Question Number 32499 by 7991 last updated on 26/Mar/18
$${proof}:\:{a}\left(−{b}\right)=\left(−{a}\right){b}=−\left({ab}\right) \\ $$
Answered by $@ty@m last updated on 26/Mar/18
$${We}\:{have}: \\ $$$$\mathrm{3}×\mathrm{2}=\mathrm{6} \\ $$$$\mathrm{3}×\mathrm{1}=\mathrm{3} \\ $$$$\mathrm{3}×\mathrm{0}=\mathrm{0} \\ $$$${if}\:{we}\:{continue}\:{the}\:{process}, \\ $$$${the}\:{no}.\:{next}\:{to}\:\mathrm{2},\mathrm{1},\mathrm{0}\:{would} \\ $$$${be}\:−\mathrm{1} \\ $$$$\&\:{no}.\:{next}\:{to}\:\mathrm{6},\mathrm{3},\mathrm{0}\:{would}\:{be}\:−\mathrm{3} \\ $$$${Hence} \\ $$$$\mathrm{3}×\left(−\mathrm{1}\right)=−\mathrm{3} \\ $$$$\Rightarrow{a}\left(−{b}\right)=−{ab} \\ $$
Answered by Joel578 last updated on 26/Mar/18
![We know for a, b ∈ R (−1) . a = −a and (−1) . b = −b a(−b) = a . [(−1)b] = [a(−1)] . b = (−a)b a(−b) = a . [(−1)b] = (−1) . (ab) = −ab (−a)b = [(−1)a]. b = (−1) . (ab) = −ab](Q32504.png)
$$\mathrm{We}\:\mathrm{know}\:\mathrm{for}\:{a},\:{b}\:\in\:\mathbb{R} \\ $$$$\left(−\mathrm{1}\right)\:.\:{a}\:=\:−{a}\:\:\:\:\mathrm{and}\:\:\:\left(−\mathrm{1}\right)\:.\:{b}\:=\:−{b} \\ $$$$ \\ $$$${a}\left(−{b}\right)\:=\:{a}\:.\:\left[\left(−\mathrm{1}\right){b}\right]\:=\:\left[{a}\left(−\mathrm{1}\right)\right]\:.\:{b}\:=\:\left(−{a}\right){b} \\ $$$$ \\ $$$${a}\left(−{b}\right)\:=\:{a}\:.\:\left[\left(−\mathrm{1}\right){b}\right]\:=\:\left(−\mathrm{1}\right)\:.\:\left({ab}\right)\:=\:−{ab} \\ $$$$\left(−{a}\right){b}\:=\:\left[\left(−\mathrm{1}\right){a}\right].\:{b}\:=\:\left(−\mathrm{1}\right)\:.\:\left({ab}\right)\:=\:−{ab} \\ $$