Examine, whether the following numbers are rational or irrational:(i)�

1.	Examine, whether the following numbers are rational or irrational:(i) \(\sqrt{7}\)(ii) \(\sqrt{4}\)(iii) \(2+\sqrt{3}\)(iv) \(\sqrt{3}+\sqrt{2}\)(v) \(\sqrt{3}+\sqrt{5}\)(vi) \((\sqrt{2}-2)^2\)(vii) \((2-\sqrt{2})(2+\sqrt{2})\)(viii) \((\sqrt{2}+\sqrt{3})^2\)(ix) \(\sqrt{5}-2\)(x) \(\sqrt{23}\)(xi) \(\sqrt{225}\)(xii) 0.3796 (xiii) 7.478478 (xiv) 1.101001000100001…..
Answer» (i) \(\sqrt7\) is not a perfect square root, so it is an irrational number. (ii) We have, \(\sqrt4 = 2 = \frac{2}{1}\) \(\sqrt4\) can be expressed in the form of \(\frac{p}q,\) so it is a rational number. The decimal expression of \(\sqrt4\) is 2.0 (iii) 2 ia a rational number, whereas \(\sqrt3\) is an irrational number. Because, sum of a rational number and an irrational number is an irrational number, so 2 + is an irrational number (iv) \(\sqrt2\) is an irrational number. Also \(\sqrt3\) is an irrational number. The sum of two irrational numbers is irrational. Therefore, \(\sqrt{3}+\sqrt{2}\) is an irrational number (v) \(\sqrt5\) is an irrational number. Also, \(\sqrt3\) is an irrational number. The sum of two irrational numbers is irrational. Therefore, \(\sqrt3+5\) is an irrational number. (vi) We have, \((\sqrt{2}-\sqrt{2})^2\) \(= (\sqrt{2})^2-2\times\sqrt2{\times2}+({2})^2\) \(= 2-4\sqrt2{+4}\) \(= 6 - 4\sqrt2\) Now 6 is a rational number, whereas \(4\sqrt2\) is an irrational number The difference of a rational number and an irrational number is an irrational number. So, it is an irrational number. (vii) We have, \((2-\sqrt2)(2+\sqrt2)\) \(({2})^2-(\sqrt2)^2\) [Therefore, (a – b) (a + b) = a² – b²] \(= 4 - 2 = 2 =\frac{2}{1}\) Since 2 is a rational number Therefore, \((2-\sqrt2)(2+\sqrt2)\) is a rational number (viii) We have, \((\sqrt2+\sqrt3)^2\) \(=( \sqrt{2})^2+2 \times\sqrt2\times\sqrt3+(\sqrt3)^2\) \(= 2 +2\sqrt6+3\) \(= 5+ 2\sqrt6\) The sum of a rational number and an irrational number is irrational number. Therefore, it is an irrational number. (ix) The difference of a rational number and an irrational number is an irrational number. Therefore, \(5 - \sqrt2\) is an irrational number (x) \(\sqrt23\) = 4.79583152331…. Therefore, it is an irrational number (xi) \(\sqrt225\) = 15 = \(\frac{15}{1}\) Therefore, it is a rational number as it is represented in the form of \(\frac{p}q\), where q ≠ 0 (xii) 0.3796, as a decimal expansion of this number is terminating, so it is an irrational number. (xiii) 7.478478….\(= 7.4\overline78\) As, decimal expansion of this number is non – terminating recurring so it is a rational number (xiv) 1.101001000100001….. It is an irrational number

Examine, whether the following numbers are rational or irrational:(i) \(\sqrt{7}\)(ii) \(\sqrt{4}\)(iii) \(2+\sqrt{3}\)(iv) \(\sqrt{3}+\sqrt{2}\)(v) \(\sqrt{3}+\sqrt{5}\)(vi) \((\sqrt{2}-2)^2\)(vii) \((2-\sqrt{2})(2+\sqrt{2})\)(viii) \((\sqrt{2}+\sqrt{3})^2\)(ix) \(\sqrt{5}-2\)(x) \(\sqrt{23}\)(xi) \(\sqrt{225}\)(xii) 0.3796 (xiii) 7.478478 (xiv) 1.101001000100001…..

Answer»

(i) \(\sqrt7\) is not a perfect square root, so it is an irrational number.

(ii) We have,

\(\sqrt4 = 2 = \frac{2}{1}\)

\(\sqrt4\) can be expressed in the form of \(\frac{p}q,\) so it is a rational number.

The decimal expression of \(\sqrt4\) is 2.0

(iii) 2 ia a rational number, whereas \(\sqrt3\) is an irrational number. Because, sum of a rational number and an irrational number is an irrational number, so 2 + is an irrational number

(iv) \(\sqrt2\) is an irrational number. Also \(\sqrt3\) is an irrational number. The sum of two irrational numbers is irrational.

Therefore, \(\sqrt{3}+\sqrt{2}\) is an irrational number

(v) \(\sqrt5\) is an irrational number. Also, \(\sqrt3\) is an irrational number. The sum of two irrational numbers is irrational.

Therefore, \(\sqrt3+5\) is an irrational number.

(vi) We have,

\((\sqrt{2}-\sqrt{2})^2\) \(= (\sqrt{2})^2-2\times\sqrt2{\times2}+({2})^2\)

\(= 2-4\sqrt2{+4}\)

\(= 6 - 4\sqrt2\)

Now 6 is a rational number, whereas \(4\sqrt2\) is an irrational number

The difference of a rational number and an irrational number is an irrational number. So, it is an irrational number.

(vii) We have,

\((2-\sqrt2)(2+\sqrt2)\)

\(({2})^2-(\sqrt2)^2\) [Therefore, (a – b) (a + b) = a² – b²]

\(= 4 - 2 = 2 =\frac{2}{1}\)

Since 2 is a rational number

Therefore, \((2-\sqrt2)(2+\sqrt2)\) is a rational number

(viii) We have,

\((\sqrt2+\sqrt3)^2\) \(=( \sqrt{2})^2+2 \times\sqrt2\times\sqrt3+(\sqrt3)^2\)

\(= 2 +2\sqrt6+3\)

\(= 5+ 2\sqrt6\)

The sum of a rational number and an irrational number is irrational number. Therefore, it is an irrational number.

(ix) The difference of a rational number and an irrational number is an irrational number.

Therefore, \(5 - \sqrt2\) is an irrational number

(x) \(\sqrt23\) = 4.79583152331….

Therefore, it is an irrational number

(xi) \(\sqrt225\) = 15 = \(\frac{15}{1}\)

Therefore, it is a rational number as it is represented in the form of \(\frac{p}q\), where q ≠ 0

(xii) 0.3796, as a decimal expansion of this number is terminating, so it is an irrational number.

(xiii) 7.478478….\(= 7.4\overline78\)

As, decimal expansion of this number is non – terminating recurring so it is a rational number

(xiv) 1.101001000100001…..

It is an irrational number

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