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Question Number 45014 by peter frank last updated on 07/Oct/18
6÷2(1+2)? which ans is correct between 1or 9  why do i look at the mobile phone calculator is 9 and when i check to scientific calculator the ans is 1.i want to clarify correctly where you are.
$$\mathrm{6}\boldsymbol{\div}\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right)?\:\mathrm{which}\:\boldsymbol{\mathrm{ans}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{between}}\:\mathrm{1}\boldsymbol{\mathrm{or}}\:\mathrm{9} \\ $$$$\boldsymbol{\mathrm{why}}\:\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{look}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{mobile}}\:\boldsymbol{\mathrm{phone}}\:\boldsymbol{\mathrm{calculator}}\:\boldsymbol{\mathrm{is}}\:\mathrm{9}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{check}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{scientific}}\:\boldsymbol{\mathrm{calculator}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{ans}}\:\boldsymbol{\mathrm{is}}\:\mathrm{1}.\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{want}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{clarify}}\:\boldsymbol{\mathrm{correctly}}\:\boldsymbol{\mathrm{where}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{are}}. \\ $$
Commented by peter frank last updated on 07/Oct/18
ajfour,tanmay,MJS,Mr_  w_3 .
$$\mathrm{ajfour},\mathrm{tanmay},\mathrm{MJS},\mathrm{Mr}_{\:} \underset{\mathrm{3}} {\mathrm{w}}. \\ $$
Commented by peter frank last updated on 07/Oct/18
why are the answers different.
$$\mathrm{why}\:\mathrm{are}\:\mathrm{the}\:\mathrm{answers}\:\mathrm{different}. \\ $$
Commented by MJS last updated on 07/Oct/18
there′s no priority for implied multiplication  although some think there is  6÷2(1+2)=6÷2×(1+2)=9  to get 1 you must write 6÷(2(1+2)) or  6÷(2×(1+2)) or (6/(2×(1+2)))
$$\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{priority}\:\mathrm{for}\:\mathrm{implied}\:\mathrm{multiplication} \\ $$$$\mathrm{although}\:\mathrm{some}\:\mathrm{think}\:\mathrm{there}\:\mathrm{is} \\ $$$$\mathrm{6}\boldsymbol{\div}\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right)=\mathrm{6}\boldsymbol{\div}\mathrm{2}×\left(\mathrm{1}+\mathrm{2}\right)=\mathrm{9} \\ $$$$\mathrm{to}\:\mathrm{get}\:\mathrm{1}\:\mathrm{you}\:\mathrm{must}\:\mathrm{write}\:\mathrm{6}\boldsymbol{\div}\left(\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right)\right)\:\mathrm{or} \\ $$$$\mathrm{6}\boldsymbol{\div}\left(\mathrm{2}×\left(\mathrm{1}+\mathrm{2}\right)\right)\:\mathrm{or}\:\frac{\mathrm{6}}{\mathrm{2}×\left(\mathrm{1}+\mathrm{2}\right)} \\ $$
Commented by MJS last updated on 08/Oct/18
Sir Peter Frank, which scientific calculator  do you use?
