If a rectangle has length (4+\sqrt{5}) and breadth (4-\sqrt{5}), what will be its area?
Correct answer: A
Area is ((4+\sqrt{5})(4-\sqrt{5})=16-5=11). Multiplying conjugate dimensions can give a rational area.
Mathematics
Irrational numbers
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.
TOPIC PRACTICE
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Up to 25 questions from this page. Select your focus, then start.
25 questions
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Hard · Level 4
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- (11)
- (16)
- (21)
- (8\sqrt{5})
Hard · Level 4
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- (7\sqrt{3})
- (5\sqrt{3})
- (9\sqrt{3})
- (\sqrt{183})
Hard · Level 4
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- (\sqrt{5})
- (\sqrt{3})
- (2\sqrt{5})
- (4)
Hard · Level 4
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- \((\sqrt{2},-\sqrt{2})\)
- \((\sqrt{2},\sqrt{8})\)
- \((\sqrt{3},2\sqrt{3})\)
- \((\sqrt{5},\sqrt{20})\)
Hard · Level 4
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- \(\sqrt{2},\ 3-\sqrt{2}\)
- \(\sqrt{2},\ 3+\sqrt{2}\)
- \(\sqrt{2},\ \frac{3}{\sqrt{2}}\)
- \(\sqrt{2},\ \sqrt{3}\)
Hard · Level 4
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- Dividing an irrational number by a non-zero rational number gives an irrational result
- Since 5 is a prime number, the fraction must always be irrational
- A fraction with a whole number in the denominator is always rational
- Dividing \(\sqrt{2}\) by 5 makes it a whole number
Hard · Level 4
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- (\sqrt{11})
- (4)
- (\sqrt{9})
- (\sqrt{13})
Hard · Level 4
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- \(3\)
- \(\sqrt{3}\)
- \(9\)
- \(\frac{1}{3}\)
Hard · Level 4
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- (3)
- (5)
- (\sqrt{5})
- (1)
Hard · Level 4
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- \(7+\sqrt{10}\)
- (49+10)
- \(\sqrt{17}\)
- \(7-\sqrt{10}\)
Hard · Level 4
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- \(p+q\) is always irrational
- \(pq\) is always rational
- \(p-q\) is always rational
- \(p/q\) is always rational
Hard · Level 4
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- \(\sqrt{19}-\sqrt{19}\)
- \(\sqrt{2}-\sqrt{3}\)
- \(\sqrt{5}-\sqrt{20}\)
- \(\sqrt{7}-\sqrt{28}\)
Hard · Level 4
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- योग सदैव अपरिमेय होता है।
- योग सदैव परिमेय होता है।
- योग परिमेय या अपरिमेय, दोनों हो सकता है।
- योग सदैव एक पूर्णांक होता है।
Hard · Level 4
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- (5+\sqrt{21})
- (\frac{5+\sqrt{21}}{2})
- (10+2\sqrt{21})
- (\frac{5-\sqrt{21}}{2})
Hard · Level 4
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- If \(5-\sqrt{7}\) were rational, then \(\sqrt{7}=5-(5-\sqrt{7})\) would also be rational, which is impossible.
- Since \(5\) is rational, subtracting any number from it always gives a rational result.
- The decimal expansion of \(\sqrt{7}\) is infinite, so \(5-\sqrt{7}\) must be an integer.
- \(5-\sqrt{7}\) is rational because both \(5\) and \(7\) are integers.
Hard · Level 4
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- (3\sqrt{5})
- (5\sqrt{5})
- (7\sqrt{5})
- (9\sqrt{5})
Hard · Level 4
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- (128)
- (64)
- (16\sqrt{2})
- (200)
Hard · Level 4
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- (\sqrt{11}+\sqrt{7})
- (\frac{\sqrt{11}-\sqrt{7}}{2})
- (\sqrt{11}-\sqrt{7})
- (2\sqrt{77})
Hard · Level 4
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- (12)
- (15)
- (17)
- (19)
Hard · Level 4
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- (\frac{\sqrt{75}}{\sqrt{3}})
- (\frac{\sqrt{125}}{\sqrt{5}})
- (\frac{\sqrt{45}}{\sqrt{9}})
- (\frac{\sqrt{108}}{\sqrt{3}})
Hard · Level 4
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- \(5\sqrt{2}\) cm; it is irrational
- \(5\sqrt{2}\) cm; it is rational
- \(\sqrt{25}\) cm; it is rational
- \(25\sqrt{2}\) cm; it is irrational
Hard · Level 4
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- \(\sqrt{2},\,-\sqrt{2}\)
- \(\sqrt{2},\,2\sqrt{2}\)
- \(\sqrt{3},\,\sqrt{12}\)
- \(\sqrt{5},\,3\sqrt{5}\)
Hard · Level 4
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- (\frac{23+8\sqrt{7}}{9})
- (\frac{23-8\sqrt{7}}{9})
- (9+8\sqrt{7})
- (4+\sqrt{7})
Hard · Level 4
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- \(0.\overline{3}\)
- \(\sqrt{2}\)
- \(\pi\)
- \(0.101001000100001\ldots\)
Hard · Level 4
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- (5\sqrt{3})
- (7\sqrt{3})
- (9\sqrt{3})
- (3\sqrt{3})
What is the simplified form of (\sqrt{300}-\sqrt{192}+\sqrt{75})?
