Use division algorithm to show that the square any position integer is of the form 9m, 9m+1 or 9m+8
Answers
Answer:
EUCLID'S ALGORITHM: a = bq + r (0 ≤ r < b)
a = 9q + r (0 ≤ r < 9) --- (1)
Given: To prove that any cube positive integer is of the form 9m, 9m+1 or 9m+8.
b = 9, r = 0,1,2,3,4,5,6,7
If r = 0, put in Eq (1),
a = bq + r
a = 9q + r
If r = 1, put in Eq (1),
a = 9q + 1
If r = 2, put in Eq (1)
a = 9q + 2
∴ We have proved Cube of an Integer is in the form of 9m (or) 9m + 1 (or) 9m + 8
We get, a = 9q, a = 9q + 1, a = 9q + 8
= a = 9q³
= a³= (9q³)
= a³ = 9³q³ -----> a (9²q²) when 9²q³ ∈
= a³ = 9m
= a³ = 9m + 1
= a³ = (9m + 1)³ (a + 3)³ - a² - b² + 3a²b + 3ab
= a³ = (9m)³ + 1³ + 3 (9m)² + 39 m²
= a³ = 9³m³ + 1 + 39m² + 3.9
= a³ = 9³m³ + 6.9m² + 3.9m + 1
= a³ = 9 (9²m³ 1.3m²) + 1
= a²m³ + 3m² + 3m ∈
= a³ = 9m + 1
= a = 9m + 2 | a³ = (9m + 2)³
We know,
a = 9m + 2
a³ = (9m + 2)³ (a+b)³ = a³ + b³ + 3a²b + 3ab²
a³ = 9³m³ + 8 + 3 (9m) 2 + 3 (9m) 2²
a³ = 9³m³ + 8 + 3 (9m) + 2 + 3 (9m) 4
a³ = 9³m³ + 6 x 9²m² + 12 (9m) + 8
a³ = 9m (9m² + 6.9m + 12) + 8
∴ 9m² + 54. + 12 ∈ i
a³ = 9m + 8
∴ The Cube of any positive integer are in the form of 9m (or) 9m + 1 (or) 9m + 8.
This Question is done using b = 9, if you want to do your Sum using b = 3 then, Check this Link below:
https://brainly.in/question/10344818