which is larger number? $\sqrt{7}-\sqrt{6}$ or $\sqrt{6}-\sqrt{5}$

Solution 1:

Hint. $$\sqrt{a+1}-\sqrt{a}=\frac{1}{\sqrt{a+1}+\sqrt{a}}$$

Solution 2:

Consider the shape of the graph of the root function. It is monotonously rising but getting flatter all the time. Hence the differences between two values ($1$ apart on the x-axis) near the y-axis is greater than between two values ($1$ apart on the x-axis) further away from the y-axis.

Solution 3:

As $(\sqrt7+\sqrt5)^2=12+2\sqrt{35}$ and $(\sqrt6+\sqrt6)^2=12+2\sqrt{36}$

$$(\sqrt7+\sqrt5)^2<(\sqrt6+\sqrt6)^2$$

$$\implies \sqrt7+\sqrt5<\sqrt6+\sqrt6\text{ as } a^2>b^2\iff a>b\text{ for }a,b>0$$

$$\implies \sqrt7-\sqrt6<\sqrt6-\sqrt5$$

Solution 4:

Consider the function $f(x) = \sqrt{x}$. Then the second derivative is $$f''(x) = -\frac{1}{4}x^{-3/2} < 0$$

Hence this function is concave down, so we see that $\sqrt{7} - \sqrt{6}$ is the smaller of the two numbers.