True or False. For all $x \in \mathbb{R}_{>0}$ there exists a y $ \in \mathbb{R}_{>0}$ such that $|2x+y| = 5$

Question

True or False: For all $x \in \mathbb{R}_{>0}$ there exists a y $ \in \mathbb{R}_{>0}$ such that $|2x+y| = 5$

I have already verified the sentence and I know it is false, but I have done it by giving a value to x:

$x=4$ Then

$ |2x+y| = 5 \ \implies \ |8+y| = 5 \ \implies \ y=-13 \text{ or } y = -3. $

Hence, $y<0$ in either case and so $y \notin \mathbb{R}_{>0}.$

How can I prove it in a general way, that is, without giving a value to x or y?

Answer 1

since $x >0, y>0$, we have $|2x+y|=2x+y$

For any $x > \frac52$ and $y>0$, $2x+y > 5$.

Remark: counter example is a proof.

Answer 2

Actually what you did is a general way and it is called proof by a contradiction. Since it says for all $x \in \mathbb{R}_{>0}$, all you need to do is to find an $x\in \mathbb{R}_{>0}$ value such that $y \notin \mathbb{R}_{>0}$, which contradicts to the claim. Then you can argue by contradiction that the claim is false.

Answer 3

The statement is false.

For positive numbers $x$ and $y$, $$|2x+y| = 2x+y =5\implies y=5-2x$$

Therefore if $ x> 2.5$ then $y$ will be negative.

That is a contradiction.

Answer 4

Simply giving a specific counter example, is a formal proof.

But if you want.

$x>0$ and $y>0 $ so $2x+y>0$ and $|2x+y|=2x+y $.

So if $|2x+y|=5$

$2x+y=5$

$y=5-2x $

If $2x > 5$ then $y <0$. So for any $x> \frac 52$ the statement fails.

If $y>\frac 52$ then $x <0$.