Number System Irrational Number Most Asked Questions HP Board Class 9th
Question 1. Prove that √2 is an irrational number.
Solution:
Assume that √2 is rational.
Then √2 = p/q, where p and q are coprime integers and q ≠ 0.
Squaring both sides:
2 = p²/q²
p² = 2q²
Therefore, p² is even, so p is even.
Let p = 2k, where k is an integer.
Substituting:
(2k)² = 2q²
4k² = 2q²
q² = 2k²
Therefore, q is also even.
This contradicts the fact that p and q are coprime.
Hence, √2 is irrational.
Question 2. Prove that √3 is an irrational number.
Solution:
Assume that √3 is rational.
Then √3 = p/q, where p and q are coprime integers and q ≠ 0.
Squaring both sides:
3 = p²/q²
p² = 3q²
Therefore, p² is divisible by 3, so p is divisible by 3.
Let p = 3k.
Substituting:
9k² = 3q²
q² = 3k²
Therefore, q is also divisible by 3.
This contradicts the assumption that p and q are coprime.
Hence, √3 is irrational.
Question 3. Prove that √5 is an irrational number.
Solution:
Assume that √5 is rational.
Then √5 = p/q, where p and q are coprime integers and q ≠ 0.
Squaring both sides:
5 = p²/q²
p² = 5q²
Therefore, p² is divisible by 5, so p is divisible by 5.
Let p = 5k.
Substituting:
25k² = 5q²
q² = 5k²
Therefore, q is also divisible by 5.
This contradicts the assumption that p and q are coprime.
Hence, √5 is irrational.
Question 4. Prove that 2 + √3 is irrational.
Solution:
Assume that 2 + √3 is rational.
Let 2 + √3 = r, where r is rational.
Then:
√3 = r − 2
Since r and 2 are rational, r − 2 must also be rational.
This means √3 is rational, which contradicts the fact that √3 is irrational.
Hence, 2 + √3 is irrational.
Question 5. Prove that 5 − √2 is irrational.
Solution:
Assume that 5 − √2 is rational.
Let 5 − √2 = r, where r is rational.
Then:
√2 = 5 − r
Since 5 and r are rational, 5 − r must be rational.
This contradicts the fact that √2 is irrational.
Hence, 5 − √2 is irrational.
Question 6. Prove that 3√2 is irrational.
Solution:
Assume that 3√2 is rational.
Let 3√2 = r, where r is rational.
Dividing both sides by 3:
√2 = r/3
Since r and 3 are rational, r/3 is rational.
This contradicts the fact that √2 is irrational.
Hence, 3√2 is irrational.
Question 7. Prove that √7 is irrational.
Solution:
Assume that √7 is rational.
Then √7 = p/q, where p and q are coprime integers and q ≠ 0.
Squaring both sides:
7 = p²/q²
p² = 7q²
Therefore, p² is divisible by 7, so p is divisible by 7.
Let p = 7k.
Substituting:
49k² = 7q²
q² = 7k²
Therefore, q is also divisible by 7.
This contradicts the fact that p and q are coprime.
Hence, √7 is irrational.
Question 8. Find three irrational numbers between 2 and 3.
Solution:
We know that:
2 = √4 and 3 = √9
The numbers 5, 6 and 7 lie between 4 and 9.
Taking their square roots:
√5, √6 and √7
Since:
4 < 5 < 9
4 < 6 < 9
4 < 7 < 9
Therefore:
2 < √5 < 3
2 < √6 < 3
2 < √7 < 3
Hence, three irrational numbers between 2 and 3 are √5, √6 and √7.
Question 9. Determine whether √16 is irrational or rational.
Solution:
We know that:
√16 = 4
Since 4 is an integer, it can be written as 4/1.
Therefore, √16 is a rational number.
Answer: √16 is rational, not irrational.
Question 10. Prove that √2 + √3 is irrational.
Solution:
Assume that √2 + √3 is rational.
Let:
√2 + √3 = r, where r is rational.
Squaring both sides:
(√2 + √3)² = r²
2 + 3 + 2√6 = r²
5 + 2√6 = r²
Therefore:
2√6 = r² − 5
√6 = (r² − 5)/2
Since r is rational, the right-hand side is rational.
This implies that √6 is rational, which contradicts the fact that √6 is irrational.
Hence, √2 + √3 is irrational.