Prove or disprove For any nonempty bounded subset S of Q sup

Prove or disprove: For any non-empty bounded subset S of Q, sup S is rational. For any non-empty subset S of Q, if S has a maximum, then max S is rational.

Solution

let Q is a set of all ratinal numbers

i.e Q = { a / a=p/q , p not= q, (p,q)=1}

given that S is a non empty subset of Q

then the eliments of S are also rational numbers

then

S={b / b=p/q ,(p,q)=1} contained in Q

   we know that sup (S) containedin S

then sup(S) is also a rational number

2) Given S is contained in Q

and S has a maximal

we know that maximal always contains in a set

then max(S) contains in S

S is a set which has rational numbers only

then

max(S) is also rational number