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

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

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

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

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

$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{priority}\mathrm{for}\mathrm{implied}\mathrm{multiplication}$$$$\mathrm{although}\mathrm{some}\mathrm{think}\mathrm{there}\mathrm{is}$$$$6\text{boldsymbol}÷2(1+2)=6\text{boldsymbol}÷2\times (1+2)=9$$$$\mathrm{to}\mathrm{get}1\mathrm{you}\mathrm{must}\mathrm{write}6\text{boldsymbol}÷\left(2(1+2)\right)\mathrm{or}$$$$6\text{boldsymbol}÷\left(2\times (1+2)\right)\mathrm{or}\frac{6}{2\times (1+2)}$$

Commented by MJS last updated on 08/Oct/18

$$\mathrm{Sir}\mathrm{Peter}\mathrm{Frank},\mathrm{which}\mathrm{scientific}\mathrm{calculator}$$$$\mathrm{do}\mathrm{you}\mathrm{use}?$$

Commented by peter frank last updated on 08/Oct/18

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

Commented by MJS last updated on 08/Oct/18

$$\mathrm{I}\mathrm{just}\mathrm{read}\mathrm{the}\mathrm{manual}\mathrm{online}\mathrm{and}\mathrm{obviously}$$$$\mathrm{the}\mathrm{calculator}\mathrm{is}\mathrm{programed}\mathrm{to}\mathrm{first}\mathrm{calculate}$$$$\mathrm{terms}\mathrm{like}2\pi \mathrm{or}5\sqrt{3}\mathrm{which}seems\mathrm{to}\mathrm{be}\mathrm{a}$$$$\mathrm{simplification}\mathrm{but}\mathrm{it}\mathrm{leads}\mathrm{to}\mathrm{mistakes}.$$$$\mathrm{what}\mathrm{does}\mathrm{this}\mathrm{mean}:$$$$1/2\pi +1/3\pi$$$$\mathrm{standard}\mathrm{math}\mathrm{is}\mathrm{definite}\mathrm{and}\mathrm{precise}$$$$1/2\pi +1/3\pi =\frac{1}{2}\pi +\frac{1}{3}\pi =\frac{5}{6}\pi$$$$1/\left(2\pi \right)+1/\left(3\pi \right)=\frac{1}{2\pi }+\frac{1}{3\pi }=\frac{5}{6\pi }$$$$1/2\sqrt{3}+5=\frac{1}{2}\sqrt{3}+5\mathrm{not}\frac{1}{2\sqrt{3}}+5\mathrm{and}\mathrm{not}\frac{1}{2\sqrt{3}+5}$$$$1/2(\pi +k)=\frac{1}{2}(\pi +k)$$$$\mathrm{sorry}\mathrm{to}\mathrm{say}\mathrm{that}\mathrm{casio}\mathrm{calculators}\mathrm{are}\mathrm{often}$$$$\mathrm{wrong}.\mathrm{buy}\mathrm{a}\mathrm{texas}\mathrm{instruments}\mathrm{or}\mathrm{a}\mathrm{hp}\mathrm{unit}$$$$\mathrm{or}\mathrm{use}\mathrm{a}\mathrm{good}\mathrm{app}\mathrm{like}\mathrm{HiPER}\mathrm{calc}\mathrm{pro}$$

Answered by Joel578 last updated on 07/Oct/18

$$\mathrm{Bracket},\mathrm{then}\mathrm{division},\mathrm{then}\mathrm{multiplication}$$$$6\text{boldsymbol}÷2\times 3$$$$=3\times 3$$$$=9$$

Commented by MJS last updated on 08/Oct/18

$$\mathrm{sorry}\mathrm{but}{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{standard}\mathrm{mathematic}$$$$\mathrm{where}\mathrm{have}\mathrm{you}\mathrm{learned}\mathrm{this}?$$

Commented by Rasheed.Sindhi last updated on 08/Oct/18

$$6\text{boldsymbol}÷2(1+2)\mathrm{and}6\text{boldsymbol}÷2\times (1+2)\mathrm{are}\mathrm{different}.$$$$\mathrm{In}6\text{boldsymbol}÷2(1+2),2(1+2)\mathrm{is}\mathrm{like}\mathrm{a}\mathrm{single}$$$$\mathrm{term}\mathrm{which}\mathrm{should}\mathrm{be}\mathrm{calculated}\mathrm{first}.$$$$(\mathrm{In}\mathrm{other}\mathrm{words}\mathrm{x}\text{boldsymbol}÷\mathrm{yz}\ne \mathrm{x}\text{boldsymbol}÷\mathrm{y}\times \mathrm{z}$$$$\ast \mathrm{x}\text{boldsymbol}÷\mathrm{yz}=\frac{\mathrm{x}}{\mathrm{yz}}$$$$\ast \mathrm{x}\text{boldsymbol}÷\mathrm{y}\times \mathrm{z}=\frac{\mathrm{x}}{\mathrm{y}}\times \mathrm{z}=\frac{\mathrm{xz}}{\mathrm{y}})$$$$\mathrm{So},$$$$6\text{boldsymbol}÷2(1+2)=6\text{boldsymbol}÷2\left(3\right)=6\text{boldsymbol}÷6=1$$$$(9\mathrm{is}\mathrm{not}\mathrm{correct}.)$$

