When can we multiply inequalities without changing their directions?

This is part of a series on common misconceptions

True or False?

If $a>b$ and $c>d,$ then $ac>bd.$

Why some people say it's true: Well, it just makes sense! How could it not be true?

Why some people say it's false: You shouldn’t just trust your intuition, it can’t be that simple!

The statement is $\color{#D61F06}{\textbf{false}}$.

Short answer: proving with one counterexample

Let $a=0, b=-5, c=1,$ and $d=0.$. This satisfies the conditions that $a>b$ and $c>d,$ but since $ac=bd=0,$ the hypothesis is false.

Long answer:

Multiple things aren’t taken into account if you go with your intuition. Were it just $a+c>b+d,$ it would be fine, but multiplication is more complex. The most obvious thing your intuition overlooks is the signs of numbers. In this case, there are 16 possibilities for the signs of $a, b, c,$ and $d.$

However, we can divide them into 5 cases:

Case 1: Exactly one of $a$ or $c$ is 0 $(ac=0),$ and exactly one of $d$ or $b$ is negative and the other is positive $(db$ is negative$).$ Then $ac>db.$

Case 2: Exactly one of $a$ or $c$ is 0 $(ac=0),$ and exactly one of $d$ or $b$ is 0 $(db=0).$ Then $ac=db.$

Case 3: At least one of $a$ or $c$ is 0 $(ac=0),$ and both $d$ and $b$ are negative $(db$ is positive$).$ Then $db>ac.$

Case 4: Both $a$ and $c$ are positive $(ac$ is positive$).$ $ac>db$ is sometimes true in this case, but not always. Since negative times negative equals positive, it is possible that $db>ac.$ However, even if $d$ and $b$ are both negative, this is not always the case — in that case, it is possible that $ac=db$ or just that $ac>db$ — and if they are not both negative, $ac>db$ is always true.

Case 5: $a, b, c,$ and $d$ are all negative $(ac$ and $db$ are both positive$).$ Then $db>ac.$

We see that in cases 2, 3, 5, and sometimes 4, our hypothesis is false. There are more counterexamples, as these cases assume that $a,c,d,$ and $b$ are all real.

Rebuttal: Why should negative times negative equal positive?
Reply: It is intuitively confusing but is always defined that way in math.

Rebuttal: If $a, b, c,$ and $d$ are all negative, isn’t $ac>bd?$
Reply:No. Let $a$ and $b$ be real positive integers where $a>b.$ Then multiply both sides by -1. We have to switch the sign around, $-a<-b,$ which does make a bit of sense intuitively when you think about it.

Rebuttal: I can’t follow all those cases!
Reply: The exceptions to this rule $($if $a, b, c,$ and $d$ are all real numbers$)$ are caused because negative times negative equals positive and negative times positive equals negative. These rules complicate the multiplication. This extra complication is why we must check so many cases if we want a general idea of this rule's truth value $($if $a,b, c,$ and $d$ are all real numbers$).$ If you don’t want to go through the whole process, let $a=0, b =-1, c=0,$ and $d=-2.$ This satisfies the conditions that $a>b$ and $c>d,$ but $ac=0$ and $bd=2,$ so $bd>ac.$ Simply proving one counterexample is enough for this.

Rebuttal: This is counter-intuitive!
Reply: Keep in mind that if $a,b,c,$ and $d$ are all real numbers; the exceptions to this rule are caused by completely counter-intuitive equations (and imaginary numbers are counter-intuitive as they are), so you shouldn’t expect an intuitive answer.