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NCERT Exemplar · Class 9 Mathematics Number Systems

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EXERCISE 1.2 1–4 (part 3 of 6)

  1. Exercise 1

    Let x\displaystyle x and y\displaystyle y be rational and irrational numbers, respectively. Is x+y\displaystyle x+y necessarily an irrational number? Give an example in support of your answer.

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    NCERT’s answer
    Yes. Let $\displaystyle x=21, y=\sqrt{2}$ be a rational number. Now $\displaystyle x+y=21+\sqrt{2}=21+1.4142 \ldots=22.4142 \ldots$ Which is non-terminating and non-recurring. Hence $\displaystyle x+y$ is irrational.
    Suppose \(\displaystyle x+y \) were rational, say \[x + y = r, \quad r \in \mathbb{Q} \] \[y = r - x \] The right side is a difference of two rationals, hence rational — contradicting that \(\displaystyle y \) is irrational. So \(\displaystyle x+y \) is irrational, always.For \(\displaystyle x = 2 \), \(\displaystyle y = \sqrt{3} \): \[x + y = 2 + \sqrt{3} \] which is irrational.Answer: Yes, \(\displaystyle x+y \) is necessarily irrational.

  2. Exercise 2

    Let x\displaystyle x be rational and y\displaystyle y be irrational. Is xy\displaystyle x y necessarily irrational? Justify your answer by an example.

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    NCERT’s answer
    No. $\displaystyle 0 \times \sqrt{2}=0$ which is not irrational .
    Take \(\displaystyle x = 0 \) and \(\displaystyle y = \sqrt{2} \): \[xy = 0 \cdot \sqrt{2} = 0 \] which is rational. So \(\displaystyle xy \) is not always irrational — the claim fails when \(\displaystyle x = 0 \).Answer: No, not necessarily; \(\displaystyle x=0,\ y=\sqrt2 \) gives \(\displaystyle xy=0 \), a rational number.

  3. Exercise 3

    State whether the following statements are true or false? Justify your answer. (i) 23\displaystyle \frac{\sqrt{2}}{3} is a rational number. (ii) There are infinitely many integers between any two integers. (iii) Number of rational numbers between 15\displaystyle 15 and 18\displaystyle 18 is finite. (iv) There are numbers which cannot be written in the form pq,q≠0,p,q\displaystyle \frac{p}{q}, q \neq 0, p, q both are integers. (v) The square of an irrational number is always rational. (vi) 123\displaystyle \frac{\sqrt{12}}{\sqrt{3}} is not a rational number as 12\displaystyle \sqrt{12} and 3\displaystyle \sqrt{3} are not integers. (vii) 153\displaystyle \frac{\sqrt{15}}{\sqrt{3}} is written in the form pq,q≠0\displaystyle \frac{p}{q}, q \neq 0 and so it is a rational number.

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    NCERT’s answer
    (i)
    False. Although $\displaystyle \frac{\sqrt{2}}{3}$ is of the form $\displaystyle \frac{p}{q}$ but here $\displaystyle p$, i.e., $\displaystyle \sqrt{2}$ is not an integer. (ii) False. Between $\displaystyle 2$ and $\displaystyle 3$, there is no integer. (iii) False, because between any two rational numbers we can find infinitely many rational numbers. (iv) True. $\displaystyle \frac{\sqrt{2}}{\sqrt{3}}$ is of the form $\displaystyle \frac{p}{q}$ but $\displaystyle p$ and $\displaystyle q$ here are not integers. (v) False, as $\displaystyle (\sqrt[4]{2})^{2}=\sqrt{2}$ which is not a rational number. (vi) False, because $\displaystyle \frac{\sqrt{12}}{\sqrt{3}}=\sqrt{4}=2$ which is a rational number. (vii) False, because $\displaystyle \frac{\sqrt{15}}{\sqrt{3}}=\sqrt{5}=\frac{\sqrt{5}}{1}$ which is $\displaystyle p$, i.e., $\displaystyle \sqrt{5}$ is not an integer.
    (i)
    False — \[\frac{\sqrt2}{3}=\frac{p}{q}\ \Rightarrow\ \sqrt2=\frac{3p}{q} \] contradicts \(\displaystyle \sqrt2\) irrational.
    (ii)
    False — no integer lies between \(\displaystyle n\) and \(\displaystyle n+1\).
    (iii)
    False — \[\frac{a+b}{2} \] between two rationals gives infinitely many.
    (iv)
    True — \(\displaystyle \sqrt2\) cannot be written as \(\displaystyle \frac{p}{q}\).
    (v)
    False — for \(\displaystyle a=\sqrt[4]{2}\), \[a^2=\sqrt2 \] still irrational.
    (vi)
    False — \[\frac{\sqrt{12}}{\sqrt3}=\sqrt4=2 \] is rational.
    (vii)
    False — \[\frac{\sqrt{15}}{\sqrt3}=\sqrt5 \] is irrational, not \(\displaystyle \frac pq\) with integer \(\displaystyle p,q\).

  4. Exercise 4

    Classify the following numbers as rational or irrational with justification : (i) 196\displaystyle \sqrt{196} (ii) 318\displaystyle 3 \sqrt{18} (iii) 927\displaystyle \sqrt{\frac{9}{27}} (iv) 28343\displaystyle \frac{\sqrt{28}}{\sqrt{343}} (v) −0.4\displaystyle -\sqrt{0.4} (vi) 1275\displaystyle \frac{\sqrt{12}}{\sqrt{75}} (vii) 0.5918\displaystyle 0.5918 (viii) (1+5)−(4+5)\displaystyle (1+\sqrt{5})-(4+\sqrt{5}) (ix) 10.124124... (x) 1.010010001...

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    (i)
    \[\sqrt{196}=14 \]
    Rational.
    (ii)
    \[\sqrt{18}=3\sqrt2\ \Rightarrow\ 3\sqrt{18}=9\sqrt2 \]
    Irrational.
    (iii)
    \[\sqrt{\frac{9}{27}}=\sqrt{\frac13}=\frac{\sqrt3}{3} \]
    Irrational.
    (iv)
    \[\frac{\sqrt{28}}{\sqrt{343}}=\sqrt{\frac{4}{49}}=\frac27 \]
    Rational.
    (v)
    \[-\sqrt{0.4}=-\sqrt{\frac25}=-\frac{\sqrt{10}}{5} \]
    Irrational.
    (vi)
    \[\frac{\sqrt{12}}{\sqrt{75}}=\sqrt{\frac{4}{25}}=\frac25 \]
    Rational.
    (vii)
    \[0.5918=\frac{5918}{10000} \]
    Rational.
    (viii)
    \[(1+\sqrt5)-(4+\sqrt5)=-3 \]
    Rational.
    (ix)
    \[10.124124\ldots=10.\overline{124} \]
    Rational — recurring.
    (x)
    \(\displaystyle 1.010010001\ldots\) never recurs.
    Irrational.
    Answer: rational — (i), (iv), (vi), (vii), (viii), (ix); irrational — (ii), (iii), (v), (x).
    NCERT prints: (vi) \(\displaystyle \frac{\sqrt{12}}{\sqrt{75}}=\frac{2}{7}\) — an arithmetic slip carried over from part (iv): \(\displaystyle \frac{12}{75}=\frac{4}{25}\), whose root is \(\displaystyle \frac25\), not \(\displaystyle \frac27\).