Question

show that every positive even integer is of the form 2q and that every positive odd integer is of the form 2q + 1 ​

Answer 1

let n be an positive integer.

on dividing n by 2 let q be the quotient and r be the remainder.

Then by euclid's divison lemma we have

n=2q+r where 0 greater than or equal to r and r greater than 2.

therefore n=2q or 2q+1 for some integer q.

case1_ when n=2m

in this case n is clearly even

case2_ when n=2m+1

in this case n is clearly odd.

hence every positive odd integer is of the form 2q or 2q+1 for some integer q.

HOPE_IT_HELPS__

Answer 2

Step-by-step Answer Explanation:

Let: 'a' and 'b' be aby Positive integer

⇒ EUCLID'S ALGORITHM: a = bq + r (0 ≤ r < b)

                                         = a = 2q + r (0 ≤ r < 2) --- (1)

∴ 'a' be the Postive Integer,

   b = 2, r = 0,1

∴ If r = 0, put the values of 'r' in Eq --- (1)

  a = bq + r

  a = 2q + 0

  a = 2q

∴ If r = 1, put the values of 'r' in Eq --- (1)

  a = 2q + r

  a = 2q + 1

⇒ Hence, by the above Illustration every positive integer shall be in the form '2q' and every '2q' integer is in the form of '2q+1' of r = 1, put the values of 'r' in Eq (1).