re talking about.
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talanum1 said:
We have: i^2=j^2=k^2=ijk=-1. Now set i=j=k=sqrt(-1) in the fourth equation:

sqrt(-1)*sqrt(-1)*sqrt(-1) = -1*sqrt(-1) not = -1, so it doesn't hold. What do I do wrong?
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The bolded bit is wrong. It doesn’t follow from the definition. Even in R, a^2=b^2 doesn’t mean that a=b.
 
cguy said:
Where? Post 66 makes the same incorrect inference.
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At post 66. It does not make the same incorrect inference: we infer |i|=|j|=|k|=sqrt(-1). Note the "size of" operators. This is correct.

If a^2 = b^2 we have a = b or -a = b or a = -b or -a = -b.
 
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talanum1 said:
At post 66. It does not make the same incorrect inference: we infer |i|=|j|=|k|=sqrt(-1). Note the "size of" operators. This is correct.
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talanum1 said:
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) = sqrt(-1)
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^ this is exactly the same incorrect inference.

talanum1 said:
or |i| = |j| = |k| = sqrt(-1)
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This is wrong too. The absolute values are equal to 1, not sqrt(-1).

talanum1 said:
If a^2 = b^2 we have a = b or -a = b or a = -b or -a = -b.
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Yes, but in H, there are infinite values for sqrt(-1). Specifically, any unit pure imaginary quaternion.
 
You are telling me that in H we cannot draw the square root on both sides of an equation. Where in the axioms is this stated?
cguy said:
Yes, but in H, there are infinite values for sqrt(-1)
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List some of them.
 
talanum1 said:
You are telling me that in H we cannot draw the square root on both sides of an equation. Where in the axioms is this stated?

List some of them.
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Yes exactly that, it’s so basic bro, even @RedViking gets it and all he knows how to do is fancy cappuccino hearts with steamed milk.
 
airborne said:
Yes exactly
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Then quaternions are non-Mathematical since you can't apply the usual rules of Mathematics to them. Thinking they are Mathematical and using them my lead to accidents.

You didn't list them.
 
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cguy said:
You can draw it but it doesn’t mean what you think it imenas.
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You have to believe an obvious falsity: "sqrt not equivalent to sqrt". It just shows you how obsessed you are with your textbooks.

cguy said:
Any q=ai+bj+cj
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Are you telling me: sqrt(i^2) = ai + bj +ck?
 
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talanum1 said:
You have to believe an obvious falsity: "sqrt not equivalent to sqrt". It just shows you how obsessed you are with your textbooks.
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You are assuming sqrt has the injective property over these domains, when it cleary it does not. You are holding the obvious falsity.

talanum1 said:
Are you telling me: sqrt(i^2) = ai + bj +ck?
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Yes, and hare are 20 out of infinity examples.
  • i
  • j
  • k
  • (i + j)/sqrt(2)
  • (i - j)/sqrt(2)
  • (i + k)/sqrt(2)
  • (i - k)/sqrt(2)
  • (j + k)/sqrt(2)
  • (j - k)/sqrt(2)
  • (i + j + k)/sqrt(3)
  • (i + j - k)/sqrt(3)
  • (i - j + k)/sqrt(3)
  • (-i + j + k)/sqrt(3)
  • 0.5*i + (sqrt(3)/2)*j
  • 0.5*i + (sqrt(3)/2)*k
  • 0.5*j + (sqrt(3)/2)*k
  • sqrt(2/3)*i + (1/sqrt(3))*j
  • sqrt(2/3)*i + (1/sqrt(3))*k
  • (1/sqrt(3))*i + sqrt(2/3)*j
  • (1/sqrt(6))*i + (1/sqrt(6))*j + (2/sqrt(6))*k
 
cguy said:
Yes, and hare are 20 out of infinity examples.
  • i
  • j
  • k
  • (i + j)/sqrt(2)
  • (i - j)/sqrt(2)
  • (i + k)/sqrt(2)
  • (i - k)/sqrt(2)
  • (j + k)/sqrt(2)
  • (j - k)/sqrt(2)
  • (i + j + k)/sqrt(3)
  • (i + j - k)/sqrt(3)
  • (i - j + k)/sqrt(3)
  • (-i + j + k)/sqrt(3)
  • 0.5*i + (sqrt(3)/2)*j
  • 0.5*i + (sqrt(3)/2)*k
  • 0.5*j + (sqrt(3)/2)*k
  • sqrt(2/3)*i + (1/sqrt(3))*j
  • sqrt(2/3)*i + (1/sqrt(3))*k
  • (1/sqrt(3))*i + sqrt(2/3)*j
  • (1/sqrt(6))*i + (1/sqrt(6))*j + (2/sqrt(6))*k
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If of those numbers are in "or" statements, then how do you select the actual answer? And if they are in "and" statements then all of them are equal to each other, which means at least i = j = k.
 
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