Question

I am given these 2 t/f questions: If you could please help explain them I would appreciate it!

a.) Let R denote the set of real numbers. For real numbers a and b, a does not equal 0, the assignments x --> ax+b is a permutation R --> R. Do these permutations form a group under composition of functions?

b.)Let H be the set consisting of the following permutations of the plane: All translations and all reflections in any line of the plane. Is H a group under composition of mappings?

Answer 1

So for something to be a group it needs a set and an operation.

Answer 2

In this case the set is \(\mathbb R\). What would the operation be?

Answer 3

composition?

Answer 4

I wil give a medal please someone help

Answer 5

anyone? Ill pay $5.00 if I have to.. I have a quiz coming up that is similar to this and would like to understand whats going on..

Answer 6

@phi @tkhunny could any of you guys help me?

Answer 7

Does the suggested operation fit all the requirements of a group? I'd start with Closure.

Answer 8

yes its closed:
let h(x) = ax+b g(x)=cx+d,

So g(h(x)) = g(cx+d)=a(cx+d)+b=(ac)x+cb+d ?

Answer 9

Well, that's a little funny. Seems like a negative is sneaking in there. Also, you should be MUCH more formal about parameters you introduce.

For Real Numbers e, f, c, d, with neither e, nor c equal to zero (0).
h(x) = ex+f
g(x) = cx+d

g(h(x)) = c(h(x)) + d = c(ex+f) + d = ecx + fc + d = (ec)x + (fc+d)
h(g(x)) will be identical.

If we agree that Real Numbers have Closure, then we are done, since:
a = ec and
b = fc + d

Okay, we're done with Closure

The official list is:
1) Closure
2) Associativity
3) Existence of an Identity Element under the Operation
4) Existence of an Inverse for each element.

Prove all four? Group!!! Yea!

(Okay, it can be a little tedious.)

Answer 10

so for Associativity: f . g . h= f . (g . h) where . is composition?

Answer 11

Not quite. (fg)h = f(gh) which may have been what you meant. It's just more clear.