# When can we multiply inequalities without changing their directions?

This is part of a series on common misconceptions.

Is this 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 $\color{#D61F06}{\textbf{false}}$.

Explanation:

Short Answer:Proving one counter exampleLet 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 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 counter-examples, 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 it is always defined that way in maths

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 rules 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 counter example 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 it is), so you shouldn’t expect an intuitive answer.

**Cite as:**When can we multiply inequalities without changing their directions?.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/when-can-we-multiply-inequalities/