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Is m < m^2 < m^3 , for all real values of m?

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Is m < m^2 < m^3 , for all real values of m?  [#permalink]

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New post 25 Nov 2018, 11:10
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A
B
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Difficulty:

  65% (hard)

Question Stats:

46% (01:26) correct 54% (01:12) wrong based on 82 sessions

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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?  [#permalink]

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New post 25 Nov 2018, 12:45
1
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?  [#permalink]

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New post 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?  [#permalink]

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New post 26 Nov 2018, 04:36
1
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 pre-statements 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 inequality-focused 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?  [#permalink]

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New post 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 pre-statements 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 inequality-focused 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?  [#permalink]

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New post 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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Our high-level "quant" preparation starts here: https://gmath.net
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Re: Is m < m^2 < m^3 , for all real values of m?   [#permalink] 27 Nov 2018, 19:56
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