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Is this set with the following operations a field?

  1. Jan 11, 2009 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    The title of the thread says it all. The set is (0,1). Here are the operations : addition is such that 0+0=0, 0+1=1, 1+0=1 and 1+1=0. Multiplication is as follows : 0.0=0, 0.1=0, 1.0=0 and 1.1=1.
    2. The attempt at a solution In order to be a field, the set and its operations must follow 8 rules. The first one is that (x+y)+z=x+(y+z). First I note that there are only 2 elements in the set but it doesn't matter. Second I don't know how I can show that the operations respect the rule without trying all the possibilities. It's a lot of arithmetic just for the first rule, so I guess I'm missing a simpler way to finish it off.
    If it is so, how could I start it?

    Thanks
     
  2. jcsd
  3. Jan 11, 2009 #2
    It looks to me that this set with this specific operation defined in it is not a Field at all.
    Why?

    For it to be a field, every element of the set should have a multiplicative inverse, which is not the case with this set, since 0 does not have an inverse in this set.

    Mult. identity of this set seems to be 1. so 0*1=0, 0*0=0, 1*0=0, i.e. there is no elelment a, of the set such that 0*a=1.

    It looks like this set with this operation forms a ring though.

    If you create the table of operation, you see that the only things you need to check in order to establish this is the associativity property with respect to addition and multiplication(the latter is very easy to show) and the distributivity property.

    There are not many calculations i guess.

    say we want to show that

    a+(b+c)=(a+b)+c.

    what i would do is; if a=b=c=0 then the above is obviously true from the way addition is defined, the same quick way is if a=b=c=1.

    now if say one of a,b,c is 0 while the others 1. then no matter how we ad them we will end up with 0, so it it true in this case. Now if two of a,b,c are 0 and the other 1, then no matter how we add them we will end up with 1.

    Now for: a(bc)=(ab)c is even easier.

    a similar reasoning i guess would go for a(b+c)=ab+ac.
     
  4. Jan 11, 2009 #3

    fluidistic

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    Thank you for your input. I'd like to ask you a question.
    Can you give me an example of a field in which 0 has an inverse? I don't get it well I guess... And ok for the rest!
     
  5. Jan 11, 2009 #4

    HallsofIvy

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    First, it is NOT true that every element of a field must have a multiplicative inverse. Every element except 0 must have an element. The additive identity of a field, 0, never has a multiplicative identity.

    Now, since there are only 2 elements of the field, there are only 23= 8 possible combinations of (x+ y)+ z and x+(y+ z). Go ahead and list them all and see if they are equal. No, that's NOT "a lot of arithmetic"- it's just hard work and there is no reason why you shouldn't be willing to do some hard work! (It should take you maybe 5 minutes. Probably less time than you spent waiting for an answer here!)
     
  6. Jan 11, 2009 #5

    fluidistic

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    Wow, thank you so much for this answer! Ok I just got all the arithmetic and compared both (x+y)+z and x+(y+z) and it gives the same! Unforntunately this means that I must check out the other 7 rules or maybe less if I find out that this is not a field. Well thanks once again, I'll try the rest alone.
     
  7. Jan 11, 2009 #6
    Contrary to sutupidmath, I'm pretty sure this is a field. In fact, I think I did this problem back in October (it's in Spivak).

    Sutupidmath, I think there's a good reason why 0 shouldn't have a multiplicative inverse.
     
  8. Jan 11, 2009 #7

    Dick

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    Sure it's a field. It's called Z_2, the integers mod 2. So a lot of those properties follow from the same properties being true for the integers. That doesn't mean it hurts to actually check them.
     
  9. Jan 11, 2009 #8

    fluidistic

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    Well I just finished the exercise! Indeed I found it to be a field.
    It would be long to put all the details, but I want to put the most interesting part anyway.
    It must satisfies the condition that for any element of the set, there must exist an opposite element such that x+(-x)=0 where 0 is the neutral element. What I did : if x is 0 then -x is also 0 and it works. While if x=1 then -x would be 1, because 1+1=0 in this set. Am I right? I doubt because I've the feeling I'm mixing things up when I consider the 0 of the set and the 0 as neutral element.
    Another condition is that there exist a unique element "1" such that x.1=x for all x in the set.
    Wait!! The word unique is very important... so this might not be a field. Let's see.
    If x=0 then the element "1" is 1 or 0: Indeed both would work : 0.0=0 and 0.1=0 in our case. While if x=1 then the "1" would be 1 with no exception.
    So I conclude that the set with its operations is not a field because it fails in the rule of a neutral element for the multiplication. Am I righ? Hmm I guess no, otherwise even R wouldn't be a field... What's wrong with what I did?
     
  10. Jan 11, 2009 #9

    Dick

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    You said "a unique element "1" such that x.1=x for ALL x". And 0.1=0 in fact shows that '0' is NOT a multiplicative identity.
     
  11. Jan 11, 2009 #10

    fluidistic

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    Ah yes, now I see. I get it. Thank you everyone.
     
  12. Jan 11, 2009 #11

    That is damn true, i don't know what i was thinking about.
    Sure 0 doesn't have a multiplicative inverse, in any ring/field with more than one elements, since there is no element in that ring R, say r, such that r*0=1=0*r. and this is due to the fact that for any such element 0*r=0.

    If R={0} then 0 is it's own inverse. In fact : If zero has a multiplicative inverse in a ring R if and only if R={a}, where a=0. THat is if R is a one element ring.
     
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