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Sachin9
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Range of a 1 element set? and square root doubt 19 Feb 2013, 08:09
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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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Range of a 1 element set? and square root doubt 19 Feb 2013, 08:18
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Sachin9
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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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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Range of a 1 element set? and square root doubt 19 Feb 2013, 08:26
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Sachin9
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,\(x < -\sqrt{2} or x > \sqrt{2}\)
Show more

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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Range of a 1 element set? and square root doubt 19 Feb 2013, 08:38
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Sachin9
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.
Show more
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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Range of a 1 element set? and square root doubt 19 Feb 2013, 21:30
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[quote="Sachin9"]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 2[/m]
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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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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