+4 Let x,y,z and be nonzero real numbers. Find all possible values of
\(\[\frac{x}{|x|} + \frac{y}{|y|} + \frac{z}{|z|} + \frac{xy}{|xy|} + \frac{xz}{|xz|} + \frac{yz}{|yz|} + \frac{xyz}{|xyz|}.\]\)
Thank you
+1926 We could use casework for this problem.
First, if x, y, and z are all postive, we have
\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{x}{x}+\frac{y}{y}+\frac{z}{z}+\frac{xyz}{xyz}\,=\,1+1+1+1\,=\,4\)
Now, if we have only x is negative, is 0, we have
\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{y}{y}+\frac{z}{z}+\frac{-|x|yz}{|xyz|}\,=\,-1+1+1-1\,=\,0\)
In the same way, having only y and z being negative will still be 0.
If only z is postive, we have
\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{-|y|}{|y|}+\frac{z}{z}+\frac{(-|x|)(-|y|)z}{|xyz|}\,=\,-1-1+1+1\,=\,0\)
In the same way, if only x and y are positive, the result will also be 0.
If all three x, y, and z are negative, we have
\(\frac{x}{|x|}+\frac{y}{|y|}+\frac{z}{|z|}+\frac{xyz}{|xyz|}\,=\,\frac{-|x|}{|x|}+\frac{-|y|}{|y|}+\frac{-|z|}{|z|}+\frac{(-|x|)(-|y|)(-|z|)}{|xyz|}\,=\,-1-1-1-1\,=\,-4\)
So, the three possible values are 0, -4, and 4.
Thanks! :)
