Comments on: Seven Random Things

by: ros

Sun, 08 Mar 2009 20:50:14 +0000

It’s more internet safety gone mad. On my first day at this school I was subjected to a short spree of ‘being safe on the internet’ talks relating to the kids. It seems we can’t trust them to have any common sense of their own any more.

by: Trig

Sun, 08 Mar 2009 17:02:26 +0000

Sounds like more political correctness gone mad.

by: vincent

Mon, 02 Mar 2009 20:03:01 +0000

Hello,

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by: Fëanor

Mon, 02 Mar 2009 15:50:37 +0000

Drat; ‘less than’ sign gobbled by HTML. I meant “leads to 0 < = c < = 7″

by: Fëanor

Mon, 02 Mar 2009 15:48:26 +0000

Yup, separate blog would be good I did (for the first): sqrt(b) = sqrt(2009)-sqrt(a), which upon squaring, results in b = 2009+a-2sqrt(2009a). 2009 = 7sqrt(41), So sqrt(41a) has to be an integer, so a = 41c^2 for some c, which after some manipulation leads to 0

by: Alex

Sun, 01 Mar 2009 00:29:18 +0000

Thank you so much for answering this meme and sharing with me Ike the polydactyl cat …

by: ros

Sat, 28 Feb 2009 22:59:12 +0000

I didn’tprove it that way… I subbed x=-y to get f(x^3)=-f(-x^3) and so f(A)= -f(-A)for all real A. I think your
condition suggests that the function is also increasing but frankly, with no smoothness condition
on this function, I am equally baffled on what to do.

Fot the first, I multiplied both sides by (root a)-(root b) then did a quick side proof that since
(root 2009)((root a)-(root b) had to be an integer, both root(2009a) and root(2009b) also had to
be integers. This gives that root a= k root(41) and root b= m (root 41) for some integers k and m ( from prime decomposition of 2009).
Also k+m =7. So assuming a is smaller than b we have a= 41k^2 for k=0,1,2,3 qnd b= 41(7-k)^2

Hmm.. perhaps I should start up a separate blog for this stuff……

by: Fëanor

Sat, 28 Feb 2009 22:09:15 +0000

Hmm, I have since done the first one although can’t claim any elegance. For the third, f(x^3) = xf(x^2), and hence odd, eh? Bafflement continues…

by: ros

Sat, 28 Feb 2009 17:28:46 +0000

Feanor, I found a nice elegant solution to the first and a bit of a clunky solution to the last.
I’m getting somewhere with the second but as for the third… Urgh! All I’ve managed to do is prove it’s an odd function. Any ideas anybody?

There’s nothing there I’m particularly embarrassed about, Trig, well, apart from the prostitute typecasting
The new job is good. I wish I could post more about it, especially cookery club, but there are rules about writing abut the kids which means I can’t say an awful lot.

by: Trig

Sat, 28 Feb 2009 17:05:56 +0000

Don’t tag me on this, Roshani. I don’t think I could be that painfully honest. Not in writing. I hope life in your new job is treating you well.