Question

Question:

1. Let$R$ be a ring and$a,b\in R$ . Let $mandn$be non-negative integers and prove that
   1. role="math" localid="1646833061292" $\left(m+n\right)a=ma+na$.
   2. $m\left(a+b\right)=ma+mb$.
   3. $m\left(ab\right)=\left(ma\right)b+a\left(mb\right)$.
   4. $\left(ma\right)\left(nb\right)=mn\left(ab\right)$.
2. Do part (a) when $mandn$are any integers.

Short answer

(a) It is proved for any non-negative integers $mandn$,

1. $\left(m+n\right)a=ma+na$.
2. $m\left(a+b\right)=ma+mb$.
3. $m\left(ab\right)=\left(ma\right)b+a\left(mb\right)$.
4. $\left(ma\right)\left(nb\right)=mn\left(ab\right)$.

(b) It is proved for any non-negative integers $mandn$,

1. $\left(m+n\right)a=ma+na$.
2. $m\left(a+b\right)=ma+mb$.
3. $m\left(ab\right)=\left(ma\right)b+a\left(mb\right)$.
4. $\left(ma\right)\left(nb\right)=mn\left(ab\right)$.

Step-by-step solution

Property of Rings:

If any ringis designated as$R$,such that $a,b\in R$, then:

$a=b\iff a-b=0$

Proof:

The given non-negative integers are$mandn$, in a ring$R$and $a,b\in R$. Then:

(i)$\begin{array}{rcl}\left(m+n\right)a& =& a+a+.........+a\\ & =& \left(a+a+.........+a\right)+\left(a+a+.........+a\right)\\ & =& ma+na\end{array}$

Hence it is proved:$\left(m+n\right)a=ma+na$

(ii)

$\begin{array}{rcl}m\left(a+b\right)& =& \left(a+b\right)+\left(a+b\right)+.........+\left(a+b\right)\\ & =& \left(a+a+.........+a\right)+\left(b+b+.........+b\right)\\ & =& ma+mb\end{array}$

Hence it is proved: $m\left(a+b\right)=ma+mb$

(iii)

$\begin{array}{rcl}m\left(ab\right)& =& \left(ab\right)+\left(ab\right)+.........+\left(ab\right)\\ & =& \left(a+a+.........+a\right)b=\left(ma\right)b\\ & =& a\left(b+b+.........+b\right)=a\left(mb\right)\end{array}$

Hence it is proved: $m\left(ab\right)=\left(ma\right)b=a\left(mb\right)$

(iv)

$\begin{array}{rcl}\left(ma\right)\left(nb\right)& =& \left(a+a+.........+a\right)\left(b+b+.........+b\right)\\ & =& \left(a+a+.........+a\right)b+\left(a+a+.........+a\right)b+...+\left(a+a+.........+a\right)b\\ & =& \left(ma\right)b+\left(ma\right)b+....+\left(ma\right)b\\ & =& m\left(ab\right)+m\left(ab\right)+.......+m\left(ab\right)\\ & =& nm\left(ab\right)\end{array}$

Hence it is proved: $\left(ma\right)\left(nb\right)=mn\left(ab\right)$

Proof:

The given integers are $mandn$, in a ring $R$such that:

$$\begin{gathered} a,b\in R \\ m,n\in \mathrm{\mathbb{Z}} \end{gathered}$$

Then:

(i) For $m<0\iff m_1=-m................\left\{say\right\}$, we have:

$\begin{array}{rcl}\left(m+n\right)a& =& \left(-m_1+n\right)a\\ & =& a+a+.........+a\\ & & \left(\left(-a\right)+\left(-a\right)+........\left(-a\right)\right)+\left(\left(-a\right)+\left(-a\right)+........\left(-a\right)\right)\\ & =& -m_{1}a+na\\ & =& ma+na\end{array}$

Hence it is proved:$\left(m+n\right)a=ma+na$

(ii) For, $m<0\iff m_1=-m................\left\{say\right\}$, we have:

$\begin{array}{rcl}m\left(a+b\right)& =& -m_1\left(a+b\right)\\ & =& \left(a+b\right)+\left(a+b\right)+......+\left(a+b\right)\\ & =& \left(a+a+.........+a\right)+\left(b+b+.........+b\right)\\ & =& ma+mb\end{array}$

Hence it is proved: $m\left(a+b\right)=ma+mb$

(iii) For, $m<0\iff m_1=-m................\left\{say\right\}$, we have:

$\begin{array}{rcl}m\left(ab\right)& =& -m_1\left(a+b\right)\\ & =& \left(ab\right)+\left(ab\right)+......+\left(ab\right)\\ & =& \left(a+a+.........+a\right)b=\left(-m_{1}a\right)b=\left(ma\right)b\\ & =& a\left(b+b+.........+b\right)=a\left(-m_{1}b\right)=a\left(mb\right)\end{array}$

Hence it is proved:$m\left(ab\right)=\left(ma\right)b=a\left(mb\right)$

(iv) Similarly, we have:

$\begin{array}{rcl}\left(ma\right)\left(nb\right)& =& \left(-m_{1}a\right)\left(nb\right)\\ & =& \left(a+a+.........+a\right)\left(b+b+.........+b\right)\\ & =& \left(a+a+.........+a\right)b+\left(a+a+.........+a\right)b+...+\left(a+a+.........+a\right)b\\ & =& \left(-m_{1}a\right)b+\left(-m_{1}a\right)b+....+\left(-m_{1}a\right)b\\ & =& \left(-m_1\right)\left(ab\right)+\left(-m_1\right)\left(ab\right)+...+\left(-m_1\right)\left(ab\right)\\ & =& n\left(-m_1\right)\left(ab\right)\\ & =& \left(mn\right)\left(ab\right)\end{array}$

Hence it is proved: $\left(ma\right)\left(nb\right)=mn\left(ab\right)$