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Question Number 158396 by zakirullah last updated on 03/Nov/21

Any proof or Idea about;  (4/2)÷((16)/3) = (4/2)×(3/(16))

$${Any}\:{proof}\:{or}\:{Idea}\:{about}; \\ $$$$\frac{\mathrm{4}}{\mathrm{2}}\boldsymbol{\div}\frac{\mathrm{16}}{\mathrm{3}}\:=\:\frac{\mathrm{4}}{\mathrm{2}}×\frac{\mathrm{3}}{\mathrm{16}} \\ $$

Answered by MJS_new last updated on 03/Nov/21

it′s just the definition  (a/b)÷(c/d)=(a/b)×(d/c)  a fraction is a division in brackets  (a/b)=(a÷b); (c/d)=(c÷d)  (a/b)÷(c/d)=(a÷b)÷(c÷d)  obviously b “makes a smaller” and c would  also. but first (brackets first!) d “makes c  smaller” ⇒ “d helps a”  ⇒  (a÷b)÷(c÷d)=a÷b÷c×d=a×d÷b÷c=  =a×d÷(b×c)=((ad)/(bc))=(a/b)×(d/c)  it′s similar to  (a−b)−(b−c)=a−b−c+d=a+d−b−c=  =(a+d)−(b+c)

$$\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{the}\:\mathrm{definition} \\ $$$$\frac{{a}}{{b}}\boldsymbol{\div}\frac{{c}}{{d}}=\frac{{a}}{{b}}×\frac{{d}}{{c}} \\ $$$$\mathrm{a}\:\mathrm{fraction}\:\mathrm{is}\:\mathrm{a}\:\mathrm{division}\:\mathrm{in}\:\mathrm{brackets} \\ $$$$\frac{{a}}{{b}}=\left({a}\boldsymbol{\div}{b}\right);\:\frac{{c}}{{d}}=\left({c}\boldsymbol{\div}{d}\right) \\ $$$$\frac{{a}}{{b}}\boldsymbol{\div}\frac{{c}}{{d}}=\left({a}\boldsymbol{\div}{b}\right)\boldsymbol{\div}\left({c}\boldsymbol{\div}{d}\right) \\ $$$$\mathrm{obviously}\:{b}\:``\mathrm{makes}\:{a}\:\mathrm{smaller}''\:\mathrm{and}\:{c}\:\mathrm{would} \\ $$$$\mathrm{also}.\:\mathrm{but}\:\mathrm{first}\:\left(\mathrm{brackets}\:\mathrm{first}!\right)\:{d}\:``\mathrm{makes}\:{c} \\ $$$$\mathrm{smaller}''\:\Rightarrow\:``{d}\:\mathrm{helps}\:{a}'' \\ $$$$\Rightarrow \\ $$$$\left({a}\boldsymbol{\div}{b}\right)\boldsymbol{\div}\left({c}\boldsymbol{\div}{d}\right)={a}\boldsymbol{\div}{b}\boldsymbol{\div}{c}×{d}={a}×{d}\boldsymbol{\div}{b}\boldsymbol{\div}{c}= \\ $$$$={a}×{d}\boldsymbol{\div}\left({b}×{c}\right)=\frac{{ad}}{{bc}}=\frac{{a}}{{b}}×\frac{{d}}{{c}} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{similar}\:\mathrm{to} \\ $$$$\left({a}−{b}\right)−\left({b}−{c}\right)={a}−{b}−{c}+{d}={a}+{d}−{b}−{c}= \\ $$$$=\left({a}+{d}\right)−\left({b}+{c}\right) \\ $$

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