Even professional researchers exploring common mistakes in algebra education are prone to saying "yes" to this question, for example:

But are parentheses a multiplicative operator? It seems clear that the answer is "no". Now clearly all of the following are multiplications ofMisconceptions: Bracket Usage -- Beginning algebra students tend to be unaware that brackets can be used to symbolize the grouping of two terms (in an additive situation) and as a multiplicative operator[Welder, "Using Common Student Misconceptions in Algebra to Improve Algebra Preparation", slide 7; references Linchevski, 1995; link]

*a*and

*b*:

*ab, (a)b, a(b), (a)(b)*, etc. But notice that the parentheses make no difference at all in this piece of writing. These are multiplications because of the usage of

**; any two symbols next to each other, barring some other operator, are connected by multiplication. Obviously, if there were some**

*juxtaposition**other*written operator like + - / ^, between the

*a*and

*b*it would be something different; but granted that multiplying is probably the most common operation, we read the

*absence*of a written operator to indicate multiplication.

The chief problem with telling students that parentheses indicate multiplying is that they then routinely get the order-of-operations incorrect. Assuming a standard ordering ([1] inside parentheses, [2] exponents & radicals, [3] multiply & divide, [4] add & subtract), students want to perform multiplying with any factors in parentheses in the first step, before exponents. One of the first and frequently repeated side-questions I ask in my class is, in an exercise like "Evaluate 2+3(5)^2" -- "Yes or no, is there any work to do inside parentheses?" On the first day of algebra, almost the entire class will answer "yes" to this (and want to do a multiply), at which point I explain that the answer is actually "no". If there is no simplifying

*inside the parentheses*, then the first piece of actual work will be to apply the exponent operation. And that's all that parentheses mean. (There is of course a multiplication here -- not because of the parentheses, but because of the juxtaposed 3, and it must take place after the exponent operator.) A majority of the class will pick up on this afterward, but not all -- some proportion of a class will continue to say "yes" and be confused by this particular question throughout the semester. (As another example, some students are prone to evaluate something like "(5)-2 = -10" for this and other reasons.)

Whether a factor is juxtaposed next to something in parentheses or not is irrelevant to the multiplication; parentheses are a separate and distinct issue. What say you? Have you ever said that the parentheses symbols actually mean multiplying?

I've never seen the misconception you raise, but there is a common confusion that is inherent in standard math notation: is f(g) the function f applied to value g, or f*g? That is, are parentheses just a grouping symbol or are they a marker for a function call?

ReplyDeleteThere is good reason why most computer languages do not interpret juxtaposition as multiplication.

Indeed, whenever the same symbol gets used for multiple meanings you get problems. Same could be said for a -1 superscript (reciprocal or inverse trig function), an overbar/vinculum (fraction, grouping, or repeating decimal), etc. But I suppose the real reason for no juxtaposition in programming is that it would be ambiguous with multi-character variable names?

DeleteFYI, I totally saw this misconception again today in the first class of a college algebra course. On the board: 6÷2(1+2). I ask: Is there any operation inside the parentheses? Yes. I ask: Exactly what operation is it? And while most of the class says "addition", one student audibly says "multiplying", and then looks around surprised at what the rest of the class said.

Deleteis the answer to this problem 1?

DeleteNo, it's 9. In detail: 6÷2(1+2) = 6÷2(3) = 3(3) = 9. Recall that any operations of the same precedence are read left-to-right across the expression (as for the divide-then-multiply in this case). See also Google

DeleteI'm glad to see the conversation here because I've been trying to figure something out regarding this problem.

DeleteI think the following is true, but I'm afraid that I've missed something.

6 ÷ 6 = 1

6 ÷ (2)(3) = 1

6 ÷ (2)(1+2) = 1

6 ÷ 2(1+2) = 1

Would parentheses be required around the "(2)(3)"? I know they would provide clarity. My question is about whether they're required.

It's been a long time since algebra, and I'm willing to acknowledge that I've forgotten something.

So multiplication implied by juxtaposition does not take priority over other × or ÷ operators? Because I just read a claim on purplemath.com that juxtaposition takes priority.

