$\begingroup$Question: Is the set of real numbers a group under the operation of multiplication?
My professor answered it by saying: No. There is no identity element (1*0=0).
However, isn't the identity element 1, did he mean to say there is no inverse because the number 0 does not have an inverse. Or did my professor try to mean something else? Or maybe I'm just mis-understanding what he wrote.
$\endgroup$13 Answers
$\begingroup$The collection of positive real numbers (and even real numbers without zero) is a group. However, once you append zero, the resulting set is no longer a group for exactly the reason your are suggesting.
One interesting thing about the positive real numbers, $(\mathbb{R}_+,\cdot)$, is that they are isomorphic to the reals with addition, $(\mathbb{R},+)$. This can be seen through the logarithm, $$\log(a\cdot b) = \log(a) + \log(b).$$ Note also that $\log(1)=0$, that is the logarithm identifies the identity elements between the two different groups.
$\endgroup$3$\begingroup$The set $R$ endowed with the operation [ . ] cannot be considered a group, since considering it a group means that every element in $R$ is invertable with respect to [ . ]. Now considering ($R$ , . ) a group, we all know that the inverse of an element say $x$ belonging to $R$ is $\frac{1}{x}$ but since 0 belongs to the group $R$ hence $\frac{1}{0}$ exists in $R$ ( Contradiction ). Meanwhile you can consider (R*,.) a group.
$\endgroup$$\begingroup$For all non-zero elements,there exist it's multiplicative inverse in $\mathbb{R}$, i.e., $a*(a)^{-1}=1=(a)^{-1}*a$.
But for the element $0$ in $\mathbb R$, there is no such element s exit in $\mathbb R$ s.t. $0*b$ (say) $=1=b*0$.
Hence, $0$ does not have its multiplicative inverse in $\mathbb{R}$. So, $\mathbb{R}$ is not a group w.r.t. "$*$".
