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 \( ab =ac \implies a (b-c) = 0 \implies a = 0 \text{ or } b-c = 0 \). 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 cannot divide by zero.
Want to make sure you've got this concept down? Try this problem:
See Also