What is the rationalised form of (\frac{6}{\sqrt{19}-\sqrt{10}})?
Correct answer: C
Multiplying by the conjugate makes the denominator (19-10=9). So the form is (\frac{6(\sqrt{19}+\sqrt{10})}{9}).
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.
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Expert · Level 4
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- (\sqrt{19}+\sqrt{10})
- (\frac{2(\sqrt{19}+\sqrt{10})}{3})
- (\frac{6(\sqrt{19}+\sqrt{10})}{9})
- (6\sqrt{190})
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- 0.272727...
- \(\sqrt{2}\)
- \(\pi\)
- 0.101001000100001...
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- \(13-\sqrt{42}\)
- \(\sqrt{55}\)
- (169+42)
- \(13+\sqrt{42}\)
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- (29)
- (15)
- (\sqrt{154})
- (2\sqrt{22})
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- If \(x\) is irrational, then \(2x\) is irrational.
- If \(x\) is irrational, then \(x^2\) is irrational.
- If \(x\) is irrational, then \(x-x\) is irrational.
- If \(x\) is irrational, then \(x/x\) is irrational.
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- (2+\sqrt{3})
- (1)
- (2-\sqrt{3})
- (\sqrt{3}-2)
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- \(\sqrt{2}\times\sqrt{3}=\sqrt{6}\)
- \(\sqrt{2}\times\sqrt{8}=4\)
- \(\sqrt{5}\times\sqrt{2}=\sqrt{10}\)
- \(\sqrt{7}\times\sqrt{3}=\sqrt{21}\)
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- \(2\sqrt{7}\)
- (14)
- (28)
- \(\sqrt{112}\)
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- The sum of any irrational number and any rational number is irrational.
- The sum of two irrational numbers is always irrational.
- The product of two irrational numbers is always irrational.
- The quotient of two irrational numbers, when the divisor is non-zero, is always irrational.
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- \(a+b\) is irrational
- \(a+b\) is rational
- \(ab\) is rational
- \(\frac{b}{a}\) is rational
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- (\sqrt{17.5})
- (5)
- (\sqrt{19})
- (\sqrt{21})
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- \(p+x\) is always irrational
- The sum of any two irrational numbers is always irrational
- The square of every irrational number is always irrational
- The product of any two irrational numbers is always irrational
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- \(r+x\) is irrational
- \(x^2\) is always irrational
- \(rx\) is always irrational
- \(r-x\) is rational
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- (8\sqrt{3})
- (10\sqrt{3})
- (12\sqrt{3})
- (\sqrt{288})
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- (\sqrt{7})
- (\sqrt{5})
- (2\sqrt{7})
- (2)
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- \(r+q\)
- \(r\times q\)
- \(r-r\)
- \(r/r\)
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- (12)
- (14)
- (16)
- (18)
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- (4\sqrt{91})
- (20)
- (2\sqrt{91})
- (91)
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- (\sqrt{18}+\sqrt{10})
- (8(\sqrt{18}+\sqrt{10}))
- (\frac{\sqrt{18}+\sqrt{10}}{8})
- (8\sqrt{180})
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- (7\sqrt{3})
- (9\sqrt{3})
- (11\sqrt{3})
- (13\sqrt{3})
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- (275)
- (143)
- (25\sqrt{11})
- (550)
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- \(\sqrt{12}\times\sqrt{3}\)
- \(\frac{\sqrt{50}}{\sqrt{2}}\)
- \(\sqrt{81}-\sqrt{16}\)
- \(\sqrt{12}+\sqrt{27}\)
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- The number of zeros between successive 1s keeps increasing; therefore, the decimal is non-recurring.
- Having only two digits makes a decimal recurring.
- Every non-terminating decimal is rational.
- It is sufficient to treat the initial block 101 as the repeating block.
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- (5(\sqrt{29}-2))
- (\frac{\sqrt{29}-2}{5})
- (\sqrt{29}+2)
- (5\sqrt{29})
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- The claim is correct; if \(r+\sqrt{7}\) were rational, subtracting the rational number \(r\) would make \(\sqrt{7}\) rational.
- The claim is correct because every sum containing a square root is always irrational.
- The claim is incorrect because the sum of a rational and an irrational number can be rational.
- The claim is incorrect because \(\sqrt{7}\) is irrational only because its decimal expansion is infinite.
A student claims that every non-terminating decimal is irrational. Which of the following examples disproves the claim?
Correct answer: A
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.
