I'm not sure you *are* allowed to say that. My approach is to take square roots of both sides, so we're now comparing sqrt7 ^ sqrt8 <=> sqrt8 ^ sqrt7. Then transpose the powers (take both roots of both sides) to get sqrt7th root of sqrt7 <=> sqrt8th root of sqrt8. Now we have a question about values of the Xth root of X. We know from prior calculus work that this has a smooth maximum at X=e=2.718. sqrt7 = 2.645, while sqrt8 = 2.828; this makes sqrt7 substantially closer to e, deltas of .073 versus .110. If we assume the Xth-root-of-X function has a reasonably symmetrical hump around e, then the sqrt7 expression is larger. Rich PS: Puzzle: Estimate how many times a day someone wishes that someone else knew more math? I was at the Borders bookstore over the weekend, in the science aisle. A couple of teen girls walked through the next aisle (math) and one spontaneously wondered "Why would someone buy a MATH book?" I restrained my urge to comment, but it left me wondering "Why would someone make such a useless remark?" -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com on behalf of Richard Guy Sent: Wed 12/15/2004 8:40 AM To: math-fun Subject: Re: [math-fun] Article on math education Am I allowed to say that sqrt is increasing more rapidly than log, so that sqrt 8 sqrt 7 ------ > ------ ln 8 ln 7 sqrt 8 sqrt 7 and hence 7 > 8 ? [ I've multiplied by ln 7 ln 8 and taken antilogs] R. On Wed, 15 Dec 2004, Bill Dubuque wrote:
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.
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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
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