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Let S be the set of all positive rational numbers x such that x2 < 3. This set is nonempty and bounded above. (Can you prove this clearly?) I claim that it does not have a maximum. To prove this, let us recall that if r were a maximum, we would have to have r ∈ S and x ≤ r for every x ∈ S. Let us show that this is impossible. Precisely, we shall show that if r ∈ S, then there exists x ∈ S with x > r. Here is the proof. Since r ∈ S, r2 < 3. Thus 3 − r2 is a positive (rational) number, and hence so is = 3 − r2 . We will show that there is a positive number δ such that x := r + δ ∈ S. Then x > r is the number we want. There are of course many possibilities for δ. For example, we can choose any δ which is less than min{ 1+2r , 1}. Here is why this works. x2 = = = < (r + δ)2 r2 + 2rδ + δ 2 3 − + δ(2r + δ) 3 − + δ(2r + 1) This last inequality is a consequence of the fact that δ ≤ 1. Since δ < δ(2r + 1) < , so we get x2 < 3 − + = 3, , 1+2r as required. Exercise: Prove that the set of rational numbers x such that x2 > 3 has no minimum. Solution: Actually this is trivial as stated, since we can take x to be a very negative number. I should have again said that the set of positive rational numbers such that x2 > 3 has no minimum. This can be proved the same way as the original problem, but it is a little easier. Suppose x > 0 and x2 > 3. Let := x2 − 3, which is positive. I claim that there is a positive number x0 < x such that x02 > 3. In other words, I claim there is a positive number δ < x such that (x − δ)2 > 3. Take any positive δ which is less than the minimum of x and /2x. Such a δ exists because x and /2x are both positive. Then 2δx < , and (x − δ)2 = x2 − 2δx + δ 2 1 Since δ 2 > 0, we get (x − δ)2 > x2 − 2δx > 3 + − 2δx > 0. 2