I wonder how those prospective teachers would have fared comparing 7^sqrt(8) vs. 8^sqrt(7) instead of 1/13 vs. 0.13 It seems this problem is deviously difficult. When it was posted to the newsgroup sci.math only two people offered correct proofs. I encourage fellow funsters to give it a try first before looking below to see my quick arithmetic proof [1]; it requires < 2 minutes of purely mental arithmetic (no pencil, paper or calculator needed!). In fact, only arithmetic of two-by-one digit integers is employed. SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW SPOILER BELOW Note 8*29^2 > 7*31^2 via x=30 in 8(x-1)^2-7(x+1)^2 = x^2-30x+1 so sqrt(8) > 31/29 sqrt(7) by taking sqrt of above. Thus to prove sqrt(8) sqrt(7) 7 > 8 it suffices to prove 31/29 sqrt(7) sqrt(7) 7 > 8 or 7^30/8^28 > 8/7 but (7^5/2^14)^6 > (42/41)^6 since 7^5 = 7(50-1)^2 > 7(2500-100) = 16800 and 2^14 = 2^4 2^10 < 2^4 1025 = 16400 > 1 + 6/41 since (1+x)^n > 1+nx [binomial theorem, x>0] > 1 + 1/7 since 7*6 > 41 Another proof [2] proceeds via calculus, namely the inequality (2n+1) All n: x f (a) > 0 => f(a+x) > f(a-x) Does anyone else know any other "interesting" proofs? Below are said sci.math threads. [1] Bill Dubuque, sci.math, 1996/06/08 http://google.com/groups?threadm=WGD.96Jun8060426@berne.ai.mit.edu [2] Don Davis, sci.math, 2000/11/24 http://google.com/groups?threadm=dtd-2411002306240001@ppp0c005.std.com --Bill Dubuque