27) Consider the following putative theorem.

Theorem. For every real number x, if |x 3| < 3 then 0 < x < 6

Is the following proof correct? If so, what proof strategies does it use? If not, can it be fixed? Is the theorem correct?

Proof. Let x be an arbitrary number, and suppose |x3| < 3. We consider

two cases:

Case 1. x3 0. Then |x3| = x3. Plugging this into the assumption

that |x 3| < 3, we get x 3 < 3, so clearly x < 6.

Case 2. x 3 < 0. Then |x 3| = 3 x, so the assumption |x 3| < 3

means that 3 x < 3. Therefore 3 < 3 + x, so 0 < x.

Since we have proven both 0 < x and x < 6, we can conclude that 0 <

x < 6.