Logic

Lincos: Design of a Language for Cosmic Intercourse. Part 1 by Hans FREUDENTHAL

By Hans FREUDENTHAL

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Extra resources for Lincos: Design of a Language for Cosmic Intercourse. Part 1

Example text

Wir f¨ uhren das f¨ ur die Addition der rationalen Zahlen vor. Sind (p ′ , q ′ ) ∈ (p, q) und (r ′ , s ′ ) ∈ (r, s) andere Repr¨asentanten, dann gilt p ′ q = q ′ p und r ′ s = s ′ r. Es ist zu zeigen, daß (p ′ s ′ + q ′ r ′ , q ′ s ′ ) ∈ (ps + qr, qs) gilt. Ausmultiplizieren liefert (p ′ s ′ + q ′ r ′ )(qs) = p ′ qs ′ s + q ′ qr ′ s = q ′ ps ′ s + q ′ qs ′ r = (ps + qr)(q ′ s ′ ), was zu zeigen war. 13 Wir definieren f¨ ur zwei Punkte (x, y), (x ′ , y ′ ) ∈ R2 (x, y) ∼ (x ′ , y ′ ) :⇐⇒ |x| + |y| = |x ′ | + |y ′ |.

N + 1 − k)! · k! n+1 . 15 (Binomischer Lehrsatz) Es sei K ein K¨orper, x, y ∈ K und n ∈ N, so gilt n n · xk · yn−k . k n (x + y) = k=0 Beweis: Wir f¨ uhren den Beweis durch Induktion nach n. Induktionsanfang: n = 0: Nach Definition gilt 0 0 · xk · y0−k . k 0 (x + y) = 1 = 1 · 1 · 1 = k=0 Induktionsschluß: n → n + 1: Es gilt (x + y)n+1 =(x + y)n · (x + y) = (x + y)n · x + (x + y)n · y n Ind. 14 n+1 =x + k=1 n+1 = k=0 n k=0 n · xk · yn+1−k k n · xk+1 · yn−k + k n k=1 n · xk · yn+1−k + k−1 n · xk · yn+1−k + yn+1 k n k=1 n · xk · yn+1−k + yn+1 k n+1 · xk · yn+1−k + yn+1 k n+1 · xk · yn+1−k k Die Aussage folgt damit aus dem Prinzip der vollst¨andigen Induktion.

Seien dazu x = (p, q), x = (p , q ), x = (p , q ) ∈ M gegeben:1 R1: F¨ ur die Reflexivit¨at m¨ ussen wir x ∼ x zeigen. Nun gilt aber pq = pq, woraus x = (p, q) ∼ (p, q) = x folgt. R2: F¨ ur die Symmetrie nehmen wir an, daß x ∼ x ′ gilt und m¨ ussen x ′ ∼ x folgern. Wegen x ∼ x ′ gilt aber nach Definition pq ′ = p ′ q, und folglich auch p ′ q = pq ′ . Letzteres bedeutet aber, daß x ′ = (p ′ , q ′ ) ∼ (p, q) = x. R3: F¨ ur die Transitivit¨at nehmen wir schließlich an, daß x ∼ x ′ und x ′ ∼ x ′′ gilt, und m¨ ussen daraus schließen, daß x ∼ x ′′ .

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