Commented by Rasheed.Sindhi last updated on 08/Oct/18

$$\mathrm{In}\mathrm{other}\mathrm{words}\mathrm{implicit}\mathrm{multiplication}$$$$\mathrm{has}\mathrm{priority}\mathrm{over}\mathrm{division}\mathrm{and}\mathrm{division}$$$$\mathrm{have}\mathrm{priority}\mathrm{over}\mathrm{explicit}\mathrm{multiplication}\dots .$$$$\mathrm{Although}\mathrm{at}\mathrm{the}\mathrm{moment}\mathrm{I}{\mathrm{haven}}^{\prime }\mathrm{t}\mathrm{a}\mathrm{reference}.$$

Commented by MJS last updated on 08/Oct/18

$$\mathrm{multiplication}\mathrm{and}\mathrm{division}\mathrm{have}\mathrm{equal}$$$$\mathrm{priority},\mathrm{similar}\mathrm{to}\mathrm{addition}\mathrm{and}\mathrm{subtraction}$$$$5-3+7-8=((5-3)+7)-8=1$$$$5/3\times 7/8=\left(\left(5/3\right)\times 7\right)/8=\frac{35}{24}$$$$$$$$\mathrm{implicit}\mathrm{multiplication}\mathrm{is}\mathrm{just}\mathrm{multiplication}$$$$\mathrm{where}\mathrm{we}\mathrm{left}\mathrm{the}“{\times }^{″}\mathrm{sign}\mathrm{because}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{less}$$$$\mathrm{writing}\mathrm{work}.{\mathrm{there}}^{\prime }\mathrm{s}\mathrm{no}\mathrm{extra}\mathrm{rule}\mathrm{for}\mathrm{it}.$$$$ab/cd=a\times b/c\times d\mathrm{must}\mathrm{be}\mathrm{calculated}\mathrm{from}$$$$\mathrm{left}\mathrm{to}\mathrm{right}.\mathrm{everything}\mathrm{else}\mathrm{is}\mathrm{sloppy}\mathrm{habit}$$$$\mathrm{people}\mathrm{write}ab/cd\mathrm{instead}\mathrm{of}\frac{ab}{cd},{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{the}$$$$\mathrm{initial}\mathrm{problem}.\mathrm{but}\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{have}one\mathrm{rule}$$$$\mathrm{if}\frac{termA}{termB}=termA/termB\iff \frac{4x+3}{3x-4}=4x+3/3x-4$$$$\mathrm{and}\mathrm{confusion}\mathrm{is}\mathrm{complete}$$$$\mathrm{instead}{\mathrm{it}}^{\prime }\mathrm{s}\frac{termA}{termB}=\left(termA\right)/\left(termB\right)$$$$\Rightarrow 6\text{boldsymbol}÷2(1+2)\ne \frac{6}{2(1+2)}$$

Commented by MJS last updated on 08/Oct/18

$$\mathrm{another}\mathrm{argument}:$$$$\mathrm{implied}\mathrm{multiplication}\mathrm{is}\mathrm{not}\mathrm{defined}\mathrm{for}$$$$a,b\in \mathbb{R}.\mathrm{or}\mathrm{how}\mathrm{can}\mathrm{we}\mathrm{write}ab\mathrm{for}a=7.43$$$$\mathrm{and}b=-2.56?$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{just}\mathrm{writing}\mathrm{shortcut},\mathrm{not}\mathrm{an}\mathrm{operation}$$

