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A Summer of Maths

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  • View Poll Results: Have you studied any Group Theory already?
    Yes, I did some Group Theory during/at my A-level.
    37.84%
    No, but I plan to study some before Uni.
    21.62%
    No, I haven't.
    40.54%

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    (Original post by jack.hadamard)
    What I thought is, if I labelled subset as \{ a_1, a_2, ..., a_p\}, and defined

    a_i * a_j = a_{(i \times_{p} j)}

    then I think, I get a cyclic group. I assume the operation is associative, but it follows from the associativity of multiplication modulo.

    Then, if I define f(a_i) = i, I get that

    f(a_i * a_j) = f(a_{(i \times_{p} j)}) = i \times_{p} j = f(a_i) \times_{p} f(a_j)

    and this is clearly bijective, so it is an isomorphism to \mathbb{Z}_p.


    In this case I get a cyclic group, for all sets with a prime number of elements.
    However, I need this to be true for every number of elements, and if it helps, they will be naturals.
    I see what you're doing. You need to be quite careful here: \mathbb{Z}_n is not (usually) constructed a subset of \mathbb{N}, or even of \mathbb{Z}, and it's not a subgroup of either under any conventional operations. It is normally either defined abstractly or defined concretely a quotient of \mathbb{Z} by the subgroup n\mathbb{Z}.

    Now, \mathbb{Z}_n is not a group under multiplication modulo n, even when n is prime (for instance, p has no multiplicative inverse modulo p). And it is always a group under addition, even when n isn't prime.

    However, \mathbb{Z}_p^* is a group under multiplication; its elements are \{ 1, 2, \cdots, p-1 \} with multiplication defined by a \times_p b = ab \pmod p. But \mathbb{Z}_p^* \cong \mathbb{Z}_{p-1}, and the correspondence is not as simple as mapping, say, 1 to 1 and 2 to 2 and 3 to 3 and so on... the isomorphism will mess up the ordering of the integers.

    And on this basis it doesn't make a difference that your indices are natural numbers, since if you want them to act multiplicatively then you'll need to screw up their ordering anyway in order to map them into \mathbb{Z}_p^*. So you can choose between making them act additively, so for instance a_i * a_j = a_{i +_n j}, in which case it makes no difference whether or not n is prime; or you can choose another group and group operation and choose your isomorphism a different way.
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    (Original post by nuodai)
    So you can choose between making them act additively, so for instance a_i * a_j = a_{i +_n j}, in which case it makes no difference whether or not n is prime; or you can choose another group and group operation and choose your isomorphism a different way.
    I see. Well, it was the first one that I picked! So, if I choose them additively, then it seems plausible?
    I'll have a look at it now, and check the details to see whether it works. Thanks!

    EDIT: I know what you mean by the above though, you didn't waste time to write it all out. I did it quickly, and slightly mispresented it.
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    (Original post by jack.hadamard)
    I see. Well, it was the first one that I picked! So, if I choose them additively, then it seems plausible?
    I'll have a look at it now, and check the details to see whether it works. Thanks!
    Yes. But you seem to be making it harder than you have to: by labelling the elements of your subset from 1 up to n you have already defined an isomorphism from your subset to \mathbb{Z}_n, by the method I described two posts ago, since the group structure is inherited from the identification with the elements of \mathbb{Z}_n.

    A more critical issue is that not all subsets of \mathbb{N} are finite, but I'm sure you can work out a way round that...
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    (Original post by nuodai)
    Yes. But you seem to be making it harder than you have to: by labelling the elements of your subset from 1 up to n you have already defined an isomorphism from your subset to \mathbb{Z}_n, by the method I described two posts ago, since the group structure is inherited from the identification with the elements of \mathbb{Z}_n.

    A more critical issue is that not all subsets of \mathbb{N} are finite, but I'm sure you can work out a way round that...
    Yes, that is true, but I want to play a bit more with it.

    It was just an idea, so I have to think through the details, and see how exactly I'm going to apply it.
    Well, more specifically, I'm looking at some periodic sequences, so that's not a problem for now.

