Exercise 1.3: Number Systems — NCERT Solutions

(i) $\sqrt{25} = 5$, a whole number. Rational.

(ii) $7$ is not a perfect square, so $\sqrt{7}$ has a non-terminating, non-recurring decimal expansion. Irrational.

(iii) $0.3796$ terminates. Rational.

(iv) $7.478478\dots = 7.\overline{478}$ is non-terminating but recurring. Rational.

Conclusion: Only $\sqrt{7}$ is irrational; the rest are rational.

(i) Use $(a+b)(a-b) = a^2 - b^2$: $(3+\sqrt{3})(3-\sqrt{3}) = 3^2 - (\sqrt{3})^2 = 9 - 3 = 6$ This is a whole number — rational.

(ii) Use $(a+b)^2 = a^2 + 2ab + b^2$: $(\sqrt{5}+\sqrt{2})^2 = (\sqrt{5})^2 + 2\sqrt{5}\sqrt{2} + (\sqrt{2})^2$ $= 5 + 2\sqrt{10} + 2 = 7 + 2\sqrt{10}$ Since $\sqrt{10}$ is irrational, $7 + 2\sqrt{10}$ is irrational.

Conclusion: (i) is rational ($6$); (ii) is irrational ($7 + 2\sqrt{10}$).

Convert to decimals: $\frac{5}{7} = 0.714285\dots, \qquad \frac{9}{11} = 0.8181\dots$

So we need irrational numbers between about $0.7142\dots$ and $0.8181\dots$. Construct decimals that are non-terminating and non-recurring (no repeating block):

$0.7401001000100001\dots$ $0.7502002000200002\dots$ $0.8080080008\dots$

Each lies between the two given numbers and has a non-repeating pattern, so each is irrational.

Conclusion: Three such irrational numbers are given above (many other answers possible).