The interesting issue here may be explicitly presented below:

\({y^x} < 0\,\,\,;\,\,\,x + y > 0\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,x > y\)

To prove that, let´s suppose, on the contrary, that we could have SIMULTANEOUSLY the following:

(1) \({y^x} < 0\)

(2) \(x + y > 0\)

(3) \(x \leqslant y\)

The contradiction follows:

\(\left( 1 \right)\,\,\,\, \Rightarrow \,\,\,\,\,y < 0\,\,\,\,\,\,\mathop \Rightarrow \limits^{\left( 3 \right)} \,\,\,x < 0\,\,\,\,\,\,\, \Rightarrow \,\,\,x + y < 0\,\,\,\,\, \Rightarrow \,\,\,\,\left( 2 \right)\,\,{\text{contradicted}}\)

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Fabio Skilnik :: http://www.GMATH.net (Math for the GMAT)

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