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Math Help - write the statement formally using the appropriate quantifers

  1. #1
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    write the statement formally using the appropriate quantifers

    The sum of any two irrational numbers is irrational


    This isnt really using quantifers but i dont know how to express on this message board.

    Is irrational numbers expressed as I?

    For all x, y in I,x +y is I??

    negation

    There exists x and y in I such that x + y =! I?

    Thank you
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  2. #2
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    Quote Originally Posted by Thetheorycase View Post
    The sum of any two irrational numbers is irrational


    This isnt really using quantifers but i dont know how to express on this message board.

    Is irrational numbers expressed as I?

    For all x, y in I,x +y is I??

    negation

    There exists x and y in I such that x + y =! I?

    Thank you

    Both your statements are correct, but now write them down using logical quantifiers and language!

    Tonio
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  3. #3
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    It is customary to denote the set of rational numbers by \mathbb{Q}. As far as I know, there is no equally accepted notation for the set of irrational numbers. Thus, a more standard way to denote irrational numbers is using set difference: \mathbb{R}\setminus\mathbb{Q}.

    Also, the following is probably understood, but just in case. If the atomic proposition (i.e., the one that cannot be broken into simpler propositions) is x\in\mathbb{R}\setminus\mathbb{Q}, then " x+y is in \mathbb{R}\setminus\mathbb{Q}" and " x+y\in\mathbb{R}\setminus\mathbb{Q}" are legal propositions while " x+y is \mathbb{R}\setminus\mathbb{Q}" and " x+y\ne\mathbb{R}\setminus\mathbb{Q}" are not.
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  4. #4
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    By "not legal propositions" do you mean "not well formed"? In what language? Since \in is being used, I surmise the language is the language for set theory. In that case,

    x+y\ne\mathbb{R}\setminus\mathbb{Q}

    is well formed given a few ordinary notational conventions.
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  5. #5
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    Granted, x+y\ne\mathbb{R}\setminus\mathbb{Q} may be a well-formed formula in set theory. But when we stay at the level of the questions in the OP, it is much more likely that this formula has an error, especially given the English equivalent.

    Similarly, a C compiler may produce a justifiable warning at the code "if (x = 1) then ... else ...". Even though this code is legal, it is likely to be a typo (= instead of ==).
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  6. #6
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    Sure, I agree that a formula like that would strongly suggest that the poster was off track somehow.
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