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The word positive is usually explained to mean that the number under consideration is greater than zero. Without the relation “$>$”, the positivity of (real) numbers may be defined specifying which numbers of a given number kind are positive, e.g. as follows.

In the set $\mathbb{R}$ of the real numbers, the numbers defined by the equivalence classes of nonzero decimal sequences are positive; these sequences (decimal expansions) consist of natural numbers from 0 to 9 as digits and a single decimal point (where two decimal sequences are equivalent if they are identical, or if one has an infinite tail of 9’s, the other has an infinite tail of 0’s, and the leading portion of the first sequence is one lower than the leading portion of the second).
For example, $1+1+1$ is a positive integer, $\frac{1+1}{1+1+1+1+1}$ is a positive rational and $5.15115111511115...$ is a positive real number.
If $a$ is positive and $a+b=0$, then the opposite number $b$ is negative.
The sets of positive integers, positive rationals, positive (real) algebraic numbers and positive reals are closed under addition and multiplication, so also the set of positive even numbers.
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proof
prove that
sum(a:1>n)a^b/sum(a:1>n)a
is an integer (b is odd)
Re: proof
induction
Re: proof
yeah yeah i was fast to answer, too fast actually,
but it feels like a double induction, on b and on n.
Re: proof
@ Tsurbaba
Can you please give the complete proof?
Re: proof
hi, actually i spoke too fast, i am quite lazy when coming down to manipulating equations, for me mathematics is a joyful art rather than an endless challenge.
, anyway to make a long story short, i gave this problem
to my father , who is a gifted equationtamer (like a lion tamer, but more hard), so here is the solution , hope this catches you before
your assignment is due (how i hated this endless assignments..)
I just solved the problem you gave me, not by induction: Newton's binom:
N = sum(i=1:n) i^m m odd.
In reverse order: N = sum(i=1:n) (n+1i)^m
Summing: 2N = sum(i=1:n) [i^m + (n+1i)^m]
With Newton: (n+1i)^m = (n+1)^m +... +(i)^m
m is odd so (i)^m cancels i^m. All other terms are multiple of n+1:
2N = K(n + 1)
Easy to show that 2N is also a multiple of n:
2N = 2sum(i=1:n1) i^m + 2n^m
Done.
Re: proof
hi Tsurbaba
Thanks for your proof. But i think we need to show that
N is a multiple of n and n+1 simultaneoulsy
because we need to show that N is divisible by n(n+1)
for e.g. x is divisible by 4*3 means we need to show that x is divisible by 12 and not that x is divisible by 4 and x is divisible by 3(because 12 is divisible by 4 and 12 is divisible by 6 does not mean that 12 is divisible by 24)
please correct me if I am wrong
What do you say?
Re: proof
please ignore my above post. I got your point. Thanks a lot for helping me. I need another help from you. Whats the units digit in the expansion of
(15+sqrt220)^19 + (15+sqrt220)^82
stretchy letters
The variables in this entry look all stretchy, I don't know why.
Also, I'm wondering about the thebibliography thing. After "begin{thebibiliography}" there is a "{1}" or a "{5}" or "{2}." I've never seen "{3}" or "{8}" for some reason. What do these numbers mean?
Re: stretchy letters
In the HTML mode they seem now to be such. I think it is a system malfunction, which will soon be over.
Re: stretchy letters
If Mathnerd sees it, you see it and I see it, maybe a rerender needs to be done.
Re: stretchy letters
knodeltheory writes:
> maybe a rerender needs to be done.
Done.
For future reference, the URL for rerendering is
http://planetmath.org/?op=rerender&from=objects&id=xxxx
with xxxx replaced by the object ID number.