# Simple inequality question $\sqrt{3} - 2 + \frac{\pi}{3} < \pi -2$

$\sqrt{3} - 2 + \frac{\pi}{3} < \pi -2$

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!

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$\sqrt 3<\frac{2\pi}3$ 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 $\sqrt3<\frac{2\pi}3$.

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Thanks a bunch for you comment! Excellent! –  Lukas Arvidsson Nov 2 '12 at 22:51

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. $$3 \cdot \sqrt{3} - 6 + \pi < 3\pi -6\tag{multiply through by 3}$$ $$3 \cdot \sqrt{3} -6+6 < 3\pi-\pi\tag{+ 6, -\pi on each side}$$ $$3\sqrt{3} < 2\pi\tag{yes?}$$

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Thank you so much for your comment! Very valuable to me! –  Lukas Arvidsson Nov 2 '12 at 22:51