Answer:
(b) ∀x∈ℝ ∃ y∈ℝ: xy=1
(c)∀x∈ℝ ∄ y∈ℝ: xy=1
Step-by-step explanation:
(a)For each x∈ℝ , there exists a y∈ℝ such that xy=1.
Given x=2∈ℝ
xy=1
2y=1
[TeX]y=\frac{1}{2}[/TeX]
But [TeX]y=\frac{1}{2}[/TeX]∉ℝ.
In fact, [TeX]y=\frac{1}{2}[/TeX]∈ℚ, the set of Rational Numbers.
Therefore, the statement is false.
(b) ∀x∈ℝ ∃ y∈ℝ: xy=1
(c)∀x∈ℝ ∄ y∈ℝ: xy=1
(d)For each x in the set of Real numbers , there does not exists a y in the set of real numbers such that such that xy=1.
