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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.

kero
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3 Answers3

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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.

Joel
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    The last property is of course true for any group homomorphism. – Sasho Nikolov Feb 16 '15 at 21:42
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    @SashoNikolov Indeed, I just wanted to emphasize it. The logarithm could be said the be the first homomorphism, since it was designed to change problems of multiplication to problems of addition. It's easy to forget that some of its properties are that of homomorphisms. We learn about it so long before we learn about homomorphisms properly. – Joel Feb 16 '15 at 21:46
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    Very interesting observations about the logarithm. – josh314 Feb 17 '15 at 15:04
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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.

mandez
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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. "$*$".

Henrik supports the community
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Suraj
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