What is the square of \(\sqrt{13+\sqrt{42}}\) equal to?
Correct answer: D
The square of a square root gives the number inside. So \(\left(\sqrt{13+\sqrt{42}}\right)^2=13+\sqrt{42}\).
If a rectangle has length (\sqrt{22}+\sqrt{7}) and breadth (\sqrt{22}-\sqrt{7}), what will be its area?
Correct answer: B
Area is ((\sqrt{22}+\sqrt{7})(\sqrt{22}-\sqrt{7})=22-7=15). Conjugate dimensions can give a rational area.
Let \(x\) be an irrational number. Which of the following statements is always true?
Correct answer: A
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.
Which is the simplified form of (\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}})?
Correct answer: C
Multiplying by the conjugate gives (\frac{(\sqrt{6}-\sqrt{2})^2}{4}=2-\sqrt{3}). Make the denominator rational.
A student claims that the product of two irrational numbers is always irrational. Which of the following examples disproves this claim?
Correct answer: B
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.
What is the value of \(\sqrt{7}\left(\sqrt{175}-\sqrt{63}\right)\)?
Correct answer: B
\(\sqrt{175}=5\sqrt{7}\) and \(\sqrt{63}=3\sqrt{7}\), so the bracket is \(2\sqrt{7}\). Multiplying by \(\sqrt{7}\) gives (14).
Which of the following statements is necessarily true?
Correct answer: A
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.
Let \(a\) be a non-zero rational number and \(b\) be an irrational number. Which of the following statements is always true?
Correct answer: A
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.
Which number definitely lies between (\sqrt{18}) and (\sqrt{20})?
Correct answer: C
Since (18<19<20), (\sqrt{19}) lies between them. Compare square roots using the numbers inside.
Let \(p\) be a rational number and \(x\) be an irrational number. Which of the following statements is always true?
Correct answer: A
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.
If \(r\) is a rational number and \(x\) is an irrational number, which of the following statements is always true?
Correct answer: A
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.
What is the simplified form of (\sqrt{432}-\sqrt{192}+\sqrt{48})?
Correct answer: A
(\sqrt{432}=12\sqrt{3}), (\sqrt{192}=8\sqrt{3}), and (\sqrt{48}=4\sqrt{3}). Therefore the result is (8\sqrt{3}).
What is the value of (\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{7}-\sqrt{5}})?
Correct answer: A
After rationalising the terms become (\frac{\sqrt{7}-\sqrt{5}}{2}) and (\frac{\sqrt{7}+\sqrt{5}}{2}). The sum is (\sqrt{7}).
If r is an irrational number and q is a rational number, which of the following expressions will always be irrational?
Correct answer: A
\(r+q\) 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.
What is the value of (\frac{\sqrt{200}+\sqrt{72}}{\sqrt{2}})?
Correct answer: C
(\sqrt{200}=10\sqrt{2}) and (\sqrt{72}=6\sqrt{2}), so the numerator is (16\sqrt{2}). Dividing gives (16).
If (r=\sqrt{13}+\sqrt{7}) and (s=\sqrt{13}-\sqrt{7}), what is the value of (r^2-s^2)?
Correct answer: A
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.
Which is the rationalised form of (\frac{8}{\sqrt{18}-\sqrt{10}})?
Correct answer: A
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 \((a-b)(a+b)=a^2-b^2\).
If (P=\sqrt{363}+\sqrt{108}-\sqrt{192}), what is (P) equal to?
Correct answer: B
(\sqrt{363}=11\sqrt{3}), (\sqrt{108}=6\sqrt{3}), and (\sqrt{192}=8\sqrt{3}). Therefore (P=9\sqrt{3}).
What is the value of ((\sqrt{44}+\sqrt{99})^2)?
Correct answer: A
(\sqrt{44}=2\sqrt{11}) and (\sqrt{99}=3\sqrt{11}), so the sum is (5\sqrt{11}). Its square is (275).
Which of the following expressions represents an irrational number?
Correct answer: D
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.
Riya claims that \(0.101001000100001\ldots\) is rational because only the digits 0 and 1 occur in it. Which is the correct refutation of her claim?
Correct answer: A
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.
What is the rationalised form of (\frac{5}{\sqrt{29}+2})?
Correct answer: B
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.
Reema claims that for any rational number \(r\), \(r+\sqrt{7}\) will be irrational. Which of the following arguments correctly evaluates her claim?
Correct answer: A
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.
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