1. **State the problem:** Find all real numbers $x$ such that $$|x - 1| > |x + 1|.$$ 2. **Recall the definition of absolute value:** For any real number $a$, $$|a| = \begin{cases} a & \text{if } a \geq 0 \\ -a & \text{if } a < 0 \end{cases}.$$ 3. **Approach:** To solve the inequality involving absolute values, consider the critical points where the expressions inside the absolute values change sign, which are $x=1$ and $x=-1$. We will analyze the inequality on the intervals $(-\infty, -1)$, $[-1,1]$, and $(1, \infty)$. 4. **Interval 1: $x < -1$** - Here, $x-1 < 0$ and $x+1 < 0$, so $$|x-1| = -(x-1) = 1 - x,$$ $$|x+1| = -(x+1) = -x - 1.$$ - The inequality becomes $$1 - x > -x - 1.$$ - Simplify: $$1 - x > -x - 1$$ Add $x$ to both sides: $$1 > -1$$ This is always true for all $x < -1$. 5. **Interval 2: $-1 \leq x \leq 1$** - Here, $x-1 \leq 0$ and $x+1 \geq 0$, so $$|x-1| = 1 - x,$$ $$|x+1| = x + 1.$$ - The inequality becomes $$1 - x > x + 1.$$ - Simplify: $$1 - x > x + 1$$ Subtract 1 from both sides: $$-x > x$$ Add $x$ to both sides: $$0 > 2x$$ Divide both sides by 2: $$0 > x$$ - So, for $x$ in $[-1,1]$, the inequality holds when $x < 0$. 6. **Interval 3: $x > 1$** - Here, $x-1 > 0$ and $x+1 > 0$, so $$|x-1| = x - 1,$$ $$|x+1| = x + 1.$$ - The inequality becomes $$x - 1 > x + 1.$$ - Simplify: $$x - 1 > x + 1$$ Subtract $x$ from both sides: $$-1 > 1$$ This is false for all $x > 1$. 7. **Combine the solutions:** - From interval 1: all $x < -1$ - From interval 2: all $-1 \leq x < 0$ - From interval 3: no solutions Therefore, the solution set is $$\boxed{(-\infty, 0)}.$$