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Is m < m^2 < m^3 , for all real values of m?
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25 Nov 2018, 11:10
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Is m < \(m^2 < m^3\) , for all real values of m? (1) \(m < m^3\) (2) \(m^3 > m^2\)
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Re: Is m < m^2 < m^3 , for all real values of m?
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25 Nov 2018, 12:45
rencsee wrote: Is m < \(m^2 < m^3\) , for all real values of m?
(1) \(m < m^3\) (2) \(m^3 > m^2\)
Very nice problem, congrats! (kudos!) \(?\,\,\,:\,\,\,m < {m^2} < {m^3}\) \(\left( 1 \right)\,\,\,m < {m^3}\,\,\,\,\left\{ \matrix{ \,{\rm{Take}}\,\,m = 2\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr \,{\rm{Take}}\,\,m =  {1 \over 2}\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\left( {{m^2} < {m^3}\,\,{\rm{is}}\,\,{\rm{false}}} \right)\,\, \hfill \cr} \right.\) \(\left( 2 \right)\,\,\,{m^3} > {m^2}\,\,\,\,\, \Rightarrow \,\,\,\,{m^2}\left( {m  1} \right) > 0\,\,\,\,\, \Rightarrow \,\,\,\,m > 1\,\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,\,m\,\,\left( { > \,0} \right)\,\,} \,\,\,\,{m^2} > m\,\,\,\,\left( {{\rm{and}}\,\,\left( 2 \right)} \right)\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\left\langle {{\rm{YES}}} \right\rangle\) The correct answer is therefore (B). This solution follows the notations and rationale taught in the GMATH method. Regards, Fabio.
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Re: Is m < m^2 < m^3 , for all real values of m?
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26 Nov 2018, 02:28
rencsee wrote: Is m < \(m^2 < m^3\) , for all real values of m?
(1) \(m < m^3\) (2) \(m^3 > m^2\) Please elaborate more on the solution. Confused on the approach to solve this problem
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Is m < m^2 < m^3 , for all real values of m?
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26 Nov 2018, 04:36
MrCleantek wrote: Please elaborate more on the solution. Confused on the approach to solve this problem Hi MrCleantek ! My pleasure. In (1) I have presented what we call a BIFURCATION: Two different VIABLE answers to the question asked (= each scenario satisfies both the question stem prestatements and the statement itself). This is the proper mathematical way to guarantee insufficiency. It avoids the common "it´s clear..." that permits falling into intuition traps! In (2) I have shown that \(m^2 > m\) and that was enough, because the other part of the inequalityfocused is given in the statement itself. My rationale was the following: I put \(m^2\) as a common nonzero factor (otherwise statement (2) would be contradicted: 0 > 0 is false), hence \(m^2 > 0\) and, therefore \(m > 1\). Afterwards I have multiplied both sides by \(m > 0\) (we know m>1), the inequality is mantained and we get the one we wished for! An alternate approach (quite similar) would be the following: \({m^3} > {m^2}\,\,\,\,\mathop \Rightarrow \limits_{\left( {m\, \ne \,0} \right)}^{:\,\,{m^2}\, > \,\,0} \,\,\,m > 1\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,m\,\,\left( { > \,0} \right)} \,\,\,{m^2} > m\) I hope everything is clear now. Regards and success in your studies, Fabio.
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Re: Is m < m^2 < m^3 , for all real values of m?
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27 Nov 2018, 03:05
fskilnik wrote: MrCleantek wrote: Please elaborate more on the solution. Confused on the approach to solve this problem Hi MrCleantek ! My pleasure. In (1) I have presented what we call a BIFURCATION: Two different VIABLE answers to the question asked (= each scenario satisfies both the question stem prestatements and the statement itself). This is the proper mathematical way to guarantee insufficiency. It avoids the common "it´s clear..." that permits falling into intuition traps! In (2) I have shown that \(m^2 > m\) and that was enough, because the other part of the inequalityfocused is given in the statement itself. My rationale was the following: I put \(m^2\) as a common nonzero factor (otherwise statement (2) would be contradicted: 0 > 0 is false), hence \(m^2 > 0\) and, therefore \(m > 1\). Afterwards I have multiplied both sides by \(m > 0\) (we know m>1), the inequality is mantained and we get the one we wished for! An alternate approach (quite similar) would be the following: \({m^3} > {m^2}\,\,\,\,\mathop \Rightarrow \limits_{\left( {m\, \ne \,0} \right)}^{:\,\,{m^2}\, > \,\,0} \,\,\,m > 1\,\,\,\,\mathop \Rightarrow \limits^{ \cdot \,m\,\,\left( { > \,0} \right)} \,\,\,{m^2} > m\) I hope everything is clear now. Regards and success in your studies, Fabio. I got it now. Thanks...
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Re: Is m < m^2 < m^3 , for all real values of m?
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27 Nov 2018, 19:56
MrCleantek wrote: I got it now. Thanks...
Great, MrCleantek! I am glad to know. See you in other posts! (Thanks for the kudos.) Regards, Fabio.
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Re: Is m < m^2 < m^3 , for all real values of m?
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27 Nov 2018, 19:56






