# Easy problem

DeepThought
Science 02 Dec '14 18:28
1. DeepThought
02 Dec '14 18:28
Well, it's easy if you spot the trick. Suppose a set S has a product which is anti-commutative. So a*b = -b*a, show that it is not associative. You can assume that there is at least one product where a*b != a*c (!= means not equal to).
2. 02 Dec '14 21:075 edits
Originally posted by DeepThought
Well, it's easy if you spot the trick. Suppose a set S has a product which is anti-commutative. So a*b = -b*a, show that it is not associative. You can assume that there is at least one product where a*b != a*c (!= means not equal to).
While I shall leave the pleasure of attempting your problem to other people,
I would like to suggest that it would help if you could state your problem
in more precise mathematically rigorous terms. To be honest, I think that
your problem--as stated--would be unacceptable (confusing terms) to be
given to students in an examination.

"Suppose a set S has a product which is anti-commutative."
--DeepThought

This pseudo-mathematical talk does not make sense *as stated*.
(I suspect that some mathematicians would criticize it as gibberish.)

Do you understand the important distinction between a set and a group?
A set *in itself* does not have a 'product'. If you want to talk about a
'product', then you must define an operation upon elements of the set.
A group is a set joined with an operation that satisfies the group axioms:
closure, associativity, and the existences of an identity and inverses.

I studied set theory and group theory, and I don't ever recall a formal
mathematical statement about 'a set having a product'. If you would like
to refer to the *Cartesian product* of sets, then please do that. Otherwise,
it's unclear to me whether you intend to refer to a 'product' of sets or a
'product' (based upon an undefined operation) of elements within a set.

As a general rule, whenever I encounter someone expressing something
mathematical without using standard mathematical terms or symbols, I
tend to become confused and avoid assuming that I know what's meant.
3. 02 Dec '14 21:25
Originally posted by Duchess64
Do you understand the important distinction between a set and a group?
A set *in itself* does not have a 'product'.
Not every set has a defined product, but the OP states that there is a product defined in the set. He assumes you know what 'product' means in this instance. He has not said that the set in question is a group under any given operation, and I assume it is not a requirement of the problem.

If you want to talk about a 'product', then you must define an operation upon elements of the set.
Yes, that is what he did.
Baby Gauss
02 Dec '14 21:321 edit
Originally posted by DeepThought
Well, it's easy if you spot the trick. Suppose a set S has a product which is anti-commutative. So a*b = -b*a, show that it is not associative. You can assume that there is at least one product where a*b != a*c (!= means not equal to).
Edited
5. 02 Dec '14 21:362 edits
Originally posted by twhitehead
Not every set has a defined product, but the OP states that there is a product defined in the set. He assumes you know what 'product' means in this instance. He has not said that the set in question is a group under any given operation, and I assume it is not a requirement of the problem.

[b]If you want to talk about a 'product', then you must define an operation upon elements of the set.

Yes, that is what he did.[/b]
Ignorant people (not necessarily always applying to DeepThought) may
make many claims in their pseudo-mathematical statements, but that
does *not* mean that those claims must be true, coherent, or clear.
(And I don't expect Twhitehead to be able to comprehend that.)

I suspected (though I don't know--that's why I asked) that DeepThought
might have been sloppy in writing 'set' when he meant 'group' or that
he might have overlooked the distinction between a set and a group.
In general, it's safer to assume that a non-mathematician is being sloppy
out of ignorance rather than assuming that a non-mathematician has some
profound reason for writing about mathematics in a needlessly confusing way.
In my university courses in set theory and group theory, I cannot recall
ever being given a problem that was presented in this confusing way.

What's the 'product' in a set of angels? Isn't dying anti-commutative?
6. wolfgang59
invigorated
02 Dec '14 21:55
Originally posted by Duchess64
I would like to suggest that it would help if you could state your problem
in more precise mathematically rigorous terms. To be honest, I think that
your problem--as stated--would be unacceptable (confusing terms) to be
given to students in an examination.
This is a chess site.

And I don't think the fun question is being submitted for a finals paper.
7. 02 Dec '14 22:042 edits
Originally posted by wolfgang59
This is a chess site.
And I don't think the fun question is being submitted for a finals paper.
I think that the original problem's expressed in sloppy or confusing terms.
But if you consider it a 'fun question', I would not attempt to stop you or
anyone else from attempting to solve the problem.

By the way, when I presented some recreational mathematics problems,
DeepThought sometimes complained that he did not understand them and
asked me clarify what I meant, and I did so to his satisfaction.
8. Soothfast
0,1,1,2,3,5,8,13,21,
02 Dec '14 22:072 edits
Originally posted by Duchess64
Ignorant people (not necessarily always applying to DeepThought) may
make many claims in their pseudo-mathematical statements, but that
does *not* mean that those claims must be true, coherent, or clear.
(And I don't expect Twhitehead to be able to comprehend that.)

I suspect (though I don't know--that's why I asked) that DeepThought
might have be ...[text shortened]... n this confusing way.

What's the 'product' in a set of angels? Isn't dying anti-commutative?
A group is necessarily associative, so that is why DeepThought doesn't give that the set is a group.

EDIT: While I have not given any thought to DeepThought's riddle, there is a fair chance his solution runs on the assumption that the set S is "closed" under the product. That is, if a,b ∈ S, then a*b ∈ S.
9. 02 Dec '14 22:155 edits
Originally posted by Soothfast
A group is necessarily associative, so that is why DeepThought doesn't give that the set is a group.
Yes, I noticed that. Perhaps I'm being too picky, yet I would have liked
DeepThought to have spent more time setting up the problem like this:
Let S be a set (of what?). Let an operation K be defined upon the elements
of S such that ... results in a product X. Then (state the problem).

