# Distributive property on fractions

I'm in seventh grade and my teacher wasn't able to explain this to me.

why is $\frac{1}{a+b}$ not equal to $\frac 1b +\frac 1a$?

I'm sorry if this is obvious.

EDIT: thank you to everyone who responded. I think I understand fractions a lot more now. it was good to get both intuitive and algebraic answers... that really nailed the point home for me

Check for yourself by trying some numbers! For example, if $a= b =1$, then $1/(a+b) = 1/2$, while $1/a + 1/b = 2$. Since $1/2 \neq 2$, we have that $1/(a+b) \neq 1/a + 1/b$ in this case.

So clearly $a/(b+c) \neq a/b + a/c$ in general. Why, on the other hand, does $(a+b)/c = a/c + b/c$? The answer is that this really is just using the usual distributive property! I can do the following algebraic tricks: $$\frac{a+b}{c} = (a+b) \frac{1}{c} = a \frac{1}{c} + b \frac{1}{c} = \frac{a}{c} + \frac{b}{c}$$ So really, all we've done here is distributed the factor of $1/c$ over the sum $(a+b)$.

The sum $(b+c)$ in $a/(b+c)$ isn't being multiplied by anything in this expression; in fact, it's being divided by! So trying to distribute the division over this sum would be a new distributive property, and as we observed above, this property does not actually hold.

• hi, thanks for answering. I know that it's true, but my main confusion was why a+b/c = a/c + b/c when 1/a+c does not equal 1/a + 1/b? Sorry for being unclear – bluesky777 Jun 22 '15 at 2:44
• +1 for "try examples!" This example generalizes nicely to adding $\frac1b$ to itself $b$ times. Surely this should add up to $1$ (as, for example when $b=4$, the four one-fourths of a pizza make a whole pizza, not one-sixteenth of a pizza!) – pjs36 Jun 22 '15 at 2:46
• @bluesky777 I'll edit my answer to address the second part of your question. – Alex G. Jun 22 '15 at 11:12
• While $\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b}$ is clearly not an identity which holds for all $a, b$, if you use complex numbers, you can find specific pairs $(a,b)$ for which this holds. If for example you fix one $a\in\mathbb{C}$ you can find an associated $b$ by solving a simple quadratic equation. – Jeppe Stig Nielsen Jun 22 '15 at 12:56

Since you know algebra, here is a proof that may satisfy you. Consider: $$\frac{1}{a+b} \stackrel{?}{=} \frac{1}{a} + \frac{1}{b}$$

If they are equal, then multiplying them by the same thing should keep them equal. Similarly, if they are unequal, then multiplying them by the same thing (except 0) should keep them unequal.

Now, multiply both sides by $ab$ to get $$\frac{ab}{a+b} \stackrel{?}{=} \frac{ab}{a} + \frac{ab}{b}$$ $$\frac{ab}{a+b} \stackrel{?}{=} b + a$$

Now, multiply both sides by $a + b$ to get $$\frac{ab(a+b)}{a+b} \stackrel{?}{=} (b + a)(a+b)$$ $$ab \stackrel{?}{=} (a + b)^2$$

You already know what the right-hand side is equal to: it is $a^2 + 2ab + b^2$: $$ab \stackrel{?}{=} a^2 + 2ab + b^2$$

Now subtract $ab$ from both sides: $$0 \stackrel{?}{=} a^2 + ab + b^2$$

### Here's the key:

Your claim was that this equality holds for all $a$ and $b$. (That is, $a^2 + ab + b^2 = 0$ for all $a$ and $b$.)

Since this is obviously false (you had complete freedom to choose $a$ and $b$), your claim cannot be true. Does this make sense?

• In fact, equality only holds if $a=b=0$, but we can't divide by 0, so it is never equal. – wythagoras Jun 22 '15 at 13:19
• @wythagoras: it is true for $a=b\left(\frac{-1\pm\sqrt{-3}}{2}\right)$, so there are other solutions in complex numbers – Henry Jun 22 '15 at 13:37
• Nice proof, easy to follow too! – CivilSigma Jun 23 '15 at 4:58

Why would it be equal?

Think of $1/3$ or $1/4$ as a single third or a single fourth. (I would use $1/c$ here, but then there isn't a convenient word to go with it.) We have $a/3+b/3=(a+b)/3$, since $a$ thirds plus $b$ thirds is $a+b$ thirds. However, we don't have $1/3+1/4=1/7$; a seventh is smaller than a third or a fourth.

Let $p$ and $q$ be positive integers (whole numbers). In the fraction $\frac{p}{q}$, the numbers $p$ and $q$ play distinct roles:

• $q$ measures the "size" of (or denominates) $\frac{1}{q}$. Intuitively, $\frac{1}{q}$ is the quantity obtained by dividing one unit into $q$ equal portions. For example, $\frac{1}{2}$ represents the result of dividing $1$ into two equal portions; $\frac{1}{3}$ results from dividing $1$ into three equal portions; and so forth. Sometimes we even read "$\frac{1}{q}$" as "$1$ divided by $q$", as in "one unit divided into $q$ equal portions".

