Debates - Potential Faulty Science

talanum1 said:
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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They're ALL solutions to q^2 = -1, which I suppose matches to "or" - there is no one answer (that would require an injective function), the answer is a family of solutions. In R where you have x^2 = 1, and you have a family of two solutions: -1 or 1, and in C you have z^2 = -1, having solutions -i and i. In H, as illustrated above, there are an infinite amount of solutions (all the points on the imaginary unit sphere).

Just like x^2 = 1 doesn't imply that -1 is 1, i^2=j^2=-1, doesn't imply that i=j.
 
If the list is "or" statements, I can choose:

sqrt(i^2) = i

sqrt(j^2)=j

and

sqrt(k^2) = k

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

and the contradiction follows.
 
I guess you can't say "sqrt(i^2) = i" because that's not the whole truth.

However we can make identical lists for j and k, which means i can be = j can be = k which contradicts the proof that they are always distinct.
 
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talanum1 said:
If the list is "or" statements, I can choose:

sqrt(i^2) = i

sqrt(j^2)=j

and

sqrt(k^2) = k

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

and the contradiction follows.
Click to expand...
talanum1 said:
I guess you can't say "sqrt(i^2) = i" because that's not the whole truth.
Click to expand...
Correct: i just one sqrt(i^2) out of many.

talanum1 said:
However we can make identical lists for j and k, which means i can be = j can be = k which contradicts the proof that they are always distinct.
Click to expand...
Just because a number a and a number b have the same square, doesn't mean that they're the same number. I.e., even in R, a^2 = x, and b^2 =x, doesn't mean that a = b. You can list the roots for a (sqrt(x), -sqrt(x)), and for b (sqrt(x), -sqrt(x)), but this doesn't mean that sqrt(x)=-sqrt(x). This is how you're finding equivalence. In H there are just a lot more than 2 categories of solution.

You're doing the mathematical equivalent of saying a Chihuahua is dog (f(c)=d), and a German Shepard is a dog (f(g)=d), so therefore a Chihuahua is a German Shepard (c=d). In general c=d is only true if f() is injective. It's not.
 
You must agree that the same numbers can be written down for j and k. Then sqrt(i^2) can = i, sqrt(j^2) can = j and sqrt(k^2) can = k. Since LS's are equal by axiom, RS's must read i can = j can = k.
 
talanum1 said:
You must agree that the same numbers can be written down for j and k. Then sqrt(i^2) can = i, sqrt(j^2) can = j and sqrt(k^2) can = k. Since LS's are equal by axiom, RS's must read i can = j can = k.
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In the same way the sqrt(1) can be -1 or 1 doesn’t imply -1 can be 1, the above doesn’t imply i can be j or be k. Just as -1 and 1 are distinct, so are i and j and k. They’re not variables.
 
I'm not saying i = j = k because sqrt(i^2)= i or j or k. See previous post for the reason.
 
Sorry, not because of axioms is LS equal but all LS can be equal by the numbers listed.
 
@talanum1 , what will you do if cguy gets tired of showing you your faulty thinking, ignore that fact again?

Weird how you still cling to your confusion dude.
 
talanum1 said:
Sorry, not because of axioms is LS equal but all LS can be equal by the numbers listed.
Click to expand...
They are all different and unique numbers. The only thing they have in common is that their squares are -1.
 
They can be identical: what's going to enforce them to be different?
 
I can assign them values that makes them identical. How does the multiplication axioms prevent me from doing this?

The axioms are inconsistent with the fact that they can have any of those values (as other Mathematics says they can have).
 
I gave them different values and still didn't get ijk = -1:

i(i + j)/sqrt(2)(i - j)/sqrt(2) = (i/2)(-1 + 1) = 0 not = -1!
 
talanum1 said:
I can assign them values that makes them identical. How does the multiplication axioms prevent me from doing this?

The axioms are inconsistent with the fact that they can have any of those values (as other Mathematics says they can have).
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How are you assigning them values? They're fixed entities.

The multiplication axioms are: ij=k, ji=-k, and ij=-ji.

If i and j were the same, then ij=k AND ij=-k, which would be a contradiction unless k=0, which it isn't since k^2=-1, not 0. A similar argument can be made for j!=k, and k!=i.

talanum1 said:
I gave them different values and still didn't get ijk = -1:

i(i + j)/sqrt(2)(i - j)/sqrt(2) = (i/2)(-1 + 1) = 0 not = -1!
Click to expand...
You've taken 3 values from that list I gave, right? That list is just a list of numbers in H that can be squared to be equal to -1.

i^2 = -1 (by definition)

((i+j)/sqrt(2))^2 = ((i+j)/sqrt(2))((i+j)/sqrt(2)) = (i+j)(i+j)/2) = (i^2+ji + ij +j^2)/2 = (-1+k-k+-1)/2 = -2/2 = -1

((i-j)/sqrt(2))^2 = ((i-j)/sqrt(2))((i-j)/sqrt(2)) = (i-j)(i-j)/2) = (i^2-ji - ij +j^2)/2 = (-1-k+k+-1)/2 = -2/2 = -1

The numbers in the list are all distinct values, including i, j, and k. i.j.k=-1, but that's not necessarily true for all selections of three values in that list - the only thing that is true is that they all square to -1 (and i.j.k is not a square).
 
There just has to be one value of i^2 = j^2 = k^2 = -1 that doesn't satisfy ijk = -1, to say that the axioms are inconsistent with the rest of mathematics.
 
talanum1 said:
There just has to be one value of i^2 = j^2 = k^2 = -1 that doesn't satisfy ijk = -1, to say that the axioms are inconsistent with the rest of mathematics.
Click to expand...
i, j and k have values i, j and k. They’re not variables. It’s like saying you need one value of 5 such that 5^2!=25.
 
Problem With a Particle Being an Excitation in a Field:

A quark is an excitation in the quark field. Then where is the specific quark's momentum recorded?
 
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