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Math Help - Relation symmetric or transitive?

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    Relation symmetric or transitive?

    Let Q denote the relation on the set Z of integers, where integers x
    and y satisfy xQy if and only if x2 + 5y2 is divisible by 6.
    Determine whether or not the relation Q is symmetric and whether or not
    the relation Q is transitive.
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  2. #2
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    Re: Relation symmetric or transitive?

    Quote Originally Posted by Cira View Post
    Let Q denote the relation on the set Z of integers, where integers x
    and y satisfy xQy if and only if x2 + 5y2 is divisible by 6.
    Determine whether or not the relation Q is symmetric and whether or not the relation Q is transitive.

    Do you plan on showing some work of your own ?
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  3. #3
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    Re: Relation symmetric or transitive?

    Quote Originally Posted by Cira View Post
    Let Q denote the relation on the set Z of integers, where integers x
    and y satisfy xQy if and only if x2 + 5y2 is divisible by 6.
    Determine whether or not the relation Q is symmetric and whether or not
    the relation Q is transitive.
    Hints: (x^2+5y^2)+(y^2+5x^2)=6(x^2+y^2) and (x^2+5y^2)+(y^2+5z^2)=(x^2+5z^2)+6y^2.
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    Re: Relation symmetric or transitive?

    Thank you Plato and Nehushtan!
    I think I now understand?

    If we suppose xQy then x2 + 5y2 is divisible by 6.
    Then (x2 +5y2) + (y2 + 5x2) = 6(x2 + y2)
    and then (y2 + 5x2) = 6(x2 + y2) - (x2 +5y2)

    6(x2 + y2) - (x2 +5y2) is divisible by 6 since the difference of 2 integers divisible by 6 is itself an integer divisible by 6.
    Thus yQx, showing that Q is symmetric.

    If we suppose xQy and yQz then (x2 +5y2) and (y2 + 5z2) are both integers divisible by 6.
    Then (x2 + 5z2) = (x2 +5y2) + (y2 + 5z2) - 6y2
    It follows that (x2 + 5z2) is divisible by 6.
    Thus xQz, showing that Q is transitive.
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