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Find Particular Set, 0 and n+3 belong to it and are N

  1. Sep 23, 2013 #1
    1. The problem statement, all variables and given/known data
    Find a set T of natural numbers such that 0 ∈ T, and whenever n ∈ T,
    then n + 3 ∈ T, but T ≠ S, where S is the set defined:
    Define the set S to be the smallest set contained in N and satisfying the following two properties:
    1. 0 ∈ S, and
    2. if n ∈ S, then n + 3 ∈ S.

    2. Relevant equations
    I can only think that this set T is not the smallest set contained in N and satisfying the properties above.

    3. The attempt at a solution
    I have no clue how to find this. I can only say that T is not S, because it is given.
    Please help. :/
     
  2. jcsd
  3. Sep 23, 2013 #2

    Dick

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    T is any set satisfying those properties. The simplest one is ALL natural numbers. There is a smaller one that also satisfies it. Can't you think what that might be?
     
  4. Sep 23, 2013 #3
    A set containing multiples of 3 would be the other choice?
     
  5. Sep 23, 2013 #4

    Dick

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    Yes!
     
  6. Sep 23, 2013 #5
    : > Thanks!
     
  7. Sep 24, 2013 #6
    S is the set of all multiples of 3.

    T also has to contain all multiples of 3, but has to be unequal to S, so it should contain at least one number that's not a multiple of 3.
     
  8. Sep 24, 2013 #7
    You are right. So, S could be ={1, 3, 6, 9, ...}
    Thanks again.
     
  9. Sep 24, 2013 #8

    Dick

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    No, it can't. The definition says 0 has to be in S. And if 1 is in S then 1+3 has to be in S. Etc.
     
  10. Sep 24, 2013 #9
    But, then why do you say the T can be all N numbers?
     
  11. Sep 24, 2013 #10

    Dick

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    I assume you are defining the natural numbers to include 0. Otherwise the whole problem makes no sense.
     
  12. Sep 24, 2013 #11
    I am confused. Even if 0 is included, then we could have that T is all N. Because the constraint that n+3 also belongs to S is there. So, T is not all natural numbers?
     
  13. Sep 24, 2013 #12

    Mark44

    Staff: Mentor

    The set has to have 0 in it. N as defined above has 0 in it.
    Pick any natural number n. Then n + 3 will also be a natural number. For example, if you picked 1, then 1 + 3 = 4, which is also a natural number. If you picked, say 7, then 7 + 3 = 10 is also a natural number.

    The problem, though, is that N is not the smallest set that works.
     
  14. Sep 24, 2013 #13

    Dick

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    There are many sets satisfying the definition of T. There is only one set satisfying the definition of S. S={0,3,6,9,...}. Just name an example of T that isn't equal to S.
     
  15. Sep 24, 2013 #14
    So, T={0,4 , 3 , 6, 9, ...}?
     
  16. Sep 24, 2013 #15
    Nope. Just go back to the requirements and check carefully and you will see why it can't be T.
     
  17. Sep 24, 2013 #16

    Dick

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    If 4 is in T then 4+3 must be in T. Read the definition of T. That doesn't work. You need to add a lot more numbers.
     
  18. Sep 24, 2013 #17
    Ok. I understand that 4 alone cannot be in T.
    A valid T set can be T={0, 3, 4, 6, 7, 9, 10, 12, 13,...}
    I think I got it now.
     
  19. Sep 24, 2013 #18

    Dick

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    Exactly. I think you've got it now too.
     
  20. Sep 24, 2013 #19
    Thank you all.
     
  21. Sep 25, 2013 #20
    Thinking back, I have one more doubt.
    Why do ALL N satisfy?

    It would be all ## \forall \mathbb{N} \geq 3 ##. Right?
     
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