$$\mathrm{Sir}\:\mathrm{Peter}\:\mathrm{Frank},\:\mathrm{which}\:\mathrm{scientific}\:\mathrm{calculator} \\ $$$$\mathrm{do}\:\mathrm{you}\:\mathrm{use}? \\ $$
Commented by peter frank last updated on 08/Oct/18
casio fx Ms 991
$$\mathrm{casio}\:\mathrm{fx}\:\mathrm{Ms}\:\mathrm{991} \\ $$
Commented by MJS last updated on 08/Oct/18
I just read the manual online and obviously  the calculator is programed to first calculate  terms like 2π or 5(√3) which seems to be a  simplification but it leads to mistakes.  what does this mean:  1/2π+1/3π  standard math is definite and precise  1/2π+1/3π=(1/2)π+(1/3)π=(5/6)π  1/(2π)+1/(3π)=(1/(2π))+(1/(3π))=(5/(6π))  1/2(√3)+5=(1/2)(√3)+5 not (1/(2(√3)))+5 and not (1/(2(√3)+5))  1/2(π+k)=(1/2)(π+k)  sorry to say that casio calculators are often  wrong. buy a texas instruments or a hp unit  or use a good app like HiPER calc pro
$$\mathrm{I}\:\mathrm{just}\:\mathrm{read}\:\mathrm{the}\:\mathrm{manual}\:\mathrm{online}\:\mathrm{and}\:\mathrm{obviously} \\ $$$$\mathrm{the}\:\mathrm{calculator}\:\mathrm{is}\:\mathrm{programed}\:\mathrm{to}\:\mathrm{first}\:\mathrm{calculate} \\ $$$$\mathrm{terms}\:\mathrm{like}\:\mathrm{2}\pi\:\mathrm{or}\:\mathrm{5}\sqrt{\mathrm{3}}\:\mathrm{which}\:{seems}\:\mathrm{to}\:\mathrm{be}\:\mathrm{a} \\ $$$$\mathrm{simplification}\:\mathrm{but}\:\mathrm{it}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{mistakes}. \\ $$$$\mathrm{what}\:\mathrm{does}\:\mathrm{this}\:\mathrm{mean}: \\ $$$$\mathrm{1}/\mathrm{2}\pi+\mathrm{1}/\mathrm{3}\pi \\ $$$$\mathrm{standard}\:\mathrm{math}\:\mathrm{is}\:\mathrm{definite}\:\mathrm{and}\:\mathrm{precise} \\ $$$$\mathrm{1}/\mathrm{2}\pi+\mathrm{1}/\mathrm{3}\pi=\frac{\mathrm{1}}{\mathrm{2}}\pi+\frac{\mathrm{1}}{\mathrm{3}}\pi=\frac{\mathrm{5}}{\mathrm{6}}\pi \\ $$$$\mathrm{1}/\left(\mathrm{2}\pi\right)+\mathrm{1}/\left(\mathrm{3}\pi\right)=\frac{\mathrm{1}}{\mathrm{2}\pi}+\frac{\mathrm{1}}{\mathrm{3}\pi}=\frac{\mathrm{5}}{\mathrm{6}\pi} \\ $$$$\mathrm{1}/\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{5}=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}}+\mathrm{5}\:\mathrm{not}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}}+\mathrm{5}\:\mathrm{and}\:\mathrm{not}\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{5}} \\ $$$$\mathrm{1}/\mathrm{2}\left(\pi+{k}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\pi+{k}\right) \\ $$$$\mathrm{sorry}\:\mathrm{to}\:\mathrm{say}\:\mathrm{that}\:\mathrm{casio}\:\mathrm{calculators}\:\mathrm{are}\:\mathrm{often} \\ $$$$\mathrm{wrong}.\:\mathrm{buy}\:\mathrm{a}\:\mathrm{texas}\:\mathrm{instruments}\:\mathrm{or}\:\mathrm{a}\:\mathrm{hp}\:\mathrm{unit} \\ $$$$\mathrm{or}\:\mathrm{use}\:\mathrm{a}\:\mathrm{good}\:\mathrm{app}\:\mathrm{like}\:\mathrm{HiPER}\:\mathrm{calc}\:\mathrm{pro} \\ $$
Answered by Joel578 last updated on 07/Oct/18
Bracket, then division, then multiplication  6÷2×3  =3×3  =9
$$\mathrm{Bracket},\:\mathrm{then}\:\mathrm{division},\:\mathrm{then}\:\mathrm{multiplication} \\ $$$$\mathrm{6}\boldsymbol{\div}\mathrm{2}×\mathrm{3} \\ $$$$=\mathrm{3}×\mathrm{3} \\ $$$$=\mathrm{9} \\ $$
Commented by MJS last updated on 08/Oct/18
sorry but that′s not standard mathematic  where have you learned this?
$$\mathrm{sorry}\:\mathrm{but}\:\mathrm{that}'\mathrm{s}\:\mathrm{not}\:\mathrm{standard}\:\mathrm{mathematic} \\ $$$$\mathrm{where}\:\mathrm{have}\:\mathrm{you}\:\mathrm{learned}\:\mathrm{this}? \\ $$
Commented by Rasheed.Sindhi last updated on 08/Oct/18
6÷2(1+2) and 6÷2×(1+2) are different.  In 6÷2(1+2),  2(1+2) is like a single   term which should be calculated first.  (In other words  x÷yz≠x÷y×z    ∗ x÷yz=(x/(yz))   ∗  x÷y×z=(x/y)×z=((xz)/y) )  So,  6÷2(1+2)=6÷2(3)=6÷6=1  (9 is not correct.)
