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Question: If A = {3, 5, 7, 9, 11, 12}, determine the truth value of each of the following. i) $\exists x \in ...

If A = {3, 5, 7, 9, 11, 12}, determine the truth value of each of the following.

i) xA\exists x \in A such that x8=1x - 8 = 1

ii) xA,x2+x\forall x \in A, x^2 + x is an even number

iii) xA\exists x \in A such that x2<0x^2 < 0

iv) xA,x\forall x \in A, x is an even number

v) xA\exists x \in A such that 3x+8>403x + 8 > 40

vi) xA,2x+9>14\forall x \in A, 2x + 9 > 14

Answer

(i) True (ii) True (iii) False (iv) False (v) True (vi) True

Explanation

Solution

Let A={3,5,7,9,11,12}A = \{3, 5, 7, 9, 11, 12\}.

  1. (i) xA\exists x \in A such that x8=1x - 8 = 1.

    • Solve: x=9x = 9. Since 9A9 \in A, it is True.
  2. (ii) xA,x2+x\forall x \in A, \, x^2 + x is even.

    • For any integer xx:
      • If xx is odd: odd2^2 is odd, and odd + odd = even.
      • If xx is even: even2^2 is even, and even + even = even.
    • Since every xAx \in A (odd or even) yields an even number, it is True.
  3. (iii) xA\exists x \in A such that x2<0x^2 < 0.

    • Squares of real numbers are never negative. Hence, False.
  4. (iv) xA,x\forall x \in A, \, x is even.

    • Most elements are odd. So, False.
  5. (v) xA\exists x \in A such that 3x+8>403x + 8 > 40.

    • Solve: 3x+8>403x + 8 > 403x>323x > 32x>32310.67x > \frac{32}{3} \approx 10.67.
    • Elements 1111 and 1212 satisfy this, so True.
  6. (vi) xA,2x+9>14\forall x \in A, \, 2x + 9 > 14.

    • Solve: 2x+9>142x + 9 > 142x>52x > 5x>2.5 x > 2.5.
    • Every element in AA (all >2.5>2.5) satisfies this, so True.