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Let the universe of discourse of x and y be the set of real numbers.Let P(x) denote "x is even." Let Q(x) denote "x is an integer."Then, "Every integer is odd." can be expressed asQuestion 1Answera.ForEvery x [ NOT Q(x) OR NOT P(x) ].b.ForEvery x NOT P(x).c.None of the Given Choices.d.ForEvery x [ P(x) AND Q(x) ].e.ForEvery x [ P(x) -> Q(x) ].

Question

Let the universe of discourse of x and y be the set of real numbers.Let P(x) denote "x is even." Let Q(x) denote "x is an integer."Then, "Every integer is odd." can be expressed asQuestion 1Answera.ForEvery x [ NOT Q(x) OR NOT P(x) ].b.ForEvery x NOT P(x).c.None of the Given Choices.d.ForEvery x [ P(x) AND Q(x) ].e.ForEvery x [ P(x) -> Q(x) ].

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Solution

Let's analyze the given statements and the logical expressions step by step.

  1. The statement "Every integer is odd" can be interpreted as: For every x, if x is an integer, then x is odd.
  2. In logical terms, this can be written as: ForEvery x [ Q(x) -> NOT P(x) ].
    • Q(x) denotes "x is an integer."
    • P(x) denotes "x is even."
    • NOT P(x) denotes "x is not even," which means "x is odd."

Now, let's match this interpretation with the given choices:

a. ForEvery x [ NOT Q(x) OR NOT P(x) ].

  • This expression means: For every x, either x is not an integer or x is not even.
  • This is logically equivalent to: For every x, if x is an integer, then x is not even (i.e., x is odd).
  • This matches our interpretation.

b. ForEvery x NOT P(x).

  • This expression means: For every x, x is not even.
  • This does not match our interpretation because it does not consider whether x is an integer.

c. None of the Given Choices.

  • This is not correct because we have found a matching choice.

d. ForEvery x [ P(x) AND Q(x) ].

  • This expression means: For every x, x is even and x is an integer.
  • This does not match our interpretation.

e. ForEvery x [ P(x) -> Q(x) ].

  • This expression means: For every x, if x is even, then x is an integer.
  • This does not match our interpretation.

Therefore, the correct answer is:

a. ForEvery x [ NOT Q(x) OR NOT P(x) ].

This problem has been solved

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