I’m inclined to say “them” refers to the hotdogs without buns, and it is inferred (but not stated) that they are served in buns, simply because that’s the most common serving method and no serving method is indicated. Still, it’s an inference, and what’s directly stated is only that the meat was served.

]]>Frank McQuarry: One of the requirements in Abstract Algebra for a set to be a group under a binary operator is that the group is closed under that operator. That is to say when the operator is applied to any two members of a set, the result is also a member of the set.

Thus {hotdog, hotdog, bun} is a group under +.

Julian Lander: Well, no. While I realize that the joke is hotdog + bun = hotdog, you only get a group if hotdog + hotdog is a member of the set. It does work as a group if buns are considered to be zero, so that what you’ve got is the plain old group of integers with hotdogs as the item being counted. Once you start including buns explicitly, you lose uniqueness of the identity element: both 1 hotdog + 0 hotdog = 1 hotdog and 1 hotdog + 1 bun = 1 hotdog. That implies (if we have a group), that a bun is the zero. (Yes, I like proofs by contradiction, although I feel that constructive proofs have more value.)

It’s hard to disentangle this discussion, since it’s not clear what the expression *hotdog* stands for. If the elements of the group are two, HOTDOG and BUN, then HOTDOG + HOTDOG must be an element, either HOTDOG (in which case + is idempotent, and that’s fine) or BUN, but in either case the results require a non-standard interpretation of +.

The alternative is that the elements of the group are hotdogs and buns, in which case hotdog-x + bun-y is a hotdog, either hotdog-x (in which case bun-y is a zero) or hotdog-z (different from hotdog-x); hotdog-z is at least consistent with the intended interpretation of +. But then what are hotdog-x + hotdog-x and hotdog-x + hotdog-z? This is where the mind starts boggling.

Group theory is not a good place to go.

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