Number System HPBOSE Class 9th Questions
Question 1
Prove that √5 is irrational.
Answer:
Assume that √5 is rational.
Then √5 = a/b, where a and b are coprime integers and b ≠ 0.
Squaring both sides:
5 = a²/b²
a² = 5b²
Therefore, 5 divides a², so 5 divides a.
Let a = 5k.
Substituting:
25k² = 5b²
b² = 5k²
Therefore, 5 divides b.
This contradicts the fact that a and b are coprime.
Hence, √5 is irrational.
Question 2
Prove that 3 + 2√2 is irrational.
Answer:
Assume that 3 + 2√2 is rational.
Let 3 + 2√2 = r, where r is rational.
Then:
2√2 = r − 3
√2 = (r − 3)/2
Since r is rational, (r − 3)/2 is rational.
This means √2 is rational, which is a contradiction.
Hence, 3 + 2√2 is irrational.
Question 3
Simplify: 1/(√7 − √5) + 1/(√7 + √5).
Answer:
= (√7 + √5)/((√7 − √5)(√7 + √5)) + (√7 − √5)/((√7 + √5)(√7 − √5))
= (√7 + √5)/2 + (√7 − √5)/2
= (2√7)/2
= √7
Question 4
Express 0.2363636... in the form p/q.
Answer:
Let x = 0.2363636...
1000x = 236.363636...
10x = 2.363636...
Subtracting:
990x = 234
x = 234/990
x = 13/55
Hence, 0.2363636... = 13/55.
Question 5
Determine whether the decimal expansion of 77/480 is terminating or non-terminating recurring.
Answer:
480 = 2⁵ × 3 × 5
The fraction 77/480 is in its simplest form.
A rational number has a terminating decimal expansion only when the denominator in its simplest form has prime factors 2 and/or 5.
Since the denominator contains the prime factor 3, the decimal expansion is non-terminating recurring.
Question 6
Simplify: (2³ × 2⁻²) × (3²/3³) ÷ 2⁻¹.
Answer:
= 2^(3 − 2) × 3^(2 − 3) ÷ 2⁻¹
= 2 × 1/3 ÷ 1/2
= 2/3 × 2
= 4/3
Question 7
Simplify: 3√12 + 2√27 − √75.
Answer:
= 3 × 2√3 + 2 × 3√3 − 5√3
= 6√3 + 6√3 − 5√3
= 7√3
Question 8
Rationalise the denominator: 1/(√5 + √3 + √2).
Answer:
Multiply the numerator and denominator by √5 + √3 − √2.
= (√5 + √3 − √2)/((√5 + √3)² − 2)
= (√5 + √3 − √2)/(6 + 2√15)
Multiply the numerator and denominator by 3 − √15.
= (√5 + √3 − √2)(3 − √15)/(18 − 30)
= (√30 − √5 − 3√3 − 2√2)/(-12)
= (√5 + 3√3 + 2√2 − √30)/12
Question 9
Find three irrational numbers between 2 and 3.
Answer:
Since 4 < 5 < 9, we get 2 < √5 < 3.
Since 4 < 6 < 9, we get 2 < √6 < 3.
Since 4 < 7 < 9, we get 2 < √7 < 3.
Hence, the three irrational numbers are √5, √6 and √7.
Question 10
If x = √3 + √2, find x² + 1/x².
Answer:
x = √3 + √2
x² = (√3 + √2)²
x² = 5 + 2√6
Since (√3 + √2)(√3 − √2) = 1, we have:
1/x = √3 − √2
Therefore:
1/x² = (√3 − √2)² = 5 − 2√6
x² + 1/x² = (5 + 2√6) + (5 − 2√6)
= 10