Weird question about inverses

This probably seems like a really stupid question. I am no math expert; I am in an Algebra II high school class. Here it goes:

addition can be interchanged such that

a + b = c
b + a = c

but why not subtraction?

a - b = c
b - a does not equal c

I don't quite understand this and with division and multiplication i thought that:

a*b/c = d

and a*(b/c) = d

I have always been told division and subtraction are interchangeable in school. I just looked at this problem and realized Ive been fooled:

6/3*2 = 4
6/(3*2) = 1

I am so confused. Can someone just clear up this whole order of operations mumbo jumbo for me. I would like to know exactly what division and multiplication are and what arithmetic works in relation with one another. I am in Algebra II and I can't believe I don't understand this.

• Yup, the way order of operations is sometimes taught (at least in the US), PEMDAS, where division comes after multiplication, doesn't quite work: if there are no parentheses, what matters is which of them comes first when reading from left to right. (Although if you think of division as a kind of multiplication as in Asydot's answer below, you do have commutativity.) – coldnumber Sep 19 '15 at 0:59
• "Order of operations" is not a mathematical concept, it is notational. It is referring to how to interpret written arithmetic when there is possible ambiguity. Other orders of operations could be used instead (but most of the other ways of doing it are less useful). – Matt Samuel Sep 19 '15 at 1:07
• What I meant is that the way order of operations is sometimes taught conflicts with the way the operations are defined algebraically. This is probably tangential, but what do you mean that it's not mathematical? It is a way of interpreting mathematical notation commonly encountered at that level (and even in college algebra; a class I'm TAing for has exercises on this, and they do trip students up). – coldnumber Sep 19 '15 at 1:23
• @coldnumber It is not intrinsic to the operations, only to the way we choose to interpret what is written. Aliens could possibly do it differently, but they would still agree that $2+3=5$. – Matt Samuel Sep 19 '15 at 1:42
• @coldnumber ...(to make my point stronger) even if they say that $3/(2/ 5)=(1.5/0.6)$ because parentheses are always applied last. – Matt Samuel Sep 19 '15 at 1:46

You may notice that subtraction can be seen as a special case of addition, i.e. $$a-b=a+(-b)$$ Thus by commutative law, $$a-b=a+(-b)=(-b)+a=-b+a$$ It is certainly not $b-a$.

Similarly, division is as a special case of multiplication, i.e. $$\frac{a}{b}=a\cdot(\frac{1}{b})$$ Where $\frac{1}{b}$ means the inverse element of $b$ with respect to multiplication.

Therefore, $$\frac{a}{b}\cdot{c}=a\cdot\frac{1}{b}\cdot{c}$$ However, $$\frac{a}{b\cdot{c}}=a\cdot\frac{1}{b\cdot{c}}=a\cdot\frac{1}{b}\cdot\frac{1}{c}$$ And they are certainly not equal.

Edited: Since you are motivated to pursue advanced math, I have a little idea to share.

I assume you have intuitively understood the arithmetic about nature numbers. It is based on this simple idea: the one-to-one correspondence between (abstract) nature numbers and (concrete) things. Then you easy find that:

$(+, 1)$ The commutative law of addition $$a+b=b+a$$

$(+, 2)$ The associative law of addition $$a+(b+c)=(a+b)+c$$

$(\times, 1)$ The commutative law of multiplication $$a\cdot{b}=b\cdot{a}$$

$(\times, 2)$ The associative law of multiplication $$a\cdot(b\cdot{c})=(a\cdot{b})\cdot{c}$$

$(\times, 3)$ The identity element of multiplication, i.e. $1$ $$a \cdot 1=1 \cdot a=a$$

$(+, \times)$ the distribution law of multiplication to addition $$a \cdot (b+c)=a\cdot {b}+a \cdot {c}$$

A possible intuitive approach has been shown by Paul Sinclair. And you also have noticed that addition and multiplication can be defined as recursive operation. That is, $$a+b:=a+\underbrace{1+1+1+...+1}_{b \text(terms)}$$ And $$a\cdot{b}:=\underbrace{a\cdot{a}...\cdot{a}}_{b \text(terms)}$$

You also have a nature idea to introduce the inverse operation, i.e. substitution and division. But you may find that $3-5$ is illegal in this stage (it has no definition), and neither is $\frac{3}{5}$.