$$\mathrm{6}\boldsymbol{\div}\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right)\:\mathrm{and}\:\mathrm{6}\boldsymbol{\div}\mathrm{2}×\left(\mathrm{1}+\mathrm{2}\right)\:\mathrm{are}\:\mathrm{different}. \\ $$$$\mathrm{In}\:\mathrm{6}\boldsymbol{\div}\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right),\:\:\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right)\:\mathrm{is}\:\mathrm{like}\:\mathrm{a}\:\mathrm{single}\: \\ $$$$\mathrm{term}\:\mathrm{which}\:\mathrm{should}\:\mathrm{be}\:\mathrm{calculated}\:\mathrm{first}. \\ $$$$\left(\mathrm{In}\:\mathrm{other}\:\mathrm{words}\:\:\mathrm{x}\boldsymbol{\div}\mathrm{yz}\neq\mathrm{x}\boldsymbol{\div}\mathrm{y}×\mathrm{z}\right. \\ $$$$\:\:\ast\:\mathrm{x}\boldsymbol{\div}\mathrm{yz}=\frac{\mathrm{x}}{\mathrm{yz}} \\ $$$$\left.\:\ast\:\:\mathrm{x}\boldsymbol{\div}\mathrm{y}×\mathrm{z}=\frac{\mathrm{x}}{\mathrm{y}}×\mathrm{z}=\frac{\mathrm{xz}}{\mathrm{y}}\:\right) \\ $$$$\mathrm{So}, \\ $$$$\mathrm{6}\boldsymbol{\div}\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right)=\mathrm{6}\boldsymbol{\div}\mathrm{2}\left(\mathrm{3}\right)=\mathrm{6}\boldsymbol{\div}\mathrm{6}=\mathrm{1} \\ $$$$\left(\mathrm{9}\:\mathrm{is}\:\mathrm{not}\:\mathrm{correct}.\right) \\ $$
Commented by Rasheed.Sindhi last updated on 08/Oct/18
In other words  implicit multiplication  has priority over division and division  have priority over explicit multiplication....  Although at the moment I haven′t a reference.
$$\mathrm{In}\:\mathrm{other}\:\mathrm{words}\:\:\mathrm{implicit}\:\mathrm{multiplication} \\ $$$$\mathrm{has}\:\mathrm{priority}\:\mathrm{over}\:\mathrm{division}\:\mathrm{and}\:\mathrm{division} \\ $$$$\mathrm{have}\:\mathrm{priority}\:\mathrm{over}\:\mathrm{explicit}\:\mathrm{multiplication}…. \\ $$$$\mathrm{Although}\:\mathrm{at}\:\mathrm{the}\:\mathrm{moment}\:\mathrm{I}\:\mathrm{haven}'\mathrm{t}\:\mathrm{a}\:\mathrm{reference}. \\ $$
Commented by MJS last updated on 08/Oct/18
multiplication and division have equal  priority, similar to addition and subtraction  5−3+7−8=((5−3)+7)−8=1  5/3×7/8=((5/3)×7)/8=((35)/(24))    implicit multiplication is just multiplication  where we left the “×” sign because it′s less  writing work. there′s no extra rule for it.  ab/cd=a×b/c×d must be calculated from  left to right. everything else is sloppy habit  people write ab/cd instead of ((ab)/(cd)), that′s the  initial problem. but we have to have one rule  if ((termA)/(termB))=termA/termB ⇔ ((4x+3)/(3x−4))=4x+3/3x−4  and confusion is complete  instead it′s ((termA)/(termB))=(termA)/(termB)  ⇒ 6÷2(1+2)≠(6/(2(1+2)))