Correct answer: A
(\sqrt{300}=10\sqrt{3}), (\sqrt{192}=8\sqrt{3}), and (\sqrt{75}=5\sqrt{3}). Therefore the result is (7\sqrt{3}).
What is the value of (\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{5}-\sqrt{3}})?
Correct answer: A
The first term becomes (\frac{\sqrt{5}-\sqrt{3}}{2}) and the second becomes (\frac{\sqrt{5}+\sqrt{3}}{2}). Their sum is (\sqrt{5}).
Reena claims, “The sum of two irrational numbers is always irrational.” Which of the following pairs disproves her claim?
Correct answer: A
Both \(\sqrt{2}\) and \(-\sqrt{2}\) are irrational, but \(\sqrt{2}+(-\sqrt{2})=0\), which is rational. Hence Reena’s “always” claim is false. Exam tip: one valid counterexample is enough to disprove an “always” statement.
In which option are both numbers irrational, but their sum is rational?
Correct answer: A
\(\sqrt{2}\) is irrational, and \(3-\sqrt{2}\) is also irrational; otherwise subtracting it from 3 would make \(\sqrt{2}\) rational. Their sum is \(3\), which is rational. Exam tip: check the sum first in such pairs.
Ravi claims that \(x=\frac{3+\sqrt{2}}{5}\) is rational because its denominator is 5. What is the error in Ravi’s claim?
Correct answer: A
\(3+\sqrt{2}\) is irrational. Dividing it by the non-zero rational number 5 keeps the result irrational. Exam tip: if \(x\) were rational, then \(5x-3=\sqrt{2}\) would be rational, which is a contradiction.
Which number definitely lies between (\sqrt{10}) and (\sqrt{12})?
Correct answer: A
Since (10<11<12), (\sqrt{11}) lies between them. For positive square roots compare the numbers inside.
A student claims that \(\frac{\sqrt{18}}{\sqrt{2}}\) must be irrational because both the numerator and denominator are irrational. What is the correct conclusion?
Correct answer: A
\(\frac{\sqrt{18}}{\sqrt{2}}=\sqrt{\frac{18}{2}}=\sqrt{9}=3\), which is rational. The quotient of two irrational numbers need not be irrational. Exam tip: combine the radicals before deciding the number type.
What is the value of (\frac{\sqrt{125}-\sqrt{20}}{\sqrt{5}})?
Correct answer: A
(\sqrt{125}=5\sqrt{5}) and (\sqrt{20}=2\sqrt{5}), so the numerator is (3\sqrt{5}). Dividing gives (3).
What is the square of \(\sqrt{7+\sqrt{10}}\) equal to?
Correct answer: A
The square of a square root gives the number inside. So \(\left(\sqrt{7+\sqrt{10}}\right)^2=7+\sqrt{10}\).
If \(p\) is an irrational number and \(q\) is a non-zero rational number, which of the following statements is always true?
Correct answer: A
If \(p+q\) were rational, then \(p=(p+q)-q\) would be the difference of two rational numbers and hence rational, a contradiction. Thus \(p+q\) is irrational. Exam tip: adding or subtracting a rational number from an irrational number remains irrational.
In which option is the difference of two irrational numbers rational?
Correct answer: A
In option A, \(\sqrt{19}\) and \(\sqrt{19}\) are both irrational, and their difference is \(\sqrt{19}-\sqrt{19}=0\). Zero is a rational number. In option B, \(\sqrt{2}-\sqrt{3}\) is irrational; option C simplifies to \(\sqrt{5}-\sqrt{20}=-\sqrt{5}\), and option D to \(\sqrt{7}-\sqrt{28}=-\sqrt{7}\), both irrational. Exam tip: the difference of identical irrational numbers is zero, which is rational.
Which of the following statements is always true about the sum of two irrational numbers?
