Recent content by Muzza

A finite field clearly has a characteristic (among the elements 1, 1 + 1, 1 + 1 + 1, ... there must be two that equal one another, since we have only finitely many elements in the field). Let p be the least number of ones we need to add up in order to get 0. Suppose p = nm with 1 < n, m < p...

Let X have N children, and Y have N children. For starters, how many ways are there to distribute the 3 tickets among the 2N children?

You have to take into consideration that Swedish universities don't require you to take any courses outside of your major. I don't really know the answer (although I am also in a Swedish mathematics program, hoping to get a Swedish MSc), but I'd say the most important determinant is this: do...

x^3 - a^3 = (x - a)(x^2 + ax + a^2).

n!/(n + 1)! = 1/(n + 1). If you write out the factorials for n = 1, 2, 3 you should see why this is the case.

This is problem 43 in chapter 1, and it actually wants you to prove that a|bc <=> a/(a, b)|c. Also, (a, b) denotes the gcd of a and b. The lcm is often written [a, b]. This problem can be solved by (for example) writing things out in terms of the prime factorizations of a, b, c.

Again, the limits in my original post were just examples. There are situations where it's difficult or even impossible to write down the iterated integral, hence the need for descriptions like those in my first post.

It's an improper integral. But this is all beside the point, I don't actually want to compute this particular integral. It was just an example I pulled out of thin air to illustrate my point.

You're right. I'm probably better off just defining some set S = {(x, y, z); x >= 3, blah} and taking the integral over S.

How do you make the limits in a triple integral look okay? I need to write something like: \iiint_{x \geq 3, y \geq 4, z \geq 5, 2z - x \geq 5} f(x, y, z)\, dx\,dy\,dz but it looks kind of silly right now.

For the first problem, you can actually prove something much stronger. Suppose you were given a set S of n + 1 numbers between 2 and 2n. Write the numbers 2, ..., 2n in "matrix", like so: 2, 3 4, 5 ... (2n - 1), (2n - 2) 2n. Now, put some kind of mark on the numbers which are in S...

You can drop the conditions that r, s are positive (but I suppose you still need r, s != 0) and r > s. Let d = gcd(rs, r + s). d divides both rs - r(r + s) and rs - s(r + s).

Real matrices are those matrices which have real entries. They are not necessarily 1x1.

If there were only one American and one Indian, could you solve this problem?

No, 10c is n with the last digit replaced with a zero. You want to consider c - 2d. This proof can be done like so (using various elementary divisibility properties): c - 2d is divisible by 7 <=> 10c - 20d is divisibly by 7 <=> 10c - 20d + 21d is divisibly by 7 <=> 10c + d is divisible by 7...