Let n be in integer Explain what is wrong with the following

Let n be in integer. Explain what is wrong with the following argument which \"shows\" that if n is both a multiple of 2 and a multiple of 3, then n is a multiple of 6: Suppose n is a multiple of 6. Then n = 6k for some integer k. Since 6 = 2 times 3, we have that n = 2 times (3k), so it is a multiple of 2, and n = 3 times (2k), so it is a multiple of 3. (b) Give a correct proof of the assertion.

Solution

a)

The problem is to assume n is a multiple of 2 and 3 and show that it is a multiple of 6

But the proof starts with assuming n is a multiple of 6 and proves that n is a multiple of 2 and 3

b)

If n is a multiple of 2 and 3

We can write, n=2k

But n is a multiple of 3 also hence, 3|2k

BUt 2 is prime

So, 3|k. so k=3p

So, n=2*3p=6p

Hence, n is a multiple of 6.

Hence proved