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So I have given the following question; Is ($\mathbb{Q}, \cdot)$ a group?

A group is defined by the monoid $(\mathbb{Q}, \cdot)$ and the inverse for $\cdot$. Since these (afaik) hold:

  • Associativity is true: Take $a, b, c \in \mathbb{Q}$: $a(bc) = (ab)c$
  • There is a neutral identity: $1$
  • $\forall a \in \mathbb{Q}: \exists a^{-1}: a \cdot a^{-1} = 1$

Why isn't $(\mathbb{Q}, \cdot)$ a group?

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    $\begingroup$ What about the inverse of $0$? $\endgroup$ – Mark May 5 at 18:43
  • $\begingroup$ It is the same argument for all fields $K$. There is no multiplicative inverse for $a=0$. $\endgroup$ – Dietrich Burde May 5 at 18:45
  • $\begingroup$ so if it were for a set: $\mathbb{Q}$ \ $0$ it would be a group? $\endgroup$ – Matthias K. May 5 at 18:56
  • $\begingroup$ @CoveredInChocolate MathJax in the title is preferred here on MSE over the sort of thing you attempted to edit in. No need to suggest edits replacing MathJax with such characters in the future. $\endgroup$ – KReiser May 5 at 19:22
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Zero doesn't have a multiplicative inverse.

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