That's how I liked to write problems and how I prefer problems be written.
I find it harder to comprehend--without making too many asssumptions--
what's meant by mathematical problems presented in a popularized way.
Whenever I hear a non-mathematician speak of 'the group of people in
this room', then I don't assume that all the group axioms must hold here!

I suppose that I believe the challenge in a problem should be in tackling
the concepts involved rather than in struggling to figure out the problem's
needlessly obscure, sloppy, or confusing terms. Or perhaps, as I grow
older, I have increasingly less patience with wasting my time deciphering
what a problem means when the writer easily could have made it clearer.

Yet I know that DeepThought did not quite understand some of my recent
recreational mathematical problems, so he asked me to explain what I
meant, and I did so to his satisfaction. I thought that much of what he
wrote in his attempts to solve these problems was too confusing for me
to follow easily (without doing more work than I would like in a hurry), so
I skipped to the parts that I could quickly check. I suppose that I generally
have some difficulties with DeepThought's style of mathematical exposition.
10. 02 Dec '14 22:58
Originally posted by Soothfast
A group is necessarily associative, so that is why DeepThought doesn't give that the set is a group.

EDIT: While I have not given any thought to DeepThought's riddle, there is a fair chance his solution runs on the assumption that the set S is "closed" under the product. That is, if a,b ∈ S, then a*b ∈ S.
"...the assumption that the set S is 'closed' under the product."
--Soothfast

I would rather have DeepThought state *explicitly* that the set S is closed
rather than have me assume it (why play guessing games?), rightly or wrongly.
At first sight, my first thought was "this is not how a serious problem by
a real mathematician tends to be presented--it just doesn't look right."
Your mileage may vary.
11. DeepThought
02 Dec '14 23:18
Originally posted by Soothfast
A group is necessarily associative, so that is why DeepThought doesn't give that the set is a group.

EDIT: While I have not given any thought to DeepThought's riddle, there is a fair chance his solution runs on the assumption that the set S is "closed" under the product. That is, if a,b ∈ S, then a*b ∈ S.
I think technically it's a module since I've implicitly got addition defined (otherwise -a*b doesn't make sense). I didn't want to give any more information in the question than was necessary. The stuff about at least one case where a*b != a*c was to rule out a trivial case (where a*b = 0 for all elements in the set, when it is trivially associative).

Yes I think I'd better make it closed under the operation (as I'm not certain of the effects of not doing that, and certainly the specific example I generalised from has that property) - although I don't think that that is necessary for the answer.
12. DeepThought
02 Dec '14 23:465 edits
Drat, didn't think it through enough I need to add two extra conditions. It's best if I restate the problem:

Let S be a set which is closed under the product * and has the following properties:

Anti-commutivity a*b = -b*a
If a and b are different elements then a*b != 0
a*b != a and a*b !=b for all a and b.

Show that S is not associative under '*'

Sorry, I tried to abstract away from a specific example and worried about trivial cases instead of whether I'd given enough information. The extra two conditions may overspecify it, but I wanted to make it different to the specific case I based it on to prevent people just looking the answer up on Wikipedia.
13. 03 Dec '14 00:241 edit
Originally posted by DeepThought to Soothfast
I think technically it's a module since I've implicitly got addition defined (otherwise [b]-a*b doesn't make sense). I didn't want to give any more information in the question than was necessary. The stuff about at least one case where a*b != a*c was to rule out a trivial case (where a*b = 0 for all elements in the set, when it ...[text shortened]... eralised from has that property) - although I don't think that that is necessary for the answer.[/b]
"I *think* technically it's a module"
--DeepThought

But a module is an additive abelian group. Soothfast concluded that you
wanted to deny that your set S is a group because a group already has
(known) associativity. So your claim now still makes no sense to me.
Do you understand what mathematical terms mean before you throw them around?

"I didn't want to give any more information in the question than was necessary."
--DeepThought

You can make a problem harder by making it about a challenging concept.
Or you can make it harder by withholding essential information and expressing
it in misleading, confusing, or contradictory terms. Which is easier?
14. DeepThought
03 Dec '14 00:30
Originally posted by Duchess64
"I *think* technically it's a module"
--DeepThought

But a module is an additive abelian group. Soothfast concluded that you
wanted to deny that your set S is a group because a group already has
(known) associativity. So your claim now still makes no sense to me.

"I didn't want to give any more information in the question than was necessary."
...[text shortened]... formation and expressing
it in misleading, confusing, or contradictory terms. Which is easier?
It's a module under addition and multiplication by a scalar, with a product operation which is anti-commutative. In the restated problem - I didn't give enough information - I'm not relying on the multiplication by a scalar so it need only be an abelian group under addition.

The problem is now easy, where before hand it's probably not possible.
15. 03 Dec '14 00:36
Originally posted by DeepThought
Drat, didn't think it through enough I need to add two extra conditions. It's best if I restate the problem:

Let S be a set which is closed under the product * and has the following properties:

Anti-commutivity a*b = -b*a
If a and b are different elements then a*b != 0
a*b != a and a*b !=b for all a and b.

Show that S is not associative under ...[text shortened]... nt to the specific case I based it on to prevent people just looking the answer up on Wikipedia.
"It's best if I restate the problem."
--DeepThought

Thanks, that's what I have been attempting to persuade you to do, even
though Twhitehead was ignorant enough to believe there was nothing that
needed to be added or changed in your original statement of the problem.

"...to prevent people just looking the answer up on Wikipedia."
--DeepThought

I would not worry about people doing that. While I was not thrilled that
you seemed to turn some of my recreational mathematics problems into
minor research projects (searching Wikipedia and asking other people)
rather than only using your brain--which was in the spirit of the problems--
I was glad that they touched your interest in mathematics.