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Prove that its a field

  1. Nov 20, 2008 #1
  2. jcsd
  3. Nov 20, 2008 #2


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    A field must be closed under addition and multiplication and every non-zero number must have a multiplicative inverse.

    Is [itex](a+ b\sqrt{5})+ (c+ d\sqrt{5})[/itex] in that same set?
    Is [itex](a+ b\sqrt{5})(c+ d\sqrt{5})[/itex] in that same set?

    Is [tex]\frac{1}{a+b\sqrt{5}}[/tex] in that same set? (Hint: rationalize the denominator.)
  4. Nov 21, 2008 #3
    the problem is that i remember these proofs as

    but here the input the root of 5 instead of sum variable


    i dont know how to construct these equations because there is no variable input?
    Last edited: Nov 21, 2008
  5. Nov 21, 2008 #4


    Staff: Mentor

    Halls has shown you what you need to do. It seems pretty clear to me.
  6. Nov 21, 2008 #5
    i clearly remember building these equations:


    but in my function there is no variable in the input

  7. Nov 21, 2008 #6
    What is a field, you probbably already know. However, In order to show that it is a ring, alll we need to show is that it is closed under subtraction, and under multiplication. Like Halls already told you.

    What you are doing there, is basically a hommomorphism, that is the perservation of the operation, but we are not dealing with mapps between two fields in here.

    So let x, and y be any two elements on that set, then obviously they will have the form

    [itex]x=(a+ b\sqrt{5}); y= (c+ d\sqrt{5})[/itex]


    [itex]x-y=(a+ b\sqrt{5})- (c+ d\sqrt{5})=(a-c)+(b-d)\sqrt{5}[/itex] so [tex]x-y \in Q(\sqrt{5})[/tex]

    now also

    [tex]xy=(a+ b\sqrt{5})(c+ d\sqrt{5})=ac+ad\sqrt{5}+bc\sqrt{5}+bd5=(ac+bd5)+(ad+bc)\sqrt{5}[/itex] which again, for obvious reasons is in [tex] Q(\sqrt{5})[/tex]

    Now,in order for that to be a field, like Halls already stated,we need to show that every el. in that set has an inverse.

    let x be in that set. so it will obviously have the form [tex]=x(a+ b\sqrt{5})[/tex] let
    s suppose that there is another el, y, such that

    [tex]xy=1 => y=\frac{1}{x}=\frac{1}{a+b\sqrt{5}}=\frac{a-b\sqrt{5}}{a^2-5b^2}[/tex] SO, the same question still holds, is this in the same set?

    I'll give u one more hint: rewrite the last expression as

  8. Nov 21, 2008 #7
    you don't need to construct these, because you are not dealing with a mapping between two rings or fields, and thus you are not asked to show whether that kind of mapping is a homomorphism. because, those expressions are used when we want to show that a particular mapping say between two rings is a homomorphism.
  9. Nov 21, 2008 #8
    you meant under subtraction here, right?
  10. Nov 21, 2008 #9
    yes if we split it:

    a/(a^2-5b^2) -(b/a^2-5b^2)*5^(1/2)

    is that correct?
  11. Nov 21, 2008 #10
    Well, now the problem breaks down into showing whether
    [tex]\frac{a}{a^2-5b^2}\in Q; and\frac{-b}{a^2-5b^2}\in Q ????[/tex]

    You can either prove this one too, or take it for granted. since a, b are rational nr. then by adding, subtracting, multiplying, and dividing them, we will still end up having rational numbers.
  12. Nov 21, 2008 #11
    hallsofIvy said to prove that that inverse of the formula is of the same set
    a/(a^2-5b^2) -(b/a^2-5b^2)*5^(1/2) => a/(a^2-5b^2) +(-b/a^2+5b^2)*5^(1/2)

    its of the same set now(it looks like the original formula)?
  13. Nov 26, 2008 #12
    how to prove that 0 or 1 are members
  14. Nov 26, 2008 #13


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    If you want to write 0 or 1 as "[itex]a+ b\sqrt{5}[/itex]", what would a and b be equal to?
  15. Nov 26, 2008 #14
    [tex]0=0+0\sqrt{5}, 1=1+0\sqrt{5}[/tex] fair enough!!

    P.S this is what Halls already said!...lol...
  16. Dec 16, 2008 #15
    how to recognize a ring proof
    and field prove
    i cant see the difference between them?
  17. Dec 16, 2008 #16
  18. Dec 17, 2008 #17
    what is the form of a field?

    what is the form of a ring?

    (how each of their formula looks?)
  19. Dec 17, 2008 #18


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    What are the definitions of "field" and "ring"?
  20. Dec 17, 2008 #19
    ring is like x=(a+b,c,g+1)
    and i need to
    prove that it stays the same type by addition
    and multiplication by scalar
    and existence of zero.

    field is like a formula

    is this correct
  21. Dec 17, 2008 #20
    Read the wiki article I linked, what you wrote is not very mathematical in nature i.e. a field is definitely not "like a formula f(7) = a + 7*b". These are pretty advanced mathematical ideas and you should have at the very least the definition of each ingrained in your brain before you go out and try to prove things about them.
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