Problems from Arrangement of the real numbers

Development:

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: $$4 > \dfrac{1}{4}$$.

Solution:

• If $$0 < a < b$$ then $$a^2 < b^2$$.
• If $$a < b < 0$$ then $$a^2 > b^2$$.
• But if $$a < 0 < b$$ then we cannot affirm anything with only this information (we have seen that both things can happen).

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