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: $-{\displaystyle \frac{1}{2}}<2$, then, $(-{\displaystyle \frac{1}{2}}{)}^2$ and $2^2=4$ and the inequality does not change: ${\displaystyle \frac{1}{4}}<4$.

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

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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