# Determine the truth value of the statement if the universe of each variable consists of (i) all real numbers, (ii) all integers.

∀x (x > 0 → ∃y ((√x)/y = 3))

I've learned about different types of proofs, and I'm thinking here that I would first negate the statement to then prove the negation. Would the negation be: ∃x(x ≯ 0 → ∀y((√x)/y ≠ 3)) ?

If it's as easy as finding a number for x in which the statement is false, I could say:

let x = 25

(√25)/y = 5/y ≠ 3

Is this sufficient proof?

In either case, how would the proof be different with integers than with real numbers? Thanks ahead.

• In your counterexample, what if $y=\frac53$? You can't pass negation through implication like that; instead consider writing your implication $P\implies Q$ in the form $Q\vee \neg P$ and then negating that. Oct 9, 2014 at 23:08

You shouldn't need to use a contradiction proof in either case. For the real number case, it is easier to use a constructive existence proof for the value of y. For the integer case, a counter example will suffice.

The field you choose changes the truth value of the proposition.

In the case of real numbers, you need to show that there exists some real number y such that ${\frac{\sqrt{x}}{y}=3}$ whenever x>0.

It is easy to see that letting y = ${\frac{\sqrt{x}}{3}}$ for any value of x>0 will yield:
${\frac{\sqrt{x}}{y} = \frac{3\sqrt{x}}{\sqrt{x}} = 3}$.

so the proposition is true in the real number field.

However, as your counter example shows it isn't true for the integers since letting ${x=25}$ causes ${\frac{\sqrt{25}}{y} = \frac{5}{y}}$ which does not equal 3 for any integer y.

Thus, the proposition is false among the integers.