    EDIT:

    (Original post by nuodai)
    More specifically, if A \subset \mathbb{N} is any subset and (G,*) is a group with \left| A \right| = \left| G \right| then any bijection f : A \to G induces a group structure (A, \cdot) on A by defining a \cdot b = f^{-1}(f(a) * f(b)).
    This is quite nice! I think I have seen it already used for the Isomorphism theorem, but obviously didn't process it well enough!
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    (Original post by nuodai)
    A more critical issue is that not all subsets of \mathbb{N} are finite, but I'm sure you can work out a way round that...
    Found some links: Does every non-empty set admit a group structure on Google. Interesting!
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    (Original post by jack.hadamard)
    Found some links: Does every non-empty set admit a group structure on Google. Interesting!
    It's often the case that statements of the form "every set has [such and such a structure]" are equivalent to the axiom of choice (or a weaker variant of it), a famous example being "every set has a well-ordering". Most people accept the axiom of choice and get on with their lives, but it is quite interesting to know when it's being used I thought that the assertion that every set has a group structure depended on the axiom of choice, but I didn't know they were equivalent!
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    (Original post by nuodai)
    .. a famous example being "every set has a well-ordering".
    So, that's the final ingredient of the argument! I have heard this axiom so many times that one may think I know what it means, but I find it interesting already!
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    Sorry to break up the Group Theory discussion - would like some thoughts on this...

    If f:\mathbb{R}\to\mathbb{R} is continuous everywhere and f(x)=C\;\;\forall x\in\mathbb{Q} can I conclude that f(x)=C\;\;\forall x\in\mathbb{R}\; ?

    My thoughts were along the lines of:

    Spoiler:
    Show

    Any irrational number y can satisfy q<y<q' with (q,q')\in\mathbb{Q}, and where y-q and y-q' can be made arbitrarily small. Since f is continuous necessarily f(q)=f(y)=f(q') (assuming f is constant for all rational numbers). Does that work?
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    One of the books given for Numbers and Sets is Numbers, Sets and Axioms by A. Hamilton.
    Nobody suggested books on Set Theory so far? Any comments?


    Usually, two sets are defined to be equal if A \subseteq B and B \subseteq A, or equivalently, if x \in A \iff x \in B.

    Why is this called the Axiom of ``Extensionality", expressed as: \forall x (x \in A \equiv x \in B) \Rightarrow A = B? It does not 'extend' anything.
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    (Original post by Lord of the Flies)
    Does that work?
    Yup, that's fine, but the wording is slightly awkward and I can't put my finger on why. When dealing directly with the definition of continuity it's good to be more specific with \varepsilon and \deltas, so something like:

    Spoiler:
    Show
    Let \varepsilon > 0, let y be irrational and let \delta > 0 be such that \left| f(x) - f(y) \right| < \delta whenever \left| x-y \right| < \delta. Since there is some q \in \mathbb{Q} with \left| q - y \right| < \delta, we have \left| C - f(y) \right| < \varepsilon. But \varepsilon > 0 was arbitrary, so f(y)=C.


    It sounds like more of a mouthful, but that way you're not skipping over details.
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    (Original post by jack.hadamard)
    Why is this called the Axiom of ``Extensionality", expressed as: \forall x (x \in A \equiv x \in B) \Rightarrow A = B? It does not 'extend' anything.
    It doesn't mean 'extension' as in 'make something bigger', it means extension as in 'the ability to include more things'. The reasoning goes a bit deeper into relations.

    Background:
    A relation R on a set X can be thought of in a few ways (all are equivalent). One way is as a subset R \subseteq X \times X, which means it's a collection of ordered pairs (x,y) where x,y \in X. (An example of a relation is the set X \times X itself, which can be thought of as the set of all 'coordinates'.) Relations generalise functions: a function f is a relation for which for each x there is exactly one y such that (x,y) \in f, and we denote this y by f(x). In this sense, we can think of a relation as a 'multi-valued function'. (For instance, the set containing all (x,y) where y=\sqrt{x} or y=-\sqrt{x} is a relation, but not a function.) But the nitty-gritty aside, just think of a relation R as being a set containing ordered pairs of elements of some set.

    Now, if (x,y) \in R then we say x is an 'R-predecessor' of y. So for instance if f is a function then you could say (although most people don't) that x is an f-predecessor of f(x) for each x in the domain of f.

    The relevant bit:
    A relation R is called 'extensional' if each element y \in X is determined by its R-predecessors. That is, if for all x \in X we have (x,y) \in R \Leftrightarrow (x,y') \in R, then y=y'. The axiom of extensionality, therefore, is just the assertion that \in is an extensional relation.
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    (Original post by nuodai)
    It doesn't mean 'extension' as in 'make something bigger', it means extension as in 'the ability to include more things'. The reasoning goes a bit deeper into relations.