Correct answer: C
The sum of two irrational numbers need not have a fixed type. For example, \(\sqrt{2}+(-\sqrt{2})=0\) is rational, whereas \(\sqrt{2}+\sqrt{3}\) is irrational. Hence, the sum can be rational or irrational. In exams, test claims using “always” with a counterexample.
Which is the simplified form of (\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}})?
Correct answer: B
Multiplying by the conjugate gives numerator (10+2\sqrt{21}) and denominator (4). So the simplified form is (\frac{5+\sqrt{21}}{2}).
A student claims that \(5-\sqrt{7}\) is a rational number. Which argument correctly proves that the claim is wrong?
Correct answer: A
\(\sqrt{7}\) is irrational. If \(5-\sqrt{7}\) were rational, subtracting it from 5 would make \(\sqrt{7}\) rational, a contradiction. Exam tip: rational ± irrational is always irrational.
What is the simplified form of (\sqrt{245}-\sqrt{180}+\sqrt{80})?
Correct answer: C
(\sqrt{245}=7\sqrt{5}), (\sqrt{180}=6\sqrt{5}), and (\sqrt{80}=4\sqrt{5}). So the result is (7\sqrt{5}-6\sqrt{5}+4\sqrt{5}=5\sqrt{5}).
What is the value of ((\sqrt{18}+\sqrt{50})^2)?
Correct answer: A
The direct answer is A, 128. First simplify each radical by taking out the largest perfect-square factor: √18 = √(9×2) = 3√2 and √50 = √(25×2) = 5√2. Therefore their sum is 8√2. Squaring gives (8√2)^2 = 64×2 = 128. Option A is correct because it equals this result. Option B, 64, forgets the factor 2 produced by squaring √2. Option C, 16√2, is not the square of the sum; it is an unsimplified or incorrect form. Option D, 200, comes from an incorrect expansion or multiplication. A useful check is to use (a+b)^2: 18+50+2√(18×50) = 68+2√900 = 68+60 = 128. Memory cue: simplify radicals first, then square the common radical carefully.
What is the rationalised form of (\frac{2}{\sqrt{11}+\sqrt{7}})?
Correct answer: B
Multiplying by the conjugate makes the denominator (11-7=4). So (\frac{2(\sqrt{11}-\sqrt{7})}{4}=\frac{\sqrt{11}-\sqrt{7}}{2}).
If (s=\sqrt{98}+\sqrt{200}), what is the value of (\frac{s}{\sqrt{2}})?
Correct answer: C
(\sqrt{98}=7\sqrt{2}) and (\sqrt{200}=10\sqrt{2}), so (s=17\sqrt{2}). Dividing by (\sqrt{2}) gives (17).
Which option will remain irrational?
Correct answer: C
The first, second, and fourth options give (5), (5), and (6) respectively. The third gives (\sqrt{5}), which is irrational.
Which conclusion about the side of a square tile with an area of 50 cm² is correct?
Correct answer: A
Side = \(\sqrt{50}=\sqrt{25\times2}=5\sqrt2\) cm. Since \(\sqrt2\) is irrational, multiplying it by non-zero rational 5 keeps it irrational. Exam tip: take the square root of the area.
A student claims, “The sum of two irrational numbers is always irrational.” Which of the following pairs disproves this claim?
Correct answer: A
\(\sqrt{2}+(-\sqrt{2})=0\), and 0 is rational. Hence, the sum of two irrational numbers need not be irrational. In option B, the sum is \(3\sqrt{2}\), which is irrational. Exam tip: one counterexample is enough to disprove an “always” statement.
Which is the simplified form of (\frac{4+\sqrt{7}}{4-\sqrt{7}})?
Correct answer: A
Multiplying by the conjugate gives denominator (16-7=9) and numerator ((4+\sqrt{7})^2=23+8\sqrt{7}). So the correct form is (\frac{23+8\sqrt{7}}{9}).
A student says that every non-terminating decimal number is irrational. Which of the following numbers shows the error in this statement?
Correct answer: A
\(0.\overline{3}=1/3\), so it is non-terminating but repeating and hence rational. In contrast, \(\sqrt{2}\) and \(\pi\) are irrational. Exam tip: a non-terminating repeating decimal is always rational.
What is (\sqrt{147}-\sqrt{75}+\sqrt{27}) equal to?
Correct answer: A
(\sqrt{147}=7\sqrt{3}), (\sqrt{75}=5\sqrt{3}), and (\sqrt{27}=3\sqrt{3}). Therefore the result is (5\sqrt{3}).
QUIZ COMPLETE