$$\mathrm{multiplication}\:\mathrm{and}\:\mathrm{division}\:\mathrm{have}\:\mathrm{equal} \\ $$$$\mathrm{priority},\:\mathrm{similar}\:\mathrm{to}\:\mathrm{addition}\:\mathrm{and}\:\mathrm{subtraction} \\ $$$$\mathrm{5}−\mathrm{3}+\mathrm{7}−\mathrm{8}=\left(\left(\mathrm{5}−\mathrm{3}\right)+\mathrm{7}\right)−\mathrm{8}=\mathrm{1} \\ $$$$\mathrm{5}/\mathrm{3}×\mathrm{7}/\mathrm{8}=\left(\left(\mathrm{5}/\mathrm{3}\right)×\mathrm{7}\right)/\mathrm{8}=\frac{\mathrm{35}}{\mathrm{24}} \\ $$$$ \\ $$$$\mathrm{implicit}\:\mathrm{multiplication}\:\mathrm{is}\:\mathrm{just}\:\mathrm{multiplication} \\ $$$$\mathrm{where}\:\mathrm{we}\:\mathrm{left}\:\mathrm{the}\:“×''\:\mathrm{sign}\:\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{less} \\ $$$$\mathrm{writing}\:\mathrm{work}.\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{extra}\:\mathrm{rule}\:\mathrm{for}\:\mathrm{it}. \\ $$$${ab}/{cd}={a}×{b}/{c}×{d}\:\mathrm{must}\:\mathrm{be}\:\mathrm{calculated}\:\mathrm{from} \\ $$$$\mathrm{left}\:\mathrm{to}\:\mathrm{right}.\:\mathrm{everything}\:\mathrm{else}\:\mathrm{is}\:\mathrm{sloppy}\:\mathrm{habit} \\ $$$$\mathrm{people}\:\mathrm{write}\:{ab}/{cd}\:\mathrm{instead}\:\mathrm{of}\:\frac{{ab}}{{cd}},\:\mathrm{that}'\mathrm{s}\:\mathrm{the} \\ $$$$\mathrm{initial}\:\mathrm{problem}.\:\mathrm{but}\:\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{have}\:{one}\:\mathrm{rule} \\ $$$$\mathrm{if}\:\frac{{termA}}{{termB}}={termA}/{termB}\:\Leftrightarrow\:\frac{\mathrm{4}{x}+\mathrm{3}}{\mathrm{3}{x}−\mathrm{4}}=\mathrm{4}{x}+\mathrm{3}/\mathrm{3}{x}−\mathrm{4} \\ $$$$\mathrm{and}\:\mathrm{confusion}\:\mathrm{is}\:\mathrm{complete} \\ $$$$\mathrm{instead}\:\mathrm{it}'\mathrm{s}\:\frac{{termA}}{{termB}}=\left({termA}\right)/\left({termB}\right) \\ $$$$\Rightarrow\:\mathrm{6}\boldsymbol{\div}\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right)\neq\frac{\mathrm{6}}{\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right)} \\ $$
Commented by MJS last updated on 08/Oct/18
another argument:  implied multiplication is not defined for  a, b ∈R. or how can we write ab for a=7.43  and b=−2.56?  it′s just writing shortcut, not an operation
$$\mathrm{another}\:\mathrm{argument}: \\ $$$$\mathrm{implied}\:\mathrm{multiplication}\:\mathrm{is}\:\mathrm{not}\:\mathrm{defined}\:\mathrm{for} \\ $$$${a},\:{b}\:\in\mathbb{R}.\:\mathrm{or}\:\mathrm{how}\:\mathrm{can}\:\mathrm{we}\:\mathrm{write}\:{ab}\:\mathrm{for}\:{a}=\mathrm{7}.\mathrm{43} \\ $$$$\mathrm{and}\:{b}=−\mathrm{2}.\mathrm{56}? \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{writing}\:\mathrm{shortcut},\:\mathrm{not}\:\mathrm{an}\:\mathrm{operation} \\ $$
Commented by ajfour last updated on 08/Oct/18
where have you been,all this while,   Rasheed Sir?
