Comments on: 17=evil. http://reallyhardsums.wordpress.com/2007/11/21/17evil/ The eternal comprehensibility of beauty Thu, 10 Sep 2009 21:17:53 +0000 http://wordpress.com/ hourly 1 By: Discount or Surcharge? « The Magic Weblog of EDGARS http://reallyhardsums.wordpress.com/2007/11/21/17evil/#comment-537 Discount or Surcharge? « The Magic Weblog of EDGARS Sun, 25 Nov 2007 23:25:47 +0000 http://reallyhardsums.wordpress.com/2007/11/21/17evil/#comment-537 [...] P.S. This is my 17th post in this blog, so let’s remember that thought about the 17 being an evil number! [...] [...] P.S. This is my 17th post in this blog, so let’s remember that thought about the 17 being an evil number! [...]

]]> By: Phil Wilson http://reallyhardsums.wordpress.com/2007/11/21/17evil/#comment-534 Phil Wilson Sun, 25 Nov 2007 22:04:36 +0000 http://reallyhardsums.wordpress.com/2007/11/21/17evil/#comment-534 Hi edgarsr. I like your proof! (I would also like to be pedantic for a moment and insert a couple of lines near the beginning that show that b=2 leads to a contradiction, so that dividing by b-2 is OK.) It takes the now-classic idea of replacing a function on the integers with a function on the reals. This would seem to make things harder, but in fact makes it easier (much like moving from the reals to the complex numbers can often help). Can you - or anyone else - think of a more geometric proof? Hi edgarsr. I like your proof! (I would also like to be pedantic for a moment and insert a couple of lines near the beginning that show that b=2 leads to a contradiction, so that dividing by b-2 is OK.) It takes the now-classic idea of replacing a function on the integers with a function on the reals. This would seem to make things harder, but in fact makes it easier (much like moving from the reals to the complex numbers can often help).

Can you – or anyone else – think of a more geometric proof?

]]> By: edgarsr http://reallyhardsums.wordpress.com/2007/11/21/17evil/#comment-495 edgarsr Fri, 23 Nov 2007 19:14:21 +0000 http://reallyhardsums.wordpress.com/2007/11/21/17evil/#comment-495 OK, I guess, I proved this! So, if the length and width of a rectangle is a and b, then the area of the rectangle is a*b, and the perimeter of the rectangle is 2(a+b). It is said these two numbers are the same: a*b=2(a+b) Or: a=2b/(b-2) Both a and b are positive integers, so the expression 2b/(b-2) is a positive integer. It is easy to see that if b>2 (if b=1 or b=2, then a is not a positive integer), then the function f(b)=2b/(b-2) is a diminishing function, and: lim 2b/(b-2) = 2 (b->infinity) So the only possible integers being able to be the result of this function are the ones within the interval (2;f(3)=6], or: f(3)=6 f(4)=4 f(6)=3 So, there are actually only two different rectangles possible - one with a=b=4, and the other one with a=3, and b=6 (or vice versa). Q.E.D. OK, I guess, I proved this!

So, if the length and width of a rectangle is a and b, then the area of the rectangle is a*b, and the perimeter of the rectangle is 2(a+b). It is said these two numbers are the same:
a*b=2(a+b)
Or: a=2b/(b-2)
Both a and b are positive integers, so the expression 2b/(b-2) is a positive integer. It is easy to see that if b>2 (if b=1 or b=2, then a is not a positive integer), then the function f(b)=2b/(b-2) is a diminishing function, and:
lim 2b/(b-2) = 2
(b->infinity)
So the only possible integers being able to be the result of this function are the ones within the interval (2;f(3)=6], or:
f(3)=6
f(4)=4
f(6)=3

So, there are actually only two different rectangles possible – one with a=b=4, and the other one with a=3, and b=6 (or vice versa). Q.E.D.

]]> By: Phil Wilson http://reallyhardsums.wordpress.com/2007/11/21/17evil/#comment-474 Phil Wilson Thu, 22 Nov 2007 20:19:28 +0000 http://reallyhardsums.wordpress.com/2007/11/21/17evil/#comment-474 Good luck, then, edgarsr. What I love about that theorem is how it reveals that in the whole infinite universe of numbers, the only two with this property are 16 and 18 - small, familiar numbers. Good luck, then, edgarsr. What I love about that theorem is how it reveals that in the whole infinite universe of numbers, the only two with this property are 16 and 18 – small, familiar numbers.

]]> By: edgarsr http://reallyhardsums.wordpress.com/2007/11/21/17evil/#comment-469 edgarsr Thu, 22 Nov 2007 16:26:53 +0000 http://reallyhardsums.wordpress.com/2007/11/21/17evil/#comment-469 Nice task about those numbers 16 and 18, I'll try to prove it later! Nice task about those numbers 16 and 18, I’ll try to prove it later!

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