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A decision method for elementary algebra and geometry by Alfred Tarski

By Alfred Tarski

In a call approach for hassle-free algebra and geometry, Tarski confirmed, through the tactic of quantifier removal, that the first-order concept of the genuine numbers below addition and multiplication is decidable. (While this outcome seemed basically in 1948, it dates again to 1930 and was once pointed out in Tarski (1931).) this can be a very curious outcome, simply because Alonzo Church proved in 1936 that Peano mathematics (the idea of typical numbers) isn't really decidable. Peano mathematics can also be incomplete via Gödel's incompleteness theorem. In his 1953 Undecidable theories, Tarski et al. confirmed that many mathematical structures, together with lattice concept, summary projective geometry, and closure algebras, are all undecidable. the speculation of Abelian teams is decidable, yet that of non-Abelian teams is not.

In the Nineteen Twenties and 30s, Tarski frequently taught highschool geometry. utilizing a few principles of Mario Pieri, in 1926 Tarski devised an unique axiomatization for aircraft Euclidean geometry, one significantly extra concise than Hilbert's. Tarski's axioms shape a first-order idea without set concept, whose people are issues, and having in simple terms primitive relatives. In 1930, he proved this idea decidable since it might be mapped into one other conception he had already proved decidable, specifically his first-order conception of the true numbers.

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6. Beweis: Rechnen modulo n entspricht dem Rechnen im Ring Z/nZ. 8. 3 (Folgerung aus dem Satz von Lagrange) (a + nZ)ϕ(n) = 1 + nZ, was gerade der Satz von Euler ist. (a) ist dann der Spezialfall n = p. 6 [Anwendung: Primzahltests I] Ist eine gegebene Zahl n ∈ N eine Primzahl? √ n, ob d ein Teiler von n ist. 1. Methode: Pr¨ ufe f¨ ur alle 1 < d Vorteil: gibt sichere Antwort. Nachteil: dauert zu lange. 2. h. w¨ahle a < n und u ufe, ob an−1 ≡ 1 ¨berpr¨ (mod n). Vorteil: geht schnell. Nachteil: nur bei negativer Antwort weiß man, dass n keine Primzahl ist.

Angenommen nicht, und ein Primelement p teilt Cg˜·h˜ . Man pr¨ uft ohne Schwierigkeiten nach, dass der nat¨ urliche Epimorphismus R[X] → (R/pR)[X] den Kern pR[X] hat und somit R[X]/pR[X] ∼ = (R/pR)[X] ein Integrit¨atsbereich ist. Im Widerspruch ˜ + pR[X] = 0 ∈ R[X]/pR[X], andererseits dazu gilt nun aber nach Annahme einerseits g˜ · h ˜ g˜ + pR[X] = 0 = h + pR[X]. 2 In der Literatur werden verschiedene Ergebnisse dieses Abschnittes als Lemma von Gauss“ bezeichnet. 4 Sei f ∈ R[X] primitiv. Dann ist f genau dann irreduzibel in R[X], wenn f irreduzibel in K[X] ist.

A9 ) nur an einer Stelle ab oder unterscheidet sich durch Vertauschen zweier Ziffern, so ¨ andert sich die Pr¨ ufziffer. h. f¨ ur das Bild von a ∈ Z unter dem nat¨ urlichen Epimorphismus Z → Z/pZ. (Z/pZ)∗ → (Z/pZ)∗ , x → x2 ist ein Gruppenhomomorphismus. F¨ ur Primzahlen p > 2 ist der Kern gerade {1, −1} = {1, p − 1}, besteht also aus zwei Elementen. Also gilt |Bild(x → x2 )| = p−1 alfte der Elemente von (Z/pZ)∗ sind Quadrate. h. genau die H¨ F¨ ur a ∈ Z, p a, definiert man nun das Legendre–Symbol a ( ) := 1 falls a ¯ Quadrat in Z/pZ p a ( ) := −1 sonst p Der Beweis des folgenden Satzes wird zeigen, dass ( p.

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