$${where}\:{have}\:{you}\:{been},{all}\:{this}\:{while},\: \\ $$$${Rasheed}\:{Sir}? \\ $$
Commented by MJS last updated on 08/Oct/18
there are some “bad spellings”, most of them  sloppiness. but we learned in school things  like  ((35)/4)=8(3/4)  this is just for children to make it visible for  them  standard math for adults ;−)  8(3/4)=8×(3/4)=6  because otherwise: confusion. solve  x(3/4)=6  hopefully it′s x=8 not x=((21)/4)  also it′s not possible to calculate with these  mixed fractions  8(3/4)±3(5/6)=  8(3/4)×3(5/6)=  8(3/4)÷3(5/6)=  we have to use standard fractions and standard  maths
$$\mathrm{there}\:\mathrm{are}\:\mathrm{some}\:“\mathrm{bad}\:\mathrm{spellings}'',\:\mathrm{most}\:\mathrm{of}\:\mathrm{them} \\ $$$$\mathrm{sloppiness}.\:\mathrm{but}\:\mathrm{we}\:\mathrm{learned}\:\mathrm{in}\:\mathrm{school}\:\mathrm{things} \\ $$$$\mathrm{like} \\ $$$$\frac{\mathrm{35}}{\mathrm{4}}=\mathrm{8}\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{just}\:\mathrm{for}\:\mathrm{children}\:\mathrm{to}\:\mathrm{make}\:\mathrm{it}\:\mathrm{visible}\:\mathrm{for} \\ $$$$\mathrm{them} \\ $$$$\left.\mathrm{standard}\:\mathrm{math}\:\mathrm{for}\:\mathrm{adults}\:;−\right) \\ $$$$\mathrm{8}\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{8}×\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{6} \\ $$$$\mathrm{because}\:\mathrm{otherwise}:\:\mathrm{confusion}.\:\mathrm{solve} \\ $$$${x}\frac{\mathrm{3}}{\mathrm{4}}=\mathrm{6} \\ $$$$\mathrm{hopefully}\:\mathrm{it}'\mathrm{s}\:{x}=\mathrm{8}\:\mathrm{not}\:{x}=\frac{\mathrm{21}}{\mathrm{4}} \\ $$$$\mathrm{also}\:\mathrm{it}'\mathrm{s}\:\mathrm{not}\:\mathrm{possible}\:\mathrm{to}\:\mathrm{calculate}\:\mathrm{with}\:\mathrm{these} \\ $$$$\mathrm{mixed}\:\mathrm{fractions} \\ $$$$\mathrm{8}\frac{\mathrm{3}}{\mathrm{4}}\pm\mathrm{3}\frac{\mathrm{5}}{\mathrm{6}}= \\ $$$$\mathrm{8}\frac{\mathrm{3}}{\mathrm{4}}×\mathrm{3}\frac{\mathrm{5}}{\mathrm{6}}= \\ $$$$\mathrm{8}\frac{\mathrm{3}}{\mathrm{4}}\boldsymbol{\div}\mathrm{3}\frac{\mathrm{5}}{\mathrm{6}}= \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{to}\:\mathrm{use}\:\mathrm{standard}\:\mathrm{fractions}\:\mathrm{and}\:\mathrm{standard} \\ $$$$\mathrm{maths} \\ $$
Commented by Rasheed.Sindhi last updated on 08/Oct/18
Thanks Sir ajfour for missing me!!!  Actually unfortunately I lost my interest  for mathematics! I′ll try to participate  in the forum again.   Thanks again!
$$\mathcal{T}{hanks}\:{Sir}\:{ajfour}\:{for}\:{missing}\:{me}!!! \\ $$$${Actually}\:{unfortunately}\:{I}\:{lost}\:{my}\:{interest} \\ $$$${for}\:{mathematics}!\:{I}'{ll}\:{try}\:{to}\:{participate} \\ $$$${in}\:{the}\:{forum}\:{again}.\: \\ $$$$\mathcal{T}{hanks}\:{again}! \\ $$
Commented by MJS last updated on 08/Oct/18
there′s an international standard, is all I′m  saying.  (1/2) is an entity, I can write 1/2  2π is an entity, I can write (1/(2π)) or 1/2π  in 1/2π, which entity has priority?  solve this:  x^2 +1/2x+1/4=0  is it obviously a 2^(nd)  degree polynome? or is  1/2x a division by  2x?  and this: ∫dx/x^2 +1/2x+1/4  or sin x+1/2  or (√)2x−3  I′ve seen similar expressions in this forum  and I refused to solve them sometimes  because it′s never obvious when it comes  to mathematics. that′s the reason I plead  in favour of the standard spelling of maths
