Using Direct Proof : Let’s assume 7|4a where ‘a’ is some integer.
So, 4a = 7x for some integer x. Here LHS is even and 7 in RHS is odd. So, x must be an even integer. Let x = 2y, for some integer y.
=> 4a = 7(2y)
=> 2a = 7y. Now also LHS is even and 7 in RHS is odd. So, y must be an even integer. Let y = 2z, for some integer z.
=> 2a = 7(2z)
=> a = 7z
So 7 is a factor of ‘a’. Therefore 7|a.