Chapter 1: Real Numbers

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 integer is either of the form 4q or 4q + 1 for some integer q.

Hence Proved.

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 values of 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 m is 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        [where m 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 in the form of 4m, 4m+1 or 4m+3.

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 + 4 where q = 5m² + 4m

When a = 5m + 3

a² = (5m + 3)² = 25m² + 30m + 9

a² = 5 (5m² + 6m + 1) + 4

a² = 5q + 4 where q = 5m² + 6m + 1

When a = 5m + 4

a² = (5m + 4)² = 25m² + 40m + 16

a² = 5 (5m² + 8m + 3) + 1

a² = 5q + 1 where q = 5m² + 8m + 3

Therefore, square of any positive integer cannot be of the form 5q + 2 or 5q + 3.

Hence Proved.

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), we get

a² = 6 (6q²) = 6m, where, m = 6q² is an integer.

When r = 1, substituting r = 1 in Eq.(i), we get

a² = 6 (6q² + 2q) + 1 = 6m + 1, where, m = (6q² + 2q) is an integer.

When r = 2, substituting r = 2 in Eq(i), we get

a² = 6(6q² + 4q) + 4 = 6m + 4, where, m = (6q² + 4q) is an integer.

When r = 3, substituting r = 3 in Eq.(i), we get

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) we get

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), we get

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 integer cannot be of the form 6m + 2 or 6m + 5 for any integer m.

Hence Proved

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, as they 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 the form 4q + 1, for some integer q.

Hence Proved.

We know that any odd positive integer n can be 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 clear that, if n is an odd positive integer

n² – 1 is divisible by 8.

Hence Proved.

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 is not divisible by 4

Therefore, if x and y are odd positive integer, then x² + y² is even but not divisible by four.

Hence Proved