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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

 smiley

 Jun 2, 2024
 #1
avatar+806 
+1

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! :) 

 Jun 2, 2024

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