# Is the division property of equality just a special case of the multiplication property?

My textbook clearly states after the lesson on Transforming Equations: Addition and Subtraction

"Notice that the subtraction property of equality is just a special case of the addition property, since subtracting the number c is the same as adding -c."

But, following the next lesson which is on Transforming Equations: Multiplication and Division there is no equivalent statement such as

The division property of equality is just a special case of the multiplication property, since dividing by the number c is the same as multiplying by the reciprocal of c.

Is division just a special case of multiplication, with the restriction of not having 0 in the denominator?

If so, why would they not include such a natural parallel statement?

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Yes. ${}{}{}{}$ – Gerry Myerson Dec 11 '12 at 12:15
A reason why you are down-voting this question would be appreciated. – Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ Dec 11 '12 at 12:17
@GerryMyerson If so why would they not include such a statement? – Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ Dec 11 '12 at 12:23
@skullpatrol: A book, however long, cannot contain all things on the subject that are true. Probably there was no deliberate exclusion. – André Nicolas Dec 11 '12 at 17:14
@skull, that is a question of psychology, or possibly of pedagogy, not of mathematics. If you really really want to know, I suggest you ask the people best placed to tell you, viz., the author(s). – Gerry Myerson Dec 11 '12 at 23:21

So in order to understand why this must be true, we have to consider what "equals" means in this context.

Assume you were talking to someone who had never heard of division before in his life. How would you explain why you can write $$\frac{1}{5}=\frac{5}{25}?$$ These fractions are equal precisely because $$25\cdot 1 = 5 \cdot 5.$$ In abstract algebra, this is how we define fractions: the equivalency classes of pairs of numbers $(a,b)$ with $b\not= 0$ under the relation $(a,b)\sim(c,d)\Leftrightarrow ad=cb$. (Note: I use the word "numbers" here loosely; see the last paragraph for more info.) We write these pairs $(a,b)$ as $a/b$ or $\frac{a}{b}$ simply because it's convenient.

So, you want to know whether the division property is a special case of the multiplicative property. Let's see if it is.

Assume for the remainder of the answer that $x\not= 0$. The multiplicative property is that $$\frac{a}{b}=\frac{c}{d} \hspace{10pt}\Leftrightarrow \hspace{10pt}\frac{xa}{b}=\frac{xc}{d}.$$ First, let's check if that's true, based on the definition of fractions above. $$\frac{a}{b}=\frac{c}{d}\hspace{10pt} \Leftrightarrow \hspace{10pt}ad=cb \hspace{10pt}\Leftrightarrow\hspace{10pt} xad=xcb\hspace{10pt} \Leftrightarrow \hspace{10pt}\frac{xa}{b}=\frac{xc}{d}.$$ Alright, I buy that. Now what about the division property? $$\frac{a}{b}=\frac{c}{d} \hspace{10pt}\Leftrightarrow \hspace{10pt}\frac{a}{xb}=\frac{c}{xd}.$$ How do we check if this is true? $$\frac{a}{b}=\frac{c}{d}\hspace{10pt} \Leftrightarrow \hspace{10pt}ad=cb \hspace{10pt}\Leftrightarrow\hspace{10pt} axd=cxb\hspace{10pt} \Leftrightarrow \hspace{10pt}\frac{a}{xb}=\frac{c}{xd}.$$ You can see that this is exactly the same thing. The only difference in the proofs are that $xad=xcb$ and $axd=cxb$, so the two properties are equivalent precisely because multiplication is a commutative operation (that is, $ab=ba$ always). In fact, the division property implies the multiplication property too: $$\frac{xa}{b}=\frac{xc}{d}\hspace{10pt}\Leftrightarrow \hspace{10pt}xad=xcb\hspace{10pt} \Leftrightarrow \hspace{10pt}axd=cxb\hspace{10pt} \Leftrightarrow \hspace{10pt}\frac{a}{xb}=\frac{c}{xd}.$$

Optional Note: All this comes from the concept of a field of fractions, which you can read about on Wikipedia if you're feeling ambitious. In short, any time you have a set of elements that can be added and multiplied, where every $a,b$ satisfy $ab=ba$ and such that no nonzero $a,b$ satisfy $ab=0$, you can define fractions of that set which work exactly as they do with regular numbers. The word for making fractions like this is "localization," and you can do it with all kinds of stuff other than numbers, like functions, for example.

