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# Algebra

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Let $x$, $y$, and $z$ be nonzero real numbers. Find all possible values of
\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|}

Dec 9, 2023

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We can consider each of the 8 possible cases for the signs of x,y,z, and x+y+z.

Case 1: x,y,z,x+y+z are all positive. In this case, the fractions ∣x∣x​,∣y∣y​,∣z∣z​,∣x+y+z∣x+y+z​ all reduce to 1, so the value is 4​.

Case 2: x,y,z are all positive and x+y+z is negative. In this case, ∣x∣x​=∣y∣y​=∣z∣z​=∣x+y+z∣x+y+z​=−1, so the value is −4​.

Case 3: x,y,z are all positive and only one of them is negative. We can say without loss of generality that x is negative. In this case, ∣x∣x​=∣y∣y​=∣z∣z​=1 while ∣x+y+z∣x+y+z​=−1, so the value is 1​.

Case 4: x,y,z are all negative and x+y+z is positive. In this case, ∣x∣x​=∣y∣y​=∣z∣z​=−1 while ∣x+y+z∣x+y+z​=1, so the value is −1​.

Case 5: x and y are positive and z is negative. In this case, ∣x∣x​=∣y∣y​=1 while ∣z∣z​=−1 and ∣x+y+z∣x+y+z​=x+y+zx+y−z​, where without loss of generality we can assume y>z, so x+y+zx+y−z​<1. Therefore, the value is between 1​ and 2​.

Case 6: x and z are positive and y is negative. In this case, ∣x∣x​=∣z∣z​=1 while ∣y∣y​=−1 and ∣x+y+z∣x+y+z​=x+y+zx+z−y​, where without loss of generality we can assume x>y, so x+y+zx+z−y​<1. Therefore, the value is between 1​ and 2​.

Case 7: y and z are positive and x is negative. In this case, ∣y∣y​=∣z∣z​=1 while ∣x∣x​=−1 and ∣x+y+z∣x+y+z​=y+z+xy+z−x​, where without loss of generality we can assume y>x, so y+z+xy+z−x​<1. Therefore, the value is between −1​ and 0​.

Case 8: x and y are negative and z is positive. In this case, ∣x∣x​=∣y∣y​=−1 while ∣z∣z​=1 and ∣x+y+z∣x+y+z​=z+x+yz−x−y​, where without loss of generality we can assume y>x, so z+x+yz−x−y​<1. Therefore, the value is between −1​ and 0​.

Dec 9, 2023