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I am quite confused with this, if I multiply both sides with -1, the sign should be reversed; if I square it first, the sign would not be reversed. May I know what operation should I do?

Thank you so much for your reply.

Henry Cai
  • 633

3 Answers3

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The function $x \mapsto -x$ reverses order, so you need to reverse the inequality. The squaring function does not preserve order unless you restrict it to either the positive numbers (preserves) or the negative numbers (reverses).

ncmathsadist
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If I square it first, the sign would not be reversed.

That is not true. For example, consider $p = 4, q = 3$. It is true that $-p < -q$. However, if you square and don't flip the sign, you get that $16 < 9$ which is absurd.

The safest thing to do it to multiply with $-1$ and change signs as that will always be correct. The sign's orientation after squaring depends on the signs of $p$ and $q$.


Case 1. $p \ge 0, q \ge 0$. In this case, you have that $-p \le 0, -q \le 0$.
Note that the function $x \mapsto x^2$ is strictly decreasing on $(-\infty, 0]$ and thus, squaring changes the sign. (This was the example above.)

Case 2. $p \le 0, q \le 0$. In this case, you have that $-p \ge 0, -q \ge 0$.
Note that the function $x \mapsto x^2$ is strictly increasing on $[0, \infty)$ and thus, squaring does not change the sign.

Case 3. $p$ and $q$ are of different signs. The first inequality forces $p$ to be positive and $q$ negative. However, nothing more can be concluded. For example, $p = 2, q = -3$ preserves sign whereas $p = 3, q = -2$ reverses it. In fact, $p = 1, q = -1$ doesn't even preserve the fact that they're unequal.

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Always multiply both sides by $-1$. The behaviour of the inequality under this operation is always well-defined, unlike squaring which destroys the relationship with negative numbers.

Parcly Taxel
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