No matter what positive real number x we choose, there exists some positive real number y
such that yz2 > xz + 10 for every positive integer z.
Translate the above statement to predicate logic and prove it using a direct approach.
I don't believe that there are relevant equations for this problem.
The Attempt at a Solution
Let Q (x, y, z) = yz2 > xz + 10
[itex]\forall[/itex]x ∈ ℝ+ [itex]\exists[/itex]y ∈ ℝ+ [itex]\forall[/itex]z ∈ [itex]Z[/itex]+ Q(x, y, z)
Before I attempted to prove the theorem, I wanted to make sure that I got the predicate logic translation right. I don't think that the above translation is right, but I hope I'm on the right track. I've never translated into predicate logic with 3 variables. It's usually just x and y, so should it be (x, y, z)? Also, does [itex]\forall[/itex]z ∈ [itex]Z[/itex]+ come after Q (x, y, z) since it does in the statement?