DeleteDerek: 6 ÷ (2)(3) = 9 (not 1). Recall that multiplying and division have the same precedence; ties are broken by reading left-to-right; so in this case the division occurs before multiplying. In more detail: 6 ÷ (2)(3) = 3(3) = 9.

DeleteIf you wanted the result to be 1, then yes, you would need additional parentheses to make the multiplying happen first.

See: Wolfram Alpha

jeverettk: These is some small historical difference of opinion on that. I've never seen it explicated in any professional math text. Can you provide a link to purplemath.com claiming that juxtaposition has higher precedence?

DeleteDelta, since I posted my comment, I've also spoken with a PhD in mathematics. He solved the problem to 1 on first glance, and immediately followed with "of course, it could also be 9." He then pointed out that, since the problem suggested two possible solutions, it was a poorly written problem.

DeleteFrom that discussion, the confusion comes from the experience that, as mathematics gets more complex, we start referring to things that are juxtaposed as if they were a single term. For example, "1 ÷ 3x" is read as "one divided by 3x," not "1 divided by 3, times x." The juxtaposed terms are treated as a single value which must be solved first before the term can be divided into 1.

So the prof told me he first read the problem as "six divided by the product of 2 & 3." And that's how you get 9. He said that the confusion here comes from the fact that we generally stop using an obelus to note division when we start multiplying by juxtapostion. And the latter involves conventions where juxtaposed items are treated as a single term, as the "product" of things rather than as a list of things.

In roughly the words of my professor friend, the author of the problem does not adequately convey the conventions from which the problem is to be approached. He said that, if he were to put this problem on an exam, he'd have to give credit for both answers because both were justifiable mathematically.

Derek: I'm going to broadly guess that your acquaintance was speaking from initial instinct, and may rethink this on later reflection.

DeleteI'm certainly happy to remove the obelus symbol because I don't think that's germane to the question (some people think that using a slash makes it more ambiguous, which is why I used an obelus there). So, how should the expression 6/(2)(3) be simplified? How will it be interpreted by a computer algebra system? It's not possible for such systems to interpret it more than one way.

In short, I've never seen it documented in any text that juxtaposition has higher precedence than other operations. Even if many people fall into the "habit" of reading so visually.

DeleteI'm not sure he's thought about it since then. He only thought about it then because I bought him lunch.

DeleteDerek, your PHD prfoessor is a moron to say there are 2 possible answers for a simple algebra question. Your professor has made the same mistake as the students in the last paragraph of the above article. There is no simplification within any brackets therefore you proceed to evaluate D and M of BEDMAS.

Deletegiven:

a(b) = a times b = a x b

a/b(c) = a / b x c

let:

a=6, b=2, c=3

a/b(c) = a / b x c = 6 / 2 x 3 = 3 x 3 = 9

"Evaluate: 2+3(5)^2". If solving this problem myself, I would start by rewriting it (at least in my head if not on paper) as: "Evaluate: 2+3(5^2)", and perhaps further as: "Evaluate: 2+(3*(5^2))". This makes it clear that the operation inside the (innermost) parenthesis is exponentiation.

ReplyDeleteSimilarly for "Evaluate: (5)-2=?", I would recognize this as "Evaluate: 5+(-2)=?", and rewrite it as simply: "Evaluate: (5-2)=?" (with or without the parenthesis).

Delta, I don't agree with your answer. My daughters text clearly states the the number beside the parentheses goes with it therefore needs to be solved 1st. So, the answer is 1.

ReplyDeleteKrystal: Your daughter's text is probably wrong. It's not a completely rare misunderstanding. Can you provide a citation (author's name, title, section, short quote?).

DeleteThe 6/2(1+2) example is very common on Facebook and create huge amounts of angry debate. Much of it due to acronym confusion ("m before d") but also a lot due to the idea of the "implied multiplication by juxtaposition" super priority that sadly is supported by some scientific calculators (notably the Casio's). There seems to be no consensus or formal (none that I've found at least) on this (I personally reject it).

ReplyDeleteWolfram, Texas Instruments, and Hewlett Packard all treat juxtaposition as the same as multiplication with no precedence for the juxtaposition over other notations for multiplication. The "1 divided by (3x)" example is a scenario where I tell my students it is written poorly. I also tell my students to never buy Casios.

