Arrangement of the real numbers

Let's suppose first that $a$ and $b$ are positive numbers, that is, $0 < a < b$, then, we multiply the inequality $a < b$ by $a$, and we obtain: $$a\cdot a < a \cdot b \Rightarrow a^2 < a\cdot b $$

Next we multiply the inequality by $b$ and we obtain: $$a\cdot b < b \cdot b \Rightarrow a\cdot b < b^2$$

And if we join both inequalities, we have : $$a^2 < a\cdot b < b^2 \Rightarrow a^2 < b^2$$

Let's suppose now that $a$ and $b$ are negative numbers, that is, $a < b < 0$, and we repeat the process. Multiplying the inequality by $a$, we obtain: $$a\cdot a > a \cdot b \Rightarrow a^2 > a\cdot b $$

And multiplying it by $b$ we obtain: $$a\cdot b > b \cdot b \Rightarrow a\cdot b > b^2$$

And if we join both inequalities, we have that: $$a^2 > a\cdot b > b^2 \Rightarrow a^2 > b^2$$

But let's see what happens if one of the numbers is positive and the other negative. Namely if we have $a < 0 < b$. Proceeding in the same way, we obtain by multiplying the inequality $a < b$ by $a$: $$a\cdot a > a \cdot b \Rightarrow a^2 > a \cdot b$$

And multiplying it by $b$ we obtain: $$a\cdot b < b \cdot b \Rightarrow a \cdot b < b^2 $$

In such a way that we cannot join both results.

In fact we can find examples of all kinds:

If we choose: $-\dfrac{1}{2} < 2$, then, $\Big(-\dfrac{1}{2}\Big)^2$ and $2^2=4$ and the inequality does not change: $\dfrac{1}{4} < 4$.

But if we choose $-2 < \dfrac{1}{2}$, then $(-2)^2=4$ and $\Big(\dfrac{1}{2}\Big)^2=\dfrac{1}{4}$ and the inequality changes: $4 > \dfrac{1}{4}$.