(1) Remember that S={x= root number 2*q: q belongs to a set of rational numbers}, so that x belongs to s is equivalent to (x/ root number 2) being a rational number.

To take any real number A, it is necessary to prove that the sequence {a_n} existing in S is close to A.

Firstly, take the rational number sequence {b_n} to approximate the real number (a/ root number 2);

Remember that a_n= radical number 2*b_n, then a_n/ radical number 2=b_n is a rational number, so a_n belongs to S;

On the other hand, {b_n} approaches (a/ radical number 2), so a_n=b_n* radical number 2 approaches a.

To sum up, the sequence {a_n} in S approaches A, which shows that S is dense on the real number set.

(2) Note that m = {x = n+1(3m): n is an integer and m is a natural number}.

Let x=n+ 1/(3m), obviously x is not equal to 2/3. Consider x > separately; 2/3 and x

If x = n+ 1/(3m) >: 2/3, then n >；; 2/3- 1/(3m)> 1/3 (because m= 1, 2,3,4, ...),

So there must be n= 1, 2, 3, 4, ..., so x-2/3 = n+1/(3m)-2/3 >; = 1+ 1/(3m)-2/3 & gt； 1/3;

If x = n+ 1/(3m) < 2/3, then n

Then 2/3-x = 2/3-n-1(3m) >: 2/3-1(3m) >1/3 (because m= 1, 2, 3, 4, ...

To sum up, in either case, there is |x-2/3| > 1/3.

This shows that the sequence in set M cannot approach 2/3, so it is not dense in real number set.