• $p$ tells you "how many portions" of size $\frac{1}{q}$ make up $\frac{p}{q}$. That is, $p$ numerates the fraction.

A rule such as $\frac{2}{7} + \frac{3}{7} = \frac{5}{7}$ holds because two of some quantity (here $\frac{1}{7}$) added to three of that same quantity gives five units of the quantity.

A rule such as $\frac{4}{14} = \frac{2}{7}$ holds for a reason a little complicated to say in words: If fourteen units of some quantity make one unit and seven units of another quantity make one unit, then four units of the first quantity equal two units of the second quantity. Here, we have twice as many portions of a quantity half as large.

Yogi Berra was once asked, "Do you want your pizza cut into six pieces or eight?" He replied, "Better make it six. I could never eat eight pieces of pizza." Now you can spoil the joke by explaining it mathematically: $\frac{6}{6} = \frac{8}{8}$.

Finally we come to the question: Why is $\frac{1}{a + b}$ not equal to $\frac{1}{a} + \frac{1}{b}$? As user2357112 notes, a "better" first question to ask might be the skeptical one, "Why are they (or why should they be) equal?" The preceding paragraphs explain why, if $p$, $p'$, $q$, and $k$ are positive integers, then $$\frac{p}{q} + \frac{p'}{q} = \frac{p + p'}{q},\qquad \frac{kp}{kq} = \frac{p}{q}.$$ Each rule (or "theorem") comes down to counting, possibly counting different numbers of units of different size.

Now, the fraction $\frac{1}{a + b}$ represents an amount, $(a + b)$ portions of which make one unit.

The expression $\frac{1}{a} + \frac{1}{b}$ represents an agglomeration of two quantities: An amount, $a$ portions of which make one unit, and an amount, $b$ portions of which make one unit.

These cannot represent the same quantity if $a$ and $b$ are positive integers: The larger the denominator, the smaller the denominated portion, because a larger number of portions are needed to make one unit (Yogi Berra's pizza slices). In symbols, we might express the argument as $$\frac{1}{a + b} < \frac{1}{a} < \frac{1}{a} + \frac{1}{b}.$$ And as you've noticed, $\frac{1}{a + b} \neq \frac{1}{a} + \frac{1}{b}$ in examples. It's effectively because the denominator measures the sizes of portions, while addition expresses agglomeration of portions of fixed size.

Since we've come this far: What is $\frac{1}{a} + \frac{1}{b}$ as a single fraction? If $a \neq b$, we're "adding apples and oranges", i.e., trying to agglomerate amounts in different systems of measurement, like adding $2$ inches to $4$ centimeters. To proceed, we need to express each quantity in terms of a common denominator, as it were. One way to do this is to cross-multiply and then count the total number of portions: $$\frac{1}{a} + \frac{1}{b} = \frac{b}{ab} + \frac{a}{ab} = \frac{a + b}{ab}.$$

By similar reasoning, you can justify the general addition formula: $$\frac{p}{q} + \frac{p'}{q'} = \frac{pq' + p'q}{qq'}.$$

We have

$$\frac{a+b}{c} = \frac{a}{c} + \frac{a}{c}$$

But on the other hand

$$\frac{a}{b+c} \not= \frac{a}{b} + \frac{a}{c}$$

Why is that?

Consider that $x/y$ can also be written as $x \cdot (y^{-1})$. As you probably know,

$$(a+b)^2 \not= a^2 + b^2$$

And indeed, in general,

$$(a+b)^n \not= a^n + b^n$$

It just doesn't work like that. Given this fact, it's not too surprising that

$$(a+b)^{-1} \not= a^{-1} + b^{-1}$$

In other words,

$$\frac1{a+b} \not= \frac1a + \frac1b$$

Again, $x/y = x \cdot (y^{-1})$. The denominator is raised to the power $-1$. But the numerator is not, so you can add and subtract with the numerator. But not the denominator.

• Your answer seems to imply that $(a+b)^n \neq a^n + b^n$ for all $n$. At least, that's the only argument you give as to why $(a+b)^{-1} \neq a^{-1}+b^{-1}$. But of course $(a+b)^1 = a^1 + b^1$! – Alex G. Jun 22 '15 at 13:27
• @AlexG. As far as I know, 0 and 1 are the only cases where this holds. – MathematicalOrchid Jun 22 '15 at 14:23
• It doesn't hold for $n=0$.... But anyway, the point is that you can't extrapolate from $(a+b)^2 \neq a^2 + b^2$ and $(a+b)^3 \neq a^3 + b^3$ that $(a+b)^{-1} \neq a^{-1} + b^{-1}$. – Alex G. Jun 22 '15 at 14:26
• It's certainly not a rigorous proof. But it's supposed to give some intuition as to why the denominator behaves differently. – MathematicalOrchid Jun 22 '15 at 14:28

Put super intuitively;

With a=3, b=4, and (a+b)=7

Left: "cut one pie into 7 pieces and give me one"

Right: "cut one pie into 4 pieces, cut a second pie into 3 pieces, and give me a slice from each pie"

The difference is that the left side represents taking a small slice from a single pie, while the right side represents taking larger pieces from two pies, then combining them.

