shrouded1 wrote:
Is \(x > y\) ?
(1) \(\sqrt{x} > y\)
(2) \(x^3 > y\)
\(x\,\,\,\mathop > \limits^? \,\,\,y\)
\(\left( 1 \right)\,\,\,\sqrt x > y\,\,\,\,\,\left\{ \begin{gathered}
\,Take\,\,\,\left( {x;y} \right) = \left( {1\,;\,0} \right)\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,Take\,\,\,\left( {x;y} \right) = \left( {\frac{1}{4}\,;\,\frac{1}{3}} \right)\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\end{gathered} \right.\)
\(\left( 2 \right)\,\,\,{x^3} > y\,\,\,\,\,\left\{ \begin{gathered}
\,(re)\,Take\,\,\,\left( {x;y} \right) = \left( {1;0} \right)\,\,\,\,\,\left\langle {{\text{YES}}} \right\rangle \hfill \\
\,Take\,\,\,\left( {x;y} \right) = \left( {2\,;\,7} \right)\,\,\,\,\,\left\langle {{\text{NO}}} \right\rangle \hfill \\
\end{gathered} \right.\)
\(\left( {1 + 2} \right)\,\,\,\,\,\)
\(\left. {x = y\,\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}
\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\,\sqrt x > x\,\,\,\,\, \Rightarrow \,\,\,\,\,0 < \sqrt x < 1\,\,\,\,\,\,\,\,\left( {\sqrt x \geqslant 1\,\,\, \Rightarrow \,\,\,\sqrt x > x\,\,{\text{false,}}\,\,\,{\text{impossible}}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\boxed{0 < x < 1} \hfill \\
\mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,\,\,{x^3} > x\,\,\,\,\, \Rightarrow \,\,\,\,\,\boxed{0 < x < 1\,\,\,{\text{false}}}\,\,\,\,\,\,\,\,\, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x \ne y\)
\(x < y\left. {\,\,\,\, \Rightarrow \,\,\,\left\{ \begin{gathered}
\mathop \Rightarrow \limits^{\left( 1 \right)} \,\,\,\,\,x < y < \sqrt x \,\,\,\,\, \Rightarrow \,\,\,\,\,\sqrt x > x\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{above}}} \,\,\,\,\,\,\boxed{0 < x < 1}\,\,\,\, \hfill \\
\mathop \Rightarrow \limits^{\left( 2 \right)} \,\,\,\,\,x < y < {x^3}\,\,\,\, \Rightarrow \,\,\,{x^3} > x\,\,\,\,\,\mathop \Rightarrow \limits^{{\text{again}}} \,\,\,\,\,\,\,\boxed{0 < x < 1\,\,\,{\text{false}}}\,\,\,\,\,\,\,\,\, \hfill \\
\end{gathered} \right.\,\,\,\,\,\,\,} \right\}\,\,\,\,\,\, \Rightarrow \,\,\,\,\,x < y\,\,\,{\text{is}}\,\,{\text{false}}\)
Conclusion: \(\left( {1 + 2} \right)\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\text{YES}}} \right\rangle\)
The above follows the notations and rationale taught in the GMATH method.
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Fabio Skilnik ::
GMATH method creator (Math for the GMAT)
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