Real Numbers — 20 Important Questions
Class 9 • Maharashtra Board • Uses MathJax for clean surds, fractions & proofs
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Q1. Classify each as terminating or non-terminating recurring: \( \frac{13}{5},\ \frac{2}{11},\ \frac{29}{16},\ \frac{17}{125},\ \frac{11}{6}\).
Hint: Factorize denominators; only primes \(2,5\) ⇒ terminating, otherwise recurring.
Answer:
\(\frac{13}{5}\) (denom \(=5\)) ⇒ terminating.
\(\frac{2}{11}\) (denom includes \(11\)) ⇒ non-terminating recurring.
\(\frac{29}{16}\) (denom \(=2^4\)) ⇒ terminating.
\(\frac{17}{125}\) (denom \(=5^3\)) ⇒ terminating.
\(\frac{11}{6}\) (denom \(=2\cdot 3\)) ⇒ non-terminating recurring.
\(\frac{2}{11}\) (denom includes \(11\)) ⇒ non-terminating recurring.
\(\frac{29}{16}\) (denom \(=2^4\)) ⇒ terminating.
\(\frac{17}{125}\) (denom \(=5^3\)) ⇒ terminating.
\(\frac{11}{6}\) (denom \(=2\cdot 3\)) ⇒ non-terminating recurring.
Q2. Write in decimal form: \( \frac{127}{200},\ \frac{25}{99},\ \frac{23}{7},\ \frac{4}{5},\ \frac{17}{8}\).
Divide or use standard recurring forms.
Answer (to 6 d.p. where recurring):
\(127/200=0.635\) (terminating); \(25/99=0.\overline{25}\); \(23/7=3.\overline{285714}\); \(4/5=0.8\); \(17/8=2.125\).
Q3. Convert \(0.\overline{7}\) to \(p/q\) form.
Let \(x=0.\overline{7}\). Multiply suitably and subtract.
Answer:
\(x=0.\overline{7}\Rightarrow 10x=7.\overline{7}\Rightarrow 10x-x=7\Rightarrow 9x=7\Rightarrow x=\frac{7}{9}\).
Q4. Convert \(7.\overline{529}\) to \(p/q\) form.
Let \(x=7.\overline{529}\). Multiply by \(1000\) and subtract \(x\).
Answer:
\(x=7.\overline{529}\Rightarrow 1000x=7529.\overline{529}\). Subtract: \(999x=7522\Rightarrow x=\frac{7522}{999}\).
Q5. Write in \(p/q\) form: \(0.6,\ 0.37,\ 3.17,\ 15.89,\ 2.514\).
Remove decimal by multiplying by powers of 10 and reduce.
Answer:
\(0.6=\frac{6}{10}=\frac{3}{5}\); \(0.37=\frac{37}{100}\); \(3.17=\frac{317}{100}\); \(15.89=\frac{1589}{100}\); \(2.514=\frac{2514}{1000}=\frac{1257}{500}\).
Q6. Outline an indirect proof that \( \sqrt{2} \) is irrational.
Assume \( \sqrt{2}=\frac{p}{q} \) in lowest terms; derive a contradiction.
Answer (sketch):
Assume \( \sqrt{2}=\frac{p}{q}\Rightarrow 2q^2=p^2\) ⇒ \(p^2\) even ⇒ \(p\) even ⇒ \(p=2t\). Then \(2q^2=4t^2\Rightarrow q^2=2t^2\) ⇒ \(q\) even.
So \(p,q\) both even, contradicting lowest terms. Hence \( \sqrt{2}\) irrational.
Q7. Comment on decimal expansions of \( \sqrt{2} \) and \( \sqrt{3}\).
Use division method idea.
Answer:
Both are non-terminating, non-recurring decimals (e.g., \( \sqrt{2}\approx1.41421\ldots, \sqrt{3}\approx1.732\ldots\)), hence irrational.
Q8. State: (i) rational ± irrational (ii) nonzero rational × irrational.
Use properties of irrational numbers.
Answer:
(i) irrational; (ii) irrational. (Sum/difference or nonzero multiple of an irrational is irrational.)
Q9. Define a surd. Which of the following are surds: \( \sqrt[3]{7},\ \sqrt[3]{27},\ \sqrt[4]{8}\)?
Surd: irrational \(n\)th-root of a positive rational number.
Answer:
\( \sqrt[3]{7}\) is a surd (irrational). \( \sqrt[3]{27}=3\) is rational ⇒ not a surd. \( \sqrt[4]{8}\) is irrational ⇒ surd.
Q10. Write in simplest surd form: \( \sqrt{48},\ \sqrt{98}.\)
Factor out perfect squares.
Answer:
\( \sqrt{48}=\sqrt{16\cdot3}=4\sqrt{3};\ \sqrt{98}=\sqrt{49\cdot2}=7\sqrt{2}.\)
Q11. Are \( \sqrt{45}\) and \( \sqrt{80}\) like surds?
Reduce each to simplest form and compare radicands.
Answer:
\( \sqrt{45}=3\sqrt{5},\ \sqrt{80}=4\sqrt{5}\Rightarrow\) both multiples of \( \sqrt{5}\) ⇒ like surds.
