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48÷2(9+3) and similar

  1. Apr 29, 2011 #1

    Hurkyl

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    An issue of mathematical grammar circulating the internet lately has been how to read a mathematical expression like
    [tex]48 \div 2 (9 + 3)[/tex]​
    that involves a combination of division and implied multiplication.

    The standard way to read arithmetic expression (i.e. order of operations) involves dealing with parentheses first, then you do all division and multiplication operations from left to right, then all addition and subtraction operations from left to right.

    So, this expression is computed as
    [tex]48 \div 2 (12)[/tex]
    [tex]24 (12)[/tex]
    [tex]288[/tex]

    It doesn't matter whether or not the multiplication is made explicit, as in
    [tex]48 \div 2 \times (9+3) = 288[/tex]​
    or if division is represented by a slash, as in
    [tex]48 / 2 (9+3) = 288[/tex]​
    all of these variations mean the same thing.



    One thing to keep in mind is that not everybody follows the standard. Some people prefer to do implied multiplication before other multiplication and division operations. Some people prefer to do all multiplications before division with /. Some people even prefer to do addition before division with /.

    So, when you are reading math from an unfamiliar source, make sure you know what convention they are adopting. And no matter what convention you prefer, you really ought to write things in an unambiguous fashion -- e.g. you should avoid
    [tex]48 / 2 (9+3)[/tex]​
    and instead use the crystal clear
    [tex](48 / 2) (9+3)[/tex]​
     
  2. jcsd
  3. May 20, 2016 #2
    I have a friend who argues for CPMD (commutative property of multiplication over division) when using PEMDAS. Basically, left to right, you solve for multiplication first, then go back and do division. PEMDAS came into being around 1986 and argues that multiplication and division hold the same priority, so for the famous equation 6/2(2+1), pre-1986 as he was taught you'd get 1; post 1986 using PEMDAS as I was taught, you'd get 9.

    This is a quandary as it does make a difference as evidenced by the two different answers.

    1. What happened to CPMD and why is it no longer important?
    2. Does it really matter as long as people read from the same book?
    3. Is 6/2(2+1) just a poorly written equation?
    4. Is the answer 1 or 9 as understood by modern experts?

    Published references would be appreciated as I've found much conflict in my internet searching on this. Thanks!
     
  4. May 20, 2016 #3

    micromass

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    That is not true. What is true is that there was a lot of confusion. There was no unique convention and all books recommended different things. So it is definitely not true that pre-1986, you would get 1 as answer. It is true that pre-1986, some books would tell you that 1 is the answer while some books would give you 9. Nowadays, there is a universal convention about this issue.

    http://www.math.ucdenver.edu/~jloats/Student pdfs/4_Order of OperationsSass.pdf

    2. No, it does not matter much.
    3. Yes it is very poorly written. Nobody would write something like this is current books.
    4. 9
     
  5. Dec 16, 2016 #4

    jack action

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    Since the thread started by @Averagesupernova was closed and the discussion was redirected here, I will continue with my 2 cents.

    About why ##6/2(1+2)## should equal ##9## instead of ##1##, here's how I see it:

    Everybody agrees ##10 - 3 + 4 = 11##. Nobody thinks that addition should be performed before the subtraction and get ##3## as the answer. The reasoning behind it, is that we are actually adding a negative number, i.e. ##10 + (- 3) + 4 = 11##.

    I can't see why this would be different for multiplication and division: When you do a division, you actually multiply the inverse of the following number, variable or parenthesis.

    The implied operator before the parenthesis is necessarily multiplication, so any of the following should be true:

    ##6/2 \times (1+2)##
    ##= 6 \times 2^{-1} \times (1+2)##
    ##= 6 \times \left(\frac{1}{2}\right) \times (1+2)##
    ##= 6 \times 0.5 \times (1+2)##

    Or even when converting multiplication to division:

    ##= 6 / 2 / (1+2)^{-1}##

    If you want to keep doing multiplication before division, then it would mean that:

    ##6/2 \times (1+2)##
    ##= 6 \times \left(2 \times (1+2)\right)^{-1}##

    Why should we need adding an extra set of parenthesis? Since when, an operator between two numbers has the «power» to include a third number, separated by another operator? It doesn't make any sense.

