tag:blogger.com,1999:blog-37148773.post4864636172892829891..comments2024-03-27T14:50:47.345-04:00Comments on <center>Sandwalk</center>: Math ChallengeLarry Moranhttp://www.blogger.com/profile/05756598746605455848noreply@blogger.comBlogger12125tag:blogger.com,1999:blog-37148773.post-5995074487760160062010-03-18T23:26:43.665-04:002010-03-18T23:26:43.665-04:00all you have to do then is prove that root(N^2 -4)...<i> all you have to do then is prove that root(N^2 -4) is irrational for all whole number values of N =3 or larger.</i><br /><br />Apparently, Dedekind and others have well-established that the square roots of non-square natural numbers are irrational. See: http://en.wikipedia.org/wiki/Quadratic_irrational for one such proof.<br /><br />Since the distance between perfect squares of size 3^2 or larger is >4, N^2-4 is not a perfect square for all values of N >=3, therefore 1 + 1/x has no integer solutions for rational values of x other than 1, -2. QED :)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-37148773.post-78507365193699082352010-03-18T20:12:55.020-04:002010-03-18T20:12:55.020-04:00Thanks Eamon. Guess I need to study more maths. :)...Thanks Eamon. Guess I need to study more maths. :)Briannoreply@blogger.comtag:blogger.com,1999:blog-37148773.post-85967410521260500772010-03-18T19:49:56.214-04:002010-03-18T19:49:56.214-04:001) is false.
1) x + 1/x = N ( N is an iNteger)...1) is false.<br /><br />1) x + 1/x = N ( N is an iNteger)<br /><br />multiplying both sides by x gives:<br /><br />2) x^2 +1 = Nx<br /><br />rearranging:<br /><br />3) x^2 -Nx +1 = 0<br /><br />solving by quadratic formula:<br /><br />4) x = ( N +/- root(N^2 -4) ) / 2<br /><br />since x, N, 2 are all rational, root(N^2 -4) must also be rational<br /><br />this is trivially true when N=2 or N=-2 and correspondingly x=1 or x=-1.<br /><br />N is an integer, and root(N^2-4) is imaginary for values of N = -1, 0 or +1.<br /><br />all you have to do then is prove that root(N^2 -4) is irrational for all whole number values of N =3 or larger. :)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-37148773.post-61033307695290993492010-03-18T11:15:59.527-04:002010-03-18T11:15:59.527-04:00I don't think this works:
P(k) = z^k = x^2 + y...I don't think this works:<br />P(k) = z^k = x^2 + y^2<br />P(k+1) = z^(k + 1) = x^2 + y ^ 2.<br />= z(z^k) = x^2 + y^2.<br /><br />In induction, you have to prove that IF the proposition is true for P(k) THEN it is also true for P(k+1). You have simply assumed that there exist x and y such that P(k=1) is true -- but they are presumably not the same as the x and y for P(k). Something has to happen to the RHS to demonstrate that the identity still works.Eamon Knighthttps://www.blogger.com/profile/04262012749524758120noreply@blogger.comtag:blogger.com,1999:blog-37148773.post-38302665269686342672010-03-18T11:10:54.296-04:002010-03-18T11:10:54.296-04:00Brian:
+/-(x + 1/x + sqrt(2))
is not the sqrt of
x...Brian:<br />+/-(x + 1/x + sqrt(2))<br />is not the sqrt of<br />x^2 + 1/x^2 + 2<br /><br />(I think #1 is false, but mostly by intuition)Eamon Knighthttps://www.blogger.com/profile/04262012749524758120noreply@blogger.comtag:blogger.com,1999:blog-37148773.post-24894217972661442792010-03-18T06:49:26.747-04:002010-03-18T06:49:26.747-04:00A short guide to the Bernoullis by your truly ;)A short guide to the <a href="http://thonyc.wordpress.com/2009/08/06/a-confusion-of-bernoullis/" rel="nofollow">Bernoullis</a> by your truly ;)Thony C.http://thonyc.wordpress.comnoreply@blogger.comtag:blogger.com,1999:blog-37148773.post-82498169182014961152010-03-17T23:35:50.245-04:002010-03-17T23:35:50.245-04:00Regarding 15, how can a rectangle have 3 distinct ...Regarding 15, how can a rectangle have 3 distinct side lengths? Seems illogical.Briannoreply@blogger.comtag:blogger.com,1999:blog-37148773.post-90209605562091301482010-03-17T23:33:40.528-04:002010-03-17T23:33:40.528-04:00n = 1: z^1 = x^2 + y^2 is true.
let z = 2, x = 1, ...n = 1: z^1 = x^2 + y^2 is true.<br />let z = 2, x = 1, y = 1. (2 = 1 + 1)<br /><br />n = 2: z^2 = x^2 + y^2 is true.<br />let z = 10, x = 6, y = 8 (100 = 36 + 64)<br /><br />n = 3: z^3 = x^2 + y^2 is true.<br />let z = 2, x = 2, y = 2 (8 = 4 + 4)<br /><br />Perhaps induction can be used?<br />P(k): z^n = x^2 + y^2 where n,x,y,z are positive integers.<br />P(1) is true. <br />P(k) = z^k = x^2 + y^2<br />P(k+1) = z^(k + 1) = x^2 + y ^ 2.<br />= z(z^k) = x^2 + y^2.<br />As z is an integer and z^k is an integer (integer to power of some integer) then P(k+1) is true. There exist positive integers x,y,z such that z^n = x^2 + y^2 is true for each positive integer n.<br /><br />Does that work?Briannoreply@blogger.comtag:blogger.com,1999:blog-37148773.post-89411949507474360452010-03-17T23:06:19.199-04:002010-03-17T23:06:19.199-04:00Difficult. Trying #1.
(x + 1/x) = n, x not equal +...Difficult. Trying #1.<br />(x + 1/x) = n, x not equal +/- 1.<br />square both sides<br />(x + 1/x)^2 = n^2<br />L.H.S. = (x + 1/x)(x + 1/x)<br />= x^2 + 1 + 1 + 1/x^2<br />= x^2 + 1/x^2 + 2<br />= R.H.S <br />= n^2<br />Take square root of both sides<br />n = +/-(x + 1/x + sqrt(2))<br />which is irrational. So, nope, no integer that (x + 1/x) = n.<br /><br />That work?Briannoreply@blogger.comtag:blogger.com,1999:blog-37148773.post-33002779651722530522010-03-17T19:15:18.600-04:002010-03-17T19:15:18.600-04:00I'm sure that Prof. Moran is well aware that t...I'm sure that Prof. Moran is well aware that the Bernoulli brothers were naturalists who were contemporary with the most important scientist who ever lived, Issac Newton. As he was wont to do, Newton carried on a feud with them, as he did with virtually every other contemporary naturalist.SLCnoreply@blogger.comtag:blogger.com,1999:blog-37148773.post-62424683218638652072010-03-17T19:01:36.405-04:002010-03-17T19:01:36.405-04:00I think 4 is false (1234567879 contain 9 numbers, ...I think 4 is false (1234567879 contain 9 numbers, then the numbers 10-19 contain 10 numbers =/= 1/9), 10 is false (it's too specific), and 11...I'm not sure about but intuitively I would say its false.karmaikohttps://www.blogger.com/profile/06210845244650948955noreply@blogger.comtag:blogger.com,1999:blog-37148773.post-78021628311859808362010-03-17T12:34:37.240-04:002010-03-17T12:34:37.240-04:00I think I got #1 and #3. Other than that....I think I got #1 and #3. Other than that....Eamon Knighthttps://www.blogger.com/profile/04262012749524758120noreply@blogger.com