1 - just a line if it's part of a number, the hook/base version if it's alone.
2 - non loop
4 - box
7 - crossed, I'm not sure why I started doing it or why it's even done in the first place
0 - usually uncrossed
z - crossed
Well so far all I've done is run a program from it, which was inefficient as I could have just clicked three times rather than writing the whole path out in the prompt. I have no doubt it's very powerful, it's just way over my head right now as I have no clue about what you guys are talking...
Hey, thanks for the replies.
I used Eclipse for Java, but when I tried to find a way to program C++ in Eclipse I couldn't figure out how to get it to work for some reason. I'd like to use Eclipse for both Java and C++ and be able to swap languages between the two easily. I don't remember...
Hey all, I decided to start learning some C++ and my book mentions compiling and running programs from the command prompt. I have taken a few classes in computer science, but those dealt mainly with java, along with some algorithms and data structures. Even though I have used computers since I...
Hey, sorry for the delayed response. I'm not sure if this works though. We assume that o(a-1) is infinite, and for the proof by contradiction we assume that o(a) is finite, but the proof in the OP requires that both be finite for it to work.
I've been working on this a bit though and I think...
Homework Statement
Let G be a group and let a be an element in G. Show that a and a-1 generate the same cyclic subgroup <a> = <a-1> and have the same order o(a) = o(a-1).
Homework Equations
This is in the second section of the book I'm using, so there aren't really any theorems that...
I tried the inverse part earlier in the thread. First, using the cancellation laws, we let a = b. Then:
It seems correct, and it's for any arbitrary element a in G so I think this is all we need to show.
Ok, I was multiplying ax = a by y and was getting nowhere. So,
ax = a >>> axb = ab >>> axb = ayb >>> xb = yb >>> x = y.
Hopefully that's right, I think it is but I've been looking at this problem too long and it's been confusing me since the start.
I'm not sure why I'm having difficulty with this, but I am. Right now I'm trying to show:
For any a,b in G, there exists identity elements x,y in G such that ax=xa=a and by=yb=b. Show x=y.
Is this the right path? We've already shown that such elements x and y exist, now we just need to...
Ahhh I understand the cancellation part now, I never understood that fully I guess. So now we have :
For each element a in G, there exists an element x such that ax = xa = a.
Now I'm a little confused here, please don't think I'm trolling or something, I just want to be 100% sure. I feel I'm...
Thank you, I haven't done this in a while...but that's really no excuse for something like this. I'm actually still confused about the hint you gave though, but it helped me get the inverse part. Isn't left/right "canceling" the same as multiplying by an inverse, which we don't know exists yet...
Homework Statement
Let G be a nonempty finite set closed under an associative operation such that both the left and right cancellation laws hold. Show that G under this operation is a group.
Homework Equations
My book defines the left and right cancellation laws as :
"For any a,b in...
After thinking about it more I realize where I went wrong. Basically a reading comprehension failure, although I think it's a slightly confusing problem to begin with.
Just took the Math Subject GRE today, and at the risk of making a fool of myself I believe I came across an error in one of the questions. I'm not sure how much I can mention about the question, but it was probably the easiest question on the entire test, which really makes me wonder if I read...
Thank you both for the help! Unfortunately, I am just not understanding this. After looking at your responses and the question for about an hour, I haven't gotten too far.
Ok so we assume k! ends in 99 zeros. We can write k! as n * 1099 = n * 299 * 599, where n is some number that comes...