OK if the question is how many elements, i.e ordered pairs have order 4, i.e can be run 4 times to bring back the identity. I have to say 0.

looking at Z_4 X Z_6

I have

{(0,0),(1,1),(2,2),(3,3),(0,4),(1,5),(2,0),(3,1),( 0,2),(1,3),(2,4),(3,5)}

now each one of these is an element right?
So
(3,3) is of order 4 {(0,0),(3,3),(2,0),(1,3)}
(1,3) is of order 4 {(0,0),(1,3),(2,0),(3,3)}

this appears to be all

for Z_2 X Z_12

{(0,0),(1,1),(0,2),(1,3),(0,4),(1,5),(0,6),(1,7),( 0,8),(1,9),(0,10),(1,11),(0,12)}

(1,9)={(0,0),(1,9),(0,6),(1,3)}


theres gotta be a quicker way. but this is what i got.

There's not a quicker way. But why are you only looking in the subgroup generated by (1,1)?

Actually I guess there is a quicker way. You can prove that the order of an element (a,b) in Z/mZ x Z/nZ is the lcm of the order of a in Z/mZ and the order of b in Z/nZ.

So for example, to construct an element of order 4 in Z/4Z x Z/6Z, you'd need elements whose orders in Z/4Z and Z/6Z have lcm 4.

Maybe it's easier to find all elements of order 12 in each group.

I think I got it:

Z_4 X Z_6

(1,1)=(1,1)
2(1,1)=(1,1)+(1,1)=(2,2)
3(1,1)=(1,1)+(1,1)+(1,1)=(3,3)
4(1,1)=3(1,1)+(1,1)=(0,4)
5(1,1)=4(1,1)+(1,1)=(1,5)
6(1,1)=5(1,1)+(1,1)=(2,0)
7(1,1)=6(1,1)+(1,1)=(3,1)
8(1,1)=7(1,1)+(1,1)=(0,2)
9(1,1)=8(1,1)+(1,1)=(1,3)
10(1,1)=9(1,1)+(1,1)=(2,4)
11(1,1)=10(1,1)+(1,1)=(3,5)
12(1,1)=11(1,1)+(1,1)=(0,0)

Z_2 X Z_12

(1,1)=(1,1)
2(1,1)=(1,1)+(1,1)=(0,2)
3(1,1)=2(1,1)+(1,1)=(1,3)
4(1,1)=3(1,1)+(1,1)=(0,4)
5(1,1)=4(1,1)+(1,1)=(1,5)
6(1,1)=5(1,1)+(1,1)=(0,6)
7(1,1)=6(1,1)+(1,1)+(1,7)
8(1,1)=7(1,1)+(1,1)=(0,8)
9(1,1)=8(1,1)+(1,1)=(1,9)
10(1,1)=9(1,1)+(1,1)=(0,10)
11(1,1)=10(1,1)+(1,1)=(1,11)
12(1,1)=11(1,1)+(1,1)=(0,0

The two dont match so not isomprohic

I have no idea what you think you're doing here.

At a glance, your method looks as good as any.

What is the order of the centralizer in Sn of (1,2), where Sn dentoes the Symmetric group of n elements

(n-2)!

Now that you know the answer, think about why.

my guess is that (1,2)x=x(1,2), when 1,2 are not in x, of which there are (n-2)! combinations.

Good guess. Can you prove it?

Edit: Ooops! My answer's not right, though certainly
|Z_{S_n}((1,2))| >= (n-2)! by what we just noted.

Do a small example-- say S_3. See how many guys are in there.

You should be able to prove this by thinking about what values x can send 1 and 2 to.

-BW

You were half right.

so the answer is (n-2)!/2?

I was thinking (n-2)! + 1 since (1,2)(1,2)=(1,2)(1,2), but (1,2) will not be included in the permutations that map 1,2 to themselves

In S_4, what permutations are in the centralizer of (1,2)?

There's only 24 total, so you can check this by hand.

Edit: you already noticed that there are (n-2)! permutations that commute with (1,2), namely, those that fix 1 and 2 (i.e. the "x" does not involve 1 or 2). But you've also shown a different permutation that commutes with (1,2), namely (1,2).

