Does cross multiply always work for inequalities?

True or False?

For all real numbers $a,b,c,d$ with $b$ and $d$ being non-zero, if $\dfrac ab > \dfrac cd$, then $a \times d > c \times b$ must be true.

Why some people say it's true:
It looks intuitive: for this problem, cross multiply is the same as multiplying both sides by $b\times d$. Then $$\dfrac a{\cancel b} \times \cancel bd > \dfrac c{\cancel d} \times b \cancel d ,$$ which implies $a \times d > c\times b$.

Why some people say it's false:
There are other cases to consider besides positive real numbers. In those other cases, this "identity" might fail with one of $b,d$ being negative.

This statement is $\color{#D61F06} {\textbf{false}}$. The claim is true if and only if the denominators are both positive or negative. In particular, for $a=b=c=1$ and $d= -1$, while the constraint of $\dfrac ab > \dfrac cd$ is indeed fulfilled, the claim of $a\times d > c\times b$ is false because $-1 > 1$ is not true.

The reason our initial claim fails is because once we multiply both sides of an inequality by a negative number, the inequality sign must be flipped. As an explicit example, the inequality $-1 > -2$ is obviously true. But if we multiply both sides by $-1$, while keeping the inequality sign the same, we have $1 > 2,$ which is obviously false.

Rebuttal: Because $a= b= c=1$ and $d=2$ makes both $\dfrac ab > \dfrac cd$ and $a\times d > b\times c$ true.

Reply: We have only shown that it's true when $a=b=c=1$ and $d=2$. In fact, we should prove (or disprove) that it's true in general. That is, we did not prove that the claim holds for all reals $a,b,c$ and $d$ with $b,d\ne 0$.

Rebuttal: If we square both sides of the inequality, we get $\dfrac{a^2}{b^2} > \dfrac{c^2}{d^2}$. Then we can cross multiply both sides by $b^2 d^2$, which is a positive number, so the inequality sign does not need to be changed. Thus $a^2 d^2 > b^2 c^2$. Taking the square roots gives the desired claim, $a \times d > b \times c$.

Reply: Recall that $\sqrt{x^2} = |x|$. So taking the square roots only yields $|ad | > |bc |$, instead of $ad > bc$.

Want to make sure you've got this concept down? Try these problems:

See Also

• Multi-step Linear Inequalities
• List of Common Misconceptions