The proof begins by assuming the non-existence of a real number z for which P(x,y,z) is true. By analyzing the given statement, a contradiction is reached, establishing the existence of such z.
Proof by contradiction:
Assume, for the sake of contradiction, that there does not exist a real number z such that P(x, y, z) is true for all x > 10 and y ∈ ℕ.
This implies that there exists some pair (x, y) such that P(x, y, z) is false for all real numbers z.
Now, consider the given statement: ∀x ∈ ℝ, ∀y ∈ ℕ, x > 10 ⇒ y < x ⇒ ∃z ∈ ℝ, P(x, y, z).
Since x > 10 and y < x, according to the given statement, there must exist a real number z such that P(x, y, z) is true. However, we have assumed the opposite.
This contradiction arises from assuming that there does not exist a real number z satisfying P(x, y, z) for all x > 10 and y ∈ ℕ.
Therefore, our assumption is false, and it follows that there exists a real number z such that P(x, y, z) is true for all x > 10 and y ∈ ℕ. The proof is complete.
The question probable may be:
Given the statement:
∀x ∈ ℝ, ∀y ∈ ℕ, x > 10 ⇒ y < x ⇒ ∃z ∈ ℝ, P(x, y, z).
Prove the statement by contradiction:
Assume that there does not exist a real number z such that P(x,y,z) is true for all x>10 and y ∈ N. This implies that there exists some pair (x,y) for which P(x,y,z) is false for all real numbers z.
Consider the given statement and show that the assumption leads to a contradiction. Conclude that there must exist a real number z such that P(x,y,z) is true for all x>10 and y ∈ N.