# Predicate Logic and Proofs

## Homework Statement

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.

## Homework Equations

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

$\forall$x ∈ ℝ+ $\exists$y ∈ ℝ+ $\forall$z ∈ $Z$+ 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 $\forall$z ∈ $Z$+ come after Q (x, y, z) since it does in the statement?

Mark44
Mentor

## Homework Statement

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.

## Homework Equations

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

$\forall$x ∈ ℝ+ $\exists$y ∈ ℝ+ $\forall$z ∈ $Z$+ 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 $\forall$z ∈ $Z$+ come after Q (x, y, z) since it does in the statement?
I would write it this way.

$\forall (x ∈ R^+, z ∈ Z^+) \exists y ∈ R^+ \ni Q(x, y, z)$

In addition to other changes, I also replaced ℝwith R, since ℝis so tiny I can barely tell it's a version of the letter R.

There's a nicer one that you can get with mathbb{R}, as in
$\mathbb{R}$.