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

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

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19 Feb 2013, 07:09
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, 07:18
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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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Re: Range of a 1 element set? and square root doubt [#permalink]

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19 Feb 2013, 07:26
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.
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Re: Range of a 1 element set? and square root doubt [#permalink]

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19 Feb 2013, 07:38
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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, 20:30
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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, 20:30
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# Range of a 1 element set? and square root doubt

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