# Order of operations when using the ÷ symbol.

I always write equations using fractions to avoid confusion, but my textbook uses the ÷ symbol. This question asks to simplify and write the equation using positive exponents:

$$x^{-1/3}y÷ xy^{-1/3}$$

If I work from left to right I can simplify to this:

$$\frac{y^{2/3}}{ x^{4/3}}$$

However if instead I split the equation into a fraction at the ÷ sign, I get the following result:

$$\frac{y^{4/3}}{x^{4/3}}$$

Shouldn't the first answer be the correct one?

• If it is ambiguous then "correct" becomes a difficult concept. I would assume that your second choice was more likely to be the intended result, but I could easily be wrong – Henry Oct 18 '16 at 10:42
• Some years ago there was a flurry of duplicates of similar questions (with integers), this one for example: math.stackexchange.com/questions/33215/what-is-48%c3%b7293 – Hans Lundmark Oct 18 '16 at 14:46

Personally, I would interpret your formula as $$\frac{x^{-1/3}y}{xy^{-1/3}},$$ because when multiplication is written by juxtaposition, I think of it as binding more strongly than $\div$. However, it is true that the notation is somewhat ambiguous, and ought to be clarified by the source. If you cannot tell from context which one the source of the formula expects, I recommend going with my/your second interpretation.
• The standard convention is that multiplication and division have the same order of precedence, because $a \div b = a \cdot \frac{1}{b}$. In this case, the computation from left to right would yield a different result. – Björn Friedrich Oct 18 '16 at 11:00
The standard convention is that multiplication and division have the same order of precedence, because $$a \div b = a \cdot \dfrac{1}{b} = a \cdot b^{-1} \;,$$ i.e., division is the same as multiplication with the inverse. And whenever you have operations with the same order of precedence, you conventionally compute from left to right. So, in an expression such as $$a \cdot b \div c \cdot d$$ we would compute $$(((a \cdot b) \cdot c^{-1}) \cdot d) \;.$$ Applied to your example, we would have $$\begin{array}{rclr} x^{-\frac{1}{3}}y \div x y^{-\frac{1}{3}} &=& \big(x^{-\frac{1}{3}}\big)y \big(x^{-1}\big) \big(y^{-\frac{1}{3}}\big) & \text{(precedence of exponentiation)} \\ &=& \big(x^{-1-\frac{1}{3}}\big)\big(y^{1-\frac{1}{3}}\big) & \text{(exponential laws)} \\ \end{array}$$