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#101 Re: Help Me ! » Inequality..help(2) » 2008-10-08 06:19:11
Clear the denominators on the left inequality and then use AM-GM.
#102 Re: Help Me ! » Inequality..help » 2008-10-07 07:02:35
Tips: evaluate 2 of the parenthesises and use cauchy shcwartz inequality
#103 Re: Help Me ! » Any suggestions » 2008-10-07 06:01:19
tony123 wrote:
Indeed, but the rule logx+logy=logxy holds in that interval too so shouldnt it be possible to prove it anyway?
#104 Re: Help Me ! » Proof » 2008-09-30 08:12:12
you could also do like this:
since exactly one of the three consecutive integers 2^n-1,2^n,2^n+1 must be divisible by 3, and 2^n isnt, 3|(2^n-1) or 3|(2^n+1)->3|(2^n-1)(2^n+1)
#105 Re: Help Me ! » Challenge Question » 2008-09-23 05:32:05
Hint: clear the denominator, factor a^3+b^3 and simplify. Do you recognize the equation?
#106 Re: Help Me ! » Is this rubbish? » 2008-09-19 20:54:41
I dont really understand how your teacher did, but i think i can solve this using vectors:
start by choosin a vector w=(x1,y1,0) such that |w|=a. This will be the center of any of the circles that created the torus. We now want all vectors u in the plane spanned by w and the z axis such that |u|=r, because then adding u and w yields a point on the torus' surface. u can then be written as a linear combination of w and e= (0,0,1), ie u=pw+qe=(px1,py1,q), with:
or equivalently, since
(1)
so we get the vector u+w= ((p+1)x1,(p+1)y1,q)= (x,y,z).
so z=q, x= (p+1)x1, y= (p+1=y1.
now we want to express a relation between x,y and z.
if we multiply the equation by (p+1)^2 we get
inserting this into (1) yields:
This is a necessary condition, I dont think I proved it to be sufficient though since I guess I used non equivalent transformations, but it may be possible to show that all points satisfying this actually lies on the torus surface.
and mikau, no, I dont think the sign function can be zero. That would be a bit weird since all we want is the sign + or -, if the entry is zero then it doesnt matter which sign we use and probably we can define sgn(0)=1.
#107 Re: Help Me ! » divisible by 2008 » 2008-09-17 06:26:58
This is easy proved by bruteforce using congruences.
2008=8*251.
5+3=8, divisible by 8.
46 and 1098 ar both divisible by 8 for n>0 (since they are even), for n=0, their sum is 1144=8*143. Thus the whole expression is divisible by 8.
now lets look at 251:
Thus the whole expression is divisible by 251, and since its divisible by 8 its also divisible by 8*251=2008. q.e.d.
(I can admit I used the computer to find the congruences for the larger numbers 
)
#109 Puzzles and Games » Hourglasses » 2008-09-16 04:56:27
- Kurre
- Replies: 2
Given two hourglasses, on a respectively b minutes, can we always measure c minutes? If not, what criteria must hold for it to be possible?
#110 Re: Exercises » How many » 2008-09-14 08:22:54
Let the triangle have sides a,b,c, with a≥b≥c. We want to solve the equation a+b+c=2007 in positive integers.
we have first that 669≤a≤1003, because if a is less that 2007/3=669, then b or c must be greater than a, if a is larger than 1003, then a>b+c contradicting the triangle inequality.
for a=669, then b=c=669 we have the equilateral triangle. Assume that a is greater than 669. If a is odd, then we start with the triangle b=c=(2007-a)/2. Now we may increase b and decrease c up to b=a, ie for any odd a, there are a-(2007-a)/2+1=3a/2-2005/2 triangles. For even a, we start with b=(2008-a)/2, c=(2006-a)/2. Now there are a-b+1=a-(2008-a)/2+1=(3a/2-2005/2)-1/2 different triangles. Summing these two yields number of triangles to:
the 167/2 is the -1/2 i pulled out from all the even numbers in the summation, altough im a bit tired so I may have calculated wrong somewhere
#111 Re: This is Cool » Algebraic numbers » 2008-09-08 23:25:40
the problem was rather trivial after all
Which problem?
uhm the question
#112 Re: This is Cool » Algebraic numbers » 2008-09-08 04:47:19
I don't think dealing in general with numbers like
will be easy.
But when we take the exponents to be rational, you may take some common divisor of the rationalexponents and after a suitable substitution, you're again in the good old algebraic field.
O yes thats true about rational exponents. Altough if we let the exponents be real there must be many more solutions since the set of such functions is not countable, thus the set of the roots cant be countable either.
Actually, considering the following expression:
taking logaritms yields
thus for each there is a such that X is a root of
altough we now restricted X to be positive...
Same applies for complex numbers, with 1/r=b+ci,
And its obvious we can choose b and c such that there is a function with X as root for all X.
very well, the problem was rather trivial after all
#113 Re: This is Cool » I just thought of something » 2008-09-07 00:13:45
more generally, let d and k be positive integers such that d|k, then N is divisible by d iff the sum of the digits in base k+1 is divisible by d.
#114 Re: This is Cool » Algebraic numbers » 2008-09-05 22:57:26
So, the algebraic number field is algebraically closed, this is a foundamental property of the field, and means exactly that every root of a polynomial with algebraic coefficients is algebraic too. This comes from basic factorisation properties of plynomials and using the fact that every first degree polynomial over the algebraic numbers has an algebraic solution.
oyea, silly me that didnt think of factorization thanks. But still what happens if we let the exponents of the terms in the "polynomial" vary?? 
ie, an expression of the form
where K is either Q, A, R or C, and S is either Q or A.
How is the set of the roots of such functions related to Q,A,R or C?
