The infinite set of numbers of the form n*sqrt(2) where n is any integer.
However, since sqrt(2) is one of the factors, there is no obvious reason to restrict the other factor to an integer. In that case, the answer is, at least the infinitely many numbers of the form x*sqrt(2) where x is any real number. But why restrict to real numbers? Why not include complex multiples?
It's not. 4 is a multiple of 2. 2 is a factor of 4 because it can divide into 4 evenly with no remainder.
Since 4 is a multiple of 2, all the multiples of 4 will be common.
Multiples of 2 are even numbers. All even numbers end in 0, 2, 4, 6 or 8.
If ordered by size, only every other multiple of 2 is a multiple of 4. All multiples of 4 are multiples of 2.
Yes, for example the multiples of 2 are: 2, 4, 6, 8, 10, 12, 14, ......etc. (all even numbers are multiples of 2) or multiples of 3 are 3, 6, 9, 12, 15, 18, 21, ....... etc.
All multiples of 8 are also multiples of 2, but not all multiples of 2 are multiples of 8.
All multiples of 16 are also multiples of 2.
All the even numbers are multiples of 2
They are all multiples of 8.
They are all the multiples of 4.
Yes.
ALL even numbers are multiples of 2... so if there is an uneven multiples of six than they aren't multiple of two.
Yes. In fact all even numbers (not just 4) will share their multiples with 2.
Since 2 is a multiple of 6, all multiples of 6 are also multiples of 2.
All multiples of 4 are multiples of 2 and 4.
It's not. 4 is a multiple of 2. 2 is a factor of 4 because it can divide into 4 evenly with no remainder.
Both of them.