But that is not hard for you, just noticed that you can define $\frac{3}{5}$ as a ratio: There certainly can be a thing $k$ such that $k\cdot{5}=3$, then we can denote $k$ by $\frac{3}{5}$. Here come the rational numbers! And you may just find

$(\times, 4)$ The reverse element of multiplication $$\text{For every number a there is a number b such that a\cdot{b}=b\cdot{a}}$$

And we can easily find that by definition, $b=\frac{1}{a}$. (the "number" here means nature number and positive rational number.)

However you are not satisfied. Sometimes you need to find a way to denote "nothing", so you need

$(+, 3)$ The identity element of addition, i.e. $0$ $$\text{For all a, a+0=0+a=a}$$

As far as I'm concerned, historically, in a relative long time, $0$ or similar notations were initially but for this kind of convenience.

A similar idea for convenience is negative number, it was originally used to express debt. With introducing negative numbers, we can derive:

$(+, 4)$ The reverse element of addition $$\text{For every number a there is a number b such that a+b=b+a=0}$$

And we can also easily find that by definition, $b=-a$.

Indeed, It took people a lot of time to accept the philosophical aspect of "nothingness", or to understand how debt times debt will be income (this was not philosophical confusion initially, but a result of wrong metaphor; however, it finally became one, but that is another story.)

But for mathematics, the most horrible thing is division by 0, it is certainly not legal. There is no help for it, so we have to ban it. Indeed, this is the only difference between addition and multiplication in abstract sense. If you consider about a more general case such as real numbers, the original, recursive idea will failed, or, at least, no more intuitive. Thus you have to define addition and multiplication in a more abstract way, that is, introducing the axioms of addition and multiplication, i.e. $(+,1)-(+,4),(\times,1)-{(\times,4)},(+,\times)$ I've mentioned before. If you compare $(+,1)-(+,4)$ with $(\times,1)-(\times,4)$, you will find they are nearly the same.

A set equipped with two operation satisfying those axioms will be the field which trb456 mentioned. (To say it clearly, the result of those operations must always within that set, and the identity element of addition can't be the same with that of multiplication.) We need this term for other reason: because there are a lot of other structures satisfying this definition. And modern algebra, though I am totally not familiar with it yet, seems like a study of mathematical structures.

This story is obviously not the real history of numbers and their arithmetic, and I just want to introduce a half-intuitive approach to understand those things. I omitted a lot of details, or maybe I just got things wrong. But I hope you enjoy it.

• Thanks it was interesting and a little rough to understand but I kind of get it. Is a^b essentially equal to aa..... b times .... *a? – coder guy Sep 19 '15 at 3:34
• and a square root is just a fractional exponent? – coder guy Sep 19 '15 at 3:36
• @banana If $b$ is a positive integer, that is OK (no matter what $a$ is). But if you want a rigor idea in mathematical sense, no. You may learn it, as well as how to construct real numbers in a rigorous way, in analysis courses in the future. – Asydot Sep 19 '15 at 3:40
• Regarding $a+b=b+a$ and $a\cdot b=b\cdot a$, it's commutative, not communicative. – Glen O Sep 19 '15 at 5:59
• @GlenO That was a spelling mistake - English is not my native language. Fixed, thanks :) – Asydot Sep 19 '15 at 6:02

I can believe you don't understand it even though you are in Algebra II. The problem is not in you, but is a weakness of the educational system. Most people, including most of your teachers, view math as little more than a collection of problem-solving techniques that must be memorized. Only a few genuinely try to understand it. And what your teachers don't really understand themselves, they cannot adequately explain to you.

A basic way of seeing intuitively that addition and multiplication ought to be commutative, associative, and that multiplication distributes over addition is by counting. If you have a set of $n$ objects and a set of $m$ objects, with nothing in common, then if you combine them, you have a set of $n + m$ objects. This is how addtion is defined for natural numbers. Does it matter which collection you dump into the combined pile first? No. So it doesn't matter what order addition is done in. If you add in a third collection of $k$ objects, you get $(n + m) + k$, but again, it doesn't matter what order you combined them in, so this is the same as $n + (m + k)$, associativity.

