Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Imagine two coins on a table at positions say A and B. Suppose we play a game wi

ID: 3168864 • Letter: I

Question

Imagine two coins on a table at positions say A and B. Suppose we play a game with the following 8 possible moves: M:Flip over the coin at A M:Flip over both coins Ms:Flip over the coin at A and then switch M7:Flip over both coins then switch M2:Flip over the coin at B M Switch the coins M6:Flip over the coin at B and then switch M:Do nothing Let G = {M,M2, M3, 4, Ms, M6, Mr. Ms) and let * be the operation defined by doing moves successively. For instance, 4 * Mi = M2 * M4-M6. (Here we are reading left to right, so we first do M4 then Mi. This is the same as doing M2 followed by M4) Granting associativity, is (G,) a group? Explain why or why not. If so, is the group commutative? Explain why or why not

Explanation / Answer

It is given that there are two coins at A and B

now the game starts and the given possible moves are

M1: Flip over the coin at A

M2: Flip over the coin at B

M3 : Flip over both coins

M4 : switch the coins

M5 : Flip over the coin at A and then switch

M7 : Flip over the coin at B and then switch

M8 : Do nothing

G={M1,M2,M3, M4, M5, M6,M7,M8 }

the operations are read left to right it means from row to column

I is the identity element of G such that I*Mn=Mn*I =Mn where n= 1,2,3,....7

now we must be understood that M1*M2= M3 since flipping over coin at A and coin at B results in Flip over both the coins

in a similar way, we must do all the moves

Granting associativity (G,*) for example

M3*M6=M3*(M2*M4)=(M3*M2)*M4=M1*M4=M5 (Associativity)

I is shown in each row it means

M1*M1= I, M2*M2=I, M3*M3=I, M4*M4=I

M5*M6=I ( since flip over coin at A and coin at B results to initial position)

M7*M7=I

M8*M8=I

Granting associativity (G,*) a group is shown

finally, G is closed since the result of every operation is an element of the group, these properties along with associativity, shows that (G,*) is a group

(G,*) is not commutative because,

M1*M5=M4 but M5*M1=M7 and M4 not equal to M7

* I M1 M2 M3 M4 M5 M6 M7 I I M1 M2 M3 M4 M5 M6 M7 M1 M1 I M3 M2 M5 M4 M7 M6 M2 M2 M3 I M1 M6 M7 M4 M5 M3 M3 M2 M1 I M7 M6 M5 M4 M4 M4 M6 M5 M7 I M2 M1 M3 M5 M5 M7 M4 M6 M1 M3 I M2 M6 M6 M4 M7 M5 M2 I M3 M1 M7 M7 M5 M6 M4 M3 M1 M2 I
Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote