- #1

SolarMidnite

- 22

- 0

## Homework Statement

No matter what positive real number x we choose, there exists some positive real number y

such that yz

^{2}> xz + 10 for every positive integer z.

Translate the above statement to predicate logic and prove it using a direct approach.

## Homework Equations

I don't believe that there are relevant equations for this problem.

## The Attempt at a Solution

Let Q (x, y, z) = yz

^{2}> 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?