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Equation 1 · Part 7 · What a Proof Assistant Changes

Symbol r^2

∀V. packing⁡(V)  ⟹  ∃c. ∀r. 1≤r  ⟹  card⁡ ⁣(V∩B(0,r))≤πr318+c r2\forall V.\ \operatorname{packing}(V) \implies \exists c.\ \forall r.\ 1 \le r \implies \operatorname{card}\!\left(V \cap B(0,r)\right) \le \frac{\pi r^{3}}{\sqrt{18}} + c\,r^{2}
r2r^{2}

What this part means

The square of r: multiply r by itself.

Its job in the formula

r2r^2 is squared: multiply the underlying quantity by itself. The square is a mathematical operation, not a second independent variable.

The passage around this formula

Hales made the same move on the Kepler conjecture and describes the choices involved. The formal statement was deliberately engineered for readability: all mention of measure was removed, and the theorem was formulated as a claim about finite packings inside a large ball, which introduces a boundary error term but avoids importing measure theory into the statement. Cleaned up, it reads: ∀V. packing⁡(V)  ⟹  ∃c. ∀r. 1≤r  ⟹  card⁡ ⁣(V∩B(0,r))≤πr318+c r2\forall V.\ \operatorname{packing}(V) \implies \exists c.\ \forall r.\ 1 \le r \implies \operatorname{card}\!\left(V \cap B(0,r)\right) \le \frac{\pi r^{3}}{\sqrt{18}} + c\,r^{2}. Every symbol here is an invitation to a fidelity question. What does packing unfold to? Is the error term doing more work than it appears? Hales notes that the constant is not explicit in the statement, and that the theorem includes no uniqueness claim [ 4 ] . None of this is concealed —…

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Learn the underlying idea

An exponent tells how a base is used in multiplication. In x³, x is the base and 3 is the exponent: x³ = x × x × x.

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Sources cited in the surrounding passage

These citations provide research context; check each source for the exact claim it supports.