    Or, when converting multiplication to division, it is even worse:

    ##= 6 / 2 / (1+2)##

    Why don't we inverse ##(1+2)## ? Again, it doesn't make any sense. We apply one set of rules one way and another one when we go the other way.

    This is why I can't see any justification to support doing division before multiplication.
     
  6. Dec 16, 2016 #5
    Probably has something to do with my fav topic, peano arithmetic. :rolleyes::cool:
     
  7. Dec 16, 2016 #6

    Averagesupernova

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    I will admit that for the expression 6 / 2(1 + 2) to equal 1 it would be more properly written: 6 / (2(1 +2)).
    -
    So you could say:
    6 / 2 (1 + 2) is this:
    $$\left(\frac 6 2\right) \left(\frac {2(1+2)}1\right)$$
    and 6 / (2(1 + 2)) is this:
    $$ \frac 6 {2(1+2)} $$
    Obviously answers of 9 and 1 respectively. So the question is if it were written like this: 6 / 2(1 + X) = 1, what would one say X is? I would feel very safe in saying X = 2 if it were presented like this: 6 / (2(1 + X)) = 1. But, thinking about the alternative of 6 / 2(1 + X) = 1 this afternoon, I cannot say that without the outer parenthesis that X = 2. Yes, LaTex is new to me so I am cheating using a slant bar instead of the division symbol. It's been so long since I looked in an algebra text book that I can't say if I ever solved for X in an equation that was written out in a single line other than something like 5 - X = 2. So, given that it could be a requirement to include the extra set of parenthesis in single line form, why is it then acceptable to write it this way:
    $$ \frac 6 {2(1+2)} $$
    instead of this way:
    $$ \frac 6 {(2(1+2))} $$

    About the only time this matters to me at all is writing software. For the record, I am big on parenthesis when writing code. More than anything it is to keep it straight in my head. Some people cram it all into one line of code where I use parenthesis liberally and break it up into several lines. So judging by my habits, I guess I would be in agreement with Hurkyl in the first post of this thread to make it obvious what the expression means.
    -
    Edit: Fixed missing parenthesis.
     
    Last edited: Dec 16, 2016
  8. Dec 16, 2016 #7

    Mark44

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    No.
    You can't use that factor of 2 twice. You have it in the denominator here: ##\frac 6 2## and you have it in the numerator here: ##\frac{2(1 + 2)}1##.
    You're missing a right parenthesis above.
    You're again missing a right parenthesis. The equation should be 6 / (2(1 + X)) = 1
    The outer parentheses in the second example are unnecessary.
    One parenthesis; two parentheses.
    Using parentheses and splitting complicated lines are generally a good idea. Too many people writing code think that writing dense code somehow makes it faster. What really happens is that it makes it much tougher for a maintenance programmer to follow the logic and find whatever bugs might be lurking there.
     
  9. Dec 16, 2016 #8

    Averagesupernova

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    Hmmm. Correct. I missed that. The way I have it the answer would be 18. Darn. I knew better. LOL.
    Yep. Missed it and copied and pasted so the second equation ended up the same way.
    Obviously. My whole point.
     
  10. Dec 17, 2016 #9

    jack action

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    The point I'm trying to make is about this concept that an operator can have an influence over a following operator. If I write:

    ##abcd##

    It means:

    ##a \times b \times c \times d##

    And I know it is the same as:

    ##\frac{a}{1} \times \frac{b}{1} \times \frac{c}{1} \times \frac{d}{1}##

    But when I do this:

    ##ab/cd##

    It means:

    ##a \times b / c \times d##

    and apparently becomes:

    ##\frac{a}{1} \times \frac{b}{1} / \frac{c}{1} / \frac{d}{1}##

    The last multiplication is transformed into a division. So that last operator is in an undetermined state, as its definition depends on the presence or absence of another operator.

