July 2nd update (2)

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July 2nd update (2)

Article:

I’ve planned 300 hours for the details of Wiles’ Proof of Fermat’s Last Theorem.

Update:

from

(a^d + b^d)^(1/d) = (z^d)^(1/d) is a norm to a side of projection to the isometric space.
The cases are (the right side : the left side), to consider
(1) rational number : rational number
(2) irrational number : rational number
(3) rational number : irrational number
(4) irrational number : irrational number
E.g. x = y = 2,
(x^d + y^d)^(1/d) = (2x^d)^(1/d) = (z^d)^(1/d),
(x^(d+1))^(1/d) = (z^d)^(1/d),
x^2 = z in the inverse square space, (1) or (4) of T.
E.g. x = y ≠ 2,
(x^d + y^d)^(1/d) = (2x^d)^(1/d) = (z^d)^(1/d),
2^(1/d) × x = 2^(1/d) × y = z in an isometric space, {2^(1/d) → β | β ∈ α}, (4) of T.

to

If (a^d + b^d)^(1/d) = (z^d)^(1/d) is a norm to a side of projection to the isometric space.
The cases are (the right side : the left side), to consider
(1) rational number : rational number
(2) irrational number : rational number
(3) rational number : irrational number
(4) irrational number : irrational number
E.g. x = y = 2,
if (x^d + y^d)^(1/d) = (2x^d)^(1/d) = (z^d)^(1/d), (x^(d+1))^(1/d) = (z^d)^(1/d),
x^((d+1)/d) ≠ z^1, d=1 → x^2 is in the inverse square space, but (2) or (4) of F.
E.g. x = y ≠ 2,
if (x^d + y^d)^(1/d) = (2x^d)^(1/d) = (z^d)^(1/d),
2^(1/d) × x^1 = 2^(1/d) × y^1 ≠ z^1 in an isometric space, {2^(1/d) → β | β ∈ α}, (2) or (4) of F.

URL:

I’ve planned 300 hours for the details of Wiles’ Proof of Fermat’s Last Theorem. – Kyo’s Blog (kyonissho.com)

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