# Search results

1. ### Median, average question

nevermind i found it. 25/4 was the last one.
2. ### Median, average question

Let f(x) be the mean of five numbers: 4, 5, 7, 9 and x. Let g(x) be the median of the same numbers. For how many values of x, a real number, is f(x) = g(x)? I only got 2. x = 0, 10 There are 3 though. Perhaps someone can help me find the other one.
3. ### Infinite number of units

Thanks to Petek I was able to clear up my misunderstanding.
4. ### Infinite number of units

Yes Petek you hit the nail on the head, actually the book I'm reading seems to have no details and the few resources I looked up online just said we can derive some particular unit that if you raise it to the nth power, it's still a unit, from the pell's equation. I know the fundamental unit for...
5. ### Infinite number of units

How do you find an infinite number of units of \mathbb{Q}(\sqrt{21}) using the \sqrt(21)? I saw one example using continued fractions but do not know how to apply it in this case. I do have the periodic form of the continued fraction of \sqrt(21).
6. ### Unique factorization domain

I guess it could be. So you are saying that you take some random element a + b\sqrt{5} \in \mathbb{Q}(\sqrt{5}) and claim there are two distinct prime factorizations and show they actually differ by a unit?
7. ### Unique factorization domain

So you see it all over the place, \mathbb{Q}(\sqrt{-5}) is not a UFD by finding an element such that it has two distinct prime factorizations...but what about showing that \mathbb{Q}(\sqrt{5}) is a UFD? I'm only concerned with this particular example, I might have questions later on regarding a...
8. ### Gauss lemma extension?

Ok, the book I'm reading states Gauss's lemma as such: If f(x) is a monic polynomial with integral coefficients that factors into two monic polynomials with coefficients that are rational, f(x) = g(x)h(x), then g(x), h(x) \in \mathbb{Z}[x]. Now one of the exercises says to prove that: If...
9. ### Continued fractions

:blushing: that's embarassing.
10. ### Continued fractions

Ok I need to know which is the right answer for evaluating the continued fraction \langle 1, 2, 1, 2, \ldots \rangle? Here's my work: x = 1 + \frac{1}{2+x} \Rightarrow x^2 + x - 3 = 0 and by quadratic formula, we get x = \frac{-1 \pm \sqrt{13}}{2} but we only want the positive root so I...
11. ### Mobius inversion?

Now you've given me something to think about... very interesting way of proving it as I thought being in the same section as that of Mobius inversion, was supposed to follow direct from that or something...
12. ### Mobius inversion?

\sigma _0 (n) = \sum_{d \mid n} 1 and \omega (n) = \sum_{p \mid n} 1 where p is a prime and d is a divisor. I thought the notation was standard for arithmetic functions.
13. ### Mobius inversion?

Would like to show \sum_{d \mid n} \mu (d) \sigma_0 (d) = (-1)^{\omega (d)}. This proof is just left out of text I'm looking at and I can't seem to piece how F(n/d) = \sigma_0 (d), where F(x) = \sum_{s \mid x} f(x).
14. ### Completely multiplicative

So you are saying that F(4) - F(2)F(2) is not 0 as it should be? I was hoping for more an explicit function, f, such that F is not completely multiplicative.
15. ### Completely multiplicative

If f is completely multiplicative, then \sum_{d \mid n} f(d) is completely multiplicative is not true. There must be an easy counterexample for this yet I cannot come up with one.

I have edited the first post....i left out p is prime and we want to see show that the congruence is not solvable.

Homework Statement Show x^2 + (p+1)/4 \equiv 0 (\mod p) where p \equiv 3 (\mod 4) and p is prime is not solvable. Homework Equations Legendre's and Jacobi symbol, congruences The Attempt at a Solution Noticing that x^2 \equiv -(k+1) (\mod p) when p = 4k + 3 ? Now (-1/p)(k+1/p) should tell...
18. ### Inverse image of phi (totient)

I don't think there should be any confusion in my terminology but in case a refresher is needed check out http://en.wikipedia.org/wiki/Image_%28mathematics%29#Inverse_image" It might also help make it clear that f: \mathbb{N} \rightarrow \phi(\mathbb{N}) where f(n) = \phi(n) cannot have an...
19. ### Inverse image of phi (totient)

So upon introduction to Euler's phi function, we can see that \phi (1) = 1 and \phi (2) = 1, where it turns out that these are in fact the only numbers in N that map to 1. Now what I'm wondering is if there is some general way to find the inverse image of numbers in the image of phi? Also...
20. ### Relatively prime and independent confusion

Not questioning the validity, but where do they get that last line from?
21. ### Relatively prime and independent confusion

so how come that wasn't your original hint, winkie?
22. ### Relatively prime and independent confusion

That's some awesome work there! Thanks. but I'm concerned by this part... if n = 1, then what is k?
23. ### Relatively prime and independent confusion

I think you missed the fact that n \in \mathbb{N} so 2100 is not considered.
24. ### Relatively prime and independent confusion

That's a lot of eye twitching there. Should probably see a doctor. I'm still confused as what the next step is.
25. ### Relatively prime and independent confusion

If a and 77 are relatively prime, show that for positive integers n, a^(10^n) modulo 77 is independent of n. I think I don't understand what this statement is asking. a^(10^n) modulo 77 independent of n means that a^(10^n) modulo 77 is always going to be the same or something?
26. ### Quick problem?

Right that's what I thought, which is why I got confused when he said that we only need odd integers.
27. ### Quick problem?

robert, i like your expansion of a^p - b^p....but I'm afraid I don't understand why exactly it has to be an odd integer. Are you saying that there will not be p terms on the right hand side something if p is even? You started with p odd prime, does that mean the original question does not work...
28. ### Quick problem?

Nevermind I figured it out using binomial theorem. But it looked like to me that p doesn't necessarily have to be prime. Am I correct?
29. ### Quick problem?

I was just browsing for some small problems the other day and came across this problem and I am unsure if it should be obvious and have a quick answer. In any case, I couldn't figure it out. If p is a prime that divides a - b, then show that p^2 divides a^p - b^p, where a, b \in \mathbb{Z}.
30. ### Tensor product exercise

Thanks for the help. I noticed you mentioned that you said all tensor elements (finite sums of basis elements of the tensor product)...how come it doesn't suffice to just show it for the pure (simple) tensor elements since these span the tensor product as a vector space?