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

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19 Feb 2013, 08:38

1

This post received KUDOS

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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