In the case where x≥0, we see that |x|=x, and since |x|=a, we can conclude that x=−a. In the case where x≥0, we know that |x|=x and so the inequality \(|x| < a\...In the case where x≥0, we see that |x|=x, and since |x|=a, we can conclude that x=−a. In the case where x≥0, we know that |x|=x and so the inequality |x|<a implies that x<a. So in both cases, we have proven that −a<x<a and this proves that if |x|<a, then −a<x<a. In Part (1) of Theorem 3.25, we proved that for each real number x, |x|<a if and only if −a<x<a.
In the case where x≥0, we see that |x|=x, and since |x|=a, we can conclude that x=−a. In the case where x≥0, we know that |x|=x and so the inequality \(|x| < a\...In the case where x≥0, we see that |x|=x, and since |x|=a, we can conclude that x=−a. In the case where x≥0, we know that |x|=x and so the inequality |x|<a implies that x<a. So in both cases, we have proven that −a<x<a and this proves that if |x|<a, then −a<x<a. In Part (1) of Theorem 3.25, we proved that for each real number x, |x|<a if and only if −a<x<a.
