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sqrtthlem is more general than resqrtthlem (sqrt section revision) #4623

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icecream17 opened this issue Feb 2, 2025 · 2 comments
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@icecream17
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https://us.metamath.org/mpeuni/sqrtthlem.html
https://us.metamath.org/mpeuni/resqrtthlem.html

sqrtthlem   | ⊒ (𝐴 ∈ β„‚           β†’ (((βˆšβ€˜π΄)↑2) = 𝐴 ∧ 0 ≀ (β„œβ€˜(βˆšβ€˜π΄)) ∧ (i Β· (βˆšβ€˜π΄)) βˆ‰ ℝ+))
resqrtthlem | ⊒ ((𝐴 ∈ ℝ ∧ 0 ≀ 𝐴) β†’ (((βˆšβ€˜π΄)↑2) = 𝐴 ∧ 0 ≀ (β„œβ€˜(βˆšβ€˜π΄)) ∧ (i Β· (βˆšβ€˜π΄)) βˆ‰ ℝ+))

This means that quite a few theorems can either be generalized to complex numbers completely, or a 0 ≀ 𝐴 condition can be removed, like in sqrt00, sqrtsq2, and sqrtmul. However, doing this is somewhat nontrivial, since sqrtthlem comes about 100 theorems after resqrtthlem.

@benjub
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benjub commented Feb 3, 2025

We should probably first prove a theorem that comes before that section on square roots: the analogous generalization of https://us.metamath.org/mpeuni/sq11.html that proves injectivity of the square function on that half-plane.

@jkingdon
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jkingdon commented Feb 4, 2025

For what it is worth, the decision to develop the real square root first and only later the complex square root is deliberate, per #2142 (comment)

Speaking from an iset.mm point of view, square root on nonnegative reals is a very different beast from complex square roots (with the latter being pretty problematic constructively and even a little weird with excluded middle - the hard part being the selection of one of the two roots). But I suppose you should do what makes sense for set.mm.

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