# Multiplying Decimals seemingly violating Order of Operations

I'm on Khan Academy, learning how to multiply decimals, and in this video (around the 1:10 mark) Sal shows that if you're multiplying a decimal by a decimal, first you convert both into natural numbers divided by exponents of 10 and then multiply, ending up with this:

2.91 x 3.2 = 291 / 100 x 32 /10


So far, so good. But then he goes and reorders the expression to the right of the equals sign into this:

2.91 x 3.2 = 291 / 100 x 32 / 10
= 291 x 32 / 100 / 10


This is the point where I get lost, because in a much earlier video about the Order of Operations around 5:40 (and more prominently around 7:40), he explains that when there's multiple operations on the same level (which multiplication and division are), you have to complete those operations from left to right. So by that logic, the above then violates Order of Operations, right?

After that, he then further seems to violate it by clearly performing the operations out of order, by doing the following:

2.91 x 3.2 = 291 / 100 x 32 / 10
= 291 x 32 / 100 / 10
= 291 x 32 / 1000


He divided 100 by 10, which is at the far right of the expression! How does that work, if you're supposed to perform your operations from left to right?

Now, I did both calculations (the way he did it and just taking the original expression and resolving it from left to right - 291 / 100 x 32 / 10 ) on my calculator and came to the exact same answer of 9.312, which tells me that the Order of Operations rule is wrong and you actually don't have to perform them from left to right?

So to sum up my question: Do you have to perform operations on the same level, like multiplication and division, from left to right, or not?

UPDATE 1: I've now done some more experiments on the Order of Operations' left to right rule, which I think might have given me more clarity on exactly how exactly it works. I tried the expressions 1 + 3 - 5 and 2 x 3 / 4 (and a couple others, unimportant) and I have found that I can re-arrange the numbers and wind up with the same answer, so long as I move their symbol (+/-) with them. For example:

1+3-5 = 1-5+3 = -5+1+3 = -1
2x3/4 = 2/4x3 = 3x2/4 = 1.5 // although I'm not sure how I would move "/4" to the left-most position of this expression.


Based on that, I want to update the conclusion of my question: Can you rearrange the numbers in an expression, so long as you move the numbers' symbols with them, thus not breaking the left to right rule?

UPDATE 2: It seems that my problem was caused when I confused the Order of Operations' left to right rule with moving numbers around in an expression (which was my suspicion in UPDATE 1). Thanks to everyone that helped me come to this realization!

• multiplication is commutative and dividing a number A by another number B is really a short hand of multiplying A by the multiplicative inverse of B. – achille hui Dec 19 '18 at 12:53

## 4 Answers

So to sum up my question: Do you have to perform operations on the same level, like multiplication and division, from left to right, or not?

thus it seems that either PEMDAS is wrong or I'm not understanding it correctly.

The PEMDAS rule should be understood as PE(MD)(AS) or even better PE(DM)(AS), where the operations inside brackets have equal priorities and should be performed from left to right.

For example: $$16\cdot 8:4=(16\cdot 8):4=16\cdot (8:4)=32.$$ However: $$8=16:8\cdot 4=(16:8)\cdot 4\ne 16:(8\cdot 4)=\frac12.$$

One way to avoid confusion (or violation of PEDMAS) is to consider "multiplication" and "division" as mutually inverse operations and remember commutativity of multiplication. For example: $$16\cdot 8:4=16\cdot 8\cdot \frac14=16\cdot \left(8\cdot \frac14\right)=16\cdot \frac14\cdot 8=16:4\cdot 8;\\ 16:8\cdot 4=16\cdot \frac18\cdot 4=16\cdot 4\cdot \frac18=16\cdot 4:8.$$

• Awesome, thanks! So does that mean that I can basically freely move around the numbers in an expression, so long's as I move their leading symbols with them, and thus not breaking PEMDAS' left to right rule, i.e. 1-5 = -5+1? – PrintlnParams Dec 19 '18 at 14:12
• Yep, check commutativity out. – farruhota Dec 19 '18 at 14:13

Consider the product in this form:

$$2.91 \cdot3.2 = \frac{291}{100}\cdot \frac{32}{10}$$

$$= \frac{291\cdot32}{100\cdot10}$$

Hope this helps.

