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Answers
Solution 1. Let a be any positive integer which may be even or odd. Then due to Eulcid's Lemma we know that,
⇒ a = 2q + r where 0 ≤ r < 2
Possible values of r = 0 and 1
Therefore,
⇒ a = 2q + (0) and a = 2q + (1)
⇒ a = 2q (for even) and a = 2q + 1 (for odd)
Q.E.D
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Solution 2. Let any positive odd integer be a. By Euclid's Lemma we know that,
a = 4q + r where 0 ≤ r < 4
So possible values of r = 0, 1, 2 and 3
Notong that here only 1 and 3 are odd integers.
Therefore a = 4q + 1 and a = 4q + 3
Q.E.D
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Solution 3. Let x be the final answer of 4ⁿ then,
x = 4ⁿ = (2²)ⁿ = 2²ⁿ
This means that, x has no factor as like 5. For an integer to end with "0", the integer should have 5 as a factor.
Therefore, 4ⁿ can never end with 0.
Q.E.D
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Solution 4. L.C.M and H.C.F by prime factorisation method
L.C.M. :
2 | 12, 18
2 | 6, 9
3 | 3, 9
3 | 1, 3
L.C.M = 2 × 2 × 3 × 3 = 36
H.C.F :
12 = 2² × 3
18 = 2 × 3²
H.C.F = 2 × 3 = 6
Answer : 36 (LCM) and 6 (HCF)
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Solution 5. (a) 16/135 = 16/(5 × 3³)
Thus dinominator is not in form of 2ᵃ × 5ᵇ therefore 16/135 is non terminating.
(b) 25/32 = 25/2⁵ = (25 × 5⁵)/(2⁵ × 5⁵)
Since dinominator is in form of 2ᵃ × 5ᵇ therefore 25/32 is terminating.
(c) 100/81 = 100/3⁴
Thus dinominator is not in form of 2ᵃ × 5ᵇ therefore 16/135 is non terminating.
(d) 41/75 = 41/(5² × 3)
Thus dinominator is not in form of 2ᵃ × 5ᵇ therefore 16/135 is non terminating.
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Solutiom 6. (a) 36/50 = 35/(2 × 5²)
= (35 × 2)/(2² × 5²) = 70/(2 × 5)² = 0.7
(b) 31/25 = 31/5² = (31 × 2²)/(5² × 2²)
= 1.24
(c) 7/8 = 7/2³ = (7 × 5³)/(2 × 5³)
= (7 × 125)/1000 = 0.875
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Solution 7. Let √2 be a rational number p/q in simplest form,
⇒ √2 = p/q
⇒ 2 = p²/q²
⇒ 2q² = p²
Thherefore, 2 | p
Now, let any natural mumber m be factor of p
⇒ 2m = p
⇒ 2²m² = p²
⇒ 4m² = 2q²
⇒ 2m² = q²
Therefore, 2 | q
This contradicts our theory that p/q is in simplest form because they have comman factor 2. Therefore √2 is irrational.
Q.E.D
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Solution 8. Let x be 5 - √3 then,
⇒ x = 5 - √3
⇒ x² = (5 - √3)²
⇒ x² = 25 + 3 - 10√3
⇒ (x² - 28)/(-10) = √3
There √3 is rational number since it is in form of p/q. But it is a fact that √3 is irrational number. Therefore our assumotion is wrong and 5 - √3 is irrational.
Q.E.D
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Solution 9. Let x be 3√2, then
⇒x = 3√2
⇒ x/3 = √2
There √2 is rational number since it is in form of p/q. But it is a fact that √2 is irrational number. Therefore our assumotion is wrong and 3√2 is irrational.
Q.E.D
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Solution 10. Let x be √2 + √3
⇒ x = √2 + √3
⇒ x² = (√2 + √3)²
⇒ x² = 2 + 3 + 2√6
⇒ (x² - 5)/2 = √6
There √6 is rational number since it is in form of p/q. But it is a fact that √6 is irrational number. Therefore our assumotion is wrong and √2 + √3 is irrational.
Q.E.D