# If AB=AC, does B=C?

This is part of a series on common misconceptions.

True or False:For real numbers \(a\), \(b\) and \(c\), if \(ab=ac\), then \(b=c\).

**Why some people say it's true:** Just divide both sides of the equation by \(a\).

**Why some people say it's false:** We cannot simply divide by \(a\), right?

The statement is \( \color{red}{\textbf{false}}\).

We can prove that the statement is false using a simple counter-example:

Proof:Consider \( a = 0, b = 1, c = 2 \). We have \( ab = 0 \times 1 = 0 \) and \( ac = 0 \times 2 = 0 \) so they are equal. However, \( b \neq c \).More generally, by the zero product property, we know that \( ab =ac \Rightarrow a (b-c) = 0 \Rightarrow a = 0 \text{ or } b-c = 0 \). Always remember that if we want to "divide both sides by \(a\)," we have to check that we are not dividing by 0.

Rebuttal:But the statement is true if \( a = 1 \). Then \( ab = ac \) gives us \( 1 \times b = 1 \times c \).

Reply:Yes, the statement is true when \( a = 1 \). However, the question is for all possible values of \(a\), \(b\) and \(c\). In particular, the statement need not be true when \( a = 0 \).

Rebuttal:But by the rules of algebra, you can divide by a.

Reply:Yes, but that implies that a is not equal to 0. After all, you can not divide by zero.

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

**See Also**

**Cite as:**If AB=AC, does B=C?.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/when-can-we-cancel-common-factors/