State which of the following are irrational numbers.
(i) 3 – √(7/25)
(ii) -2/3 + ∛2
(iii) 3/√3
(iv) -2/7 ∛5
(v) (2 – √3) (2 + √3)
(vi) (3 + √5)2
(vii) (2/5 √7)2
(viii) (3 – √6)2
Solution:
(i) 3 – √(7/25)
Let us simplify,
3 – √(7/25) = 3 – √7/√25
= 3 – √7/5
Hence, 3 – √7/5 is an irrational number.
(ii) -2/3 + ∛2
Let us simplify,
-2/3 + ∛2 = -2/3 + 21/3
Since, 2 is not a perfect cube.
Hence it is an irrational number.
(iii) 3/√3
Let us simplify,
By rationalizing, we get
3/√3 = 3√3 /(√3×√3)
= 3√3/3
= √3
Hence, 3/√3 is an irrational number.
(iv) -2/7 ∛5
Let us simplify,
-2/7 ∛5 = -2/7 (5)1/3
Since, 5 is not a perfect cube.
Hence it is an irrational number.
(v) (2 – √3) (2 + √3)
Let us simplify,
By using the formula,
(a + b) (a – b) = (a)2 (b)2
(2 – √3) (2 + √3) = (2)2 – (√3)2
= 4 – 3
= 1
Hence, it is a rational number.
(vi) (3 + √5)2
Let us simplify,
By using (a + b)2 = a2 + b2 + 2ab
(3 + √5)2 = 32 + (√5)2 + 2.3.√5
= 9 + 5 + 6√5
= 14 + 6√5
Hence, it is an irrational number.
(vii) (2/5 √7)2
Let us simplify,
(2/5 √7)2 = (2/5 √7) × (2/5 √7)
= 4/ 25 × 7
= 28/25
Hence it is a rational number.
(viii) (3 – √6)2
Let us simplify,
By using (a – b)2 = a2 + b2 – 2ab
(3 – √6)2 = 32 + (√6)2 – 2.3.√6
= 9 + 6 – 6√6
= 15 – 6√6
Hence it is an irrational number.
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