Recent content by jhson114

R is just a program language kind of like matlab. "More generally, is your question whether the "sample vs. population" formula (sample stat = population stat/sqrt(n)) applies to median and std. dev. in addition to the mean?" This is exact what I'm asking.

i have a set containing 10000 data. i took 1000 samples of size 4, 16, 64, and 1024 and took the medians, means, and stadard deviations of each size. i graphed them sd of medians vs sample size, sd of mean vs sample size, and sd of s.d. vs sample size. for sample mean, i know from a textbook...

for b, 2m +1 = 3n +2 => 2m = 3n+1. does this mean n is even and m is a multiple of 6 PLUS 1; therefore P must be... i don't know.. T.T I know what you did with a, but having that extra 1 totally confused me on how to go about after setting the two equations equal to each other.

Using this formula: a \equiv b\left( {\bmod 6} \right),b \in \left\{ {\mathop 0\limits^\_ ,\mathop 1\limits^\_ ,\mathop 2\limits^\_ ,\mathop 3\limits^\_ ,\mathop 4\limits^\_ ,\mathop 5\limits^\_ } \right\},a \in Z which states that all whole numbers can be represented as either, 6n, 6n+1...

Tide, by first assuming what you said (p>3 is prime then both p-1 and p+1 are even), 1. Assume p>3, p+-1 are even 2. P+-1 = 2n , n is an integer 3. p+-1 = 2n congruent to 0 mod 2 4. p+-1 = 2(3n) = 6n congruent to 0 mod 6 5. p = 6n +- 1 6. p = 6n -1 congruent to 5 mod 6 7. p = 6n +1...

I'm trying to prove that any prime number bigger than 3 is congruent to 1 or 5 modulo 6. I started out by saying that that is the same as saying all prime numbers bigger than 3 are in the form 6n +- 1, n is an integer since 1 or 5 mod 6 yields either 1 or -1 and if you divide 6n+-1 by 6, you...

thank you very much for all your help Tide and devious. :)

oh i see. 2^(1/2) is irrational number, which disproves the above statement. however like Tide said, 1^1 is a rational number. but since there's a one statement that made it false, it makes the entire statement false, right?

i don't understand what you mean by "consider square-root of two". can you be more specific?

Im trying to either prove or disprove that if a and b are rational numbers, then a^b is also rational. I tried doing it with a contradiction, but i can't seem to correctly arrive at a solution. this is how i started the problem defn of rational number: a,b = {m/n: m,n are all nonzero...

I ran into this logic puzzle and have been working on it for couple hours now but i can't seem to explain clearly why the answer i came up with is the answer. Heres the question: The police have three suspects for the murder of Mr. Cooper: Smith, Jones, and Williams. Smith, Jones, and...

1). suppose that y1, y2, y3 are the eigenvalues of a 3 by 3 matrix A, and suppose that u1, u2,u3 are corresponding eigenvectors. Prove that if { u1, u2, u3 } is a linearly independent set and if p(t) is the characteristic polynomial for A, then p(A) is the zero matrix. I thought...

ah ha! that makes more sense. and that's so simple too. thanks a lot!

i know that det(A') = det(A) = -det(A). i think i also mentioned this in the earlier post. But i still don't get why all the determinants are zero. oh yeah. i figured out number 2. thanks

det(A') and det(A) are equal. so that means the two determinants are zero?? man i suck