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I have difficulty understanding following expression: $$(64x^3÷27a^{-3})^\frac{-2}{3}$$

Should I interpret the division sign as follows: $$\left(\frac{64x^3}{27a^{-3}}\right)^\frac{-2}{3}$$

Or as I originally interpreted it: $$\left[\left(\frac{64x^3}{27}\right)\times a^{-3}\right]^\frac{-2}{3}$$

The reason for my confusion is the order of operations as we can write a division as a multiplication (which is how I came to my original interpretation):

$$\left[64x^3\times\frac{1}{27}\times a^{-3}\right]^\frac{-2}{3}$$

I know this is wrong as I cannot get the correct answer this way. This leaves me with the conclusion that the division sign they use is the same as everything after it should be on the bottom of the fraction?

The link to the exercise including the solution: Mathopolis exercise

user3051847
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    Division signs aren't used in any serious mathematics as they are ambiguous, as in your example above. Both interpretations are possibly correct. – Kaynex Aug 07 '17 at 15:22
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    I would've definitely said the whole thing is a fraction if the 'a' wasn't to a negative power. – Jordan Abbott Aug 07 '17 at 15:23
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    @Kaynex Is right about the nonuse of division signs. In your case I think the first interpretation is what's intended. – Ethan Bolker Aug 07 '17 at 15:24
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    According to the order of operations, your original interpretation is correct. Judging by the answers posted on the site, the author intended it to be interpreted as $$\left(\frac{64x^3}{27a^{-3}}\right)^{-\frac{2}{3}}$$ – N. F. Taussig Aug 07 '17 at 15:25
  • We have many questions on the site about the order of operations. I answered on here and another here and another here I think your original interpretation is technically correct but it is likely the author meant the other. A good reason for parentheses, which I rag on frequently – Ross Millikan Aug 07 '17 at 16:17
  • In some contexts, multiplication between an explicit number and a variable (maybe with a possible exponent) binds stronger than other operation symbols: usually $\sin2x$ is interpreted as “sine of twice $x$” and not “sine of $2$ times $x$”. But in this case it's very ambiguous. – egreg Aug 07 '17 at 17:57
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    Related – Stephen S Aug 07 '17 at 18:46

5 Answers5

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This is a good question. Without realizing my assumptions, I interpret $(64x^3 \div 27a^{-3})^\frac{-2}{3}$ as $(\frac{64x^3}{27a^{-3}})^\frac{-2}{3}$. I realize now there is an implicit pair of parentheses around the expression $27a^{-3}$. Clearly, the authors of that question assume parentheses around that term as well.

It is always good to be aware of the assumptions you bring to a problem, so thank you for bringing this up. In general, however, I wouldn't expect to see a lot of problems with division signs like that. Most rational functions (in calculus, for example) are written as fractions; higher math seems to eschew the grade-school division sign.

scott
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The "right" answer, in the sense that it is the generally accepted convention, is that you are correct; multiplication and division are performed in a single pass left to right, so

$$64x^3÷27a^{-3}$$

parses as

$$((64 x^3) \div 27 )a^{-3}$$

However, many (most?) people don't really learn the convention, and write what they think "looks" right. There are also some people that learn the convention wrongly, thinking that multiplication and division are to happen in separate passes.

Thus, it is unfortulately common for people to write such an expression when they actually mean for it to be parsed as

$$(64x^3)÷(27a^{-3})$$

So how should you interpret this expression? Unfortunately, there is no rule here: you have to guess what the author intended. Sometimes, the surrounding context (e.g. the previous or next step in a calculation) can give you clues as to which is meant.

And you should never write such expressions if you can avoid it, since it's prone to having the reader misunderstand.

