Math Proof (∀ x ∈ R) (∃ y ∈ R)[(-y > x^2)]?
Prove the following statement:
(∀ x ∈ R) (∃ y ∈ R)[(-y > x^2)]
Your proof should start with \Let x be an arbitrary real number. . . " and go on to produce a y in terms
of x which satisfies the desired property.
So I have no idea how to do this, I started with:
Let x in R be arbitrary.
We know x^2 ≥ 0
so, -y ≥ 0
dividing by -1 we get,
y ≤ 0 Therefore there exists a y for all x such that --y > x^2
Is that a valid answer?
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Verified answer
"so, -y ≥ 0 "
What is y? You haven't said.
Since x^2 >=0, all you need is any negative y, like -1.
It doesn't have to depend on x.
-1 works for all x.
It's that simple.
The problem with your proof is that it says "so, -y >= 0" without introducing what y actually is. The problem is to show, for that given x, that there _is_ some y that makes something happen, and so the proof should be very clear on where this y comes from (ie, this y should depend on x). Your "we know x^2 >= 0, so, -y >= 0" is, in more detail, probably something like "We know that x^2 >= 0, so if we *can* find a y with -y > x^2, we would need to have -y > 0." This is definitely true, but it's not very helpful (ie, it doesn't actually show us that we can find a y with -y > x^2, it just tells us a property that that y would have to have --- and this property, by itself, isn't enough to ensure that -y > x^2."
Here is one way of doing it.
Let x in R be arbitrary. Let y = -(x^2 + 1). We have
-y = -(-(x^2 + 1)) = x^2 + 1.
Since 1 > 0, we know that x^2 + 1 > x^2 + 0. From the facts that -y = x^2 + 1 and x^2 + 1 > x^2 we deduce that -y > x^2. This shows that there is a real number y --- namely -(x^2 + 1) --- with the property that -y > x^2. End of proof.