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I cannot speak for the authors, but my guess is that the relation between addition and subtraction is sometimes subtly different from that between multiplication and division, while sometimes the difference is more pronounced. For example, when dealing with real equations, division is simply multiplication by the inverse, except when dividing by $0$. However, in the case of $\mathbb{Z}$ and $\mathbb{R}[x]$, there are multiplicative inverses of only a few elements, whereas there are additive inverses of all of the elements.

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Thanks for the perspective. – Sᴋᴜʟʟ ᴘᴇᴛʀᴏʟ Dec 19 '12 at 22:01

If the book is not very "symmetric" in general, it's possible that they just kind of forgot it or something. However, my guess is that the book is aimed at beginners, and it's easy for them to forget that zero is a special case, especially when variables are used. For me it's easier to remember to check for zero when I'm thinking of dividing instead of multiplying by the reciprocal. So, suppose I have $$ac=bc.$$ If I learn that I can just multiply by the reciprocal of $c$ I would get $$a=b$$ which is not necessarily true. However, if I think that I could divide by $c$, I'll probably remember that zero is always an exception and I would get $$a=b\quad\text{or}\quad c=0$$ which is correct. Of course you can learn that zero is an exception in the multiply-by-reciprocal case, too, but I think that the division-by-zero problem is more deeply planted already.

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I'd write that "the subtraction property can be subsumed under the addition property, since subtracting $c$ is the same as adding $-c$." ( A "special case" of addition would be adding $1$.)

It is certainly not the same thing with multiplication and division. Of course, dividing by $5$ is the same as multiplying by ${1\over5}$. But one has to take more complicated situations into account. Assume that $\Phi=\Phi(\ldots)$ and $\Psi=\Psi(\ldots)$ are complicated expressions containing all sorts of variables and parameters playing a rôle in the problem at hand, and that $c=c(\ldots)$ is another, maybe simpler, such expression. Then for all values of the variables hidden in $(\ldots)$ you have the valid conclusions $$\Phi=\Psi\quad\Longrightarrow\quad \Phi+c=\Psi+c\ ,\quad \Phi-c=\Psi-c$$ and $$\Phi=\Psi\quad\Longrightarrow\quad c\cdot \Phi=c\cdot \Psi\ ;$$ but you are not allowed to make use of $$\Phi=\Psi\quad\Rightarrow\quad {\Phi\over c} = {\Psi\over c}\qquad(*)$$ (which might result in a simplification of the expressions involved) without a discussion, under which circumstances $c$ might be zero and whether such circumstances could occur in the case at hand.

If you use $(*)$ carelessly nevertheless it may happen ten minutes later that you fix some variables in a way such that $c$ (not even appearing in your formula anymore) evaluates to zero; so that for this particular instance the application of $(*)$ is illegal, and $1=0$ follows easily $\ldots$

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Short answer: Yes, but vice-versa applies too.

Long answer with a little philosophy:

Yes, in terms of algebra and pre-calculus, division is a special case of multiplication if dealing with real numbers ($\mathbb R$). For example, dividing by $2$ is the same as multiplying by ${1 \over 2}$. But, in turn, we also see that multiplication is a special case of division: dividing by $1 \over 2$ is the same as multiplying by $2$.

THE POINT TO BE NOTED$\ \ \ \$ Both multiplication and division are significant in their own uses. There is no questioning on their existence, just in case you think that division is just an extra operator...

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