How do I show that $x+(1-\alpha)y < 0$ for all $0<\alpha < 1$?

Question

Suppose that $x\in\mathbb{R}^{-}$, $y\in\mathbb{R}^{+}$, and $0<\alpha <1$. I would like to construct an argument to show that the expression $x+(1-\alpha)y$ is negative. My take on it is as follows: $$ x+(1-\alpha)y\leq x-|\alpha-1|y<0. $$ I think that this does it. However, I'm worried about what I did with $(1-\alpha)$, knowing that $0<\alpha<1$. Am I thinking correctly, or is there something I'm missing here?

Answer 1

This result is not true for all $x \in \mathbb{R}^-$, $y \in \mathbb{R}^+$ and $\alpha < 1$. For example, take $\alpha = 1/2 < 1$. Then you would like for $x + (1-\alpha)y = x + y/2$ to be always negative. However, it cannot be true for any $x > 0$ and $y < 0$ since you can take $y > -2x$, and hence $x + y/2 > x - 2x/2 = 0$.

To see what's wrong in your proof, remember that by the definition of the absolute value, for all $x \in \mathbb{R}$ we have $-|x| \leq x \leq |x|$, so it is certain that $\alpha - 1 \leq |\alpha - 1|$. However, since $y$ is positive, the inequality you used here is true if $(1-\alpha) \leq -|\alpha - 1|$, which is equivalent to $|\alpha - 1| \leq \alpha - 1$. So this can be true only if $|\alpha - 1| = \alpha - 1$, that is if $\alpha - 1 > 0$, or $\alpha > 1$. It obviously contradicts your hypothesis.