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SubjectsMathematics
अपरिमेय संख्याएँ
In this Class 9 Mathematics topic from the Number Systems chapter, students learn that irrational numbers cannot be written in the form p/q, where p and q are integers and q is not zero. They explore familiar examples such as √2 and π, understand their non-terminating, non-repeating decimal expansions, and distinguish them from rational numbers. The topic also develops skills for representing irrational numbers on the number line and understanding their place within the real number system.
Correct answer: C. (\frac{6(\sqrt{19}+\sqrt{10})}{9})
Explanation: Multiplying by the conjugate makes the denominator (19-10=9). So the form is (\frac{6(\sqrt{19}+\sqrt{10})}{9}).
Correct answer: A. 0.272727...
Explanation: 0.272727... is a recurring decimal. If \(y=0.272727...\), then \(100y-y=27\), so \(y=27/99=3/11\). Therefore, it is rational. Exam tip: convert recurring decimals into fractions to check rationality.
Correct answer: D. \(13+\sqrt{42}\)
Explanation: The square of a square root gives the number inside. So \(\left(\sqrt{13+\sqrt{42}}\right)^2=13+\sqrt{42}\).
Correct answer: B. (15)
Explanation: Area is ((\sqrt{22}+\sqrt{7})(\sqrt{22}-\sqrt{7})=22-7=15). Conjugate dimensions can give a rational area.
Correct answer: A. If \(x\) is irrational, then \(2x\) is irrational.
Explanation: If \(2x\) were rational, then \(x=(2x)/2\) would also be rational, a contradiction. Hence \(2x\) is irrational. But \(x^2\) need not be irrational: for \(x=\sqrt{2}\), \(x^2=2\). In exams, test claims using a counterexample.
Correct answer: C. (2-\sqrt{3})
Explanation: Multiplying by the conjugate gives (\frac{(\sqrt{6}-\sqrt{2})^2}{4}=2-\sqrt{3}). Make the denominator rational.
Correct answer: B. \(\sqrt{2}\times\sqrt{8}=4\)
Explanation: Both \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, but \(\sqrt{2}\times\sqrt{8}=\sqrt{16}=4\), which is rational. Hence the claim is false. Exam tip: multiply radicands and check for a perfect square.
Correct answer: B. (14)
Explanation: \(\sqrt{175}=5\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the bracket is \(2\sqrt{7}\). Multiplying by \(\sqrt{7}\) gives (14).
Correct answer: A. The sum of any irrational number and any rational number is irrational.
Explanation: Let \(x\) be irrational and \(r\) be rational. If \(x+r\) were rational, then \((x+r)-r=x\) would be rational, a contradiction. Option B fails since \(\sqrt{2}+(-\sqrt{2})=0\). Exam tip: test “always” claims with a counterexample.
Correct answer: A. \(a+b\) is irrational
Explanation: If \(a+b\) were rational, subtracting the rational number \(a\) would make \(b\) rational, a contradiction. Thus \(a+b\) is irrational. Exam tip: test an “always” statement using contradiction.
Correct answer: C. (\sqrt{19})
Explanation: Since (18<19<20), (\sqrt{19}) lies between them. Compare square roots using the numbers inside.
Correct answer: A. \(p+x\) is always irrational
Explanation: If \(p+x\) were rational, subtracting the rational number \(p\) would make \(x\) rational, a contradiction. B fails because \(\sqrt2+(-\sqrt2)=0\). Exam tip: use contradiction for such properties.
Correct answer: A. \(r+x\) is irrational
Explanation: If \(r+x\) were rational, subtracting rational \(r\) would make \(x\) rational, a contradiction. But \((\sqrt{2})^2=2\). Exam tip: use closure of rational numbers under subtraction.
Correct answer: A. (8\sqrt{3})
Explanation: (\sqrt{432}=12\sqrt{3}), (\sqrt{192}=8\sqrt{3}), and (\sqrt{48}=4\sqrt{3}). Therefore the result is (8\sqrt{3}).
Correct answer: A. (\sqrt{7})
Explanation: After rationalising the terms become (\frac{\sqrt{7}-\sqrt{5}}{2}) and (\frac{\sqrt{7}+\sqrt{5}}{2}). The sum is (\sqrt{7}).
