REAL NUMBERS
Exercise: 1.2
1). Prove that √5 is an irrational number.
Solution: Let’s assume that √5 is a rational number
Therefore, there exists two integers r and s such that
√5 = r/s (s ≠ 0)
s√5 = r
squaring on both sides
(s√5)2 = r2
5s2 = r2 . . . . . . . . . . . . . . (I)
Thus r2 is divisible by 5 so r is also divisible by 5
Let r = 5k for some value of k
Substituting in eqn (I) we get,
5s2 = (5k)2
s2 = 5k2
it means s is divisible by 5.
Clearly r and s are not co-primes.
Our assumption √5 is a rational number is not correct
Hence, √5 is an irrational number.
2). Prove that 3 + 2√5 is an irrational number.
Solution: Let assume that 3 + 2√5 is a rational number
Therefore, there exists two integers r and s such that
3 + 2√5 = r/s (s ≠ 0)
2√5 = (r/s) -3
√5 =1/2 (r/s) -3
But 1/2 (r/s) -3 is a rational number
Therefore, √5 is also a rational number
But it contradicts the fact that √5 is an irrational number
Our assumption 3 + 2√5 is a rational number is not correct
Hence, 3 + 2√5 is an irrational number.
3). Prove that the following are irrational.
i). 1/√2
ii). 7√5
iii). 6 + √2
Answers:
i). 1/√2
Solution: Let assume 1/√2 is a rational number
Therefore, there exists two integers r and s such that
1/√2 = r/s (s ≠ 0)
√2 = r/s
Since r and s are integers
Therefore, √2 is also a rational number
But it contradicts the fact that √2 is an irrational number
Our assumption 1/√2 is a rational number is not correct
Hence, 1/√2 is an irrational number.
ii). 7√5
Solution: Let’s assume 7√5 is a rational number
Therefore, there exists two integers r and s such that
7√5 = r/s (s ≠ 0)
√5 = r/7s
Since r and s are integers
Therefore, √5 is also a rational number
But it contradicts the fact that √5 is an irrational number
Our assumption 7√5 is a rational number is not correct
Hence, 7√5 is an irrational number.
iii). 6 + √2
Solution: Let assume that 6 + √2 is a rational number
Therefore, there exists two integers r and s such that
6 + √2 = r/s (s ≠ 0)
√2 = (r/s) -3
But (r/s) -3 is a rational number
Therefore, √2 is also a rational number
But it contradicts the fact that √2 is an irrational number
Our assumption that 6 + √2 is a rational number is not correct
Hence, 6 + √2 is an irrational number.
