Real Numbers Ex 1.3 Solutions
Question - 1 : - Show that the square of any positive integeris either of the form 4q or 4q + 1 for some integer q.
Answer - 1 : -
Solution:
According to Euclid’s division lemma,
a=bq+r
According to the question,
When b = 4.
a = 4k + r, 0 < r < 4
When r = 0, we get, a = 4k
a2 = 16k2 =4(4k2) = 4q, where q = 4k2
When r = 1, we get, a = 4k + 1
a2 = (4k + 1)2 =16k2 + 1 + 8k = 4(4k + 2) + 1 = 4q + 1, where q = k(4k + 2)
When r = 2, we get, a = 4k + 2
a2 = (4k + 2)2 =16k2 + 4 + 16k = 4(4k2 + 4k + 1) = 4q, where q= 4k2 + 4k + 1
When r = 3, we get, a = 4k + 3
a2 = (4k + 3)2 =16k2 + 9 + 24k = 4(4k2 + 6k + 2) + 1
= 4q + 1, where q = 4k2 + 6k +2
Therefore, the square of any positive integeris either of the form 4q or 4q + 1 for some integer q.
Hence Proved.
Question - 2 : - Show that cube of any positive integer is of the form 4m, 4m + 1 or 4m +3, for some integer m.
Answer - 2 : -
Solution:
Let a be any positive integer and b = 4.
According to Euclid Division Lemma,
a = bq + r [0 ≤ r < b]
a = 3q + r [0 ≤ r < 4]
According to the question, the possible valuesof r are,
r = 0, r = 1, r = 2, r = 3
When r = 0,
a = 4q + 0
a = 4q
Taking cubes on LHS and RHS,
We have,
a³ = (4q)³
a³ = 4 (16q³)
a³ = 4m [where mis an integer = 16q³]
When r = 1,
a = 4q + 1
Taking cubes on LHS and RHS,
We have,
a³ = (4q + 1)³
a³ = 64q³ + 1³ + 3 × 4q × 1 (4q + 1)
a³ = 64q³ + 1 + 48q² + 12q
a³ = 4 (16q³ + 12q² + 3q) + 1
a³ = 4m + 1 [wherem is an integer = 16q³ + 12q² + 3q]
When r = 2,
a = 4q + 2
Taking cubes on LHS and RHS,
We have,
a³ = (4q + 2)³
a³ = 64q³ + 2³ + 3 × 4q × 2 (4q + 2)
a³ = 64q³ + 8 + 96q² + 48q
a³ = 4 (16q³ + 2 + 24q² + 12q)
a³ = 4m [where m is an integer =16q³ +2 + 24q² + 12q]
When r = 3,
a = 4q + 3
Taking cubes on LHS and RHS,
We have,
a³ = (4q + 3)³
a³ = 64q³ + 27 + 3 × 4q × 3 (4q + 3)
a³ = 64q³ + 24 + 3 + 144q² + 108q
a³ = 4 (16q³ + 36q² + 27q + 6) + 3
a³ = 4m + 3 [where m is an integer =16q³ +36q² + 27q + 6]
Hence, the cube of any positive integer is inthe form of 4m, 4m+1 or 4m+3.
Question - 3 : - Show that the square of any positive integercannot be of the form 5q + 2 or 5q + 3 for any integer q.
Answer - 3 : -
Solution:
Let the positive integer = a
According to Euclid’s division lemma,
a = bm + r
According to the question, b = 5
a = 5m + r
So, r= 0, 1, 2, 3, 4
When r = 0, a = 5m.
When r = 1, a = 5m + 1.
When r = 2, a = 5m + 2.
When r = 3, a = 5m + 3.
When r = 4, a = 5m + 4.
Now,
When a = 5m
a2 = (5m)2 =25m2
a2 = 5(5m2) = 5q,where q = 5m2
When a = 5m + 1
a2 = (5m + 1)2 =25m2 + 10 m + 1
a2 = 5 (5m2 +2m) + 1 = 5q + 1, where q = 5m2 + 2m
When a = 5m + 2
a2 = (5m + 2)2
a2 = 25m2 +20m + 4
a2 = 5 (5m2 +4m) + 4
a2 = 5q + 4where q = 5m2 + 4m
When a = 5m + 3
a2 = (5m + 3)2 =25m2 + 30m + 9
a2 = 5 (5m2 +6m + 1) + 4
a2 = 5q + 4where q = 5m2 + 6m + 1
When a = 5m + 4
a2 = (5m + 4)2 =25m2 + 40m + 16
a2 = 5 (5m2 +8m + 3) + 1
a2 = 5q + 1where q = 5m2 + 8m + 3
Therefore, square of any positive integercannot be of the form 5q + 2 or 5q + 3.
Hence Proved.
Question - 4 : - Show that the square of any positive integer cannot be of the form 6m +2 or 6m + 5 for any integer m.
Answer - 4 : -
Solution:
Let the positive integer = a
According to Euclid’s division algorithm,
a = 6q + r, where 0 ≤ r < 6
a2 = (6q + r)2 =36q2 + r2 + 12qr [∵(a+b)2 = a2 + 2ab + b2]
a2 = 6(6q2 +2qr) + r2 …(i), where,0 ≤ r < 6
When r = 0, substituting r = 0 in Eq.(i), weget
a2 = 6 (6q2) = 6m,where, m = 6q2 is an integer.
When r = 1, substituting r = 1 in Eq.(i), weget
a2 = 6 (6q2 +2q) + 1 = 6m + 1, where, m = (6q2 + 2q) is an integer.
