MHB Understanding Squaring Inequalities

[tmt1]

I have

$$-1 \le \cos\left({2x}\right) \le 1 $$

If everything is squared, it goes to

$$0 \le \cos^2\left({2x}\right) \le 1 $$

and I'm not sure how $(-1)^2$ turns into $0$

 

[topsquark]

I have

$$-1 \le \cos\left({2x}\right) \le 1 $$

If everything is squared, it goes to

$$0 \le \cos^2\left({2x}\right) \le 1 $$

and I'm not sure how $(-1)^2$ turns into $0$

It doesn't. If we squared both sides and did it mechanically we'd get [math]1 \leq cos^2(2x) \leq 1[/math], which is a ridiculous statement. (It is unwise to square both sides of an inequality. Weird things happen.) Instead we set the lower bound by noting that the square of any real quantity is greater than or equal to 0.

-Dan

 

[soroban]

I have: $$-1 \le \cos\left({2x}\right) \le 1 $$

If everything is squared, it goes to: $$0 \le \cos^2\left({2x}\right) \le 1 $$

and I'm not sure how $(-1)^2$ turns into $0$. . . Actually, it doesn't.

In the first statement, we have a small quantity between -1 and +1.
. . (Think of a proper fraction, positive or negative.)

If we square that quantity, we have a positive value (greater than zero).

(Ah, Dan beat me to it.)