let s be the set of real numbers & let x be any element of s such that x< b then show that suprimum of s =b

Answer 1

First, you know the supremum is <= b since b > s for all s in the set.

So, let the supremum be b - e, where e > 0.

If e > 0 then b - e/2 < b so b - e/2 is also in s, but b - e/2 > b - e, which contradicts the choice of e. So e <= 0 but also e >= 0, so e = 0, sup{s} = b.

Answer 2

Take any real number a Than take (c+b)/2= d/2^2 ( I called a+b=d)and substact it from b you get e= b-d/2^3
After doing so n times you´ll get a number z with z = b-d/2^n less than b but as near as you wish because you can make d/2^n less than any positive small number

Answer 3

(one million+sqrt(a))^2>=0 implies one million+a>=2sqrt(a) that's merely am-gm. Now prod[one million+a_i]>=prod[2sqrt{a_i}]=2^n. end. -------- @Steiner, you're suitable, and that i've got favorite lots your paintings (i spent various time to comprehend thoroughly what's precisely a Lagrange multiplier some month in the past). in spite of the undeniable fact that that's significantly a olimpic question (as maximum of inequalities..), i think of (in spite of the undeniable fact that that's merely a opinion) that's preferible a olimpic answer.. you additionally are suitable once you assert that there are different style of answer: (yet another answer) prod{a_i+one million} = prod{a_i} + (n over one million)prod{a_ia_j} + (n over 2)prod{a_ja_ja_k} +... and so forth. yet each product is <=one million nevertheless by using Am-gm. And so LHS<= (n over 0)+(n over one million)+...+(n over n)=2^n.