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.
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...
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).
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?
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...
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...
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...
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...
\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.
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).
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.
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.