ReplyDeleteInteresting!

DeleteI teach astrobiology at the university level and deal with mathematical expressions daily. Mainly, I must agree the ongoing problem '6/2(1+2)' is very poorly written.

ReplyDeleteHere is the defeat of the logic for those saying the answer is 9

6÷2(2+1)

Left to right, correct? With PEMDAS in mind. I will follow your logic here:

6÷2(3)

PEMDAS, so parenthesis first

6÷2(3)

PEMDAS, so parenthesis first

6÷2(3)

PEMDAS, so parenthesis first

6÷2(3)

PEMDAS, so parenthesis first

6÷2(3)

....on and on forever....

THAT IS WHY AFTER SOLVING THE PARENTHESIS YOU REMOVE THEM! AND SINCE YOU CANT JUST HAVE A FLOATING NUMBER YOU APPLY IT IMMEDIATELY TO THE 2

6÷2(2+1)

6÷6= 1

THE ANSWER IS 1

I hate to say it, but -- this is totally incoherent. You're not even using the word "solving" correctly.

DeleteCan I ask where you teach? A Google search for "Edward Starski astrobiology" doesn't turn up anything. Pretty skeptical at the moment.

Garbage, off-topic replies from new profiles being deleted here.

DeleteI'm having the same discussion on FB now

Delete6^2/2(3)+4. Some are saying 36/2*3+4 is 58 because you divide the 36 by two before multiplying in the 3. Others suggest that since it is a juxtaposition using a parenthesis that you multiply the 2(3) first and divide the 36 by 6 and add the 4 giving you 10 as the answer. The PurpleMath page explaining this as mentioned above is here:http://www.purplemath.com/modules/orderops2.htm

That's a nice link and example of the problem. The PurpleMath page says at the bottom that, "Note that

Deletedifferent software will process this differently; even different models of Texas Instruments graphing calculators will process this differently." So we all agree that there's some confusion around this issue.But I really think that PurpleMath's answer to the question must be in error. Consider that their justification is that "though multiplication and division are at the same level (so the left-to-right rule should apply), parentheses outrank division". This argument -- that a multiplication juxtaposed outside parentheses follows the ordering of "parentheses" -- falls into exactly the trap I wrote about above, e.g., that 3(5)^2 = (15)^2 = 225. We all know that's not correct.

Now, if PurpleMath had written that they believe juxtapositions should have ordering above other multiplications, then that would at least be coherent, if nonstandard, and not according to any written definitions I've ever seen. But the argument that they follow the ordering as "parentheses" leads immediately to all kinds of insanity.

Why isn’t distribution apart of the solution? 2(1+2) is in the form of a(b+c), which by property equals ab + ac. Therefore

ReplyDelete2(1+2) = 2*1 + 2*2 = 2+4 = 6

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

That move violates the left-to-right precedence of the operations. In other words: Division is non-associative.

DeleteIt's the same as arguing that that since 2 - 1 = 1, it must be the case that 3 - 2 - 1 = 3 - 1 = 2.

I was tought in UK school 70s + 80s that implied multiplication is shorthand for the standard explicit mulitplication in algebra because it could become confused with the curly x variable. So it was just implied that it was a multiply where no operator was present next to a factor.

ReplyDeleteThus x4 = x Multiply 4, abc = a Multiply b Multiply c, 2(3) = 2 x (3) = 2 x 3. I passed my high school exams with that knowledge.

So is x=2/ab the same as x=2b/a?

ReplyDeleteCertainly if one types 2/a*b into a computer or spreadsheet system, it will be evaluated left-to-right, and you'll get a similar value as 2*b/a.

DeleteI think everyone agrees it's best to avoid writing stuff like that, in case any readers are unclear about that. Recent StackExchange question.

There is absolutely multiplication implied by the parenthesis...I know of nobody who disagrees. The question is when to resolve that multiplication.

ReplyDeleteAs a mathematician by training, I shudder at the idea that some multiplication being a higher order of magnitude than other multiplication. While a(b) resolves to ab, 2(3) does not resolve to 23 of course, it is 2*3, but when is it done.

