# Binary operation

1. Oct 7, 2006

In the definition of a group on mathworld, http://mathworld.wolfram.com/Group.html" [Broken], implies closure, so, isn't it unnecessary to talk about the property closure in the definition of a group?

Last edited by a moderator: May 2, 2017
2. Oct 7, 2006

### AKG

What do you mean "isn't it sufficient to talk about the property of closure in the definition of a group?" It is not necessary to talk about the property of closure in the defintion of a group. But "not necessary" is not the same thing as "sufficient". Anyways, although it is not necessary, in theory, to talk about the property of closure, you are often just given a set S with a function * with domain SxS, and you have to verify that * is indeed an operation, that is, that closure does indeed hold.

3. Oct 7, 2006

Yes, I corrected that, I meant 'not necessary'. OK, I get it, it's more general, since we don't always know if * is a binary operation.

4. Oct 7, 2006

### matt grime

There is often redundancy in how axioms. The statements are usually made so as to appear as clean as possible. One can, and often does, define a group without invoking the strict definition of binary operation, and just takes it to mean some operation that takes two objects and gives a third without necessarily saying where the third lies.