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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|}

 Apr 30, 2024
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Note that for any non-zero number a, the quantity \(\dfrac a{|a|}\) can only take two values: -1 when a is negative, and 1 when is positive.

 

Therefore, \(\dfrac a{|a|}\) serves as an indicator of whether a is positive. So, there are only 5 cases:

 

Case 1: x, y, z are all positive, making x + y + z also positive. Then \(\displaystyle {\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|}} = 4\).

Case 2: One of x, y, z, x + y + z is negative (for example, x = -1, y = 3, z = 4). Then \(\displaystyle\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|} = 2\).

Case 3: Two of x, y, z, x + y + z are negative (for example, x = -1, y = -2, z = 4). Then \(\displaystyle\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|} = 0\).

Case 4: Three of x, y, z x + y + z are negative (for example, x = -7, y = -5, z = 1). Then \(\displaystyle\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|} = -2\).

Case 5: x, y, z are all negative, making x + y + z also negative. Then \(\displaystyle\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|} = -4\).

 

Therefore, the possible values of \(\displaystyle\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{x + y + z}{|x + y + z|}\) are \(-4,-2,0,2,4\).

 Apr 30, 2024

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