Commented by ajfour last updated on 08/Oct/18

$$wherehaveyoubeen,allthiswhile,$$$$RasheedSir?$$

Commented by MJS last updated on 08/Oct/18

$$\mathrm{there}\mathrm{are}\mathrm{some}“\mathrm{bad}{\mathrm{spellings}}^{″},\mathrm{most}\mathrm{of}\mathrm{them}$$$$\mathrm{sloppiness}.\mathrm{but}\mathrm{we}\mathrm{learned}\mathrm{in}\mathrm{school}\mathrm{things}$$$$\mathrm{like}$$$$\frac{35}{4}=8\frac{3}{4}$$$$\mathrm{this}\mathrm{is}\mathrm{just}\mathrm{for}\mathrm{children}\mathrm{to}\mathrm{make}\mathrm{it}\mathrm{visible}\mathrm{for}$$$$\mathrm{them}$$$$\mathrm{standard}\mathrm{math}\mathrm{for}\mathrm{adults};-)$$$$8\frac{3}{4}=8\times \frac{3}{4}=6$$$$\mathrm{because}\mathrm{otherwise}:\mathrm{confusion}.\mathrm{solve}$$$$x\frac{3}{4}=6$$$$\mathrm{hopefully}{\mathrm{it}}^{\prime }\mathrm{s}x=8\mathrm{not}x=\frac{21}{4}$$$$\mathrm{also}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{not}\mathrm{possible}\mathrm{to}\mathrm{calculate}\mathrm{with}\mathrm{these}$$$$\mathrm{mixed}\mathrm{fractions}$$$$8\frac{3}{4}\pm 3\frac{5}{6}=$$$$8\frac{3}{4}\times 3\frac{5}{6}=$$$$8\frac{3}{4}\text{boldsymbol}÷3\frac{5}{6}=$$$$\mathrm{we}\mathrm{have}\mathrm{to}\mathrm{use}\mathrm{standard}\mathrm{fractions}\mathrm{and}\mathrm{standard}$$$$\mathrm{maths}$$

Commented by Rasheed.Sindhi last updated on 08/Oct/18

$$\mathcal{T}hanksSirajfourformissingme!!!$$$$ActuallyunfortunatelyIlostmyinterest$$$$formathematics!I^{\prime }lltrytoparticipate$$$$intheforumagain.$$$$\mathcal{T}hanksagain!$$

Commented by MJS last updated on 08/Oct/18

$${\mathrm{there}}^{\prime }\mathrm{s}\mathrm{an}\mathrm{international}\mathrm{standard},\mathrm{is}\mathrm{all}{\mathrm{I}}^{\prime }\mathrm{m}$$$$\mathrm{saying}.$$$$\frac{1}{2}\mathrm{is}\mathrm{an}\mathrm{entity},\mathrm{I}\mathrm{can}\mathrm{write}1/2$$$$2\pi \mathrm{is}\mathrm{an}\mathrm{entity},\mathrm{I}\mathrm{can}\mathrm{write}\frac{1}{2\pi }\mathrm{or}1/2\pi$$$$\mathrm{in}1/2\pi ,\mathrm{which}\mathrm{entity}\mathrm{has}\mathrm{priority}?$$$$\mathrm{solve}\mathrm{this}:$$$$x^2+1/2x+1/4=0$$$$\mathrm{is}\mathrm{it}\mathrm{obviously}\mathrm{a}{2}^{\mathrm{nd}}\mathrm{degree}\mathrm{polynome}?\mathrm{or}\mathrm{is}$$$$1/2x\mathrm{a}\mathrm{division}\mathrm{by}2x?$$$$\mathrm{and}\mathrm{this}:\int dx/x^2+1/2x+1/4$$$$\mathrm{or}\sin x+1/2$$$$\mathrm{or}\sqrt{}2x-3$$$${\mathrm{I}}^{\prime }\mathrm{ve}\mathrm{seen}\mathrm{similar}\mathrm{expressions}\mathrm{in}\mathrm{this}\mathrm{forum}$$$$\mathrm{and}\mathrm{I}\mathrm{refused}\mathrm{to}\mathrm{solve}\mathrm{them}\mathrm{sometimes}$$$$\mathrm{because}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{never}\mathrm{obvious}\mathrm{when}\mathrm{it}\mathrm{comes}$$$$\mathrm{to}\mathrm{mathematics}.{\mathrm{that}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{reason}\mathrm{I}\mathrm{plead}$$$$\mathrm{in}\mathrm{favour}\mathrm{of}\mathrm{the}\mathrm{standard}\mathrm{spelling}\mathrm{of}\mathrm{maths}$$

Commented by MrW3 last updated on 08/Oct/18

$$SirRasheed:$$$$Youaremissedbymetoo.Welcome$$$$back!$$

Commented by Rasheed.Sindhi last updated on 09/Oct/18

$$\mathcal{T}hanks\text{boldsymbol}sir\text{boldsymbol}MrW3.$$$$\text{boldsymbol}Sir\text{boldsymbol}MJS$$$$Samediscussionispresentonthe$$$$followinglink\left(seecommentsbelow\right)$$$$whichsuggestthat{there}^{\prime }snoagreed$$$$internationalstandardinthisconnection.$$$$Bothconventionsarecontinuesimultaneously!!$$

Commented by Rasheed.Sindhi last updated on 09/Oct/18

Commented by Rasheed.Sindhi last updated on 09/Oct/18

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