Question

Check, please

Answer 1

I'm here.

Answer 2

thanks a lot!!!

Answer 3

the last part is ok, we checked already

Answer 4

so part A?

Answer 5

Part1

Answer 6

Part One in attachment 1 first. Can you read it well? my computer gives out the bad paper

Answer 7

I can read it...

Answer 8

It all boils down to what fr = ?

Answer 9

to the least , no need to get fr if it's not the least length

Answer 10

You need it so that you can simplify the long expressions involving fr...

Answer 11

From your drawing, it would seem
\[\huge fr = r^{-1}f\]

Answer 12

not inverse, TJ

Answer 13

Are you sure? Start at any point, you'll find that travelling fr is the same as going back one reverse \(\large (r^{-1})\) and then switching lanes (f)

Answer 14

I suggest starting here instead, for clarity...

Answer 15

for i) if I start at node 1 , go to the right 5 times, and then moving along f 5 times, and then go to the left 2 times, and then up along f 1 time. I stop at node 4. so, my least length to go from 1 --> 4 is r^3

Answer 16

Hang on... thinking :D

Answer 17

okay,
\[\huge fr = r^{-1}f\]\[\huge rf = fr^{-1}\]

Answer 18

ok, I can go on that way to check , it's quite easier and more logic than mine

Answer 19

Maybe it is easier to derive a formula first for this..
\[\huge f^nr^n\]

Answer 20

we meet there. apply yours i got the same answer

Answer 21

You can check, but it seems
\[\huge f^n r^n = r^{(-1)^nn}f^n\]

Answer 22

can you tell me something about the order in the stuff, something like fr is order2 ....
Quite confuse

Answer 23

Well of course :)
Simply because
\[\huge (fr)^2=frfr = ffr^{-1}r=1\]
Thus, fr = 2, since \(\large (fr)^2=1\)

Answer 24

so, it will correspond to a if a^2 =1, right?

Answer 25

I guess so, if you're looking for isomorphisms... but we haven't considered any other group besides this one \(\Large <f,r>\)

Answer 26

yes, I think about isomorphisms, not this. because you give out the formula already and that works well, I can self checking .

Answer 27

about partB in paper 1 do you have a rule for checking the inverse?

Answer 28

Wait, just remember these rules for part 1
\[\huge fr = r^{-1}f\]\[\huge rf = fr^{-1}\]
\[\huge f^nr^n = r^{(-1)^nn}f^n\]\[\huge r^nf^n = f^nr^{(-1)^nn}\]

\[\huge (fr)^n=(fr)^{n(\mod 2 \ )}\]\[\huge (rf)^n=(rf)^{n(\mod 2 \ )}\]

Answer 29

Now working on part B...

Answer 30

you mean the remainder in (mod2) , for example if I have (fr)^5 = (fr)^5*1 =

Answer 31

No, 5(mod 2) = 1

Answer 32

Just the remainder, you do not multiply it by anything :)

Answer 33

no, n(mode2) is the whole number, is the remainder only, right? misunderstood, only. so far so good
so (fr)^5 = fr since 2|5 =1 mode2

Answer 34

Yes, so (fr)^5 = (fr)^1 = fr
ok? :)

Answer 35

yeap

Answer 36

Okay, part B...inverses
any particular item you want to do?

Answer 37

general formula, and check mine, hihihi..

Answer 38

hey,TJ for f^nr^n I don't get, give me example

Answer 39

I have to copy every thing into my note. this stuff never known before. My prof said very fast and just go over them in 1hour and a half

Answer 40

It's not that hard...
You can think of it this way...
\[\huge f^nr^n = \left\{\begin{array}{11}r^{-n}f^n \quad \text{n is odd}\\r^nf^n \quad \ \ \ \text{n is even}\end{array}\right.\]

Answer 41

So for example
\[\huge f^8r^8 = r^8f^8\]\[\huge f^9r^9=r^{-9}f^9\]

Answer 42

yes, it's better from your previous note, quite clear

Answer 43

you know why? because you confused me when r^(-1)n*n f^n

Answer 44

It's more compact that way. I don't like piecewise functions :D

Answer 45

ok, inverse

Answer 46

Okay, it would only make sense to put up the basic inverses, some of these you may have already figured out...
\[\huge r^{-n}=fr^nf\]
\[\huge f^{-1} = f\]

Answer 47

Now, when taking inverses, you take them in reverse. Let me show you what I mean...

Answer 48

got it

Answer 49

I cannot read part B, number (v)
could you write it down?

Answer 50

fr^3 f

Answer 51

Okay...
\[\huge fr^3f\]
There may be faster ways to get inverses, but here's the method that would work every time...
Get the inverses, one part at a time, STARTING FROM THE RIGHT.

Answer 52

\[\huge (fr^3f)^{-1} = ?\]

Answer 53

r^3

Answer 54

So... inverse, starting from the right
\[\huge fr^3\color{red}f\rightarrow \color{blue}f\]\[\huge f\color{red}{r^3}f\rightarrow f\color{blue}{fr^3f}\]\[\huge \color{red}fr^3f\rightarrow ffr^3f\color{blue}f\]

Answer 55

And yes, \[\huge ffr^3ff = r^3\]