• Thank you for your answer, it does make sense! But I'm now still confused about the Order of Operations rule of left to right, which seems to be wrong (or more likely I'm just misunderstanding it). How does going from 291 / 100 x 32 / 10 to 291 x 32 / 100 / 10 not break that rule? You're literally scrambling up the expression and doing the operations out of order. To me, 291 / 100 x 32 / 10 seems like (((291 / 100) x 32) / 10) is implied, is that not so? – PrintlnParams Dec 19 '18 at 12:58
• When rewriting the sum you need to make use of brackets correctly,>2.91 x 3.2 = (291/100) x (32/10) – PolynomialC Dec 19 '18 at 13:23
• Right, that's what I thought too, at first. But then, that should also mean that you have to do each division first, before multiplying each division's answer, right? Because, writing it the way you did now - (291/100) x (32/10) - also seems to break the left to right rule when you reorder it as 291 x 32 / 100 / 10. Actually, now it seems to break the whole concept of Order of Operations because parentheses gets resolved before multiplication and division. – PrintlnParams Dec 19 '18 at 13:28

The order of operations is an attempt to explain what is actually meant.

The reality is there are two multiplies and two divides. The outer one is seperated by × and ÷ signs, which run left to right. The inner one is close up symbols like $$ab/cd=(ab)/(cd)$$. Here the multiplication happens before the division, and both of these before the outer ones.

Calculators only use the outer rule, so the example is $$a×b÷c÷d$$, which is not what is written.

• I don't understand. Where's the two multiplies? I only see one in 291 / 100 x 32 /10, or are you referring to another expression? I also don't know what "close up symbols" are or any of the ab/cd = (ab)/(cd) stuff. My apologies, but I'm only now starting to learn pre-algebra and have never dealt with any of that before. Also, the calculator part (in my question) is not really what I meant to put emphasis on, I simply meant that both 291 / 100 x 32 /10 and 291 x 32 / 100 / 10 gets me the same answer and thus it seems that either PEMDAS is wrong or I'm not understanding it correctly. – PrintlnParams Dec 19 '18 at 13:06
• In calculator or outer mode, you can take everything to the left of a given sign as being multiplied by the next factor. So 291/100=2.91, then 2.91×32 =93.12 then 93.12/10=9.312. The order is PE(MD)(AS), where the brakets are current to the cumulative left. – wendy.krieger Dec 19 '18 at 13:15

The "left-to-right rule" is actually just a convention that allows us to handle strings of operations including the $$\div$$ symbol from basic calculators. It doesn't necessarily make any difference. For example, in the early video you mentioned, he takes the left-to-right approach to get $$10\times 4\div 2=40\div 2=20,$$ but right-to-left would get us $$10\times 4\div 2=10\times 2=20,$$ as well. However, left-to-right yields $$12\div 6\times 2=2\times 2=4,$$ while right-to-left yields $$12\div 6\times 2=12\div 12=1.$$

This ambiguity is something we'd like to avoid, so instead of $$\div,$$ we begin to use fractions and rational notation, making it clear whether we want $$\frac{12}{6}\times2=4$$ or $$\frac{12}{6\times 2}=1.$$ Having adopted this new notation, instead, we see in the latter case an implied parenthetical, so we're effectively doing $$12\div(6\times 2),$$ which standard order of operations tells us how to handle.

If we're dealing with fractions, though, divisions and multiplications can be performed in any order. For example, let's consider $$\frac23\times\frac9{10}.$$ We could think of this as a multiplication and two divisions, or as simply a multiplication of two non-whole numbers. Perhaps we could think of this as finding the area of a rectangle that's $$\frac23$$ of a unit by $$\frac9{10}$$ of a unit, or some other equivalent way, but ultimately, we can perform the operations in any order we choose. We should end up with $$\frac35,$$ regardless.

