Mathematics

Example: Show that any positive integer which is of the form 6q + 1 or 6q + 3 or 6q + 5 is odd, where q is some integer.

Solution. If a and b are two positive integers such that a is greater than b; then according to Euclids division algorithm; we have
a = bq + r; where q and r are positive integers and 0 ≤ r < b.
Let b = 6, then
a = bq + r ↠ a = 6q + r; where 0 ≤ r < 6.
When r = 0 ↠ a = 6q + 0 = 6q;
which is even integer
When r = 1 ↠ a = 6q + 1
which is odd integer
When r = 2 ↠ a = 6q + 2 which is even.
When r = 3 ↠ a = 6q + 3 which is odd.
When r = 4 ↠ a = 6q + 4 which is even.
When r = 5 ↠ a = 6q + 5 which is odd.
This verifies that when r = 1 or 3 or 5; the integer obtained is 6q + 1 or 6q + 3 or 6q + 5 and each of these integers is a positive odd number.
Hence the required result.

Example: Use Euclid's Division Algorithm to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m.

Solution. Let a and b are two positive integers such that a is greater than b; then: a = bq + r; where q and r are also positive integers and 0 ≤ r < b
Taking b = 3, we get:
a = 3q + r; where 0 ≤ r < 3
↠ The value of positive integer a will be 3q + 0, 3q + 1 or 3q + 2
i.e., 3q, 3q + 1 or 3q + 2.
Now we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m, or 3m + 1 for some integer m.
↠ Square of 3q = (3q)²
= 9q² = 3(3q²) = 3m; 3 where m is some integer.
Square of 3q + 1 = (3q + 1)²
= 9q² + 6q + 1
= 3(3q² + 2q) + 1 = 3m + 1 for some integer m.
Square of 3q + 2 = (3q + 2)²
= 9q² + 12q + 4
= 9q² + 12q + 3 + 1
= 3(3q² + 4q + 1) + 1 = 3m + 1 for some integer m.
↠ The square of any positive integer is either of the form 3m or 3m + 1 for some integer m. Hence the required result.

Example: Use Euclids Division Algorithm to show that the cube of any positive integer is either of the 9m, 9m + 1 or 9m + 8 for some integer m.

Solution. Let a and b be two positive integers such that a is greater than b; then:
a = bq + r; where q and r are positive integers and 0 ≤ r < b.
Taking b = 3, we get:
a = 3q + r; where 0 ≤ r < 3
↠ Different values of integer a are
3q, 3q + 1 or 3q + 2.
Cube of 3q = (3q)³ = 27q³ = 9(3q³) = 9m; where m is some integer.
Cube of 3q + 1 = (3q + 1)3
= (3q)³ + 3(3q)² ×1 + 3(3q) × 12 + 13
[Formula Used : (a + b)³ = a³ + 3a²b + 3ab² + 1]
= 27q³ + 27q² + 9q + 1
= 9(3q³ + 3q² + q) + 1
= 9m + 1; where m is some integer.
Cube of 3q + 2 = (3q + 2)³
= (3q)³ + 3(3q)² × 2 + 3 × 3q × 22 + 23
= 27q³ + 54q² + 36q + 8
= 9(3q³ + 6q² + 4q) + 8
= 9m + 8; where m is some integer.
↠ Cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8. Hence the required result.