# How do I show that this fraction is smaller than 1?

I have $a>b>0$ and $z = \frac{a-\sqrt{a^2-b^2}}{b}$

As $b\to a$ we have $z \to 1$ and as $b\to 0$ we have $z\to 0$. Is this sufficient to show that $z\lt 1$? If not how can I do it?

$$\frac{a-\sqrt{a^2-b^2}}{b}\lt\frac{a-\sqrt{(a-b)(a+b)}}{b}\lt\frac{a-\sqrt{(a-b)(a-b)}}{b}\lt1$$

• The first $<$ should be a $=$. You should evaluate $\sqrt{(a-b)(a-b)}$ and then reduce the fraction. – Trevor Norton Jan 12 '16 at 22:10

$$\frac{a-\sqrt{(a-b)(a+b)}}{b} < \frac{a-\sqrt{(a-b)(a-b)}}{b}$$