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[Foglio 2] Esercizio 1 #19

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Elia-Belli opened this issue Oct 11, 2023 · 3 comments
Closed

[Foglio 2] Esercizio 1 #19

Elia-Belli opened this issue Oct 11, 2023 · 3 comments
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da revisionare Almeno una possibile soluzione pubblicata

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@Elia-Belli
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@Elia-Belli Elia-Belli added the da risolvere Nessuna soluzione pubblicata label Oct 11, 2023
@Elia-Belli
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Elia-Belli commented Oct 12, 2023

  • 1.1) Siano $e,e'\in G, e\neq e'$ per assurdo, elementi neutri per $\cdot$ e $g \in G$:
    per definizione di elemento neutro $g\cdot e = g = e \cdot g$ e $g\cdot e' = g = e' \cdot g$
    siccome $e \in G$ possiamo sostituire $g$ con $e$: $e \cdot e' = e = e' \cdot e$, analogamente per $e' \in G$: $e' \cdot e = e' = e \cdot e'$
    ottenendo dalle uguaglianze $e' \cdot e = e \cdot e' \rightarrow e = e'$ (contraddizione) $\Rightarrow$ l'elemento neutro è unico.
  • 1.2) Identico a [Lezione 9/10/23] Esercizio 1 #16
  • 1.3) Identico a [Lezione 9/10/23] Esercizio 2 #17

@Elia-Belli Elia-Belli added da revisionare Almeno una possibile soluzione pubblicata and removed da risolvere Nessuna soluzione pubblicata labels Oct 12, 2023
@CiottoloMaggico
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Analogo @Elia-Belli

@CarloDaRomadev
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-Nota-18-18_annotato.pdf

@sapienzastudentsnetwork sapienzastudentsnetwork locked and limited conversation to collaborators Dec 15, 2023
@Elia-Belli Elia-Belli converted this issue into discussion #166 Dec 15, 2023

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