CBSE Class 9 Mathematics Number System Assignment Set 09

NUMBER SYSTEM

Q1. Express \(0.\overline{32}\) as a rational number.
Answer: Let \(x = 0.323232\dots\) ------- (1)
Since there are two repeating digits, multiply both sides of equation (1) by 100:
\(100x = 32.323232\dots\) ------- (2)
Subtract equation (1) from equation (2):
\(100x - x = 32.323232\dots - 0.323232\dots\)
\(99x = 32\)
\(x = \frac{32}{99}\)
In simple words: Convert a repeating decimal into a fraction by multiplying by a power of 10 and subtracting the original number.

Exam Tip: For purely repeating decimals with two digits under the bar, directly write the repeating part in the numerator and 99 in the denominator.

Q2. Express \(0.\overline{123}\) as a rational number.
Answer: Let \(x = 0.123123123\dots\) ------- (1)
Since there are three repeating digits, multiply both sides of equation (1) by 1000:
\(1000x = 123.123123\dots\) ------- (2)
Subtract equation (1) from equation (2):
\(1000x - x = 123.123123\dots - 0.123123\dots\)
\(999x = 123\)
\(x = \frac{123}{999} = \frac{41}{333}\)
In simple words: Multiply by 1000 because three digits repeat, then subtract to eliminate the decimal part.

Exam Tip: Always remember to simplify the final fraction to its lowest terms if possible.

Q3. Express \(12.\overline{34}\) in the form of \(\frac{p}{q}\).
Answer: Let \(x = 12.343434\dots\) ------- (1)
Multiply both sides by 100 as two digits are repeating:
\(100x = 1234.343434\dots\) ------- (2)
Subtract equation (1) from equation (2):
\(100x - x = 1234.343434\dots - 12.343434\dots\)
\(99x = 1222\)
\(x = \frac{1222}{99}\)
In simple words: Use the same algebraic method for mixed or whole number decimals with repeating decimal parts.

Exam Tip: Ensure the decimal parts align perfectly before subtracting the two equations.

Q4. Write \(0.1\overline{2}\) in the form of \(\frac{p}{q}\).
Answer: Let \(x = 0.12222\dots\) ------- (1)
Multiply both sides by 10 to shift the non-repeating digit past the decimal point:
\(10x = 1.22222\dots\) ------- (2)
Multiply equation (1) by 100 to shift one repeating period past the decimal point:
\(100x = 12.22222\dots\) ------- (3)
Subtract equation (2) from equation (3):
\(100x - 10x = 12.22222\dots - 1.22222\dots\)
\(90x = 11\)
\(x = \frac{11}{90}\)
In simple words: For mixed recurring decimals, multiply by powers of 10 to align the repeating parts so they cancel out on subtraction.

Exam Tip: The denominator has one 9 for the repeating digit and one 0 for the non-repeating decimal digit.

Q5. Express \(0.25\overline{62}\) in the form of \(\frac{p}{q}\).
Answer: Let \(x = 0.25626262\dots\) ------- (1)
Multiply both sides by 100 to bring the non-repeating digits before the decimal point:
\(100x = 25.626262\dots\) ------- (2)
Multiply equation (1) by 10000 to shift two more digits (the repeating block):
\(10000x = 2562.626262\dots\) ------- (3)
Subtract equation (2) from equation (3):
\(10000x - 100x = 2562.626262\dots - 25.626262\dots\)
\(9900x = 2537\)
\(x = \frac{2537}{9900}\)
In simple words: Shift the decimal point past both non-repeating and repeating digits to eliminate the infinite tail through subtraction.

Exam Tip: Count the number of non-repeating decimal digits to determine the number of zeros in the denominator (two 9s and two 0s).

Q6. Show that \(\sqrt{2}\) is not a rational number.
Answer: Let us assume to the contrary that \(\sqrt{2}\) is a rational number.
Then, there exist coprime integers \(a\) and \(b\) (\(b \neq 0\)) such that:
\(\sqrt{2} = \frac{a}{b}\)
Squaring both sides, we get:
\(2 = \frac{a^2}{b^2} \implies a^2 = 2b^2\) ------- (1)
This means that 2 divides \(a^2\). By theorem, 2 must also divide \(a\).
So, we can write \(a = 2c\) for some integer \(c\).
Substituting \(a = 2c\) into equation (1):
\((2c)^2 = 2b^2 \implies 4c^2 = 2b^2 \implies b^2 = 2c^2\)
This means that 2 divides \(b^2\), and therefore 2 divides \(b\).
Thus, both \(a\) and \(b\) have at least 2 as a common factor, which contradicts our assumption that \(a\) and \(b\) are coprime.
Hence, our assumption is false, and \(\sqrt{2}\) is an irrational number.
In simple words: Proof by contradiction shows that assuming \(\sqrt{2}\) is a fraction leads to a logical contradiction where numbers share a common factor.

Exam Tip: Clearly state the method of proof by contradiction and mention that \(a\) and \(b\) are coprime integers.

Q7. Prove that \(\sqrt{3}\) is an irrational number.
Answer: Assume to the contrary that \(\sqrt{3}\) is a rational number.
Then, there exist coprime integers \(a\) and \(b\) (\(b \neq 0\)) such that:
\(\sqrt{3} = \frac{a}{b}\)
Squaring both sides:
\(3 = \frac{a^2}{b^2} \implies a^2 = 3b^2\) ------- (1)
This implies that 3 divides \(a^2\), and hence 3 divides \(a\).
Let \(a = 3c\) for some integer \(c\).
Substituting \(a = 3c\) into equation (1):
\((3c)^2 = 3b^2 \implies 9c^2 = 3b^2 \implies b^2 = 3c^2\)
This implies that 3 divides \(b^2\), and hence 3 divides \(b\).
Thus, 3 is a common factor of both \(a\) and \(b\), which contradicts the fact that \(a\) and \(b\) are coprime.
Therefore, \(\sqrt{3}\) is an irrational number.
In simple words: Follow the standard proof template by replacing 2 with 3 to show that the assumption of rationality fails.

Exam Tip: Practice this proof format thoroughly as it frequently appears in board examinations with different prime numbers like \(\sqrt{2}\), \(\sqrt{3}\), or \(\sqrt{5}\).

Q8. Find three irrational numbers between 3 and 4. Can you find more?
Answer: Three irrational numbers between 3 and 4 are:
\(3.101001000100001\dots$
\(3.202002000200002\dots$
\(3.303003000300003\dots$
Yes, we can find infinite irrational numbers between any two distinct real numbers.
In simple words: Create non-terminating and non-recurring decimals lying between 3 and 4.

Exam Tip: Ensure that the pattern of zeros increases by one each time to guarantee that the decimal is non-recurring and non-terminating.

Q9. Find an irrational number between \(3.2220200200002\dots\) and \(3.2222020020002\dots\).
Answer: An irrational number between the given two numbers is:
\(3.222101001000100001\dots$
In simple words: Choose a non-terminating non-recurring decimal value that lies strictly between the given lower and upper bounds.

Exam Tip: Compare digits from left to right to find a suitable interval and insert a custom non-repeating tail.