2 ms·

You clearly know more than I do about these things, but: 0 < sqrt(2)^a <= a for all a in 0..2) is not true. It fails for a=1. If you can make it right, could

• albertzeyer 16y ago
  Whoops, yes you are right. Fixed, more complete version of this part: 1 < sqrt(2)^a <= 2 for all a in [1,2]. I.e. the sequence (with x=sqrt(2)) is lower-bounded by 1 and upper-bounded by 2. Also, because sqrt(2)^a >= a for all a>sqrt(2), the sequence is also monotone. By that, it follows that it converges to some value y. Now, given that it converges, you know that sqrt(2)^y = y and also y in [1,2]. And then you have y=2.