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]

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]

19 Feb 2013, 08:18

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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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Director

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

19 Feb 2013, 08:38

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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]

19 Feb 2013, 21:30

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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.

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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Director

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

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.
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Re: Range of a 1 element set? and square root doubt [#permalink] 19 Feb 2013, 08:26

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Range of a 1 element set? and square root doubt

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