Linear Algebra Proof on Composition of One-to-One Functions

jrk012
Messages
7
Reaction score
0

Homework Statement




Prove that the composition of one-to-one functions is also a one-to-one function.



Homework Equations




A function is one-to-one if f(x1)=f(x2) implies x1=x2. Composition is (f*g)(x)=f(g(x)). Proof-based question.



The Attempt at a Solution




A one-to-one function does not repeat the image. If we have two one-to-one function f(x) and g(x), then f and g do not repeat their images. Then, when then the composition, for example f(g(x)), for all x, g(x) does not repeat the image and after that applying f(x) also does not repeat the image, therefore the composition of the function is one-to-one as well.

Is this a good proof for the question?
 
Physics news on Phys.org
I would suggest a proof by contradiction. Let g be a one-to-one function from set A to set B, f a one-to-one function from set B to set C. Suppose the statement were not true- that f(g(x)) is not one-to-one. Then there exist a, b, a not equal to b, in A such that f(g(a))= f(g(b)). Since f is one-to-one, there must exist a unique x in b such that f(x)= f(g(a))= f(g(b)). Can you complete this?
 
If in doubt, proof by contradiction!
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top