Real numbers
Exercise 1.1
1. Use Euclid’s division algorithm to find the HCF of :
(i) 135 and 225.
(ii) 196 and 38220.
(iii) 867 and 255.
Ans :
(i) 135 and 225
Since 225>135, we apply the division lemma to 225 and 135 to obtain
225 = 135×1+90
Since remainder 90 is not equal to 0 , we apply the division lemma to 135 and 90 to obtain
135 = 90×1+45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain
90 = 2×45+0
Since the remainder in zero , the process stops.
Since the divisor at this stage is 45.
Therefore, the HCF of 135 and 225 is 45 .
(ii) 196 and 38220
Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196×195+0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220is 196
(iii) 867 and 225
Since 867>255, we apply the division lemma to 867 and 255 to obtain
867 = 255×3+102
Since remainder 102 is not zero, we apply the division lemma to 255 ne 102 to obtain
255 = 102×2+ 51
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
102 = 51×2+0
Since the remainder is zero, the process is stops.
Since the divisor at this stage is 51.
Therefore, HCF of 867and 255 is 51.
2. Show that any positive odd integer is of the form 6q+1, or 6q+3, or 6q+5, where q is some integer.
Ans :
Let a be any positive integer and b=6. Then, by Euclid’s algorithm,
a=6q+r for some integer q ≥ 0, and r = 0,1,2,3,4,5 because 0 ≤ r < 6.
Therefore, a = 6q or 6q +1 or 6q+2 or 6q+3 or 6q+4or 6q+5
Also, 6q+1 = 2×3q+1 = 2k¹+1, where k¹ is positive integer
6q+3 = (6q+2) + 1 = 2(3q+1) + 1 = 2k²+1, where k² is an integer
6q+5 = (6q+4) + 1 = 2(3q+2) + 1 = 2k³+1, where k³ is an integer
Clearly, 6q+1,6q+3,6q+5 are not exactly divisible by 2. Hence ,these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q+1, or 6q+3, or 6q+5
3. An army contingent of 616 members is to much behind an army band 32 members n a parade. The two groups are to much in the same number of columns. What is the maximum number of columns in which they can match?
Ans :
HCF (616,32) will give the maximum number of columns in which they can match.
We can use Euclid’s algorithm to find the HCF.
616 = 32×19+8
32 = 8×4+0
The HCF (616,32) is 8
Therefore, they can match in 8 columns each.
4. Use Euclid’s division lemma to show that the square of any positive integer is either of from 3m or 3m+1 for some integer m.
[Hint: Let x be any positive integer then it is of the from 3q, 3q+1, or 3q+2. Now square each of these and show that they can be rewritten in the form 3m or 3m+1.]
Ans:
Let a be any positive integer and b=3.
Then a=3q+r for some integer q≥0
And r=0,1,2 because 0≤r<3
Therefore, a=3q or 3q+1 or 3q+2
Or
a²=(3q)² or (3q+1)² or (3q+2)²
a²=(9q²) or 9q²+6q+1 or 9q²+12q+4
=3×(3q²) or 3(3q²+2q)+1 or 3(3q²+4q+1)+1
=3k1 or 3k2+1 or 3k3+1
Where k1,K2,and k3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the from 3m or 3m+1.
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5. Use Euclid’s division lemma to show that the cube of any positive integer is of the from 9m,9m+1 or 9m+8.
Ans:
Let a be any positive integer and b=3
a=3q+r, where q≥0 and 0≤r<3
* a=3q or 3q+1 or 3q+2
Therefore, every number can be represented as there three forms. There are three cases.
Case 1: when a=3q,
a³= (3q)³=23q³ = 9(3q³) = 9m
Where m is an integer such that m=3q³
Case 2: when a=3q+1,
a³=(3q+1)³
a³=27q³+27q²+9q+1
a³=9(3q³+3q²+q)+1
a³=9m+1
Where m is an integer such that m=(3q³+3q²+q)
Case 3: when a=3q+2,
a³=(3q+2)³
a³=27q³+54q²+36q+8
a³=9(3q³+6q²+4q)+8
a³=9m+8
Where m is an integer such that m=(3q³+6q²+4q)
Therefore, the cube of any positive integer is of the from 9m, 9m+1, or 9m+8.
