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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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25 Nov 2018, 11:10
2
6
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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) $$m^3 > m^2$$
GMATH Teacher
Status: GMATH founder
Joined: 12 Oct 2010
Posts: 936
Re: Is m < m^2 < m^3 , for all real values of m?  [#permalink]

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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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Fabio Skilnik :: GMATH method creator (Math for the GMAT)
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]

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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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Posts: 936
Is m < m^2 < m^3 , for all real values of m?  [#permalink]

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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.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Manager
Joined: 26 Apr 2011
Posts: 60
Location: India
GPA: 3.5
WE: Information Technology (Computer Software)
Re: Is m < m^2 < m^3 , for all real values of m?  [#permalink]

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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 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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GMATH Teacher
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Joined: 12 Oct 2010
Posts: 936
Re: Is m < m^2 < m^3 , for all real values of m?  [#permalink]

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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.
_________________
Fabio Skilnik :: GMATH method creator (Math for the GMAT)
Our high-level "quant" preparation starts here: https://gmath.net
Re: Is m < m^2 < m^3 , for all real values of m?   [#permalink] 27 Nov 2018, 19:56
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