    The other thing that annoys me is an equation that would leave me even more perplex with such a rule than the usual Internet equations: I'm not sure what division preceded by division means? For example:

    [tex]a / b c / d[/tex]

    Is it equal to:
    [tex]\frac{a}{\frac{bc}{d}} \left(= \frac{ad}{bc}\right)[/tex]
    or:
    [tex]\frac{\frac{a}{bc}}{d} \left(= \frac{a}{bcd}\right)[/tex]

    If one thought ##6/2(1+2)## could be confusing, this is way worst. But by considering the inverse and setting the same priority for multiplication and division, it is clear that:

    [tex]a/bc/d = a \times b^{-1} \times c \times d^{-1} = a \div b \div c^{-1} \div d = \frac{ac}{bd}[/tex]
     
  11. Dec 17, 2016 #10

    Averagesupernova

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    Hmmmm
    Hmmmmmm. Let's clarify. What differences do you see in the following:
    $$ {ab} \div {cd}$$
    $$ {ab} / {cd}$$
    $$ \frac {ab} {cd}$$

    Let's replace a, b, c, and d with 10000, 50, 20, and 10 respectively.
    My take on it is as follows:
    Case 1: Answer is 250000
    Case 2: Up for discussion.
    Case 3: Answer is 2500

    Just as you say, the operator has changed from Case 1 to Case 3.
    -
    So is the slant bar considered a fraction or division sign? I am sure that I am probably guilty of sloppy notation somewhere along the line. As I said before, mostly this only matters to me in programming. I anyone wants my perspective, I think that one thinks about it differently when creating the equation rather than trying to solve someone else's equation.
     
  12. Dec 17, 2016 #11

    jack action

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    What's the difference?

    Even assuming there is one, there wouldn't be one. A fraction is a numerator divided by a denominator. So ##ab/cd## would set the numerator to be ##b## and the denominator to be ##c##, such that it would be equivalent to ##a(b/c)d## or ##ab\div cd##. Why would anyone extend the denominator pass the next operator?

    Is ##2/3a+1## equal to ##\frac{2}{3}a + 1## OR ##\frac{2}{3a} + 1## OR ##\frac{2}{3a + 1}##? To me it is clearly the first option (the denominator stops at the first operator encountered).
     
  13. Dec 17, 2016 #12
    I dony see any difference between those, answer would be the same..o_O

    edit: yes, if we divide first before multiplying answer differes
    OR2YHpC.png
    OX5sTMr.png
    edit: slant bar is considered division sign in C#, and maybe other programming languages.
    edit: my conclusion is, you need more parenthesis, otherwise, ACCORDING to rules it is:

    a*b first...
    then divide by C, then multiplied into d.:cool:
     
    Last edited: Dec 17, 2016
  14. Dec 18, 2016 #13

    Averagesupernova

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    EXACTLY! Here's the deal. Take the standard equation for capacitive reactance in electronics. It is:
    $$ \frac 1 {2\pi fC}$$
    I NEVER see it like this:
    $$ \frac 1 {(2\pi fC)}$$
     
  15. Dec 18, 2016 #14

    Mark44

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    The bar in the fraction acts exactly the same as the parentheses do when the first fraction is written in a single line. In other words, this -- ## \frac 1 {2\pi fC} ## -- is exactly the same as this -- ##1/(2\pi fC)##.
    BTW, neither of these is an equation -- an equation always has = in it somewhere. Your two examples are expressions.
     
  16. Dec 18, 2016 #15
  17. Dec 19, 2016 #16

    NascentOxygen

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    When this has been brought up previously, I recall someone pointing out there is popular software where the expression evaluates to 1. And as if things couldn't get worse, it requires the presence of a space to the right of the first 2 to cause the software to decide this should evaluate to 9.

    https://www.physicsforums.com/attachments/holly-1756-gif.110502/
     
  18. Dec 19, 2016 #17

    Averagesupernova

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    Last edited by a moderator: May 8, 2017
  19. Dec 19, 2016 #18

    rbelli1

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    I have made an effort to add parens even when logically it is irrelevant. For example x*y*2 is relatively unambiguous as far as the answer is concerned. But this could mean double the product of x and y or double y multiplied by x. The answer is the same but understanding the logic around this expression could be made easier by knowing which it is.

    In a non-time constrained environment comments will be included that eliminate the ambiguity. But how often does that happen?

    BoB

    PS: Is that an environment that is constrained but does not include time or one that does not include a time constraint?
     
  20. Dec 20, 2016 #19

    mfb

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    "I deliberately make my notation confusing" is not mathematics, and we discussed this often enough.
     
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