$$\mathrm{there}'\mathrm{s}\:\mathrm{an}\:\mathrm{international}\:\mathrm{standard},\:\mathrm{is}\:\mathrm{all}\:\mathrm{I}'\mathrm{m} \\ $$$$\mathrm{saying}. \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{is}\:\mathrm{an}\:\mathrm{entity},\:\mathrm{I}\:\mathrm{can}\:\mathrm{write}\:\mathrm{1}/\mathrm{2} \\ $$$$\mathrm{2}\pi\:\mathrm{is}\:\mathrm{an}\:\mathrm{entity},\:\mathrm{I}\:\mathrm{can}\:\mathrm{write}\:\frac{\mathrm{1}}{\mathrm{2}\pi}\:\mathrm{or}\:\mathrm{1}/\mathrm{2}\pi \\ $$$$\mathrm{in}\:\mathrm{1}/\mathrm{2}\pi,\:\mathrm{which}\:\mathrm{entity}\:\mathrm{has}\:\mathrm{priority}? \\ $$$$\mathrm{solve}\:\mathrm{this}: \\ $$$${x}^{\mathrm{2}} +\mathrm{1}/\mathrm{2}{x}+\mathrm{1}/\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{is}\:\mathrm{it}\:\mathrm{obviously}\:\mathrm{a}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{degree}\:\mathrm{polynome}?\:\mathrm{or}\:\mathrm{is} \\ $$$$\mathrm{1}/\mathrm{2}{x}\:\mathrm{a}\:\mathrm{division}\:\mathrm{by}\:\:\mathrm{2}{x}? \\ $$$$\mathrm{and}\:\mathrm{this}:\:\int{dx}/{x}^{\mathrm{2}} +\mathrm{1}/\mathrm{2}{x}+\mathrm{1}/\mathrm{4} \\ $$$$\mathrm{or}\:\mathrm{sin}\:{x}+\mathrm{1}/\mathrm{2} \\ $$$$\mathrm{or}\:\sqrt{}\mathrm{2}{x}−\mathrm{3} \\ $$$$\mathrm{I}'\mathrm{ve}\:\mathrm{seen}\:\mathrm{similar}\:\mathrm{expressions}\:\mathrm{in}\:\mathrm{this}\:\mathrm{forum} \\ $$$$\mathrm{and}\:\mathrm{I}\:\mathrm{refused}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{them}\:\mathrm{sometimes} \\ $$$$\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{never}\:\mathrm{obvious}\:\mathrm{when}\:\mathrm{it}\:\mathrm{comes} \\ $$$$\mathrm{to}\:\mathrm{mathematics}.\:\mathrm{that}'\mathrm{s}\:\mathrm{the}\:\mathrm{reason}\:\mathrm{I}\:\mathrm{plead} \\ $$$$\mathrm{in}\:\mathrm{favour}\:\mathrm{of}\:\mathrm{the}\:\mathrm{standard}\:\mathrm{spelling}\:\mathrm{of}\:\mathrm{maths} \\ $$
Commented by MrW3 last updated on 08/Oct/18
Sir Rasheed:  You are missed by me too. Welcome   back!
$${Sir}\:{Rasheed}: \\ $$$${You}\:{are}\:{missed}\:{by}\:{me}\:{too}.\:{Welcome}\: \\ $$$${back}! \\ $$
Commented by Rasheed.Sindhi last updated on 09/Oct/18
Thanks sir MrW3.  Sir MJS  Same discussion is present on the  following link(see comments below)  which suggest that there′s no agreed  international standard in this connection.  Both conventions are continue simultaneously!!
$$\mathcal{T}{hanks}\:\boldsymbol{{sir}}\:\boldsymbol{{MrW}}\mathrm{3}. \\ $$$$\boldsymbol{{Sir}}\:\boldsymbol{{MJS}} \\ $$$${Same}\:{discussion}\:{is}\:{present}\:{on}\:{the} \\ $$$${following}\:{link}\left({see}\:{comments}\:{below}\right) \\ $$$${which}\:{suggest}\:{that}\:{there}'{s}\:{no}\:{agreed} \\ $$$${international}\:{standard}\:{in}\:{this}\:{connection}. \\ $$$${Both}\:{conventions}\:{are}\:{continue}\:{simultaneously}!! \\ $$
Commented by Rasheed.Sindhi last updated on 09/Oct/18
Commented by Rasheed.Sindhi last updated on 09/Oct/18

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