Q12. Compare: \(6\sqrt{2}\) and \(5\sqrt{5}\).
Square both sides (positive quantities) to compare: compare \(72\) and \(125\).
Answer: \(72<125\Rightarrow 6\sqrt{2}<5\sqrt{5}.\)
Q13. Simplify: \(7\sqrt{3}+29\sqrt{3}-5\sqrt{3}\).
Add coefficients of like surds.
Answer: \((7+29-5)\sqrt{3}=31\sqrt{3}.\)
Q14. Compute: (i) \( \sqrt{50}\cdot\sqrt{18}\) (ii) \( \dfrac{\sqrt{125}}{\sqrt{5}}\).
Use \( \sqrt{a}\sqrt{b}=\sqrt{ab}\) and simplify.
Answer:
(i) \( \sqrt{900}=30\) (rational); (ii) \( \sqrt{25}=5\) (rational).
Q15. Rationalize the denominator: \( \dfrac{1}{\sqrt{5}}\), \( \dfrac{3}{\sqrt{2}+\sqrt{7}}\) (set up only).
Multiply by a suitable rationalizing factor.
Answer:
\( \dfrac{1}{\sqrt{5}}\cdot\dfrac{\sqrt{5}}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}.\)
For \( \dfrac{3}{\sqrt{2}+\sqrt{7}}\), multiply by the conjugate \( \sqrt{2}-\sqrt{7}\):
\( \dfrac{3(\sqrt{2}-\sqrt{7})}{(\sqrt{2}+\sqrt{7})(\sqrt{2}-\sqrt{7})}=\dfrac{3(\sqrt{2}-\sqrt{7})}{2-7}=-\dfrac{3}{5}(\sqrt{2}-\sqrt{7}).\)
For \( \dfrac{3}{\sqrt{2}+\sqrt{7}}\), multiply by the conjugate \( \sqrt{2}-\sqrt{7}\):
\( \dfrac{3(\sqrt{2}-\sqrt{7})}{(\sqrt{2}+\sqrt{7})(\sqrt{2}-\sqrt{7})}=\dfrac{3(\sqrt{2}-\sqrt{7})}{2-7}=-\dfrac{3}{5}(\sqrt{2}-\sqrt{7}).\)
Q16. Show that \( (\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})\) is rational.
Use \( (a+b)(a-b)=a^2-b^2\).
Answer: \(5-3=2\) (rational). Conjugate pair product is always rational.
Q17. Using order properties on \(\\mathbb{R}\): If \(a0\), show \(ac
State the property and the sign reversal idea.
Answer:
If \(a0\Rightarrow acbc\) (inequality reverses).
Q18. Solve \( |x-5|=2\).
Split into two linear cases.
Answer:
\(x-5=2\Rightarrow x=7\) or \(x-5=-2\Rightarrow x=3.\)
Q19. Compare \(7\sqrt{2}\) and \(5\sqrt{3}\) without calculator.
Square and compare: \(98\) vs \(75\).
Answer: \(98>75\Rightarrow 7\sqrt{2}>5\sqrt{3}.\)
Q20. Rationalize \( \dfrac{3+\sqrt{8}}{\sqrt{2}+\sqrt{5}}\) and simplify fully.
Multiply by conjugate of denominator and simplify surds.
Answer (steps):
\(\displaystyle \frac{3+\sqrt{8}}{\sqrt{2}+\sqrt{5}}\cdot\frac{\sqrt{2}-\sqrt{5}}{\sqrt{2}-\sqrt{5}} = \frac{(3+\sqrt{8})(\sqrt{2}-\sqrt{5})}{(\sqrt{2})^2-(\sqrt{5})^2}\) \(= \dfrac{3\sqrt{2}-3\sqrt{5}+\sqrt{16}-\sqrt{40}}{2-5} = \dfrac{3\sqrt{2}-3\sqrt{5}+4-2\sqrt{10}}{-3}.\)
Hence \(= -\dfrac{1}{3}\big(3\sqrt{2}-3\sqrt{5}+4-2\sqrt{10}\big) = -\sqrt{2}+\sqrt{5}-\dfrac{4}{3}+\dfrac{2}{3}\sqrt{10}.\)
\(\displaystyle \frac{3+\sqrt{8}}{\sqrt{2}+\sqrt{5}}\cdot\frac{\sqrt{2}-\sqrt{5}}{\sqrt{2}-\sqrt{5}} = \frac{(3+\sqrt{8})(\sqrt{2}-\sqrt{5})}{(\sqrt{2})^2-(\sqrt{5})^2}\) \(= \dfrac{3\sqrt{2}-3\sqrt{5}+\sqrt{16}-\sqrt{40}}{2-5} = \dfrac{3\sqrt{2}-3\sqrt{5}+4-2\sqrt{10}}{-3}.\)
Hence \(= -\dfrac{1}{3}\big(3\sqrt{2}-3\sqrt{5}+4-2\sqrt{10}\big) = -\sqrt{2}+\sqrt{5}-\dfrac{4}{3}+\dfrac{2}{3}\sqrt{10}.\)