Backes Posted February 3, 2015 Share Posted February 3, 2015 Hey, I started to learn what groups are... I asked myself, is the multiplication in R a group? My counterargument is, that there is no identity element, because 1*e = e*1 = e, expect for e=0. And the second one is, that there is no symmetric element, because: a*b = b * a = e, where b = 1/a and e = the no identity element. This works, except for a=0... Am I wrong in my argumentation, or is the multiplication in the real set no group? Thank you! Link to comment Share on other sites More sharing options...
ajb Posted February 3, 2015 Share Posted February 3, 2015 The set of all real numbers is a monoid under standard multiplication. 2 Link to comment Share on other sites More sharing options...
mathematic Posted February 3, 2015 Share Posted February 3, 2015 Real numbers without 0 form a group under multiplication. Link to comment Share on other sites More sharing options...
ajb Posted February 27, 2015 Share Posted February 27, 2015 (edited) Shameless plug here, but I use the monoid structure of the real and comment on the use of the group of non-zero reals in my latest preprint (http://arxiv.org/abs/1502.06092). What I use is the fundamental result that any smooth action of the monoid of multiplicative reals on a manifold leads to a particularly nice N-grading of the (structure sheaf of the) manifold. Such actions are called homogeneity structures. In particular we have only functions that have a positive grading and the homogeneous functions form a bases of all functions. We have a kind of polynomial bundle known as a graded bundle. These are the commutative versions of non-negatively graded supermanifolds that you find in the literature, an example would be the so called N-manifolds. Anyway, if you now replace the homogeneity structure with the action of group is non-zero reals, the theory is similar but fundamentally geometrically different. You end up with a principal bundle whose structure group is the group of non-zero reals. Importantly, to every such structure corresponds a line bundle over the base of this principal bundle. In reverse, you can always construct a principal bundle with structure group of the non-zero reals by just forgetting the zero section. There is a one-to-one correspondence. So instead of dealing with line bundles a such, you can consider certain principal bundles and these can be described by a smooth action of the group of non-zero reals. This is something that has not yet be greatly exploited. I thought it maybe nice for Backes to know that these structures are used in modern mathematics. Edited February 27, 2015 by ajb Link to comment Share on other sites More sharing options...
Keen Posted November 16, 2015 Share Posted November 16, 2015 And just to answer the actual question: the set of real numbers is not a group under the usual multiplication law, but not because there is no neutral element, because 1 is a neutral element under multiplication, but not everyone is invertible, because 0 isn't. If you put e=0 in your original argument Backes, it still works: 1*0=0*1=0, so 1 is neutral even for 0. The real numbers are on the other hand a group under the addition law. 2 Link to comment Share on other sites More sharing options...
studiot Posted November 16, 2015 Share Posted November 16, 2015 Well pitched answer keen +1 Link to comment Share on other sites More sharing options...
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