Math Is Fun Forum

I assume you mean prime integers.  If that's the case, every integer z is in U.  Further, because the exponents can be -1, 1/z is also in U.  The result of this is that U = K.

This reminds me of something a professor of mine said.  Not exactly the same, but similar: "It's very easy to come up with invariants in mathematics.  It's much harder to come up with invariants that aren't always zero."

bobby:

This fact once becomes obvious once you think of numbers in binary form:

1111111 + 1 = 10000000

The immediate corollary is your proposition.

zetafunc: p(k) = k^7 - k^6 - k^5 - k^4 - k^3 - k^2 - k - 1 is an irreducible polynomial.  This means that if there are two polynomials f(k) and g(k) with integer coefficients such that f * g = p, then either f = ±1 or g = ±1.

Further, p(k) has only one real root, easily verified by graphing.  An elementary result from these two facts from Galois theory is that the roots of p(x) can not be expressed in terms of +, -, ×, ÷, and nth roots alone.  Such roots are called insolvable.

I'm guess that you learned this on your own?  Did you use a book, or just various web pages?  If you used a book, which?

The identities above involve:

Differential geometry
Analysis
Complex analysis
Vector calculus

Each of these merits their own study for at least a semester if not more.  In turn, to study these subjects, you should know "advanced calculus" (intro to analysis), abstract algebra, and differential equations.  So if you want to understand the identities involved above, this is fine.  But if you want to learn mathematics, then I recommend starting with a good solid foundation in calculus.

Sheesh... now I understand all your posts.  I was wondering what it was you were trying to learn.

How much math have you had zeta?

Do you know any multivariable calculus?

Umm, who looks through a phonebook sequentially?  Worst case you would need to look at 14 entries.  So it looks like he is talking about an unsorted phone book.

As part of a fictitious biography:

One of his most important discoveries, still in use today, was that you can turn your paper sideways.  This allowed mathematicians to work on even more complex formulas than they were previously capable.

Any natural number can only have a finite number of digits.  Such is not true for a real number.

(2) = (4) = (6) = (8) (this is easy to prove: 4*3 = 2, 6*2 = 2, 8*4 = 2)
(1) = (3) = (7) = (9) (3*7 = 1, 7*3 = 1, 9*9 = 1)

So there are four principal ideals: (0), (1), (2), and (5).

Each of these numbers are relatively prime, so any ideal generated by two of them must be the entire ring, (1).  Thus, there are only four ideals:

(0), (1), (2), and (5)

Ignoring (0) and (1) because these are trivial, (2) and (5) are easy to check whether they are maximal and/or prime: clearly they are both maximal (there are only 4 ideals!).  Remembering that:

(G/H)/(K/H) = G/K

If we mod out by (2), this is: (Z/10Z)/(2Z/10Z) = Z/2Z.  As this is an integral domain, this ideal is prime.  Same for (5) which would give us Z/5Z.

No.  1 is out because A_8 is a simple group, and if there is only 1 Sylow 7-subgroup, then this subgroup must necessarily be normal.

Yes, this works.  You must prove that each element of order 7 in A_8 is indeed a 7-cycle, and then counting them suffices.  No Sylow theory needed at all.

For #3: I'll give you a hint:  Certainly any element a/b where 11 does not divide a is a unit in this ring (prove it!).  So to find irreducible, it suffices to look at elements of the form:

Now there is a restriction here, in other words it isn't true for all k and b that the number above is irreducible.  Find this restriction, and then prove all the elements you've found are also prime.

(It's very important you understand how this is wayyy more meaningful than 20160 )

Now Sylow's theorem says that if n_7 is the number of Sylow 7-subgroups, then

The RHS restriction will give you 2^9 possible numbers right off the bat, and the LHS should reduce this into a manageable amount.  I'm not sure how do to do this one without aid of a computer, but the list of possibilities is:

1, 8, 15, 36, 64, 120, 288, 960

Now 1 is out immediately (why?).  The goal is to try to eliminate all but one.  This can be tricky.

For the same reason that the sequence

Does not suggest there's a finite number of primes.

bobby: Yes.  This would suffice to prove that the integers are infinite, assuming that you've been able to define them without implicitly having this property.