# Show that for $\alpha, \beta > 0$ and $0 < \delta < 1$, that $\left| \alpha\beta -1 \right| \leq 3\delta$

I am working on a pretty long proof, which is irrelevant here. But in one step, I need to prove the following:

Let $$\alpha > 0$$, $$\beta > 0$$ and $$0 < \delta < 1$$ be real values with $$\left|\alpha - 1\right| \leq \delta$$ and $$\left|\beta - 1\right| \leq \delta$$. Show that: $$\left|\alpha\beta -1 \right| \leq 3\delta$$

I'm sadly stuck and don't even know where to start here, which is holding me back in my proof. How can this be proven?

$$|ab-1| \leq |ab-a+a-1|\leq |a||b-1|+|a-1|$$ $$\leq (|a-1|+1)|b-1|+|a-1|$$ $$\leq \delta^2+\delta+\delta <3\delta$$

• Thank you, that was very helpful. I will accept your answer as soon as it is possible. Nov 19, 2019 at 16:52

From $$|\alpha - 1| \leq \delta$$ and $$|\beta - 1| \leq \delta$$, we have $$1-\delta \leq \alpha \leq 1 + \delta$$ and $$1-\delta \leq \beta \leq 1 + \delta \text{,}$$ so $$1-2\delta + \delta^2 \leq \alpha \beta \leq 1+2\delta + \delta^2 \text{.}$$ Then $$-2\delta + \delta^2 \leq \alpha \beta - 1\leq 2\delta + \delta^2 \text{.}$$ Since $$0 < \delta < 1$$, the triangle inequality gives $$|-2 \delta + \delta^2| < 2 \delta + \delta^2 < 3 \delta$$. So you have the slightly stronger $$|\alpha \beta - 1| < 2\delta +\delta^2 < 3\delta \text{.}$$

• Unless I am mistaken, the last estimate $2\delta +\delta^2 < 3\delta^2$ is wrong for $0 < \delta < 1$. Nov 20, 2019 at 7:45
• @MartinR : Oops. One too many squares. Thanks! and Fixed. Nov 20, 2019 at 14:51

(I'm using Latin letters rather than Greek because I'm lazy.)

$$(a-1)(b-1) =ab-a-b+1 =ab-1-(a-1)-(b-1)$$

so

$$ab-1 =(a-1)+(b-1)+(a-1)(b-1)$$.

Taking absolute values,

$$|ab-1| \le 2d+d^2 \lt 3d$$.