The reason behind this comes from the way we may define division. In general, if $$a$$ is a non-zero number, there is a number $$a^{-1}$$ such that $$a\times a^{-1}=1.$$ We call $$a^{-1}$$ the reciprocal of $$a,$$ and often write it as $$\frac1a.$$ In general, then, we can define $$b\div a:=b\times a^{-1},$$ or, using the fractional form, $$b\div a:=\frac{b}{a}.$$ Notice how we have one thing on top of another with a line in between? That's what the $$\div$$ symbol is meant to invoke! The number preceding it goes into the numerator, and the number following goes into the denominator.

What this means is that division is effectively just a multiplication, which simplifies things a great deal, because of some nice properties that multiplication has. One is that multiplication is commutative, meaning that for any real numbers $$a$$ and $$b,$$ we have $$a\times b=b\times a.$$ Another is that multiplication is associative, meaning that if we have to do more than one multiplication, then order doesn't matter! In particular, there's no need to go left-to-right. Put more precisely, for any real numbers $$a,$$ $$b,$$ and $$c,$$ we have $$(a\times b)\times c=a\times(b\times c),$$ so the parentheses don't actually make any difference on the outcome of the calculation, which allows us to simply leave them out.

Putting these two properties together, for any real numbers $$a,$$ $$b,$$ and $$c,$$ we have $$a\times b\times c=a\times c\times b=c\times a\times b=c\times b\times a.$$ Consequently, we can let those "implied parentheses" I mentioned earlier disappear, as follows. For any non-zero real numbers $$a$$ and $$b,$$ we can prove that $$(a\times b)^{-1}=a^{-1}\times b^{-1},$$ meaning that the reciprocal of a product is the product of the reciprocals. So, in our earlier example, we get $$\frac{12}{6\times 2}:=12\times(6\times 2)^{-1}=12\times6^{-1}\times2^{-1}.$$ Commutativity means we could instead write this as $$12\times2^{-1}\times6^{-1},$$ so we could translate it back into left-to-right terms by either $$12\div6\div2$$ or $$12\div2\div6.$$

Please let me know if there are specific parts of my answer that aren't clear to you, or if you just want to make sure you understand something correctly.

• Thanks for the awesome answer, I managed to understand up to where you defined division, after which it started looking alien again. But what stood out, and what I feel proves that the left to right rule is broken (or more likely, I'm misunderstanding it) is where you mention the implied parentheses. Back to my example, doesn't 291 / 100 x 32 / 10 have implied parentheses like so (((291 / 100) x 32) / 10) (which to me looks like exactly what's meant with the left to right rule), and thus reordering that to 291 x 32 / 100 / 10 then breaks that rule? – PrintlnParams Dec 19 '18 at 13:17
• The use of $/$ instead of $\div$ is intended to divorce us from the left-to-right notion, and lead us instead to consider fractions, instead. We can define fraction multiplication by $$\frac{a}{b}\times\frac{c}{d}:=\frac{a\times c}{b\times d},$$ and many texts do so. However, I find this better viewed as an identity--that is, both sides are equal, so we can use them interchangeably, if one is more convenient than the other--and not a rule of calculation. (cont'd) – Cameron Buie Dec 19 '18 at 13:34
• Put into multiplicative terms, $$\frac{291}{100}\times\frac{32}{10}=291\times100^{-1}\times32\times10^{-1}=291\times32\times100^{-1}\times10^{-1}.$$ Does that help? – Cameron Buie Dec 19 '18 at 13:36
• The "implied parentheses" I refer to come from having a group of numbers and operations in a numerator and/or denominator, and nowhere else. But here again, the multiplicative notation comes to our rescue. We can prove that $(a\times b)^{-1}=a^{-1}\times b^{-1}$ for any real numbers $a,b.$ So, for example, $$\frac{12}{6\times 2}=12\times(6\times 2)^{-1}=12\times 6^{-1}\times 2^{-1},$$ so we can lose the implied parentheses altogether, and return to just several multiplications. – Cameron Buie Dec 19 '18 at 13:42
• I made some corrections and clarifications to my answer. Please let me know if you have any further questions. – Cameron Buie Dec 19 '18 at 14:52