Okay I got this number problem and I don't really know what the questions are trying to say.
1) Give sufficient examples to convince someone that every multiple of 3 that is divisible by 4 is also a multiple of 6.
2) Show algebraically that this is true in general for any such numbers.
Any help would be greatly appriciated.
1) will five examples be enough?
24, 36, 72, 144, 216
2) Let x be a multiple of 3. then 3|x. (3|x means "3 divides x," that is 3 goes into x and leaves no remainder.)
since 3|x, x = 3m for some integer m.
similarly, if x is also a multiple of 4, then 4|x. so x = 4n for some integer n
=> 3m = 4n
=> 3m/4 = n, where n is some integer divisible by 3 and 4.
Now suppose 6|n, then n = 6k for some integer k. that is 3m/4 = 6k.
so 3m = 24k = 6(4k).
since 4k is an integer, and 3m = 6(4k), 6|3m. but 3m is x, and so 6|x, and therefore, x is a multiple of 6 as well.
there's something i dont like about this proof, i'll think about it and get back to you
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.