Prove that the following numbers are Irrational:-
(i) 1/√3
(ii) 1/√5
(iii) 1/√7
(iv) 1/√11
(v) √3 - √2
(vi) √3 + √5
(vii) √3 - √5
Answers
Answer:
(i) 1/√3
= Let: Assume that 1/√3 is a Rational Number
As the definition of the Rational Numbers it is expressed in the p/q form, where p,q are Co-primes and q≠0.
∴ 1/√3 is Rational
= p/q = 1/√3
= q = 1/√3
∴ By Squaring on both sides,
= q² = (p√3)²
= q² = 3p² - (1)
∴ q is divisible by 3
⇒ Again, p² = 3q²
p² is a multiple of 3
p and q aren't being Co-primes
Hence, our assumption 1/√3 is Rational is False/Wrong.
So, 1/√3 is Irrational.
(ii) 1/√5
= Let: Assume that 1/√5 is a Rational Number
As the definition of the Rational Numbers it is expressed in the p/q form, where p,q are Co-primes and q≠0.
∴ 1/√5 is Rational
= p/q = 1/√5
= q = 1/√5
∴ By Squaring on both sides,
= q² = (p√5)²
= q² = 5p² - (1)
∴ q is divisible by 5
⇒ Again, p² = 5q²
p² is a multiple of 5
p and q aren't being Co-primes
Hence, our assumption 1/√5 is Rational is False/Wrong.
So, 1/√5 is Irrational.
(iii) 1/√7
= Let: Assume that 1/√7 is a Rational Number
As the definition of the Rational Numbers it is expressed in the p/q form, where p,q are Co-primes and q≠0.
∴ 1/√7 is Rational
= p/q = 1/√7
= q = 1/√7
∴ By Squaring on both sides,
= q² = (p√7)²
= q² = 7p² - (1)
∴ q is divisible by 7
⇒ Again, p² = 7q²
p² is a multiple of 7
p and q aren't being Co-primes
Hence, our assumption 1/√7 is Rational is False/Wrong.
So, 1/√7 is Irrational.
(iv) 1/√11
= Let: Assume that 1/√11 is a Rational Number
As the definition of the Rational Numbers it is expressed in the p/q form, where p,q are Co-primes and q≠0.
∴ 1/√11 is Rational
= p/q = 1/√11
= q = 1/√11
∴ By Squaring on both sides,
= q² = (p√11)²
= q² = 11p² - (1)
∴ q is divisible by 11
⇒ Again, p² = 11q²
p² is a multiple of 11
p and q aren't being Co-primes
Hence, our assumption 1/√11 is Rational is False/Wrong.
So, 1/√11 is Irrational.
(v) √3 - √2
Let: √3 - √2 = p/q, where p,q are Co=primes and q≠0.
∴ √3 = p/q + √2
∴ By Squaring on Both Sides,
= 1(√3)² = (p/q + √2)²
= 3 = p²/q² + 2 . p/q √2 [∵(a + b)² = a² + b² + 2ab]
= 2p/q √2 = 3 - 2 - p²/q²
= 2p √2 = 1 - p²/q²
= 2p/q √2 = q² - p²/q²
= √2 = (q² - p²/q²)
= √2 = (q² - p²√/q²) x (q/2p)
= √2 = (q² - p²/2pq)
= √2 is Rational.
∵ p,q ∈ ; p² + q²/2pq is in the form p/q which denores that it is Rational number wich can't be equal to √2 which is Irrational.
Hence, our assumption that √3 - √2 Rational is False.
∴ √3 - √2 is Irrational.
(vi) √3 + √5
Let: √3 + √5 = p/q, where p,q are Co=primes and q≠0.
∴ √3 = p/q + √5
∴ By Squaring on Both Sides,
= (√3)² = (p/q - √5)²
= 3 = p²/q² + 5 + 5 - 2 p/q √5 [∵(a + b)² = a² + b² - 2ab]
= 2p/q √5 = p²/q² + 5 - 3
= 2p/q √5 = 2 + p²/q²
= 2p/q √5 = p²/2q²/q²
= √5 = (q² - p²/q²)
= √5 = (p² + 2q²) x (q/2p)
= √5 = p² + 2q²/2pq
= √5 is Rational.
∵ p,q ∈ ; p² + q²/2pq is in the form p/q which denores that it is Rational number wich can't be equal to √5 which is Irrational.
Hence, our assumption that √3 + √5 Rational is False.
∴ √3 + √5 is Irrational.
(vii) √3 - √5
Let: √3 - √5 = p/q, where p,q are Co=primes and q≠0.
∴ √3 = p/q + √5
∴ By Squaring on Both Sides,
= (√3)² = (p/q + √5)²
= 3 = p²/q² 5 - 2 . p/q √5 [∵(a + b)² = a² + b² - 2ab]
= 2p/q √3 = 3 - 5 - p²/q²
= 2p/q √3 = 2 - p²/q²
= 2p/q √3 = (2q² - p²/q²)
= √3 = (2q² - p²√/q²) x (q/2p)
= √3 = (2q² - p²/2pq)
= √3 is Rational.
∵ p,q ∈ ; p² + q²/2pq is in the form p/q which denores that it is Rational number wich can't be equal to √3 which is Irrational.
Hence, our assumption that √3 - √5 Rational is False.
∴ √3 - √5 is Irrational.