Originally Posted by

**Jhevon** here is a better proof, i think.

Background knowledge:

you should know that an integer is even if you can express it as 2n for some integer n, it is odd if you can express it as 2n + 1 for some integer n. if a number is divisible by 6 (a multiple of 6), you can write it as 6n for n an integer, if its divisible by 4, you can write it as 4n for n an integer and so on.

I'll try to tone done the logic and formal math for this proof.

We want to show that if a number x is divisible by 4 and divisible by 3 then it is divisible by 6.

Proof: assume that x is not divisible by 6. then we can write x as 6n + r, where n is any integer and r is the remainder when dividing by 6. therefore, r = 1,2,3,4, or 5.

and so we have 5 cases.

case 1: x = 6n + 1 (that is when we divide x by 6 we have 1 as a remainder).

note that x = 6n + 1 = 3(2n) + 1, since 2n is an integer, it means x is not divisible by 3 (we have a remainder 1).

case 2: x = 6n + 2

notice x = 6n + 2 = 3(2n) + 2

that means we have 2 as a remainder when we divide x by 3, so x is not divisible by 3

case 3: x = 6n + 3

note that x = 6n + 3 = 3(2n + 1), so in this case x is divisible by 3. now we will check if it is divisible by 4.

notice that x = 6n + 3 = 2(3n + 1) + 1. this means x is odd, and therefore is not divisible by 4.

case 4: x = 6n + 4

so x = 6n + 4 = 3(2n + 1) + 1, so x is not divisible by 3

case 5: x = 6n + 5

so x = 6n + 5 = 3(2n + 1) + 2, so x is not divisible by 3.

so we see in all cases that when a number is not divisible by 6 it is not divisible by either 3 or 4 as well. so we can conclude the opposite is true, by way of the contropositive.