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-

Question Number 45014 by peter frank last updated on 07/Oct/18

$$\mathrm{6}\boldsymbol{\div}\mathrm{2}\left(\mathrm{1}+\mathrm{2}\right)?\:\mathrm{which}\:\boldsymbol{\mathrm{ans}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{correct}}\:\boldsymbol{\mathrm{between}}\:\mathrm{1}\boldsymbol{\mathrm{or}}\:\mathrm{9} \\ $$$$\boldsymbol{\mathrm{why}}\:\boldsymbol{\mathrm{do}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{look}}\:\boldsymbol{\mathrm{at}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{mobile}}\:\boldsymbol{\mathrm{phone}}\:\boldsymbol{\mathrm{calculator}}\:\boldsymbol{\mathrm{is}}\:\mathrm{9}\:\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{when}}\:\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{check}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{scientific}}\:\boldsymbol{\mathrm{calculator}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{ans}}\:\boldsymbol{\mathrm{is}}\:\mathrm{1}.\boldsymbol{\mathrm{i}}\:\boldsymbol{\mathrm{want}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{clarify}}\:\boldsymbol{\mathrm{correctly}}\:\boldsymbol{\mathrm{where}}\:\boldsymbol{\mathrm{you}}\:\boldsymbol{\mathrm{are}}. \\ $$

Commented by peter frank last updated on 07/Oct/18

$$\mathrm{ajfour},\mathrm{tanmay},\mathrm{MJS},\mathrm{Mr}_{\:} \underset{\mathrm{3}} {\mathrm{w}}. \\ $$

Commented by peter frank last updated on 07/Oct/18

$$\mathrm{why}\:\mathrm{are}\:\mathrm{the}\:\mathrm{answers}\:\mathrm{different}. \\ $$

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

$$\mathrm{casio}\:\mathrm{fx}\:\mathrm{Ms}\:\mathrm{991} \\ $$

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

$${where}\:{have}\:{you}\:{been},{all}\:{this}\:{while},\: \\ $$$${Rasheed}\:{Sir}? \\ $$

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

$${Sir}\:{Rasheed}: \\ $$$${You}\:{are}\:{missed}\:{by}\:{me}\:{too}.\:{Welcome}\: \\ $$$${back}! \\ $$

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