Question

Show that any positive integer is of the form 3M or (3M + 1) or (3 m + 2) for some integer m.​

Answer 1

Let 3m be a natural positive integer. This number is divisible by 3.

The next multiple of 3 is 3m+3.

(3m+3) would be the third consecutive integer after 3m. Thus, the integer before 3m+3 is(3m+3-1), the one before that is (3m+3-1-1) and then it's (3m+3-1-1-1) that is, 3m itself.

Hence we have three numbers: 3m, 3m+1, 3m+2.

Take the next consecutive integer after 3m+2. That would be 3m+3, which can be written as 3(m+1).

This (3(m+1)) can now be written as 3n, where n is another integer. Hence, again employing the former procedure, we have another 3 successive integers after 3n.

Here it is proved that any integer can be expressed as 3m, 3m+1, 3m+2. By changing m as (m+1), we get the next integer, that is 3m+3 and so on.