Simple inequality question $\sqrt{3} - 2 + \frac{\pi}{3} < \pi -2$ Can someone please explain how this conclusion is made:
$\sqrt{3} - 2 + \frac{\pi}{3} <  \pi -2$
because $\sqrt{3} < 2 < \frac{2\pi}{3}$
I think this one is quite simple, just that I cant really get my head around it.
Thank you!
 A: When working with inequalities, remember that you can operate on each side, just as you would an equation except that when you multiply or divide by a negative number, the inequality reverses, for example:  $$ -2x < 8 \iff \left(-\frac{1}{2}\right)\cdot (-2x) > \left(-\frac{1}{2}\right) \cdot 8 \iff x> -4$$
Note the reversal of the "direction"  of the inequality from $<$ to $>$ when I multiplied each side by $(-\frac{1}{2})$. So you can operate on your inequality just as you normally would with an equation, just remember to use caution when multiplying or dividing by a negative value!
You have
$$\sqrt{3} - 2 + \frac{\pi}{3} <  \pi -2$$
and you want to confirm it is true.
$$3 \cdot \sqrt{3} - 6 + \pi < 3\pi -6\tag{multiply through by 3}$$
$$3 \cdot \sqrt{3} -6+6 < 3\pi-\pi\tag{+ 6, $-\pi$ on each side}$$
$$ 3\sqrt{3} < 2\pi\tag{yes?}$$
A: $\sqrt 3<\frac{2\pi}3$ implies $\sqrt3-2+\frac\pi3<\pi-2$ (by adding $\frac\pi3-2$ to both sides)
The number $2$ is included in the middle probably for the reasons that it is easy to see that $\sqrt3<2$ (since $3<2^2$) and $2<\frac{2\pi}3$ (since $\pi>3$). So this helps you to deduce that $\sqrt3<\frac{2\pi}3$.
