Explanation
Assume that √p+√q is rational. So,
√p+√q=ab
p+q+2√pq=a2b2
√pq=12(a2b2–p−q)
Since the RHS of the above equation is rational but √pq is an irrational number, so the assumption is wrong.
Therefore, it is true that √p+√q is irrational.
Step - 1: Solving A
Prime factorizing 8 and 40,
⇒ 8 = 2 × 2 × 2
⇒ 40 = 2 × 2 × 2 × 5
Multiplying common prime factors to get HCF,
H.C.F = 2 × 2 × 2 = 8
Step - 2: Solving B
Prime factorizing 15 and 25,
⇒ 15 = 3 × 5
⇒ 25 = 5 × 5
H.C.F = 5
Step - 3: Solving C
Prime factorizing 16 and 60,
⇒ 16 = 2 × 2 × 2 × 2
⇒ 60 = 2 × 2 × 3 × 5
H.C.F = 2 × 2 = 4
Step - 4: Solving D
Prime factorizing 14 and 35,
⇒ 14 = 2 × 7
⇒ 35 = 5 × 7
H.C.F = 7
Step - 5: Solving E
Prime factorizing 15, 25 and 60,
Step - 6: Solving F
Prime factorizing 16, 48 and 80,
⇒ 48 = 2 × 2 × 2 × 2 × 3
⇒ 80 = 2 × 2 × 2 × 2 × 5
H.C.F = 2 × 2 × 2 × 2 = 16
Hence, we have found H.C.F of different pair of numbers using prime factorization method
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