There is a number of good answers already but since this approach to mathematics is a pet peeve of mine I will add one more.

The question is very much the wrong way around and to me, as someone who has taught introductory college courses (basic calculus), embodies what is wrong with mathematical education at the elementary level.

In math the question about identities you should generally be asking is Why would so and so hold? as opposed to why doesn't so and so hold. Any identity should be assumed false until rigorously proven true. The reasons for this are multiple but the most essential one in my opinion is that whatever intuition you think you have about math is going to be wrong much more often than it's right, and moreover there are many more identities that don't hold then ones that do hold.

Your particular question is an example of a very common fallacy which I like to call Everything is linear. I have no idea where it originates, but it seems deeply ingrained and leads to identities such as $\sqrt[p]{a^p+b^p}=a+b$, $(a+b)^p=a^p+b^p$ (holds only over fields with characteristic $p$), $\ln(a+b)=\ln(a)+\ln(b)$, and also in reverse $e^{a+b}=e^a+e^b$.

All the above example can be easilly shown not to hold by plugging in a couple numbers yet time and time again students ask Why doesn't this hold? or more often just use the "identities" in their work and then are surprised when they get no points.

In conclusion I believe the right way to approach mathematical problems (particularly of this kind) is Do I know of some reason why this should hold?, believe me this will save you pain in the long run.

Addendum: This approach is usually a good idea even in higher mathematics where IMO the reason is that in general few Why doesn't so and so hold? questions have a very satisfying answer (though some do) whereas finding out why something does hold usually teaches you a lot about the structure you are working with.

Did you ever think that you would get such a response?! Are you more confused now, or less? Welcome to the Math stack exchange! I'm sure you have noticed that this is a community of very bright mathematicians. But what language are they speaking?! Right?

The field of Mathematics does have a language of it's own, which becomes more and more apparent as you progress to higher and higher levels of math. Before long you'll see equations filled with greek letters and all kinds of new symbols that mean nothing to you in 7th grade. But you have actually already learned a lot more of this language than you may think. 2 + 2 is not English, but in elementary school, if not earlier, you learned that it represents the combination of two things with two more things. And you've learned that you get four things from that combination.

Many people get lost along the way in the language of math. You are now seeing equations that include horizontal bars which represent fractions. You may also recall that the fraction bar also represents division. ($\frac{8}{4}$ is the same as 8 ÷ 4.) You are learning a set of rules about how you can move symbols around without breaking the truth of the equation. If you have two things added together on the top of a fraction bar, you can separate them into their own fractions with the same denominator, but you can't do the same if they are added together on the bottom. It can feel like an overly complex boardgame with so many rules that you will never remember them. And if you try, by the way, you will probably always hate math.

The beauty of math, and probably the reason that people in this community have always gotten so excited about it, is that the rules all reflect something far beyond the equation game. They are not made up at all, but derived from the meaning of the symbols in the language. They reflect an irrefutable truth. I never memorized the fact that you cannot distribute addition across the denominator of a fraction. I just tried to understand what a denominator was. Once you do that, all of the rules make so much sense that you can forget the rules entirely, and simply re-derive when faced with a question again.

The top of the fraction is like the numbers you have always used. If you add to them, you get bigger numbers. If you subtract from them you get smaller numbers; even negative numbers if you subtract enough. The bottom of the fraction (the denominator) is what divides the top, so think of it as the number of people you have to share with, or the number of buckets you have to split something into. If you increase that number, you have to split the same amount among more. A pie divided into 8ths rather than 4ths means that the same size pie has to be divided into twice as many pieces. So they are smaller. When you add to that number, it is the opposite (in some sense) of adding to the top number.

So now, your question: Consider what it means. $\frac{1}{a}$ is what you get if something is divided a ways. $\frac{1}{b}$ is what you get if something is divided b ways. $\frac{1}{a+b}$ is what you get if something is divided a+b ways, which is not at all the same as adding the $\frac{1}{a}$ and $\frac{1}{b}$ portions together. You are doing the opposite. You are adding to the number of ways you have to divide this portion. It's like the difference between bringing more food to the table and bringing more hungry people to the table. The food in this analogy is on top of the fraction bar; the hungry people on the bottom.

Multiply both sides by $a+b$, then use distributive property:

$$\frac{1}{a+b}\stackrel{?}=\frac{1}{a}+\frac{1}{b}$$

$$\iff 1\stackrel{?}=\frac{a+b}{a}+\frac{a+b}{b}$$

$$=\frac{1}{a}(a+b)+\frac{1}{b}(a+b)=$$

$$=\frac{a}{a}+\frac{b}{a}+\frac{a}{b}+\frac{b}{b}$$

$$=1+\frac{b}{a}+\frac{a}{b}+1$$

$$\iff \frac{a}{b}+\frac{b}{a}\stackrel{?}=-1$$

Clearly $\frac{a}{b}+\frac{b}{a}$ doesn't even have to be negative. The equality can't always hold.

For any positive $a,b$, LHS is positive and $\neq -1$.