For real numbers a and b, when is the equation |a + b| = |a – b| true? I put that the statement was true only when a = 0 and b = 0 but the correct answer was that it only held true for a = 0 OR b = 0. With 'and' I figured |0 + 0| = 0 and |0 - 0| = 0. Could someone explain why my answer was incorrect? Thank you. 
 A: If we square both sides of the equation $|a + b| = |a - b|$, we obtain
\begin{align*}
|a + b|^2 & = |a - b|^2\\
a^2 + 2ab + b^2 & = a^2 -2ab + b^2\\
4ab & = 0\\
ab & = 0
\end{align*}
which holds if $a = 0$ or $b = 0$ since a product is equal to zero if and only if one of the factors is equal to zero.  
The statement $a = 0$ or $b = 0$ means that at least one of the equations is true, that is, it means that $a = 0$ is true, $b = 0$ is true, or both $a = 0$ and $b = 0$ are true.  The statement $a = 0$ and $b = 0$ means that both $a = 0$ and $b = 0$ are true.  
As you showed, the statement $|a + b| = |a - b|$ is certainly true when $a = 0$ and $b = 0$.  However, this is not the only solution.  For instance, the equation is satisfied when $a = 1$ and $b = 0$ since 
$$1 = |1| = |1 + 0| = |1 - 0| = |1| = 1$$
It is also satisfied when $a = 0$ and $b = 1$ since
$$1 = |1| = |0 + 1| = |0 - 1| = |-1| = 1$$
More generally, the statement $|a + b| = |a - b|$ is satisfied when $a = 0$ and $b$ is any real number or $b = 0$ and $a$ is any real number.  
For the case $a = 0$ and $b \in \mathbb{R}$, observe that the equation $|a + b| = |a - b|$ holds since 
$$|b| = |0 + b| = |0 - b| = |-b|$$
which, from a geometric point of view, is valid since a number and its additive inverse are equidistant from $0$ on the real number line and, from an algebraic point of view, is true since 
$$|b| = \sqrt{b^2} = \sqrt{(-b)^2} = |-b|$$
For the case $b = 0$ and $a \in \mathbb{R}$, observe that the equation $|a + b| = |a - b|$ holds since 
$$|a| = |a + 0| = |a - 0| = |a|$$
Thus, the statement $a = 0$ or $b = 0$ gives us the general solution, while the statement $a = 0$ and $b = 0$ gives us a particular solution.
A: Because of the modulus, it suffices to check if it is true in only these two cases: (i) $a,b\geq0$ and (ii) $a>0,\  b\leq0$. It should not be difficult now.
