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Math Help - equivalence relation and equivalence classes

  1. #1
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    Question equivalence relation and equivalence classes

    Hi i need help with this following question. I am not sure on how to approach it.

    Define the relation R on \mathbb{N} by xRy\Leftarrow \Rightarrow x^{2}-y^{2} is divisible by 7 (x,y\in  \mathbb{N})
    Prove that R is an equivalence relation on \mathbb{N} and describe the equivalence classes R_{1} and R_{3}

    Any help is appreichated

    thanks

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  2. #2
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    Quote Originally Posted by rpatel View Post
    Define the relation R on \mathbb{N} by xRy\Leftarrow \Rightarrow x^{2}-y^{2} is divisible by 7 (x,y\in  \mathbb{N})
    Prove that R is an equivalence relation on \mathbb{N} and describe the equivalence classes R_{1} and R_{3}
    Is x^2-x^2 divisible by 7?

    If x^2-y^2 is divisible by 7 is y^2-x^2?

    Now what is left for you to do?
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  3. #3
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    Quote Originally Posted by rpatel View Post
    Hi i need help with this following question. I am not sure on how to approach it.

    Define the relation R on \mathbb{N} by xRy\Leftarrow \Rightarrow x^{2}-y^{2} is divisible by 7 (x,y\in  \mathbb{N})
    Prove that R is an equivalence relation on \mathbb{N} and describe the equivalence classes R_{1} and R_{3}

    Any help is appreichated

    thanks

    Since 7 is prime, then if 7\big|(x^2-y^2) we have

    7\big|(x-y) or 7\big|(x+y), such that

    xRy\Leftrightarrow x\equiv\pm y\mod 7.

    Therefore we have the equivalence classes [0]_7,[1]_7\cup[6]_7,[2]_7\cup[5]_7,[3]_7\cup[4]_7 of R.
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  4. #4
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    how did you calculate the equivalence classes because i know that some being mod 7 means it divisble by some multiple of 7 and remainder is written in the front of the mod7. hence 34=6mod7 since 34-(7x4)=6.

    but i don't understand how you've done it in terms of x and y. and also could you please explain the sqaure bracket notation for the equivalence classes. I have seen this notation before but i don't understand what the number inside the brackets and the lower case power to it means.

    thanks for your help
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  5. #5
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by rpatel View Post
    how did you calculate the equivalence classes because i know that some being mod 7 means it divisble by some multiple of 7 and remainder is written in the front of the mod7. hence 34=6mod7 since 34-(7x4)=6.

    but i don't understand how you've done it in terms of x and y. and also could you please explain the sqaure bracket notation for the equivalence classes. I have seen this notation before but i don't understand what the number inside the brackets and the lower case power to it means.

    thanks for your help
    Asssuming that hatsoff is correct with his solution, the notation \left[n\right]_k usually means \left\{n\in\mathbb{Z}:z\equiv k\text{ mod } n\right\}. Residue classes.
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  6. #6
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    Quote Originally Posted by rpatel View Post
    how did you calculate the equivalence classes because i know that some being mod 7 means it divisble by some multiple of 7 and remainder is written in the front of the mod7. hence 34=6mod7 since 34-(7x4)=6.

    but i don't understand how you've done it in terms of x and y. and also could you please explain the sqaure bracket notation for the equivalence classes. I have seen this notation before but i don't understand what the number inside the brackets and the lower case power to it means.

    thanks for your help
    Sure thing. You may wish to memorize this extremely useful theorem:

    For any two integers a,b, an integer n divides a-b if and only if a\equiv b\mod n.

    So, if 7 divides x-y, then x\equiv y\mod 7. Similarly, if 7 divides x+y, then x\equiv -y\mod 7. And so if 7 divides either x-y or x+y, then x\equiv \pm y\mod 7.

    As for the notation issue, for some integers k\in\mathbb{Z}^+ and n\in[0,k), we have the equivalence class [n]_k=\{m:m\in\mathbb{Z},m\equiv n\mod k\} (the set of integers which, when divided by k, have a remainder of n).
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  7. #7
    Senior Member Shanks's Avatar
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    I completely agree with hatsoff. He briefly shows the key point of this problem .
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