### Nutter, nutter, nutter

Today, whilst I walked from A to B, you might have just been able to hear me muttering the word nutter to myself. I was obviously referring to myself. That conclusion was due to a sequence of actions which lead to my bag somehow taking off, and landing in a painful way. My initial reaction had been to jump on the spot whilst looking around in an agitated manner, hoping to have gone unnoticed. I was in a quiet area, and thankfully no one was actually around on this occasion. After picking my bag up with a manic laugh, I resorted to calling myself Nutter. Funnily enough, it is either Mondays that have this effect on me, or the fact that I had less then seven hours sleep. I don't know whether we will ever find out the real cause for my behaviour, since I can't guarantee 7 hours sleep for Mondays!

This morning I was quite lucky because I ended up greeting all the maths buildings in the morning. I can't but help mention my intense dislike for Oxford Road. Ever since last year, I have dreaded the thought of unfortunately walking down Oxford Road more than once in a day. That road is plain horrible. Ugly. Full. Busy. I can think of many more colourful adjectives to describe it (feel free to share), but will pause! (I don't want to be complaining about something new in each post). This morning I walked the maybe slightly longer, but pleasant way to university. I smiled and nodded to the mss building, and then happily waved at the Lamb and Newman building. I can't recall my reaction to the AT building, but the discovery that both doors can open did lead to a lot of 'aahs'.

From this day forth(!) I will attempt to walk on that route every single morning. I don't care if I have to wake up slightly early, but I refuse to walk down Oxford Road at 9am. Firstly there is no grass and pleasant scenery. You also can't really have a conversation with yourself, since you are constantly maneuvering yourself around people who don't know how to walk. (Well that may be a bit harsh, but you can't peacefully walk to lectures). Whereas the other route has less people for a start, which means that you are free to talk about whatever you want (with yourself of course!) I can also say hi to more maths buildings this way, especially the mss one. I will probably eventually do the opposite of what I am saying (on occasions when I am running late), but I am hoping that for mornings at least, I am able to do this. (At the end of the day, there are more variables to consider).

Hmmm, I seem to have gotten that all out of the way. There is no sense in mentioning this poor woman looking at me weirdly, after I was found talking to myself after registration!

Did I mention that the Algebraic Structure lecture was what was keeping me going during the day? I don't think it is necessary for me to mention how much I enjoy them lectures, and today was a particularly enjoyable/memorable one too! It was one of them lectures, where your brain is telling you to switch off due to being tired, but you don't turn off. You don't listen to your brain, because it speaks a lot of waffles at times (well mine does anyway). All you need to know is that there was mention of a parrot (with an illustration), and some music mentioned. (I at least recognised who was talked about from English anthology lessons; however I think I'm more of a swan lake \sout{person} bean.) Would you believe it - I just googled Swan Lake and I am rather embarrassed at my discovery or ignorance! [You may wish to do the same...] To save myself further embarrassment, I will talk about some maths now. I wouldn't mind knowing something, but I will have to wait for Thursday to find out I suppose.

EDIT: I just couldn't resist. I present you with the Swan Lake. I find it rather... cool! But yes - note I am still embarrassed at not knowing a startling fact behind it! Well what can I say, I have never claimed to be an expert in such matters, but just show some interest.

We started groups today, and I think I may as well write my foolish thoughts to remind myself in the future about what is wrong. The identity element has to be commutative i.e. for a binary operation * defined on a non-empty set S, e is the identity if e*a=a*e= a, for all a in S. We require that to be true for the identity, but that doesn't imply that the binary operation is commutative! I was thinking that maybe we could prove that one of the exotic examples can be shown to 'not' have an identity, by showing that it is not commutative. That is very wrong.

So that means that a group does not have to be commutative, since having an identity element does not imply that the group is commutative. There is something else floating around my brain, which was my confusion over the binary operation being a 'function'. It is (I guess) technically speaking a function or a rule, but I had bijective floating about. A function does not have to be bijective! I had been thinking (stupidly one might add), that all functions are bijective. So whilst doing the problem sheet and when considering whether or not * was a binary operation defined on a set S, I had been checking the 'onto' property. Damn. The idea is that all ordered pairs (a,b) are mapped onto an element of S (and something else). If they are not then the binary operation is not defined. There is this second condition in the Fraleigh book which I have misinterpreted, and that has led to my confusion. I will have another proper look at it later, but I am glad to having hopefully stopped this train of thought here. (The problem sheet is long...)

The positives of the problem sheet: I remembered something about the power set from my first year supervision (I think). If you have the set A={1,2,3}, then P(A)= 2^|A|. (The power set of A is equal to 2^(the cardinality of A)). However all this came later, what I remembered is this: the set has three elements, so consider the line of the Chinese triangle which has the number three in it i.e. 1 3 3 1. Sum the four numbers to get 8, which is equal to 2^3. I can't really remember the motivation behind that, but I have a tendency to remembering weird things.

There is something else, but with my mind focusing on food, I have forgotten this thing. I dare not strain too much, but I was very happy during and after the algebraic structures lecture. It was a nice way to end the day.