When r = 2, substituting r = 2 in Eq(i), weget
a2 = 6(6q2 +4q) + 4 = 6m + 4, where, m = (6q2 + 4q) is an integer.
When r = 3, substituting r = 3 in Eq.(i), weget
a2 = 6(6q2 +6q) + 9 = 6(6q2 + 6a) + 6 + 3
a2 = 6(6q2 +6q + 1) + 3 = 6m + 3, where, m = (6q + 6q + 1) is integer.
When r = 4, substituting r = 4 in Eq.(i) weget
a2 = 6(6q2 +8q) + 16
= 6(6q2 + 8q) + 12 + 4
⇒ a2 = 6(6q2 + 8q + 2) + 4 = 6m +4, where, m = (6q2 + 8q + 2) is integer.
When r = 5, substituting r = 5 in Eq.(i), weget
a2 = 6 (6q2 +10q) + 25 = 6(6q2 + 10q) + 24 + 1
a2 = 6(6q2 +10q + 4) + 1 = 6m + 1, where, m = (6q2 + 10q + 4) is integer.
Hence, the square of any positive integercannot be of the form 6m + 2 or 6m + 5 for any integer m.
Hence Proved
Question - 5 : - Show that the square of any odd integer is of the form 4q + 1, for someinteger q.
Answer - 5 : -
Let a be any odd integer and b = 4.
According to Euclid’s algorithm,
a = 4m + r for some integer m ≥ 0
And r = 0,1,2,3 because 0 ≤ r < 4.
So, we have that,
a = 4m or 4m + 1 or 4m + 2 or 4m + 3 So, a =4m + 1 or 4m + 3
We know that, a cannot be 4m or 4m + 2, asthey are divisible by 2.
(4m + 1)2 = 16m2 +8m + 1
= 4(4m2 + 2m) + 1
= 4q + 1, where q is some integer and q = 4m2 +2m.
(4m + 3)2 = 16m2 +24m + 9
= 4(4m2 + 6m + 2) + 1
= 4q + 1, where q is some integer and q = 4m2 +6m + 2
Therefore, Square of any odd integer is of theform 4q + 1, for some integer q.
Hence Proved.
Question - 6 : - If n is an odd integer, then show that n2 – 1 isdivisible by 8.
Answer - 6 : -
Solution:
We know that any odd positive integer n canbe written in form 4q + 1 or 4q + 3.
So, according to the question,
When n = 4q + 1,
Then n2 – 1 = (4q +1)2 – 1 = 16q2 + 8q + 1 – 1= 8q(2q + 1), is divisible by 8.
When n = 4q + 3,
Then n2 – 1 = (4q +3)2 – 1 = 16q2 + 24q + 9 – 1= 8(2q2 + 3q + 1), is divisible by 8.
So, from the above equations, it is clearthat, if n is an odd positive integer
n2 – 1 is divisible by 8.
Hence Proved.
Question - 7 : - Prove that if x and y are both odd positiveintegers, then x2 + y2 is even but notdivisible by 4.
Answer - 7 : -
Solution:
Let the two odd positive numbers x and y be 2k+ 1 and 2p + 1, respectively
i.e., x2 + y2 =(2k + 1)2 +(2p + 1)2
= 4k2 + 4k + 1 + 4p2 +4p + 1
= 4k2 + 4p2 +4k + 4p + 2
= 4 (k2 + p2 +k + p) + 2
Thus, the sum of square is even the number isnot divisible by 4
Therefore, if x and y are odd positiveinteger, then x2 + y2 is even but not divisibleby four.
Hence Proved
Question - 8 : - Prove that √5 is irrational.
Answer - 8 : - 
Question - 9 : - Prove that 3 + 2√5 + is irrational.
Answer - 9 : -
Solutions:
Letus 3 + 2√5 is rational.
Then we can find co-prime x and y (y ≠ 0)
Then 3+ 2√5 = x/y
Rearranging, we get,
Since, x and y are integers, thus,
is a rational number.
Therefore, √5 is also a rationalnumber. But this contradicts the fact that √5 is irrational.
So, we conclude that 3 + 2√5 isirrational.
Question - 10 : - Prove that the following are irrationals:-
(i) 1/√2
(ii) 7√5
(iii) 6 + √2
Answer - 10 : -
Q (i) 1/√2
Solutions:
(i) 1/√2
Let us 1/√2 is rational.
Then we can find co-prime x and y (y ≠ 0)
Then1/√2 = x/y
Rearranging, we get,
√2 = y/x
Since, x and y are integers, thus, √2 is arational number, which contradicts the fact that √2 is irrational.
Hence, we can conclude that 1/√2 isirrational.
Q (ii) 7√5
Solutions:
(ii) 7√5
Let us 7√5 is a rational number.
Then we can find co-prime a and b (b ≠ 0)
Then 7√5= x/y
Rearranging, we get,
√5 = x/7y
Since, x and y are integers, thus, √5 is arational number, which contradicts the fact that √5 is irrational.
Hence, we can conclude that 7√5 is irrational.
Q (iii) 6 + √2
Solutions:
(iii) 6+√2
Let us 6 +√2 is a rational number.
Then we can find co-primes x and y (y ≠ 0)
Then 6 +√2 = x/y⋅
Rearranging, we get,
√2 = (x/y) – 6
Since, x and y are integers, thus (x/y) – 6 isa rational number and therefore, √2 is rational. This contradicts the fact that√2 is an irrational number.
Hence, we can conclude that 6 +√2 is irrational.