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# Range of a 1 element set? and square root doubt

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Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
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Range of a 1 element set? and square root doubt [#permalink]

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19 Feb 2013, 08:09
Bunuel/Karishma,

1) What is the range of a 1 element set?

2) say x square > 2

then root(x)> root 2
=> |x| >root 2

=>

x>root 2 and x< -ve root 2

Is this right? Kindly explain.
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Re: Range of a 1 element set? and square root doubt [#permalink]

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19 Feb 2013, 08:18
1
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Sachin9 wrote:
Bunuel/Karishma,

1) What is the range of a 1 element set?

2) say x square > 2

then root(x)> root 2
=> |x| >root 2

=>

x>root 2 and x< -ve root 2

Is this right? Kindly explain.

1. The range of a one element set is 0.

2. $$x^2 > 2$$

$$x^2 - 2 > 0$$

$$(x + \sqrt{2})(x - \sqrt{2}) > 0$$

So,$$x < -\sqrt{2} or x > \sqrt{2}$$
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Director
Status: Gonna rock this time!!!
Joined: 22 Jul 2012
Posts: 533
Location: India
GMAT 1: 640 Q43 V34
GMAT 2: 630 Q47 V29
WE: Information Technology (Computer Software)
Followers: 3

Kudos [?]: 66 [0], given: 562

Re: Range of a 1 element set? and square root doubt [#permalink]

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19 Feb 2013, 08:26
MacFauz wrote:
Sachin9 wrote:
Bunuel/Karishma,

1) What is the range of a 1 element set?

2) say x square > 2

then root(x)> root 2
=> |x| >root 2

=>

x>root 2 and x< -ve root 2

Is this right? Kindly explain.

1. The range of a one element set is 0.

2. $$x^2 > 2$$

$$x^2 - 2 > 0$$

$$(x + \sqrt{2})(x - \sqrt{2}) > 0$$

So,$$x < -\sqrt{2} or x > \sqrt{2}$$

Thanks mate but what I have deduced above is correct?

And why is the range of 1 element set 0? I know SD is zero because you get zero by substituting the number in its formula.
_________________

hope is a good thing, maybe the best of things. And no good thing ever dies.

Who says you need a 700 ?Check this out : http://gmatclub.com/forum/who-says-you-need-a-149706.html#p1201595

My GMAT Journey : http://gmatclub.com/forum/end-of-my-gmat-journey-149328.html#p1197992

Moderator
Joined: 02 Jul 2012
Posts: 1223
Location: India
Concentration: Strategy
GMAT 1: 740 Q49 V42
GPA: 3.8
WE: Engineering (Energy and Utilities)
Followers: 124

Kudos [?]: 1483 [1] , given: 116

Re: Range of a 1 element set? and square root doubt [#permalink]

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19 Feb 2013, 08:38
1
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Sachin9 wrote:
MacFauz wrote:
Sachin9 wrote:
Bunuel/Karishma,

1) What is the range of a 1 element set?

2) say x square > 2

then root(x)> root 2
=> |x| >root 2

=>

x>root 2 and x< -ve root 2

Is this right? Kindly explain.

1. The range of a one element set is 0.

2. $$x^2 > 2$$

$$x^2 - 2 > 0$$

$$(x + \sqrt{2})(x - \sqrt{2}) > 0$$

So,$$x < -\sqrt{2} or x > \sqrt{2}$$

Thanks mate but what I have deduced above is correct?

And why is the range of 1 element set 0? I know SD is zero because you get zero by substituting the number in its formula.

The range of a set is the difference between the set's highest element and lowest element.. In a single element set both are the same and hence the difference is 0.
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Re: Range of a 1 element set? and square root doubt [#permalink]

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19 Feb 2013, 21:30
1
KUDOS
Expert's post
Sachin9 wrote:
Bunuel/Karishma,

1) What is the range of a 1 element set?

2) say x square > 2

then root(x)> root 2
=> |x| >root 2

=>

x>root 2 and x< -ve root 2

Is this right? Kindly explain.

Yes, this is another way of putting it.

$$x^2 > 2$$
This is equivalent to $$|x|^2 > 2$$
Taking square root since both sides are positive,
$$|x| > \sqrt{2}$$

So $$x > \sqrt{2}$$ or $$x < - \sqrt{2}$$

Though, I feel that tackling it using inequalities is more straight forward.
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Re: Range of a 1 element set? and square root doubt   [#permalink] 19 Feb 2013, 21:30
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