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We know that $R$ is a field with the usual addition and usual multiplication. Is the set of extended real numbers also a field with the same operations? What are the additive inverse and multiplicative inverse of $\infty$ in the extended field?

(We know ($\infty-\infty$) is an indeterminate form, so $-\infty$ can't be additive inverse of $\infty$. And $0$ can't be multiplicative inverse of $\infty$ because we have $c.\infty=0$ if $c=0$.)

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    $\begingroup$ No${{{{{{}}}}}}$. $\endgroup$ – Lord Shark the Unknown Oct 1 '17 at 6:50
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    $\begingroup$ $1 + \infty = \infty$. So if $\infty$ has an additive inverse, then $1 = 0$. $\endgroup$ – Michael Oct 1 '17 at 7:12
  • $\begingroup$ Extended reals are not a field. $\endgroup$ – Oria Gruber Oct 1 '17 at 8:04
  • $\begingroup$ Multiplication is also bad ... $\infty\cdot 2 = \infty\cdot 1$ and $\infty \ne 0$, but it does not follow that $2 = 1$. $\endgroup$ – GEdgar Oct 1 '17 at 11:28
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I think you've answered your question. In a non trivial field $0 \ne 1$ and $a + q = a \iff q = 0$. So $\infty + 1 \ne \infty$ which... defeats the purpose of the extended reals.

It's not a field. (And you really shouldn't be surprised.)

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