Search All of the Math Forum:

Views expressed in these public forums are not endorsed by NCTM or The Math Forum.

Notice: We are no longer accepting new posts, but the forums will continue to be readable.

Topic: Composites, and neat relation
Replies: 18   Last Post: Sep 14, 2004 3:38 AM

 Messages: [ Previous | Next ]
 Dik T. Winter Posts: 7,899 Registered: 12/6/04
Re: Composites, and neat relation
Posted: Sep 12, 2004 7:15 PM

In article &lt;3c65f87.0409121043.5988edb8@posting.google.com&gt; jstevh@msn.com (James Harris) writes:
&gt; So someone pointed out that there's the trivial relation
&gt;
&gt; [x] + [x + 1/2] = [2x] where you're in reals,
&gt;
&gt; and I started thinking about
&gt;
&gt; [x] + [x + 1/k] = [2x]
&gt;
&gt; also in reals, with k&gt;1, and it turns out you need x&gt;1 as well, which
&gt; I was thinking about didn't put down before.

Have you a proof for that? Did you apply the null test to that proof?
What do you think of the following proof:
Suppose: [x] + [x + 1/k] = [2x]
Trivially: [x + 1] = [x] + 1
So: [x] = [x - 1] + 1,
[x + 1/k] = [(x - 1) + 1/k] + 1,
[2x] = [2(x - 1)] + 2.
And so:
[x - 1] + [(x - 1) + 1/k] = [2(x - 1)].
The conclusion is that if the statement holds for x, it also holds for
x - 1. Think a bit about the actual condition you need for that relation
to be true. As a hint, it has to do with the value of 'x - [x]', and not
of the value of x relative to some number.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; <a href="http://www.cwi.nl/~dik/">http://www.cwi.nl/~dik/</a>

Date Subject Author
9/12/04 JAMES HARRIS
9/12/04 The Last Danish Pastry
9/12/04 Jim Burns
9/12/04 Tim Smith
9/13/04 David C. Ullrich
9/12/04 Nate Smith
9/12/04 C. BOND
9/12/04 Dik T. Winter
9/12/04 Dik T. Winter
9/12/04 JAMES HARRIS
9/13/04 Paul Murray
9/13/04 JAMES HARRIS
9/13/04 C. BOND
9/13/04 Jim Burns
9/14/04 Nate Smith
9/13/04 Dik T. Winter
9/14/04 Paul Murray
9/13/04 Dik T. Winter
9/13/04 David C. Ullrich