$\forall x {\in} \Bbb R\; \exists y {\in} \Bbb R \;\,2y^2(x-5)<1$

Question

Prove or disprove that $\forall x {\in} \Bbb R\; \exists y {\in} \Bbb R \;\,2y^2(x-5)<1.$

Since $2y^2(x-5)<1 \Rightarrow y^2(x-5)<\frac{1}{2}.$

Take $y=\frac{1}{\sqrt {x-5}}.$

Then $\left(\frac{1}{\sqrt {x-5}}\right)^2(x-5)=1<\frac{1}{2},$ which is a contradiction.

So, the proposition $\forall x \in \Bbb R, \exists y \in \Bbb R: 2y^2(x-5)<1$ is false.

Is my reasoning correct?

Answer 1

Unpacking your reasoning

You have shown that for all $x$ greater than $5$, there exists a real number $y$ such that $2y^2(x-5)>1$. In other words:

$$ \forall x \in \{x\in \mathbb{R} : x>5\}, \exists y \in \mathbb{R} : 2y^2(x-5)>1 $$

by taking $y=\frac{1}{\sqrt{x-5}}$. This statement isn't equivalent to the statement you are trying to prove

If you want to prove by contradiction, you need to negate the quantifiers as well as negating the inequality. When negating quantifiers, "for all" becomes "there exists", and vice versa; therefore, the statement you are trying to prove is:

$$ \exists x \in \mathbb{R}. \forall y \in \mathbb{R} : 2y^2(x-5)\geq 1. $$

This is a valid approach, but I think it's clearer to argue directly.

Direct solution

Let $x \in \mathbb{R}$.

If $x \leq 5$, then $2y^2(x-5) \leq 0 < 1$ for all $y$ so we are done. Hence assume that $x > 5$.

Due to the order of our quantifiers, the value of $y$ is allowed to depend on the value of $x$, so we choose $y$ small enough to ensure $2y^2(x-5) < 1$. In particular, we choose any $y < \frac{1}{\sqrt{2(x-5)}}$, for example $y = \frac{1}{2\sqrt{2(x-5)}}$.

With this choice of $y$ we see

$$ 2y^2(x-5) = \frac{2(x-5)}{(2\sqrt{2(x-5)})^2} = \frac{1}{4} < 1, $$

and we are done.

Notes

Our above argument is a good approach to take when working with quantifiers and inequalities. We let $x$ be arbitrary because our statement must be true for all $x$, and then try and find a $y$ that works. As EzTheBoss mentioned in their answer, we often find this $y$ by looking at the inequality and working backwards. The value of $y$ we presented depends on the value of $x$. Alternatively, as suggested in the comments, we could have also chosen $y=0$.

Answer 2

To show that the expression is false, all that is needed is a counter example. You would need to come up with some $x \in \mathbb{R}$ $:$ $\nexists y \in \mathbb{R}, 2y^2(x-5) \lt 1$. But the expression has to be true! As for any $x$, no matter how large you can choose a small enough $y$ so that the expression is less than $1$. How we prove this is by choosing some function $y: \mathbb{R} \to \mathbb{R}$ of $x$ so that the expression holds. Here is my proof: $\DeclareMathOperator{\sgn}{sgn}$
Pf: Define $y(x) = \sqrt{\frac{1}{4}(\frac{1}{|x-5|})}$, then $$2y^2(x-5)=2(\frac{1}{4})(\frac{1}{|x-5|})(x-5)=\frac{1}{2}\sgn(x-5) \lt 1$$ $\square$

Note that for these sorts of proof it helps to work backwards, I started by solving the expression for $y$ to get a suitable $x$ instead of starting from the beginning of the proof. Also the sign function shows up because we have an expression that is $1$ when $x \gt 5$ and $-1$ when $x \lt 5$, but in either case we have $\pm\frac{1}{2}\lt 1$ which holds regardless of $x$.