Recent content by sportlover36

so (1,2,3)(1,4,5)(2,3) = (1,4,5,2)(3)?

ohh so (1,2,3)(2,3)= (1,2)(3) since 2 goes to 3 and 3 goes to 1

umm is it (1,2,3)(2,3)= (1)(2,3)? I don't really know how to do ones that don't come out perfectly

which are {(1,2,3), (1,4,5,2,3), (1,5,4,2,3), (1,2,3), (1,4,5,2,3), (1,5,4,2,3)}...and i think i am doing it wrong :/

it H={(1), (1,4,5)(2,3), (1,5,4)(2)(3), (1)(5)(4)(2,3), (1,4,5)(2)(3), (1,5,4)(2,3)}

One of my homework problems asks me to list the left coset (1,2,3)H where σ=(1,4,5)(2,3) and H=<σ>. I know that you have to take the do the permutation of (1,2,3)(1,4,5)(2,3) but i am not sure how you can do that? I got (1,2,3)H={(1,2,3)(3)(1,2,4,5)} but i do not think that is right

Thank you soooo much for your help and time!

so there is no inverse since it does not exist in the S!

umm well x times 0 equals 0 but if we are saying x*1 as the operation then the only one after plugging it into x+y=0 is -1 i don't know if that made sense

whats confusing me is the 6 in min(x+y , 6). And I am not sure how to find the inverse property using the identity element were the first 3 right?

ohsorry! i meant x*y=min(x+y,6)

ohsorry! i meant x*y=min(x+y,6)

Homework problem stated this: Let S={0,1,2,3,4,5,6} let * be defined by x*y=max(x-y,0) is it commutative, associative, does it have an identity and inverse? help please! I put that yes it is commutative and yes for associative, but I'm not to sure they are right because the 6 is...

how can i find a power series for this integral? \int cos(x^3)

I mean for it to say d/dx (1/(x-9)) sorrry