Recent content by xcr

Homework Statement Let us assume that ray XY is a limiting parallel to line l, with P*X*Y. Prove that ray PY is a limiting parallel to line l. Homework Equations Steps will go something like this: Show that ray PZ meets line lat a point V. Pick a point S such that P is between S and...

Gotcha. And I could show that abk=b2ka by doing aba-1=b2 and then raising each side to the k gives abka-1=b2k. Move the a over and get abk=b2ka. Thank you so much Deveno and I like Serena. Hope ya'll both have a great holiday season.

Yea that was a typo, it should have been 5. So does that prove that the order of b is 31 or is there more that needs to be done?

Ok, so far I have that aba-1=b2, rewritten as b-1ab=ba. Raising both sides to the 5th power gives us e=(ba)5, therefore the order of ba is 5 or m where m|5. so m=1,2,3,4. m can't be 2,3,4 because they don't divide m. 1|5 but then we have ba=e, where ba=b-1ab. So b-1ab=e, giving us ab=b, which...

If you multiply both sides by a4 on the right, you get b=a4, so I see now that there is a possibility that ba=e where a and b do not equal e. But if you do b=am, then b5=(am)5=a5m=(a5)m=em=e. Not sure if this is correct but I'd say that this is a contradiction because 5 does not divide 31 and we...

Ok, so ba≠e because then a must equal e and b must equal e, which cannot be true because of our assumptions

I thought that since 2,3, and 4 did not divide 5 that the could not be the order?

Well if (ba)=e, then (ba)2=(ba)(ba)=ee=e, which is not true. Therefore ba≠e

I would just need to show that (ab)m, where m=1,2,3,4 does not equal e. So just show that (ab)1≠e, (ab)2=a(ba)b≠e, (ab)3=a(ba)2b≠e, (ab)4=a(ba)3b≠e.

ok so I took e=(ba)5=b(ab)4a=b(ab)(ab)(ab)(ab)a. I then took ab=b2a and substituted it in. After doing this REPEATEDLY I moved all the b's to the left and all the a's to the right. I ended up with e=b31a5, but a5=e, so e=b31e, or e=b31. Therefore the order of b is 31. Felt like I was doing the...

ab=b2a b-1ab=ba (b-1ab)5=(ba)5 (b-1a5b)=(ba)5 b-1eb=(ba)5 b-1b=(ba)5 e=(ba)5 Therefore |a|=|ba| As for proving that (ba)m≠e for m<5, do I just show that when m=1,2,3,4, you get b-1amb=(ba)m, which ≠e? Then do I just show that (aba-1)n=(b2)n, which can be rearranged as b-nabn=bna...

Homework Statement Let f:R→S be a homomorphism of rings. If J is an ideal in S and I={r∈R/f(r)∈J}, prove that I is an ideal in R that contains the kernal of f. Homework Equations The Attempt at a Solution I feel like I have the problem right, but would like to have someone look...

I understand how you got to this point: abn = b2na. The steps to get there go like this: aba-1=b2 so raising both sides to the nth power would look like (aba-1)n=(b2)n. On the left side we would have n terms of aba-1, but regrouping them would cause then the aa-1 terms to cancel. yielding...

Homework Statement Let G be a group with identity e. Let a and b be elements of G with a≠e, b≠e, (a^5)=e, and (aba^-1)=b^2. If b≠e, find the order of b. Homework Equations Maybe the statement if |a|=n and (a^m)=e, then n|m. Other ways of writing (aba^-1)=b^2: ab=(b^2)a...

Ok, I think I got it now. I got the quotient is (4x^2)-3x and the remainder is 2x+1. I multiplied the quotient and (2x^2)+x+1 and added the remainder and it came to the original polynomial. Thanks for checking over my work.