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    Is the set Y of real numbers that are less than or equal to 0 closed under addition? And closed under multiplication?
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    Closure says that if a,b \in Y , then a*b \in Y . This is true for addition, but not for multiplication in the set you described.
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    (Original post by Perpetuallity)
    Yes. If a,b \in Y , then a*b \in Y . This is true for both multiplication and addition.
    But...if its two numbers that are less than 0, lets say -1 and -2, multiplying them gives 2 and thats not in the set?
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    (Original post by T13)
    But...if its two numbers that are less than 0, lets say -1 and -2, multiplying them gives 2 and thats not in the set?
    It's not true for multiplication, I'm on autopilot here, sorry.

    As you said, multiplying two negative numbers, a*b \notin Y because the set is resticted to x \leq 0
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    (Original post by Perpetuallity)
    It's not true for multiplication, I'm on autopilot here, sorry.

    As you said, multiplying two negative numbers, a*b \notin Y because the set is resticted to x \leq 0
    Ah no problem, im just glad i understand it
    The set Z of real numbers r such that −2 ≤ r ≤ 2...surely that is not closed under addition nor multiplication?
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    (Original post by T13)
    Ah no problem, im just glad i understand it
    The set Z of real numbers r such that −2 ≤ r ≤ 2...surely that is not closed under addition nor multiplication?
    Nope, that set isn't closed for either.
 
 
 
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