# If-X-A-and-X-B-then-why-X-AB-Give-me-the-Mathematical-Explanation-

Question Number 11701 by Nayon last updated on 30/Mar/17
$$\\$$$$\\$$$${If}\:{X}\propto{A}\:\:\:\:{and}\:{X}\propto{B}\:, \\$$$${then}\:{why}\:{X}\propto{AB}\:?? \\$$$${Give}\:{me}\:{the}\:{Mathematical} \\$$$${Explanation}…. \\$$$$\\$$
Commented by Nayon last updated on 30/Mar/17
$${mrw}\mathrm{1}\:{pls}\:{solve} \\$$
Commented by mrW1 last updated on 01/Apr/17
$${X}\propto{A}\:{means}\:{X}\:{and}\:{A}\:{are}\:{proportional}, \\$$$${i}.{e}.\:\frac{{X}}{{A}}={k}_{\mathrm{1}} ={constant}\neq\mathrm{0} \\$$$${but}\:{X}\:{and}\:{A}\:{are}\:{not}\:{constant}! \\$$$$\\$$$${In}\:{the}\:{same}\:{way}:\:\frac{{X}}{{B}}={k}_{\mathrm{2}} ={constant}\neq\mathrm{0} \\$$$$\\$$$$\frac{{X}}{{AB}}=\frac{{X}}{\frac{{X}}{{k}_{\mathrm{1}} }×\frac{{X}}{{k}_{\mathrm{2}} }}=\frac{{k}_{\mathrm{1}} {k}_{\mathrm{2}} }{{X}}=\frac{{constant}}{{no}\:{constant}}\neq{constant} \\$$$$\Rightarrow{X}\:{and}\:{AB}\:{are}\:{not}\:{proportional}! \\$$$${i}.{e}.\:{X}\propto{AB}\:{is}\:{not}\:{true}! \\$$$$\\$$$${Example}: \\$$$${A}=\mathrm{2}{X} \\$$$${B}=\mathrm{3}{X} \\$$$$\frac{{X}}{{A}}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow{X}\propto{A} \\$$$$\frac{{X}}{{B}}=\frac{\mathrm{1}}{\mathrm{3}}\Rightarrow{X}\propto{B} \\$$$${but} \\$$$$\frac{{X}}{{AB}}=\frac{{X}}{\mathrm{6}{X}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{6}{X}}\neq{constant} \\$$$$\Rightarrow{X}\:{not}\:\propto{AB}! \\$$
Commented by Nayon last updated on 03/Apr/17
$${But}\:{this}\:{formula}\:{we}\:{use}\:{in}\:{physics} \\$$
$${X}\propto{A}\Rightarrow{X}={k}_{\mathrm{1}} {A},\:{k}_{\mathrm{1}} \neq\mathrm{0} \\$$$$\\$$$${is}\:{the}\:{meaning}“{X}\propto{A}''\:{like}\:{this}?? \\$$
$$\\$$$$\\$$$${If}\:{my}\:{view}\:{is}\:{right}… \\$$$${then}\:{it}\:{should}\:{be}\:{written}\:{as}“{X}^{\mathrm{2}} \propto{AB}''. \\$$
$${no}\:,\:{it}\:{is}\:{X}\propto{AB} \\$$
$${Give}\:{me}\:{the}\:{Exat}\:{Reason}. \\$$
$${Well}… \\$$$${if}\:{A}\:{and}\:{B}\:{are}\:{constants} \\$$$${then}, \\$$$${X}\propto{A}\Rightarrow\exists{k}_{\mathrm{1}} \in{constant},\:{X}={Ak}_{\mathrm{1}} ,\:{k}_{\mathrm{1}} \neq\mathrm{0} \\$$$${X}\propto{B}\Rightarrow\exists{k}_{\mathrm{2}} \in{constant},\:{X}={Bk}_{\mathrm{2}} ,\:{k}_{\mathrm{2}} \neq\mathrm{0} \\$$$$\therefore{X}={Ak}_{\mathrm{1}} ={Bk}_{\mathrm{2}} \\$$$$\Rightarrow\frac{{X}}{{AB}}=\frac{{k}_{\mathrm{1}} }{{B}}=\frac{{k}_{\mathrm{2}} }{{A}}\in{constant} \\$$$$\Rightarrow{X}\propto{AB} \\$$