# 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
\[ \require{cancel}\dfrac a{\cancel b} \times \cancel bd > \dfrac c{\cancel d} \times b \cancel d ,\]
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 \( \color{red} {\textbf{false}}\). 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 forallreals \(a,b,c\) 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:

You should know that

\[ \frac{a}{b} + \frac{c}{d} \ne \frac{a+c}{b+d}.\]

However, if \(a, b, c, d \) are positive integers that satisfy \(\frac{a}{b} < \frac{c}{d},\) what can we say about \(\frac{a+c}{b+d} ?\)

\[ \frac{a}{b} < \frac{c}{d} \]

If \(a, b, c, d \) (are real variables that) satisfy the above inequality, what can we say about

\[ \frac{a+c}{b+d} ?\]

Related problem.

**See Also**

**Cite as:**Does cross multiply always work for inequalities?.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/does-cross-multiple-always-work-for-inequalities/