Correct answer: A. \(r+q\)
Explanation: \(r+q\) is always irrational; otherwise, subtracting the rational number \(q\) would make \(r\) rational, a contradiction. But \(r\times q\) can be 0 when \(q=0\). Exam tip: adding or subtracting a rational number from an irrational number remains irrational.
Correct answer: C. (16)
Explanation: (\sqrt{200}=10\sqrt{2}) and (\sqrt{72}=6\sqrt{2}), so the numerator is (16\sqrt{2}). Dividing gives (16).
Correct answer: A. (4\sqrt{91})
Explanation: The direct answer is A, \\(4\\sqrt{91}\\). Use the difference-of-squares identity \\(r^2-s^2=(r-s)(r+s)\\). Given \\(r=\\sqrt{13}+\\sqrt{7}\\) and \\(s=\\sqrt{13}-\\sqrt{7}\\), calculate \\(r-s=2\\sqrt{7}\\) and \\(r+s=2\\sqrt{13}\\). Therefore \\(r^2-s^2=(2\\sqrt{7})(2\\sqrt{13})=4\\sqrt{91}\\). Option A is correct. Option B, 20, would result from a different expression and is not obtained here. Option C, \\(2\\sqrt{91}\\), misses a factor of 2. Option D, 91, incorrectly removes the square root and the factor 4. The safest method is to factor first rather than expand both squares.
Correct answer: A. (\sqrt{18}+\sqrt{10})
Explanation: To rationalise a denominator containing \(\sqrt{18}-\sqrt{10}\), multiply the fraction by the conjugate \(\sqrt{18}+\sqrt{10}\) in both numerator and denominator. This preserves the value of the fraction while changing the denominator into a difference of squares.
The denominator becomes \((\sqrt{18}-\sqrt{10})(\sqrt{18}+\sqrt{10})=18-10=8\). Thus the expression becomes \(\frac{8(\sqrt{18}+\sqrt{10})}{8}=\sqrt{18}+\sqrt{10}\). The rationalised result is therefore the sum of the two square roots, corresponding to option A as supplied. The essential step is using the conjugate and the identity \((a-b)(a+b)=a^2-b^2\).
Correct answer: B. (9\sqrt{3})
Explanation: (\sqrt{363}=11\sqrt{3}), (\sqrt{108}=6\sqrt{3}), and (\sqrt{192}=8\sqrt{3}). Therefore (P=9\sqrt{3}).
Correct answer: A. (275)
Explanation: (\sqrt{44}=2\sqrt{11}) and (\sqrt{99}=3\sqrt{11}), so the sum is (5\sqrt{11}). Its square is (275).
Correct answer: D. \(\sqrt{12}+\sqrt{27}\)
Explanation: In D, \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so it becomes \(5\sqrt{3}\), which is irrational. A becomes \(\sqrt{36}=6\). Exam tip: extract perfect-square factors first.
Correct answer: A. The number of zeros between successive 1s keeps increasing; therefore, the decimal is non-recurring.
Explanation: Option A is correct. The 1s occur at positions 1, 3, 6, 10, …, so the zero-gaps 1, 2, 3, … keep growing and no fixed block repeats. B is false because using two digits does not ensure repetition. Exam tip: rational decimals terminate or repeat.
Correct answer: B. (\frac{\sqrt{29}-2}{5})
Explanation: For a denominator of the form \(\sqrt{29}+2\), use its conjugate \(\sqrt{29}-2\). Multiplying by the conjugate is useful because the middle terms cancel. The denominator then becomes a rational number, so no square root remains below the fraction line.
Multiply numerator and denominator by \(\sqrt{29}-2\). The denominator is \((\sqrt{29}+2)(\sqrt{29}-2)=29-4=25\). The result is \(\frac{5(\sqrt{29}-2)}{25}=\frac{\sqrt{29}-2}{5}\). This matches option B. Options that leave an irrational denominator or fail to include the correct factor do not give the simplified rationalised form.
Correct answer: A. The claim is correct; if \(r+\sqrt{7}\) were rational, subtracting the rational number \(r\) would make \(\sqrt{7}\) rational.
Explanation: The claim is correct. If \(r+\sqrt{7}\) were rational, then \(\sqrt{7}=(r+\sqrt{7})-r\) would also be rational, a contradiction. Exam tip: rational ± irrational is always irrational.
