Is G={f: R -> R : f(x)=ax+b, a ≠ 0} a Group Under Composition?

[Artusartos]

Homework Statement

What exactly does G={ f: R -> R : f(x)=ax+b, where a is not equal to zero} is a group under composition, mean? So what are the elements of G? Are they (for example) f(x)=ax+b and g(x)=a'x+b'? Or are they f(x)=ax+b and f(y)=ay+b?

Thanks in advance

Homework Equations

The Attempt at a Solution

 

[Dick]

Homework Statement

What exactly does G={ f: R -> R : f(x)=ax+b, where a is not equal to zero} is a group under composition, mean? So what are the elements of G? Are they (for example) f(x)=ax+b and g(x)=a'x+b'? Or are they f(x)=ax+b and f(y)=ay+b?

Thanks in advance

Homework Equations

The Attempt at a Solution

The first, of course, they are all linear polynomials in x.

 

[Artusartos]

The first, of course, they are all linear polynomials in x.

Thanks...

 

[Dick]

But how can you tell which one they're talking about?

I read the problem statement, the other interpretation doesn't make much sense. f(x)=ax+b and g(y)=ay+b define the same function. Some example of elements of G are 2x+1, x-1, -2x+4, etc etc.

 

[Artusartos]

I read the problem statement, the other interpretation doesn't make much sense. f(x)=ax+b and g(y)=ay+b define the same function. Some example of elements of G are 2x+1, x-1, -2x+4, etc etc.

Thanks. Yes, I figured that out...it doesn't make any sense...I don't know why I was confused. :)

 

[HallsofIvy]

On the other hand, although the group is defined as {f(x)= ax+ b}, it is not a good idea to use "f" to represent two different members. Rather, say that f(x)= ax+ b and g(x)= a'x+ b'. The group operation, "composition" would give fg= a(a'x+ b')+ b= aa'x+ (ab'+ b) and gf= a'(ax+ b)+ b'= aa'x+ (a'b+ b'). Since, in general, $ab'+ b\ne a'b+ b'$ this group is not commutative.

Of course, the function f(x)= 1x+ 0= x is the group identity: if g(x)= ax+ b then fg= 1(ax+ b)+ 0= ax+ b and gf= a(x)+ b= ax+b. What is the inverse of f(x)= ax+ b?

 

[Artusartos]

On the other hand, although the group is defined as {f(x)= ax+ b}, it is not a good idea to use "f" to represent two different members. Rather, say that f(x)= ax+ b and g(x)= a'x+ b'. The group operation, "composition" would give fg= a(a'x+ b')+ b= aa'x+ (ab'+ b) and gf= a'(ax+ b)+ b'= aa'x+ (a'b+ b'). Since, in general, $ab'+ b\ne a'b+ b'$ this group is not commutative.

Of course, the function f(x)= 1x+ 0= x is the group identity: if g(x)= ax+ b then fg= 1(ax+ b)+ 0= ax+ b and gf= a(x)+ b= ax+b. What is the inverse of f(x)= ax+ b?

The inverse is: $$f^{-1}(x) = a^{-1}x - a^{-1}b$$, right? Because...

$$a(a^{-1}x - a^{-1}b) + b = (x-b)+b = x$$

 

[Dick]

The inverse is: $$f^{-1}(x) = a^{-1}x - a^{-1}b$$, right? Because...

$$a(a^{-1}x - a^{-1}b) + b = (x-b)+b = x$$

Yes, that's exactly right.