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Math Help - Equivalence Classes

  1. #1
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    Equivalence Classes

    Hi, I need some help with this question if anyone can help me.

    For part a and b my answer is:

    There are infinitely many E-equivalence classes, and each E-equivalence class is infinite because the set of Natural numbers is infinite and there are infinitely many primes.

    Is this correct? And how do I do part c and d?

    Thanks
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  2. #2
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    Quote Originally Posted by Nguyen View Post
    Hi, I need some help with this question if anyone can help me.

    For part a and b my answer is:

    There are infinitely many E-equivalence classes, and each E-equivalence class is infinite because the set of Natural numbers is infinite and there are infinitely many primes.

    Is this correct? And how do I do part c and d?

    Thanks

    For (a) I'd rather say there are infinite E-eq. classes since there are infinite primes and we can form a

    natural number with any arbitrary number of different primes

    For (b) the answer is not since the E-eq. class of {1} has ony two elements (which natural is the other one?)

    For (c) check that for example [101] = [103] (why?), [102]=[105] (why?), etc.

    For (d) check how many different primes divide 120, and then form the 6 lowest possible numbers with those divisors

    Tonio
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  3. #3
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    Quote Originally Posted by tonio View Post
    For (a) I'd rather say there are infinite E-eq. classes since there are infinite primes and we can form a

    natural number with any arbitrary number of different primes

    For (b) the answer is not since the E-eq. class of {1} has ony two elements (which natural is the other one?)

    For (c) check that for example [101] = [103] (why?), [102]=[105] (why?), etc.

    For (d) check how many different primes divide 120, and then form the 6 lowest possible numbers with those divisors

    Tonio

    Thanks Tonio.

    For b) the E-eq. class {1} has only one element which is 1 from what I understand now. What is the second?
    And to check my understanding, what are the elements of the E-eq. class {4}?

    For c) so [101] = [103] because 101 and 103 are prime numbers?
    and [102]=[105] because 102 and 105 can be written as the product of 3 primes?

    So I get [101]=[103]=[107]=[109] , [102]=[105]=[110] and [104]=[106]=[108] ?
    So there is a total of 3 different E-eq. classes.

    for d) 120=2^3 x 3 x 5, so what do I do from here?

    Thanks heaps!
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  4. #4
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    Quote Originally Posted by Nguyen View Post
    for d) 120=2^3 x 3 x 5, so what do I do from here?
    Here is the smallest 2\cdot 3\cdot 5=30.
    Then 2\cdot 3\cdot 7=42 and 2^2\cdot 3\cdot 5=60.
    Can you finish?
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  5. #5
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    Quote Originally Posted by Plato View Post
    Here is the smallest 2\cdot 3\cdot 5=30.
    Then 2\cdot 3\cdot 7=42 and 2^2\cdot 3\cdot 5=60.
    Can you finish?
    oh yeah, thanks Plato. I fully understand now.
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  6. #6
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    Quote Originally Posted by Nguyen View Post
    Thanks Tonio.

    For b) the E-eq. class {1} has only one element which is 1 from what I understand now. What is the second?


    My bad: I thought the equivalence is defined on the integers and not on the integers.

    Tonio



    And to check my understanding, what are the elements of the E-eq. class {4}?

    For c) so [101] = [103] because 101 and 103 are prime numbers?
    and [102]=[105] because 102 and 105 can be written as the product of 3 primes?

    So I get [101]=[103]=[107]=[109] , [102]=[105]=[110] and [104]=[106]=[108] ?
    So there is a total of 3 different E-eq. classes.

    for d) 120=2^3 x 3 x 5, so what do I do from here?

    Thanks heaps!
    .
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