Disprove: For every integer a, if 8 divides a, then 6 divides a.


im still stuck. i wrote:
Disproof. Suppose 8|a. by definition of divisibility, we know 8|a means there is an integer b with a=8b. There is not value of y.

i mean value of b

it's always easy to disprove a false statement. Just find a counterexample.

8|32 but 6∤32

To disprove the statement "For every integer a, if 8 divides a, then 6 divides a," you need to show a counterexample. In other words, you need to provide one specific integer value of 'a' for which 8 divides 'a' but 6 does not divide 'a'.

Your attempt at a disproof is not quite correct. Here's how you can approach it:

Disproof: Let's find a counterexample by providing a specific value of 'a' where 8 divides 'a', but 6 does not divide 'a'.

One way to find such a counterexample is to look for an integer value of 'a' that is divisible by 8 but not divisible by 6. Let's try 'a = 16'.

Step 1: Check if 8 divides 16.
We can verify this by dividing 'a' by 8: 16/8 = 2. Since the division result is an integer (2), 8 divides 16.

Step 2: Check if 6 divides 16.
We can verify this by dividing 'a' by 6: 16/6 ≈ 2.67. Since the division result is not an integer, 6 does not divide 16.

Therefore, we have found a counterexample where 8 divides 'a' (a = 16) but 6 does not divide 'a'. This counterexample disproves the original statement.