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    Hmm... I agree that the convention says so, but I'm not sure I'm willing to say the convention is "correct" and "most people" are wrong. I think Lloyd Tao and Scott are correct in claiming the convention doesn't take coefficients into account. Coefficients (although technically mulitplication) are "glued" to their terms and can not be thought separate from their terms. I think this convention is a hammered rule that didnt take all things into account. – fleablood Aug 07 '17 at 17:34
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    I think anybody who writes $2/3x$ to mean $2x/3$ deserves to be misunderstood. – David Richerby Aug 07 '17 at 22:00
  • @fleablood: Multiplication notated by juxtaposition is the standard, not the exception that might need to be considered in an unusual fashion. –  Aug 07 '17 at 22:33
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    The reality is nobody actually writes $2/3x$ to mean $(2/3)x$, so clearly it doesn't mean that. It's not because people don't learn the "convention", it's because people learn that the rule is not the convention. – user541686 Aug 08 '17 at 09:37
  • @DavidRicherby I think anybody who writes $2/3x$ deserves to be misunderstood. – JiK Aug 08 '17 at 09:39
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    @JiK: I think everybody who writes deserves to be misunderstood ;) – user541686 Aug 08 '17 at 09:41
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    This is wrong: it is conventional for multiplication-notated-by-juxtaposition to bind tighter than any infix symbol. So $ab ⊛ cd$ always means $(ab) ⊛ (cd)$, whatever infix symbol ⊛ you use. The division operator ÷ is no different in this respect. (This seems to arise from confusing operations such as multiplication/division with symbols such as ×, ÷, ∕ and juxtaposition - it is possibly for two different symbols to denote the same operation, but have different binding priority.) – psmears Aug 08 '17 at 09:44
  • @psmears Indeed. Juxtaposition is special. Nobody ever claims that $1/56$ is ambiguous, even though it "could" refer to $\tfrac15\times 10 + 6=8$. – David Richerby Aug 08 '17 at 09:50
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I have always interpreted coefficients and variables next to each other to have an order of operation higher than MD. This is because when someone is creating an expression or equation, and they intend for that $ \div $ or $ \times $ to be used with what looks like the coefficient, they will notice the ambiguity instantly and put the parenthesis in. This is my basis for the assumption.

However, there is no real set convention for dealing with the order of operations in such a scenario. It's good to state how you interpreted the question before starting, and if you can, try to derive what the calculation should be based on the context (or even the complexity) of the calculation.

LloydTao
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Short answer:

"Should I interpret the division sign as follows:$\left(\frac{64x^3}{27a^{-3}}\right)^\frac{-2}{3}$"

Yes. That is the correct interpretation.

"Or as I originally interpreted it: $\left[\left(\frac{64x^3}{27}\right)\times a^{-3}\right]^\frac{-2}{3}$"

No. But that was a perfectly reasonable (unfortunately incorrect) interpretation.

Long (and weird) answer:

A statement $A \div B$ implies that we think of $B$ as a single "chunk" and it is conventional that $27a^{-3}$, if it is expressed is a single "thing". Why? Hmmm, it's a good question.

This probably isn't a good answer but I think of it this way: Multiplication takes precedence over addition (that is because we can distribute multiplication over addition: $a(b+c) = (b*c) + (b*c)$) and makes addition, in my mind, a fluid ongoing modification, whereas multiplication is a "gluing" bonding modification. We never consider $a - b + c$ to mean $a - (b+c)$ but always to mean $(a-b) + c$ because there is nothing "permanently bonding" about $b+c$ so that whenever we see $b+c$ we think "wow, that is solid 'chunk'".

Multiplication however "feels" different. It is a bonding. In my intuition, it feels almost "chemical" in nature. Where as $a + b - c + d +f$ seems like a liquid flowing process, $27a^{-3}$ seems like a crystalline little rock pebble.

In a way this is why prime factorization is so critical whereas sums are not. If you want to solve $n + m = 27; n,m \in \mathbb N$ we can just let $n$ be anything from $1$ to $26$ because addition is "fluid" and we can break it anywhere. But if you want so solve $n*m = 27; n, m \in \mathbb N$ it is far more finicky. Not any old natural number will divide into $27$. You must tap at it until you find a "subparticle" such as $3$ or $9$ and chisel out the rest.

(Lots of rock and liquid metaphors.)

But... obviously, if I got before a math class and tried to teach that way, my students would ... be perplexed. That's not math, that's ... impressionism.

Well, remember the rules. There's some dumb mnemonic about the orders of operation all you young kids are using these days[$*$]. I can never remember it but ... Let multiplication take precedence over the division symbol. Just... obey, and stop asking questions.... I guess.

