Question

Let C be the set $\mathbf{ℝ}\mathbf{\times }\mathbf{ℝ}$with the usual coordinatewise addition (as in Theorem 3.1) and a new multiplication given by

$\left(a,b\right)\left(c,d\right)\mathbf{=}\left(ac-bd,ad+bc\right)$

Show that with these operations C is a field.

Short answer

It is proved that C with a given operation is a field.

Step-by-step solution

Prove the first seven properties of field

Prove that $C\left(\mathrm{ℝ}\times \mathrm{ℝ}\right)$satisfies all Axioms with the given operations as:

Consider r, s, and t to be any elements of a set C, where $r=\left(a,b\right),s=\left(c,d\right),t=\left(e,f\right)$.

1. Prove that$r+s\in C$ as:
   $\begin{array}{rcl}r+s& =& \left(a,b\right)+\left(c,d\right)\\ & =& \left(a+c,b+d\right)\end{array}$
   This implies that $r+s\in Casa+c,b+d\in \mathrm{ℝ}$.
2. Prove that$C\left(\mathrm{ℝ}\times \mathrm{ℝ}\right)$ is associative as:
   $\begin{array}{rcl}r+\left(s+t\right)& =& \left(a,b\right)+\left(\left(c,d\right)+\left(e,f\right)\right)\\ & =& \left(a,b\right)+\left(c+e,d+f\right)\\ & =& \left(a+c+e,b+d+f\right)\\ \left(r+s\right)+t& =& \left(\left(a,b\right)+\left(c,d\right)\right)+\left(e,f\right)\\ & =& \left(a+c,b+d\right)+\left(e.f\right)\\ & =& \left(a+c+e,b+d+f\right)\end{array}$
   This implies that $r+\left(s+t\right)=\left(r+s\right)+t$.
3. Prove that$C\left(\mathrm{ℝ}\times \mathrm{ℝ}\right)$ is commutative as:
   $\begin{array}{rcl}r+s& =& \left(a,b\right)+\left(c,d\right)\\ & =& \left(a+c,b+d\right)\\ & =& \left(c+a,d+b\right)\\ & =& \left(c,d\right)+\left(a,b\right)\\ & =& s+r\end{array}$
   This implies that $r+s=s+r$.
4. Consider$0_c=\left(0,0\right)$, we have:
   $\begin{array}{rcl}r+0_c& =& \left(a,b\right)+\left(0,0\right)\\ & =& \left(a+0,b+0\right)\\ & =& \left(a,b\right)\\ & =& r\\ & =& \left(0+a,0+b\right)\\ & =& 0_{c+r}\end{array}$
   
5. Consider$r\text{'}=\left(-a,-b\right)$; this implies that:
   $\begin{array}{rcl}r+r\text{'}& =& \left(a,b\right)+\left(-a,-b\right)\\ & =& \left(0,0\right)\\ & =& 0_c\end{array}$
   
6. Prove that$rs\in C$ as:
   $\begin{array}{rcl}rs& =& \left(a,b\right)\left(c,d\right)\\ & =& \left(ac-bd,ad+bc\right)\\ & \in & C\end{array}$
   
7. Prove that$r\left(st\right)=\left(rs\right)t$ as:
   $\begin{array}{rcl}r\left(st\right)& =& \left(a,b\right)\left(\left(c,d\right)\left(e,f\right)\right)\\ & =& \left(a,b\right)\left(ce-df,cf+de\right)\\ & =& \left(a\left(ce-df\right)-b\left(cf+de\right),a\left(cf+de\right)+b\left(ce-df\right)\right)\\ & =& \left(ace-adf-bcf-bde,act+ade+bce-bdf\right)\\ & & \\ \left(rs\right)t& =& \left(\left(a,b\right)\left(c,d\right)\right)\left(e,f\right)\\ & =& \left(ac-bd,ad+bc\right)\left(e,f\right)\\ & =& \left(\left(ac-bd\right)e-\left(ad+bc\right)f,\left(ac-bd\right)f+\left(ad+bc\right)e\right)\\ & =& \left(ace-bde-adf-bcf,acf-bdf+ade+bce\right)\end{array}$

Thus, r (st) = (rs) t .

Prove the result

1. Prove that$r\left(s+t\right)=rs+st$ as:
   $\begin{array}{rcl}r(s+t)& =& \left(a,b\right)\left(\left(c,d\right)+\left(e,f\right)\right)\\ & =& \left(a,b\right)\left(c+e,d+f\right)\\ & =& \left(a\left(c+e\right)-b\left(d+f\right),a\left(d+f+b\left(c+e\right)\right)\right)\\ & =& \left(ac+ae-bd-bf,ad+af+bc+be\right)\\ & =& \left(ac-bd,ad+bc\right)+\left(ae-bf,af+be\right)\\ & =& \left(a,b\right)\left(c,d\right)+\left(a,b\right)\left(e,f\right)\\ & =& rs+st\end{array}$
   Thus, $r\left(s+t\right)=rs+st$.
2. Prove that rs = sr as:
   $\begin{array}{rcl}rs& =& \left(a,b\right)\left(c,d\right)\\ & =& \left(ac-bd,bd+bc\right)\\ & =& \left(ca-bd,cb+da\right)\\ & =& \left(c,d\right)\left(a,b\right)\\ & =& sr\end{array}$
   Thus, $rs=sr$.
3. Consider$1_c=\left(1,0\right)$. We have:
   $\begin{array}{rcl}r{1}_c& =& \left(a,b\right)\left(1,0\right)\\ & =& \left(a-0,b+0\right)\\ & =& \left(a,b\right)\\ & =& r\\ & =& \left(1,0\right)\left(a,b\right)\\ & =& 1_{c}r\end{array}$
   
4. Whenever $a,b\in R$and $ab=0_R$, then $a=0_R$or $b=0_R$.
5. Assume that $\left(a,b\right)\ne 0_c$and$\left(a,b\right)\left(c,d\right)=1_c$.therefore, we have:
   $\begin{array}{rcl}\left(a,b\right)\left(c,d\right)& =& \left(ac-bd,ad+bc\right)\\ & =& \left(1,1\right)\end{array}$
   This implies that:
   $$\begin{gathered} ac-bd=1 \\ bc+ad=1 \end{gathered}$$
   After the solution, we get $c=\frac{a+b}{a^2+b^2}$and$d=\frac{a-b}{a^2+b^2}$.
   For every $\left(a,b\right)\ne \left(0,0\right)$such that there exits c and d, the equation $rx=1_c$ has a solution inC.
   Hence it is proved that C with given operation is a field.