What is the correct translation of the following statement into mathematical…

2012

What is the correct translation of the following statement into mathematical logic?

“Some real numbers are rational”

Answer: C. \(\exists x (\text{real}(x) \wedge \text{rational}(x))\)Correct translation: There exists at least one x that is both real and rational. Symbolically: ∃x (real(x) ∧ rational(x)). Reasoning: The formula ∃x (real(x)…

  1. A.

    \(\exists x (\text{real}(x) \lor \text{rational}(x))\)

  2. B.

    \(\forall x (\text{real}(x) \to \text{rational}(x))\)

  3. C.

    \(\exists x (\text{real}(x) \wedge \text{rational}(x))\)

  4. D.

    \(\exists x (\text{rational}(x) \to \text{real}(x))\)

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Correct answer: C

Correct translation: There exists at least one x that is both real and rational. Symbolically: ∃x (real(x) ∧ rational(x)).

Reasoning:

  • The formula ∃x (real(x) ∨ rational(x)) says there exists an x that is real or rational. It does not require the same x to be both real and rational, so it does not express "some real numbers are rational."

  • The formula ∀x (real(x) → rational(x)) says every real number is rational. That is a universal (all) claim and is stronger than the intended existential (some) claim.

  • The formula ∃x (real(x) ∧ rational(x)) correctly asserts the existence of an x that has both properties, matching the English statement.

  • The formula ∃x (rational(x) → real(x)) is true for many x simply because the implication is true when x is not rational; it therefore does not capture that some x are both real and rational.

Note: If the domain of discourse is already all real numbers, the predicate real(x) is true for every x and the correct translation can be simplified to: ∃x rational(x).

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