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Show that the cube of any positive integer is of the

Show that the cube of any positive integer is of the

 Using Euclid’s division Lemma, show that the cube of any positive integer is of the form 9m or (9m+1) or (9m+8) for some positive integer m.

Solution: when ‘n’ is divided by 3 to get ‘q’ as a remainder and ‘r’ as a quotient. Applying Euclid’s division lemma,

            n = 3q + r, satisfy 0 ≤  r < 3.

     All possible remainder = 0, 1, 2

   Case I: If r = 0

   By Euclid’s division lemma, we get

              n = 3 x  q + 0

              n = 3 x q

              n3 = (3 x q)3

                 n3 = 27 x q3

                n3 = 9 x 3 x q

            Let m = 3 x q

              n3 = 9m

  Case II: If r = 1

 By Euclid’s division lemma, we get

            n = 3 x q + 1

           n3 = (3 x q+1)3

            n3 = 27 x q3 + 27 x q2 + 9 x q + 1

            n3 = 9 x (3 x q3 + 3 x q2 + 1) + 1

        Let m = 3 x q3 + 3 x q2 + 1

           n3 = 9m + 1

     Case III: If r = 3

  By Euclid’s division lemma, we get

         n = 3 x q + 2

         n3 = (3 x q+2)2

         n3 = 27 x q3 + 54 x q2 +36 x q + 8

         n3 = 9 x (3 x q3+ 6 x q + 4) + 8

     Let m = (3 x q3+ 6 x q + 4)

       n3 = 9m + 8

   So, the cube of any positive integer is of the form 9m or (9m+1) or (9m+8) for some integer m.

 

 

Some important question of Real Number

1. What is Euclid’s Division Lemma?

Show that the cube of any positive integer is of the form 9m

2. State Fundamental Theorem of Arithmetic.

 3. Prove that if x and y are both odd positive integers then x 2+ y 2 is even but not divisible by 4.

 

 

 4. For any positive integer n, prove that n3 – n is divisible by 6.

 Solution: when (n3 – n) is divided by 6, then (n3 – n) = n (n -1) (n +1) is also divided by 6.

          (n3 – n) = n  x (n -1) x (n +1)

 If a number is divisible by both 2 and 3, it will also be  divisible by number 6. When a number (n) is divided by 3, to get q as quotient and r as remainder.

  By Euclid’s division lemma, we get

      n = 3q + r, 0 ≤ r < 3.

   All possible remainders are 0, 1, 2.

Case I: When r = 0                

       n = 3q + r = 3q + 0 = 3q is divisible by 3.

Case II: When r = 1

       n = 3q + 1, then n-1 = 3q + 1 – 1 = 3q is divisible by 3.

Case III: When r = 2

      n = 3q + 2, then n+1 = 3q + 2 + 1 = 3q + 3 = 3(q+1) is divisible by 3.

 We can say that n, (n-1) and (n+1) is divisible by 3. Then n (n-1) (n+1) is always divisible by 3.

Similarly, when a number is divided by 2 to get q as quotient and r as remainder.

       n = 2q + r, where 0  r < 2.

   All possible remainder = 0, 1

Case I: when r = 0

         n = 2q + 0 = 2q is divisible by 2.

Case II: When r = 1

          n = 2q + 1

         n – 1 = 2q + 1 – 1 = 2q is divisible by 2.

   And n + 1 = 2q + 1 + 1 = 2q + 2 = 2(q+1) is divisible by 2.

  We can say that n, (n-1) and (n+1) is divisible by 2 then n (n-1) (n+1) is also divisible by 2.

   Since, n (n-1) (n+1) is divisible by both 2 and 3.

   Hence, n (n-1) (n+1) = n3 – n is divisible by 6.

 

 

 5. Show that any positive odd integer is of the form (6m+1) or (6m+3) or (6m+5), where m is some integer.

Solution:  when ‘n’ is divided by 6 to get ‘m’ as a quotient and ‘r’ be a remainder. Applying Euclid’s division algorithm, we get

      n = 6m + r,  0 ≤ r <6

    All possible remainder = 0, 1, 2, 3, 4, 5

    All possible odd remainder = 1, 3, 5

Case I: when r = 1

 By Euclid’s division lemma, we get

        n = 6m + 1               

  In this case, n is clearly odd.

Case II: when r = 3

  By Euclid’s division lemma, we get

         n = 6m + 3

  In this case, n is clearly odd.

Case III: when r = 5

    By Euclid’s division lemma, we get

       n = 6m + 5

     In this case, n is clearly odd.

  Hence, any positive odd integer is of the form (6m + 1) or (6m + 3) or 6m + 5), where m is some integer.

I hope you learn this concept

Show that the cube of any positive integer is of the form 9m or (9m+1) or (9m+8) for some positive integer m.

 

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