#115 Maths Is Fun - Suggestions and Comments » Preview button doesnt work » 2008-09-03 05:19:53
- Kurre
- Replies: 3
When writing on a new post, sometimes when I click preview, it posts instead. Im sure its not only me.
fix plzz
#116 This is Cool » Algebraic numbers » 2008-09-02 19:16:30
- Kurre
- Replies: 7
Algebraic numbers are numbers that are roots of polynomials with integer coefficients. But what if we let the "polynomial" have rational exponents? Algebraic exponenets? Real exponenets? maybe complex exponents? Will we then get other number sets? 
anyone knows?
My first idea was what happaned if we replaced the coeffecients with algebraic numbers, but I found on wikipedia that is has been shown that a root of a algebraic polynomial is again algebraic. Anyone knows how to prove that?
#117 Re: Help Me ! » Combinatorial proof of fermats little theorem » 2008-09-02 06:35:37
Ah, nevermind, I got it, if p wasnt prime then there are some necklaces where patterns of colored pearls can repeat, and thus doesnt have p different cyclic permutations. 
#118 Help Me ! » Combinatorial proof of fermats little theorem » 2008-09-02 00:24:47
- Kurre
- Replies: 1
This is from the book "problem solving strategies" by Arthur Engel:
We have pearls with a colors. From these we make necklaces with exactly p pearls. First we make a string of pearls and there are a^p different strings. If we throw away the a one-colored strings a^p-a strings will remain. We connect the ends of each string to get necklaces. We find that two strings that differ only by a cyclic permutation result in indistinguishable necklaces, but there are p cyclic permutations of p pearls on a string. Hence the number of distinct necklaces is (a^p-a)/p. But this number must be an integer, thus p | a^p-a. q.e.d.
I find this proof pretty elegant, but I cant see how the proof uses that p must be a prime number? It seems that this proof shows that p | a^p-a for all natural p, but this isnt true. What am I missing?
#119 Re: Help Me ! » Linear algebra » 2008-08-20 07:26:15
Yes that struck me too, that if B=-A, then A^2=B^2, but A≠B. But still, A^2 is similar to B^2, but how do I prove there is no matrix P such that A=P-¹BP=-P-¹AP ?
"1. You've already argued that Px and P-¹x are both nonzero so long as x is nonzero This means that if P-¹BPx = 0, then Bx = 0. "
Am I being stupid, or am i missing how that really is implying the other?
But anyway, I had my test today, and it was the most boring math test I have every done in my life (and my first at the university actually ). It was all about diagonalizing matrices and finding eigenvalues and other time-requiring counting exercises with loads of numbers, no "real" mathematics.. 
But I think I did well.
#120 Re: Help Me ! » Linear algebra » 2008-08-19 07:35:53
Riiiiiiiiickyyyyyyyyyyyyyyyyyyyy where are youu!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!:o
I got exam tomorrow!!!!!!!!!!
#121 Re: Help Me ! » Linear algebra » 2008-08-19 00:07:06
3. Contrapositive is the way to go on this one. Remember that if a is not equal to b, then a*a is certainly not equal to b*b.
I dont get it, how does equality between a and b have anything do with similarity?
#122 Re: Help Me ! » Linear algebra » 2008-08-18 23:31:27
2. I see now that my idea on 2 fails, so here is my second attempt:
If x is written in the basis B, then I can choose
Now multiplying this out will get a nx1 matrix, the jth column of C, on the left side and the nx1 matrix that is the jth column of D on the right showing that cij=dij for all i. And letting j be 1,2,...,n shows that cij=dij for all j as well, and thus C= D.
Is that what
means, just the nx1 matrix obtained by the coefficients when x is written in base B?#123 Re: Help Me ! » Linear algebra » 2008-08-18 22:49:12
hm I see now that me arguments for P is not enoguh for surjectivity, to prove that P is a mapping to all y in ker(B) I must prove that all y in ker(B) has the form y=Px, so far I have only proved that all Px lies in ker(B). Altough this is easy fixed by letting
and consider all y such By=0, then
multiplication from left by P^-1:
thus there is an x in ker(A) such that
Thus for all y in ker(B), there is an x in ker(A) such that y=Px and hence P is surjective.
But I dont get what you mean that they should be subsets of eachother. Then every x such that Ax=0, should also satisfy Bx=0,
so must imply that Bx=0. I dont really get it
#124 Help Me ! » Linear algebra » 2008-08-18 07:59:00
- Kurre
- Replies: 11
Need help with some problems in linear algebra:
1. Prove that the nullity is the same for two similar matrices.
My proof:
let
Multiply both sides with P:
thus
The linear transformation P is injective since P is invertible, and since P is a mapping from all x in ker(A) to all y=Px in ker(B), P must be bijective, and thus ker(B) is isomorphic to ker(A) and Null(A)=Null(B).
Is this proof completely correct?
2. Let C and D be mxn matrices, and let B=(v1,...,vn) be a basis for a vector space V. Show that if
for all x in V, then C= D.If this holds for all x, then we can choose x such that xi≠0, and multiply both sides with y= (1/x1,...,1/xn) and get C= D. What I dont get is why a different basis is introduced, does that change the problem somehow, or is my proof still correct?
3. If A^2 and B^2 are similar, must A and B be similar?
Im not sure, But I dont think so.
so det(A) could equal -det(B) which would contradict similarity, but this doesnt really prove anything.
I maybe need to find a counterexample, but I dont really see how...
#125 Re: Help Me ! » I am useless at proving inequalities » 2008-08-09 23:58:42
Well I was writing a solution and when I clicked preview It got posted instead and then I didnt manage to complete the proof. I substituted c=1/ab and the inequality for your case reduced to:
for a,b≤1