Similarly, multiplication is repeated addition of the same size set. If you have $n$ sets of $m$ objects, you can arrange the objects in $n$ rows of $m$ objects each:$$\begin{matrix} x & x & ... & x \\ x & x & ... & x \\& ...\\x & x & ... & x \end{matrix}$$ The total count is $n \times m$. But note that rotating the array by 90 degrees leaves you with $m$ rows of $n$ objects each. I.e., $n \times m = m \times n$. If you have $k$ such sets of $n \times m$ objects, you can arrange them in $k$ layers, but again, which is a row, a column, or a layer, depends only on how you look at the array, not on the number of objects in it: $(n \times m) \times k = n \times (m \times k)$. And if in the $n \times m$ array, we divide up each row into the first $a$ and last $b$ objectsm then by the definition of addition we have that $a + b = m$, and further, we have divided the entire array into an $n \times a$ array and a $n \times b$ array, so $n \times (a + b) = n \times a + n \times b$, distributivity.

But with subtraction, we don't have these nice pictures to show commutivity, associativity, or distributivity. (There are ways of picturing them, but they don't give us the properties like the pictures for addition or multiplication do.) For these what properties they have are usually obtained from the inverse relationships: $n - m = n + (-m)$ and $\frac{n}{m} = n \times m^{-1}$.

• Thanks, and I totally understand what you mean with the education system. I think teachers feel that mathematics is not something most students will use in their life, so they don't bother putting in much effort in advanced concepts. They also dumb down all of the terminology to that of a 5th grade class so that the anti-intellectual mainstream crowd of public schooling can understand. It saddens me, but I really do love mathematics, and I want to major in it. Thanks for your help. – coder guy Sep 19 '15 at 1:20
• Likely your current teachers do understand math better than most, and could explain it. But they have a certain amount of material they must cover, and this doesn't usually give them much time to go back and correct deficits left over from your previous instruction. – Paul Sinclair Sep 19 '15 at 1:26
• Especially when kids in the class don't listen so he has to move at a slower pace. Sometimes its the students fault, sometimes the teacher, sometimes both. – coder guy Sep 19 '15 at 3:23
• @banana so that the anti-intellectual mainstream crowd of public schooling can understand. I just want to (sort of) prevent you from developing an arrogant attitude towards "non-mathematicians" here: Someone having problems understanding math does not make him or her stupid nor part of an "anti-intellectual mainstream"! If this was true, this site would probably be the largest gathering of idiots on the internet. I know you just wanted to weigh in on the criticism and express your disagreement with the educational system, but are you sure you are getting to the root of the problem here? – Piwi Sep 19 '15 at 3:40
• @Piwi I'm not just speaking of maths. A large percentage of people who go to my school are nihilistic towards learning in general. – coder guy Sep 19 '15 at 22:40

Maybe the question that you should ask yourself is: Why would you expect that you can interchange them? It's not that subtraction is special in that you cannot exchange the terms, it's rather that addition is special in that you can!

If I give you $a$ dollars and then I give you $b$ dollars, you have the same amount of money as if I first give you $b$ dollars and then $a$ dollars, namely $a+b$ dollars.

If I give you $a$ dollars and then take $b$ dollars away from you, it's clearly not the same as if I give you $b$ dollars and then take $a$ dollars away from you (unless $a=b$, of course). In the first case, your money has changed by $a-b$ dollars, in the second case it has changed by $b-a$ dollars.

Okay. We could say that subtraction is interchangeable (commutative), but what you need to realize is that the sign before each number is part of the number. So $a-b=c$ would commute to $-b+a=c$. (Whenever there's no sign, that's the same as $+$, so $a=+a$.

What we've done is identify subtraction as a special form of addition, and commuted that.

Your multiplication issue might be clarified as follows.

$$\frac{6}{3}\times\frac{2}{1}=\frac{12}{3}=4$$
$$\frac{6}{3\times 2}=1$$

So, you're not really multiplying the whole fraction by $2$ in the second case.

In your central example, the equality holds because both of the left sides equal $$\frac{a}{1}\times\frac{b}{c}$$

• This also helped me understand a lot better. So subtraction basically means adding negative numbers, and division is basically multiplying numbers by their reciprocals? – coder guy Sep 19 '15 at 1:12

In fact, we have $a - b = b - a$ if and only if $a=b$ (can you see why?), so, precisely, it should be said that subtraction is not always commutative.

