This question is taken from real numbers in which we have to prove that any positive odd integer is of the form (6m+1) or (6m+3) or (6m+5), where m is some integer.
Kindly give me a detailed solution of this question
RS Aggarwal, Class 10, chapter 1A, question no 6
Let n be the given positive odd integer.
On dividing n be 6,
Assume m be the quotient
And r be the remainder.
Then,
By Euclid’s division lemma, we have
n=6m+r, where 0≤r<6
n=6m+r,
where r=0, 1, 2, 3, 4, 5
⇒ 6m or (6m+1) or (6m+2) or (6m+3) or (6m+4) and (6m+5).
But, n=6m, (6m+2), (6m+4) give even values of n.
Hence, any positive odd integer is of the form (6m+1) or (6m+3) or (6m+5) Where m is some integer.