At this stage, we're looking for

*an* upper bound for S. It doesn't have to be the

*least* upper bound. So we're looking for a number that's bigger than anything in S. The difficulty is that if x>1 then

, whereas if x<1 then

. If

and a≥1, then you can conclude that

. That doesn't work when a<1. For example (taking a=1/4), suppose that

. You can't conclude that x>1/4. (In fact, the best you can say is that x>1/2.) But if

then certainly x>1, and 1 in turn is greater than a. The device of using max{a,1} is simply a mechanism for covering the two cases a≥1 and a<1 simultaneously.

Originally Posted by

**Famboozle** We proceed to show that

. First,

, for

, since

. Next, for any

such that

we have

, so

**Also, why is there a min{a,1} here? Why not just drop the min {1,a}?**

Similar sort of reason. It is necessary to show that y>0 (because the subsequent section of the proof requires that y–ε>0, for some ε>0). So we need to find a positive number whose square is less than a. If a≥1 then 1 is such a number; if a<1 then a itself is such a number. In either case, min{a,1} will do the job.