Ex 1.1, 5 - Use Euclid’s division lemma to show that cube - Ex 1.1

Ex 1.1, 5 - Chapter 1 Class 10 Real Numbers - Part 3

The rest of the post is locked. Join Teachoo Black to see the full post.

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Ex 1.1 , 5 Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m+ 1 or 9m+ 8. As per Euclid’s Division Lemma If a and b are 2 positive integers, then a = bq + r where 0 ≤ r < b Let positive integer be a And b = 3 Hence a = 3q + r where ( 0 ≤ r < 3) r is an integer greater than or equal to 0 and less than 3 hence r can be either 0 , 1 or 2 If r = 0 Our equation becomes a = 3q + r a = 3q + 0 a = 3q Cubing both sides a3 = (3q)3 a3 = 27q3 a3 = 9(3q3) a3 = 9m Where m = 3q3 If r = 1 Our equation becomes a = 3q + r a = 3q + 1 Cubing both sides a3 = (3q + 1)3 a3 = (3q)3 + 13 + 3 × 3q × 1(3q + 1) a3 = 27q3 + 1 + 9q × (3q + 1) a3 = 27q3 + 1 + 27q2 + 9q a3 = 27q3 + 27q2 + 9q + 1 a3 = 9(3q3 + 3q2 + q) + 1 a3 = 9m + 1 Where m = 3q3 + 3q2 + q If r = 2 Our equation becomes a = 3q + r a = 3q + 2 Cubing both sides a3 = (3q + 2)3 a3 = (3q)3 + 23 + 3 × 3q × 2(3q + 2) a3 = 27q3 + 8 + 18q × (3q + 2) a3 = 27q3 + 8 + 18q2 + 6q a3 = 27q3 + 54q2 + 36q + 8 a3 = 9(3q3 + 6q2 + 4q) + 8 a3 = 9m + 8 Where m = 3q3 + 6q2 + 4q Hence, cube of any positive number can be expressed of the form 9m or 9m + 1 or 9m + 8 Hence proved

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.