Problem
A function 
is called even if
![\[ f(-x) = f(x)\]](https://www.risanp.com/notes/wp-content/ql-cache/quicklatex.com-d4d24fc5bf4e70df11bb08e30111f732_l3.png "Rendered by QuickLaTeX.com")
for all
. A function is called odd if ![\[ f(-x) = -f(x)\]](https://www.risanp.com/notes/wp-content/ql-cache/quicklatex.com-bbaabf824635d05026d9a8a357205444_l3.png "Rendered by QuickLaTeX.com")
for all
. Let 
denote the set of real-valued even functions on 
and let 
denote the set of real-valued odd functions on . Show that: ![\[ \mathbb{R}^\mathbb{R}=U_e \oplus U_o\]](https://www.risanp.com/notes/wp-content/ql-cache/quicklatex.com-ec41fd47c1ab9b8c8bf058f09ddb98fd_l3.png "Rendered by QuickLaTeX.com")
Proof. To show that
is a direct sum, it is sufficient to prove that 
. Let 
. Hence, 
for all 
, which only happens for 
. Hence, is a direct sum.Let
. The equation 
means that 
can be written as the sum of an odd function 
and an even function 
. Or in other words, for all : (1) 
(2) 
By the definition of the odd and even functions, (2) can be written as:
(3) 
The system of linear equations (1) and (3) has a unique solution
and 
. Hence, can always be uniquely written as the sum of an odd function and an even function. 
be a group and let 
. Prove the following:
, and suppose 
has a 
th root, say 
. Then 
iff 
are relatively prime.
. For 
and 
, we have 
, 
, and 
. However, 
but 
consists of all even numbers of 
. A contradiction.