So there are at least (n-2)! + 1 in the centralizer. Can you find any more, say in S_4?

Theres 20 cars and theres a 5% chance of a car being blue.

i)Whats the probability that none of the cars will be blue out of the twenty

ii)the probability that two or more of the cars will be blue out of twenty.

I think theres some simple formulae for these but i cant seem to find them anywhere,

Refer to the binomial distribution. I answered these in your original thread.

Please show work:

a + b + c = d
Find d if
a^3 x b x c = 7776
a x b^3 x c = 17496
a x b x c^3 = 3456.

--------------

Find the sum in the series and show work

[1/(sq. root of 1 + sq root of 2)] + [1/(sq. root of 2 + sq root of 3)]...... + [1/(sq root of 35 + sq root of 36) =

a) 2
b) 3
c) 4
d) 5
e) 6

---------------------------------

Show work
The radical
sq root of [(sq root of 1440) + 94] can be expressed as sq root of a + sq root of b, where a<b. What is the product ab?

Multiply these last 3 equations:

(abc)^5 = 7776*17496*3456

abc = (7776*17496*3456)^(1/5) = 216

a = sqrt(a^3*bc/abc) = sqrt(7776/216) = 6

b = sqrt(a*b^3*c/abc) = sqrt(17496/216) = 9

c = sqrt(abc^3/abc) = sqrt(3456/216) = 4

d = a + b + c = 19











This is a telescoping sum where all terms cancel except the first and the last, leaving







sqrt(1440) + 94 = [a + sqrt(b)]^2

sqrt(1440) + 94 = a^2 + 2*a*sqrt(b) + b

Trying 2*a*sqrt(b) = sqrt(1440), and a^2 + b = 94:

2*a*sqrt(b) = sqrt(1440) = 2*2*sqrt(90)

a = 2, b = 90

This works in second equation too: a^2 + b = 2^2 + 90 = 94

ab = 180

Thanks

The only thing is, for the last question when you type
sq root of [(sq root of 1440) + 94] = into the calculator you get 11.4868 while sq root of 2 plus sq root of 90 = 10.9

I misread the last question as a + sqrt(b) instead of sqrt(a) + sqrt(b). Here is the corrected solution:


sqrt(1440) + 94 = [sqrt(a) + sqrt(b)]^2

sqrt(1440) + 94 = a + 2*sqrt(ab) + b

Trying a + b = 94, and 2*sqrt(ab) = sqrt(1440)

a + b = 94
ab = 360

a(94 - a) = 360

a^2 -94a + 360 = 0

(a - 90)(a - 4) = 0

a = 90 or a = 4

Since a< b, the solution is a=4, b=90.

ab = 360.

thanks.

theres another question too, but with too many sq roots and fractions and stuff, i'd probably mess something typing it out so i'm not gonna.

How about this one (this is in the recurrence relation section of my book):

Let a_n be the number of n-digit binary numbers that do not contain 3 consecutive 1s.

I got that a_n = a_(n-1) + a_(n-2) + a_(n-3) [is this correct?]

a_0 = 1, a_1 = 2, a_3 = 4.

Is there a formula for a_n that can reduce the recurrence relation?

fyp

also solve the recurrence relation and that'll be your formula

or get the generating function if you like

R is set of real numbers. Scalar multiplication is defined at c*x=cx and addition is defined as x+y=max(x,y) (where max(0,1)=1, etc). Is R a vector space with these operations?

So it has to satisfy the 8 vector space rules, as well as being closed under scalar multiplication and vector addition.

Is this counterexample correct in showing that R is not a vector space?

Let x=(0,1/2), y =(0,2)

x + y = (0,2 1/2) !=max(x,y)

?

No, because x+y = (0,2), not (0,2.5). They are *telling you* that x "+" y = max(x,y). You've thrown away the old definition of +.

Try to show that the vector spade rules hold, and see what happens. Post back with what you've got.

Yes you are correct. I think this works:

x + (-x) = 0

let x = 1, so 1 + (-1) = 0, however x+y=max(x,y), so 1 + (-1) = max(1,-1) = 1, therefore is is not a vector space.

Thanks