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Not only is the symbol daunting, but the words are too

  1. Sep 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Check that [tex]\mathbb{F}_2[/tex] is a field



    3. The attempt at a solution

    The problem set given to us is mixed in notes, so I might have missed something because it's so messy.

    These are the two properties given.

    A field F is a set with + and * on it such that

    [tex](x,y) \to x + y \in \mathbb{F}[/tex] + : [tex]\mathbb{F} \times \mathbb{F} \to \mathbb{F}[/tex]

    [tex](x,y) \to xy \in \mathbb{F}[/tex] \cdot : [tex]\mathbb{F} \times \mathbb{F} \to \mathbb{F}[/tex]

    The notes previous saidly something about "a curious field F = {0,1}"

    What do they mean by "curious field"?
     
  2. jcsd
  3. Sep 22, 2011 #2

    HallsofIvy

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    It means, believe it or not, a field that is, in some way, "curious" (in the sense of "odd", "unusual" or "unique"). What's curious about that field is that it is the smallest possible field.
     
  4. Sep 22, 2011 #3
    I thought {0,0} is smaller...?

    How do I manipulate the properties of the fields to help me with the question?
     
  5. Sep 25, 2011 #4
    My prof said to verify A1 to A4

    [​IMG]

    [​IMG]

    and M1 to M4 and D (but not prop 2.7)

    Does that sound right? How does that relate to the "curious field"? I thought those axioms hold for all fields
     
  6. Sep 25, 2011 #5

    SammyS

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    By the set {0, 0} you must mean the set {0}

    This can't be used to form a field. It can't even be used to form a ring. There is no multiplicative identity element in {0} \ {0}, because {0} \ {0} is the empty set. (See M3.)
     
  7. Sep 25, 2011 #6

    SammyS

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    Of course they apply to all fields.

    Do you have addition and multiplication tables for the "curious field" ?
     
  8. Sep 25, 2011 #7
    I am digging, one sec.
     
  9. Sep 25, 2011 #8
    [​IMG]

    We were given this. Most of them seem okay except 1 + 1 = 0...
     
  10. Sep 25, 2011 #9

    SammyS

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    What's not OK with 1 + 1 = 0 ?
     
  11. Sep 25, 2011 #10
    Doesn't A2 say 1 + 1 = 1 + 1 and not 0...?
     
  12. Sep 25, 2011 #11

    SammyS

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    If 1+1 = 1, then 1 has no additive inverse.
     
  13. Sep 25, 2011 #12
    No big words please...
     
  14. Sep 25, 2011 #13

    SammyS

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    If 1 + 1 = 1, then there is no inverse for 1 using the addition operation. In other words, there would be no way to fulfill A4 .
     
  15. Sep 25, 2011 #14
    Let's slow down a bit and I need something clarified. Through 2.45 to 2.52, do I have to verify all of them to show that F_2 is indeed a field?

    I jsut started field so I am still a novice.

    When they say F_2 = {0,1}, they really want us to show every combination of addition and multiplication of the numbers 0 and 1?

    I still understand how that makes 1 + 1 = 1, it shuold be really, not 1 or 0.

    What does having an inverse have to do with having a unique -x? (x being 1)
     
  16. Sep 25, 2011 #15

    SammyS

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    For this field, -1 = 1 . This is because the thing you need to add to 1 to give you the identity element for addition is 1.
     
  17. Sep 26, 2011 #16
    This word "field" is being abused way too much to bypass basic addition rules...

    Let's just pick on that guy and the first one, 0 + 0 = 0 first (I am guessing := can
    be replaced by = here?)

    All I have to do is verify right?

    So by A2 0 + 0 = 0 + 0 = 0

    which is true (order does not matter)

    I can't get my way around 1 + 1 = 1 still because I still thinking this in basic addition that 1 + 1 = 2 and you are telling me that this "field" has such a property that -1 = 1, a lot of conflict in my mind.
     
  18. Sep 26, 2011 #17

    SammyS

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    There is no such element as 2 in the set {0, 1}.

    This field, F2 a.k.a. the "curious field", is the set {0,1} with the operations, + and ∙ , defined in post #8. The ":=" combination is used to indicate a definition. Thus, 2.45 through 2.52 are definitions for the field, F2.

    For this field, F2, you cannot have 1 + 1 = 2. If you did have 1 + 1 = 2, then F2 would not be closed for the + operation, would not be a field.

    Some important WORDS (Not all are big, but the ideas have big importance.):
    set
    binary operation
    closed , closure
    identity -- as in identity element, identity function, identity matrix ...
    additive identity --> the identity element for the addition operation. Your text denotes this as the numeral, 0 .
    multiplicative identity --> the identity element for the multiplication operation. Your text denotes this as the numeral, 1 .​
    group, ring, field
    negative -- otherwise known as the opposite or the additive inverse
    inverse -- your text uses this for the multiplicative inverse which is often called the reciprocal.
    The word inverse is also used to describe functions, matrices, binary operations, etc.​
    associative
    commutative
    distributive
    ...​
     
  19. Sep 26, 2011 #18

    HallsofIvy

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    What do you mean by "{0, 0}"? {0}? Most definitions of "field" require that there be at least two members.
     
  20. Sep 26, 2011 #19
    Oh okay, so to verify (the original problem of this thread) all of those properties, i must use A1 - A4 and M1 - M4, D

    But how do I do that properly? PLease help!

    Like for the first one, you have 0 + 0 := 0, that doesn't fall into any of the axioms.

    The one that even looks similiar is A2, i.e.

    0 + 0 = 0 + 0 = 0
     
  21. Sep 26, 2011 #20

    micromass

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    No, you mustn't use A1-A4, M1-M4 and D. You must verify them.
    For example, for A1, you must check that

    [tex]x+(y+z)=(x+y)+z[/tex]

    So, take 3 arbitrary elements of [itex]\mathbb{F}_2[/itex] and go check associativity.
     
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