Question

Two natural numbers are given. Prove that their sum, difference or product is a multiple of 3.​

Answer 1

Answer:

yes It is the multiple of 3

Answer 2

Answer:

let a and b be multiples of 3 , then 3a+3b=3n"

Not quite. If a is a multiple of 3 then a=3n for some integer n. We can skip the variable a altogether and say "a multiple of 3 is of the form 3n for some integer n.

So if 3n is a multiple of 3, and 3m is a multiple of 3 then you want to prove that:

To prove: 3n+3m is a multiple of 3.

And as 3n+3m=3(n+m) then 3n+3m is a multiple of 3. (As it is 3 times the integer (n + m).)

....

To be neat and to have style points:

An integer n is a multiple of 3 if n=3k for some integer k.

So if a and b are two multiples of 3, then a=3n and b=3m for some integers, n,m. Then a+b=3n+3m=3(n+m). As n+m is an integer, a+b=3(n+m) is a multiple of 3