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Mathematics

Assertion (A): (3+5)+(6−25)=9−5(3 + \sqrt{5}) + (6 - 2\sqrt{5}) = 9 - \sqrt{5}.

Reason (R): The sum of two irrational numbers is always irrational.

  1. A is true, R is false.
  2. A is false, R is true.
  3. Both A and R are true.
  4. Both A and R are false.

Rational Irrational Nos

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Answer

A is true, R is false.

Explanation

Given,

(3+5)+(6−25)=9−5(3 + \sqrt{5}) + (6 - 2\sqrt{5}) = 9 - \sqrt{5}.

Taking L.H.S.

=(3+5)+(6−25)=3+5+6−25=3+6+5−25=(3+6)+(5−25)=9+(−5)=9−5= (3 + \sqrt{5}) + (6 - 2\sqrt{5})\\[1em] = 3 + \sqrt{5} + 6 - 2\sqrt{5}\\[1em] = 3 + 6 + \sqrt{5} - 2\sqrt{5}\\[1em] = (3 + 6) + (\sqrt{5} - 2\sqrt{5})\\[1em] = 9 + (-\sqrt{5})\\[1em] = 9 - \sqrt{5}

L.H.S. = R.H.S.

∴ Assertion(A) is true.

Let the two irrational numbers be 2 + 2\sqrt2 and 4 - 2\sqrt2

Sum of two irrational numbers = (2 + 2\sqrt2) + (4 - 2\sqrt2)

= 2 + 2\sqrt2 + 4 - 2\sqrt2

= 2 + 4 + 2−2\sqrt2 - \sqrt2

= 6 (rational number)

∴ The sum of two irrational numbers is not always irrational.

∴ Reason(R) is false.

Hence, Assertion (A) is true and Reason (R) is false.

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