    Background:
    A relation R on a set X can be thought of in a few ways (all are equivalent). One way is as a subset R \subseteq X \times X, which means it's a collection of ordered pairs (x,y) where x,y \in X. (An example of a relation is the set X \times X itself, which can be thought of as the set of all 'coordinates'.) Relations generalise functions: a function f is a relation for which for each x there is exactly one y such that (x,y) \in f, and we denote this y by f(x). In this sense, we can think of a relation as a 'multi-valued function'. (For instance, the set containing all (x,y) where y=\sqrt{x} or y=-\sqrt{x} is a relation, but not a function.) But the nitty-gritty aside, just think of a relation R as being a set containing ordered pairs of elements of some set.

    Now, if (x,y) \in R then we say x is an 'R-predecessor' of y. So for instance if f is a function then you could say (although most people don't) that x is an f-predecessor of f(x) for each x in the domain of f.

    The relevant bit:
    A relation R is called 'extensional' if each element y \in X is determined by its R-predecessors. That is, if for all x \in X (x,y) \in R \Rightarrow (x,y') \in R, then y=y'. The axiom of extensionality, therefore, is just the assertion that \in is an extensional relation.
    where can I learn what all that notation means?
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    (Original post by kingkongjaffa)
    where can I learn what all that notation means?
    Most textbooks introduce it as you go along; the notation I used would all be in a typical first year textbook. I can't really think of any to suggest though. The table on Wikipedia isn't very useful but might help a little bit.
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    (Original post by nuodai)
    ...
    Merci!
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    (Original post by nuodai)
    The relevant bit:
    A relation R is called 'extensional' if each element y \in X is determined by its R-predecessors. That is, if for all x \in X (x,y) \in R \Rightarrow (x,y') \in R, then y=y'. The axiom of extensionality, therefore, is just the assertion that \in is an extensional relation.
    I see. Just for clarity, this notion of a ``relation" is as general as it can be.
    e.g. I don't attribute it any properties other than being a subset of a Cartesian product?


    I am still trying to decipher the last bit though.


    Is x \in X (x,y) \in R \Rightarrow (x,y') \in R the same as

    If for all x \in X, we have x R y \Rightarrow x R y', then y = y'?


    In this case, I get that the axiom of extensionality would be: R is \in and y = A, y' = B.

    Following this logic, would = and \equiv be extensional relations?
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    (Original post by jack.hadamard)
    I see. Just for clarity, this notion of a ``relation" is as general as it can be.
    e.g. I don't attribute it any properties other than being a subset of a Cartesian product?
    Well it's nothing more than that - a subset of a Cartesian product (of a set with itself). Relations are extremely useful (equivalence relations, partial orders and preorders are all types of relation). They can be generalised further, say to n-ary relations, which are subsets of X^n or to subsets of X \times Y, or even more general products.

    (Original post by jack.hadamard)
    Is x \in X (x,y) \in R \Rightarrow (x,y') \in R the same as

    If for all x \in X, we have x R y \Rightarrow x R y', then y = y'?
    Yes. (x,y) \in R, xRy and Rxy are three ways of writing the same thing. I prefer the former unless it's some kind of ordering or inclusion.

    (Original post by jack.hadamard)
    In this case, I get that the axiom of extensionality would be: R is \in and y = A, y' = B.
    Yes.

    (Original post by jack.hadamard)
    Following this logic, would = and \equiv be extensional relations?
    Yes. But, for instance, if we write H \le G to mean that H is a subgroup of a group G (where we consider the groups as being 'up to isomorphism'); then \le is not extensional (for instance, D_6 and C_6 have the same subgroups).
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    (Original post by nuodai)
    Yes. But, for instance, if we write H \le G to mean that H is a subgroup of a group G (where we consider the groups as being 'up to isomorphism'); then \le is not extensional (for instance, D_6 and C_6 have the same subgroups).
    Oh, yes. I wasn't very sure about the meaning of = in that case (definition), but took it as being the same object.

    Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful -- I notice most of these have no prerequisites.
    Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. )
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    (Original post by jack.hadamard)
    Oh, yes. I wasn't very sure about the meaning of = in that case, but took it as being the same object.

    Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful -- I notice most of these have no prerequisites.
    Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. )
    I didn't learn naive set theory from a book, and as far as I know there isn't a copy of the lecture notes online, so I can't really say. I've heard of that book, though, which probably means it's not too bad.

    Promise me you won't just do maths this summer? I'm a bit worried.
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    (Original post by nuodai)
    Promise me you won't just do maths this summer? I'm a bit worried.
    I won't get the chance, otherwise.. no doubt! I will be a lot more busier next month.
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    (Original post by jack.hadamard)

    Would you recommend Naive Set Theory? Or, perhaps, any other book that might be useful -- I notice most of these have no prerequisites.
    Comments suggest it is a well structured introduction that covers all axioms of ZF? (It's funny how naive is this book with this axiomatic nature. )
    Yep, great book

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