Dividing a number $a$ by a nonzero number $b$ is to multiply $a$ by $b^{-1} = 1/b$. Note that $\frac{6}{3}\times 2 = 6\times 3^{-1} \times 2$, but $\frac{6}{3\times 2} = 6\times 3^{-1} \times 2^{-1}$, so you get different outcomes.

I hope the above is plain enough for you to understand.

Good question! The mathematical word for interchange is commutative. So you have discovered that addition and multiplication are commutative, but subtraction and division are not commutative.

But when you get to more advanced math, you will discover something amazing:

Subtraction and division do not exist!

OK, that's extreme, but it's close to the truth.

When you do ordinary arithmetic, you are almost certainly working in a field. A field is just a set of numbers where you can do all "ordinary" addition, subtraction, multiplication, and division. But in the formal definition of a field, there are only two arithmetic operations: addition and multiplication.

So why no subtraction and division? Because in a field, for any element $x$, we are guaranteed that two other elements exist:

The additive inverse $-x$, where $x + -x = -x + x = 0$;

and the multiplicative inverse $1/x$, where $x \times 1/x = 1/x \times x = 1$ (except for $x=0$ of course).

And so in a field, subtraction and division can be defined in terms of addition and multiplication, respectively, the latter of which are both commutative. So subtraction is not commutative, but additive inverses are commutative, and similarly for division and multiplicative inverses.

• @MattSamuel: I guess I don't agree. The OP is asking a more advanced question than their current level of learning has taught them. Why not introduce the proper language given their advanced question? I just said a field is where you can do ordinary arithmetic. Why is that confusing? I trust that people asking good questions can handle more advanced answers. They are trying to learn, after all. – user452 Sep 19 '15 at 1:03
• @trb456 Actually I do want to pursue advanced math. I want to major in mathematics. But you are right in the fact that some of these terms are confusing to me. Are you basically saying subtraction is almost like adding negative numbers to positive numbers, and division is simply multiplying by fractions/decimals? I am still trying to grasp it all, but this has cleared a few things up for me. – coder guy Sep 19 '15 at 1:03
• @banana:Your intuition is exactly right! All I have done is given you some terminology that will serve you in the future. But you get it for sure! – user452 Sep 19 '15 at 1:05
• @banana That applies only when you do multiplication between nature numbers (positive integers). – Asydot Sep 19 '15 at 1:08
• This answer seems strange. It looks analogous to saying the natural number $1$ doesn't exist, because Peano's axioms make no direct reference to it. – Excluded and Offended Sep 19 '15 at 2:40

The commutative law or commutative property means that an operator (e.g. +, -, ×, ÷, and many others) or function has the same result regardless of the order of its operands or inputs. For example if you look at addition then the results are the same regardless of the order of the operands, $$x + y = y + x$$ and likewise for multiplication, $$x × y = y × x$$

There is also a reverse (of sorts) property called the anti-commutative property. In this case when the two operands are reversed the result is the inverted. For example $$x - y = - (y - x)$$ but this is actually shorthand for the more general solution, given operator · and the right identity $I_{(·)}$ you get the following $$x · y = I_{(·)} · (y · x)$$ and so for the example of subtraction above you get $$x - y = 0 - (y - x)$$ where $0$ is the right identity for subtraction and addition (any number that you subtract zero from is itself, i.e. $x - 0 = x$). And then for the division operator (÷) you get the following $$x ÷ y = 1 ÷ (y ÷ x)$$(likewise $1$ is the right identity for division and multiplication, i.e. $x ÷ 1 = x$).

And in fraction form it is $$\frac{x}{y} = \frac{1}{\frac{y}{x}}$$

Obviously there are limits to the anti-commutative property in that $x · y$ and $y · x$ have to exist and that a right identity for the operator $I_{(·)}$ has to exist as well.

Please note that some texts will state or imply that the anti-commutative property only applies to the additive inverse (in other words subtraction) but there is a more general form of anti-commutativity (as described above with the operator $·$ and identity $I_{(·)}$).