Anyhow, unfortunately for you $A\div B$ is really ambiguous. Fortunately "serious" mathematicians stop using it precisely for that reason and pretty soon you'll just use fraction notation $\frac AB$ and won't ever have to worry about this again.

[$*$] I was thinking of PE(M/D)(A/S) which WOULD indicate $5\div 2\times 3$ is $(5\div 2)\times 3$ and not $5\div (2\times 3)$. So basically according to that $64x^3 \div 27a^{-3}$ would be $(64x^2*\frac 1{27} a^{-3})$. But the mnemonic is, apparently wrong. I'd modify is as PE(Coeffecient terms)(M/D)(A/S) so that $5\div 2\times 3 = \frac 52\times 3$ whereas $5\div 2a^2$ is $\frac 5{2a^2}$. However what is $5\div ab$? is $ab$ a coefficient term? Hmmm, I think my assumption would be that it is. Or, at least that'd be what I'd assume if I didn't consciously think about it specifically. But it is ambiguous.

fleablood
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    Why? Typesetting. In my opinion, people parse formulas via spatial recognition moreso than actually following rules. $\div$ has wide spaces setting it off from its arguments, and juxtaposition has very little space. Thus, that's how many people read the expression, in defiance of established convention. –  Aug 07 '17 at 17:03
  • Hmmm, not bad. So $64x^2 \ \div 27\ \ a^{-3}$ will look like $\frac {64x^2}{27}a^{-3}$ whereas $64x^2 \ \div \ \ 27a^{-3}$ will look like $\frac {64x^2}{27a^{-3}}$? Perhaps. But I still think $27a^{-3}$ looks like a "solid" entity while $27 + B$ or even $27\times a^{-3}$ does not. I think both you and I would agree meaning trumps rules anyway. But the students dilemma is that recognizing meaning without rules is something one picks up by experience. Which students don't have. – fleablood Aug 07 '17 at 17:26
  • Yup, I parse $64x^3 \div 27a^{-3}$ as $(64x^3) \div (27a^{-3})$ but $64 \times x^3 \div 27 \times a^{-3}$ as $((64 \times x^3) \div 27) \times a^{-3}$. This is a concern for me as a programming language designer—you really don’t want to make it easy for a programmer to create bugs due to misleading operator precedence. – Jon Purdy Aug 08 '17 at 04:52
  • While I agree that an expression like $x/2y$ should usually be interpreted as $x/(2y)$, this answer seems to suggest that there is nothing wrong with writing $x/2y$ instead of $x/(2y)$, which is something that I cannot agree with. You also say that "serious" mathematicians will always write $\frac{A}{B}$ instead of $A/B$, which is in my experience not true in cases where space is a concern (inline equations, exponents...). – Peter Aug 08 '17 at 09:12
  • @Peter. I consider $a/b$ sign and the $a\div b$ sign to be completely different notation and $a/b$ and $\frac a b$ to be exactly the same notation. The only issue is one of typesetting. $\frac x{2y}$ can not be interpreted as $\frac x2y$ and there is no ambiguity. Visually we can not tell whether the string of symbols $x/2y$ is supposed to be $\frac x{2y}$ or $\frac x2y$. But that is a visual ambiguity. Not a symbol ambiguity. As symbols either $y$ is in the denominator or it isn't. With the notation $\div$ and is unclear whether $x\div2y$ the divisor is a "compound" $2y$ or just $2$. – fleablood Aug 08 '17 at 18:36
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$(64x^3 \div 27a^{-3})^{\frac{-2}{3}}$ should be interpreted as $(64x^3 \div 27 \cdot a^{-3})^{\frac{-2}{3}}$. Using the order of operations rules, multiplication and division have the same operator precedence, so they are evaluated from left to right. A similar question went viral a while back. The question was: "What is $6 \div 2(1+2)$". At first sight, this might seem to be $6 \div (2 \cdot 1 + 2 \cdot 2)=6 \div (6) = 1$, but in reality it should be $(3)(1+2)=3(3)=9$.

ericw31415
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    Yes. It should be so interpretted. But it wasn't. And I think very few serious mathematicians would interpret it that way. At any rate the OP's text specifically did NOT interpret it that way, while the OP did. Hence the OP's question. And thoroughly justified confusion. – fleablood Aug 07 '17 at 17:29