## 10 comments:

beans said:

"If you have the set A={1,2,3}, then P(A)= 2^|A|. (The power set of A is equal to 2^(the cardinality of A)). However all this came later, what I remembered is this: the set has three elements, so consider the line of the Chinese triangle which has the number three in it i.e. 1 3 3 1. Sum the four numbers to get 8, which is equal to 2^3. I can't really remember the motivation behind that, but I have a tendency to remembering weird things."

you mean (regardless of what set

"A" might be) |P(A)| = 2^|A|

(both equal to 8 in your example);

P(A) *itself* is of course a *set*

(not a number) ...

the "1 3 3 1" line of the

"chinese triangle", in this context,

tells us that there are

*one* subset of A

(i.e., member of P(A))

having *zero* elements;

*three* subsets having *one* element;

*three* subsets having *two* elts;

and *one* subset having *three* elts.

vlorbik

Ah yes, thanks for the correction!

And thanks for the insight about the chinese triangle. I didn't get the chance to locate my first year work. I think I remembered the Chinese triangle because I had a particularly difficult time trying to understand the power set! I didn't understand how the empty set could be a subset of a set...

Hopefully I won't forget this now!

If you can't stand Oxford Road, why did you apply to Manchester University for? Why did you not apply to "campus" universities such as Oxford, Cambridge, Warwick or York for example, where you don't have to pass a fleet of buses and 1000 people to get from A to B. You do know that Manchester University is one of the most populated Universities; if not THE most (if you exclude London and the OU). So it's a bit strange that you weren't put off by Manchester on your Open Day when you visited the place.

The power set of a set A is just the set of all subsets of A (and has cardinality 2^|A|)

so e.g. if A = {1,2} |A|=2

then P(A)= {{1},{2},{1,2},{empty set}} |P(A)| = 4

I find it easiest to state the link with binomial coefficients as follows:

Given a power set with n elements; the power set has nCk sets (as elements) with k elements.

So in the above example, P(A) has 2C1 = 2 sets with one element i.e. {1} and {2}

In terms of the binary operations, they are just a special type of function i.e. f: S^2 --> S

however, there is no restriction on them being injective (one-one)

orsurjective (onto).e.g. Let S = {1,2} and let

*: S^2 --> S

*((1,1)) -> 1

*((1,2)) -> 1

*((2,1)) -> 1

*((2,2)) -> 1

then * is a binary operation on S but * is neither surjective (nothing is mapped to 2) or injective (all elements of S^2 are mapped to the same element of S)

HTH

Hi Jake,

I always forget the proper link with the binomial co-efficients, but I think that now I have both things on one page it should be easier to remember. It is funny how when looking back at things after a long time, they make more sense!

Regarding the binary operation: that example was similar to PS's example too. Hmmm, I think I need to re-write the sentence from the book in my own words. I just have to remember what you have written, and that if *((1,1)-> 4, then that is not a binary operation. [I think my head was in the clouds whilst doing the work!]

HTH? (I have tried working out what that means, but to no avail).

Hi anonymous,

It is not that I can't stand Oxford Road, but having experienced moving around from both campuses I had gotten used to that (last year). I am not too concerned about campus universities, and on my visit to the university, I hadn't really thought much about Oxford Road! I actually didn't think about much else but maths on the visit.

However, can we agree that I am not clever enough to apply to Oxbridge! I like the idea of moving around a lot - eg from Renold to Chemistry etc. Also my "allegiance" to the mss building has something to do with this feeling too!

My point is that I can get through to my buildings without having to walk past a fleet of buses (and I aim to do so). I probably can get to my destination quicker that way too.

I have been having a humbug type of week, and it has only been two days! What more is to come, I wonder.

I just have to remember what you have written, and that if *((1,1)-> 4, then that is not a binary operation. [I think my head was in the clouds whilst doing the work!]Well, to confuse matters even further... some authors would still call that a binary operation.

E.g. Whitelaw defines a binary operation on a set S as a mapping

*: S^2 --> A where A isn't neccesarily the same set as S. He then says that a binary operation is closed on a set iff S is a subset or equal to A. So Whitelaw's

closedbinary operation is the same as our plain binary operation. So, for example, when defining a group, he has to add an axiom stating that the binary operation is closed on the underlying set of the group.HTH? (I have tried working out what that means, but to no avail).

Hope this helps.

Yes, that does confuse matters further! It took a while to register, but I guess you have to be consistent with your definitions.

I think Whitelaw's might be slightly 'jumping' the gun, and I prefer PS. The simpler a definition the easier it is to remember in my case! [I recall some first year students (inc. me) having quite some problems with the idea of something 'being closed'.]

Thanks, it did help. :D

You can actually get the full strength of a group (left and right inverses, left and right identity):

Let G be a set with an associative binary operation that has a right identity e and right inverses i.e, a*e = a and a*a^(-1)=e for all a in G.

Then (exercise :-) you can show a^(-1) is also a left inverse and that e is a left identity.

This is going round the houses, but I think it's a good exercise and it avoids having to write a*e=a=e*a so it would have avoided the commutativity confusion in your case!

Hi Graeme,

Exercise eh? :p I think something about this was mentioned in the previous lecture, so we may be told about it on Friday. (But I will surely be having a duel with the exercise tomorrow, whilst finishing the problem sheet.)

It is a bit like the idea of a limit (and possibly invertible matrices). If the right identity(e) is the same as the left identity(e) then the identity exists (i.e. e). But if they are different then the identity doesn't exist?

Thanks for that. (And you have just reminded me about sponsoring you!)

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