All of the decades of time that I have had with math tells me that it turns into regular multiplication because having a separate term in an equation is too prone to errors. Ultimately either way could be correct, but I'll go for the one that simpler and less prone to confusion. It's an ambiguous convention as to how to resolve it.

Which of course just say - you should have used more parenthesis to begin with.

I think that you're wrong. The multiplication is due to the

Deletejuxtaposition, that is, the lack of any explicit operator, not the parentheses. For example:Is there multiplication in 5 - (2)? No.

Is there multiplication in (3) + 7? No.

Is there multiplication in (x2-x1)/(y2-y1)? No.

Actually there is a multiplication to get rid of the parenthesis and it is by 1. So in your first equation you could write it as 5-1(2) but there is no need as that is implied. You could think of it as "five minus one group of 2". That is why the equation on facebook of 6÷2(2+1) resolves to 1. Again in written form this would be"Six divided by two groups of two plus one." If you wanted the answer of 9 then the equation should be 6/2*1(2+1) or just 6/2*(2+1).

DeleteI disagree with the "invisible 1" argument. 5 - (2) does not inherently mean any multiplication, it is simply showing 5 minus the quantity (2).

DeleteTo argue that there are "invisible 1"'s around, one may as well also argue that there is a multiplication on the 5, to wit, 1×5. Or why not a whole bunch of them? 1×1×1×1×5. But that's not actually inherent or necessary in the given, written expression.

But that isn't what the given, written expression actually means.

In your problem: "Evaluate 2+3(5)^2" The exponent is part of the parenthetical term and must be done before the multiplication. By multiplying the exponent of the number inside(which is 1) the parenthesis with the exponent outside(which is 2) you get the new exponent of 2. Simplify: 2+3(5^2) = 2+3(25) = 2+75 = 77

ReplyDeleteObviously. You may have missed the point of the article. The facts on the ground are that many students show errors due to being misled by the "parentheses are multiplication" claims.

DeleteDelta, I think you have the most consistent argument on the thread. That said, I think confusion comes from the notion that multiplication is implied merely through juxtaposition. The juxtaposition without parenthesis wouldn't work so they become intrinsic to the implication. That is not to say implication takes precedence. I'm coming to grips with relearning this myself. The problem with a lot of the school prealgebra textbooks is that there is a high amount of intentional ambiguity in presentation with the aim being to hallenge the students. It comes at the cost of misrepresenting how one would clearly and concisely write equations in an economical manner. As assignments near their final problems they come with a fair but present element of trickery the result being highly contrived equations.

ReplyDeleteAnyway, Delta, you've convinced me that the answer to the problem 6 / 2(1+2) is 9 with what seems to be a grounded and unbiased reasoning, whereas those clinging to an answer of 1 (I'm still curious) are doing it with an unhealthy dose of habit. Are there any resources for math, akin to the English language's OED or Webster publications, that can help put this topic to bed once and for all for those having trouble accepting your authority on the matter?

Thanks for the thoughtful post. I wish I were aware of some resource like OED/Webster's for math notation, but I don't think there is one. (Even for OED/Webster's, you quickly get into the "English is descriptive not prescriptive" argument which results in no one ever being wrong there, either.)

DeleteI would guess that the ambiguity in precalculus texts isn't intentional, but more likely an accident of the authors themselves being unclear on what their starting point is (or possibly publisher pressure to reduce page count or formality). Even after a dozen years of teaching college algebra, I'm embarrassed about how unclear I am on what the starting assumptions are.

The most helpful thing I've found this summer is to go back to my Abstract Algebra text (Hungerford) and work back through that again. My main observation is that almost all algebra/precalculus texts effectively start off presenting/assuming the (five) axioms for a field structure. I feel in a few ways that might not be the most self-evident starting point for those students, but that's what's almost universally done.

Secondly I've been reading some material by Hung-Hsi Wu intended for formally training teachers in pre-algebra and algebra which I've found quite helpful.

For links see my question on StackExchange Math Educators, "Is there a resource that formally develops the topics of elementary algebra?"