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A rational number between \sqrt{2} and \sqrt{3} is – 

(A) \frac{\sqrt{2}+\sqrt{3}}{2}

(B) \frac{\sqrt{2}.\sqrt{3}}{2}

(C) 1.5

(D) 1.8

Answers (1)

Answer: [C]

Solution.

Any number which can be represented in the form of p/q where q is not equal to zero is a rational number. So it is basically a fraction with non-zero denominator.
Irrational numbers are real numbers which cannot be represented as simple fractions.
Now, we have \sqrt{2}=1.414 and \sqrt{3}= 1.732
(A) \frac{\sqrt{2}+\sqrt{3}}{2} cannot be represented as simple fraction hence it is irrational.
(B) \frac{\sqrt{2}.\sqrt{3}}{2} =\frac{\sqrt{3}}{\sqrt{2}}=\sqrt{\frac{3}{2}}=\sqrt{1.5} cannot be represented as simple fraction hence it is irrational.
(C) 1.5 is a rational number between \sqrt{2}=1.414  and
\sqrt{3}= 1.732  
Q 1.414 < 1.5 < 1.732
(D) 1.8 is a rational number but does not lie between
\sqrt{2}=1.414  and
\sqrt{3}= 1.732.
Q 1.732 < 1.8

Therefore option (C) is correct.

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