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quick question here:

In my proofs class we had a problem that after a little work we end up with: $x(x-y)=(x+y)(x-y)$ where $ x = y $. Now, I know this is pretty basic, but my teacher said that for the next step, one cannot cancel out $(x-y)$ from both sides as $(x-y) = 0 $. Can someone explain the logic and/or the reasoning behind this?

I'm pretty sure this falls under some obscure basic algebra rule that I've forgotten over the years but I cannot find anything about this on the internet.

Edit:

To clear up some confusion here, I am not looking for how to solve this problem, but rather the why this particular rule is so.

The problem I am working on gives a proof. I am supposed to mark the errors in the proof. For this problem, the error was that they cancelled out $(x-y)$ and I am trying to understand why that's an error.

Descoladan
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    Suppose for instance we have the following silly equation $2x=1x$. This is only true for $x=0$. Suppose we had divided each side by $x$ - we would have gotten $1=2$. – SDiv Oct 21 '14 at 12:51
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    Are you working on an integral domain? – Aaron Maroja Oct 21 '14 at 12:59
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    @SDiv We really do have that silly equation here. If (x-y)=0, then x=y. So substitution in x(x−y)=(x+y)(x−y) produces x(x−x)=(x+x)(x−x). And then we just let u=x(x−x) to get u=2u. – Keen Oct 21 '14 at 21:55
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    @Cory That was a whole lot of work for little gain. Since you know that x-y equals 0 you know that both sides of the equation equal 0. And therfore that the equation is valid, given that x-y=0. – Taemyr Oct 22 '14 at 08:46
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    You'd need a case split: In case (x - y) = 0 the equation holds for any x and y. In the other case you can "cancel" this term and proceed. – usr Oct 22 '14 at 10:55
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    @Taemyr 'Tis a silly equation. – Keen Oct 22 '14 at 13:28
  • @Cory It's part of a proof. So the validity of the equation might be all that is needed. – Taemyr Oct 22 '14 at 13:52
  • @Taemyr I mean only that the equation I produced is valuable for its silliness, and the exercise was refreshing anyway, so within my decision system, it was worth producing. – Keen Oct 22 '14 at 13:58
  • See this wikipedia article for how dividing by zero can be used to produce fallacious proofs. – rlms Oct 22 '14 at 20:11
  • Also, technically the title is not correct. It should be "Can I cancel out a factor, if the factor is zero?" – nickalh Oct 24 '14 at 11:15
  • Easiest way to understand: Suppose that $a\ne b$ and $x=0$. It is true that $ax=bx$, because $0=0$, duh. But if you "cancel" the x (that is, divide by $x$, which is undefined when $x=0$), you get the contradiction $a=b$. – Larry Gritz Oct 27 '14 at 19:31

11 Answers11

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As a teacher, I would prefer to enjoin my students to eradicate the word “cancel” from their vocabulary and the supposed process of cancellation from their technique. What we may do is multiply both sides of an equation by any number, knowing that the equality will be preserved. In the example, you are trying to multiply by $1/0$, not a number.

Lubin
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  • Duly noted, thank you for that tip. Also, thank you for giving the actual answer I was looking for. – Descoladan Oct 21 '14 at 13:48
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    I have banned the word "cancel" from my classes also. – JP McCarthy Oct 21 '14 at 13:48
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    How do you explain the act of simplifying a fraction by cancelling common factors? You're not going to say both can be multiplied by the same number (possibly $0$, certainly a number...). – Marc van Leeuwen Oct 21 '14 at 17:40
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    @MarcvanLeeuwen Simplifying, say, 4/6 to 2/3 is multiplying by (1/2)/(1/2), which (because that's equal to 1) doesn't change the value. – cpast Oct 21 '14 at 18:47
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    In agreement with @cpast, let me say that my rule was for equations only; the corresponding rule for fractions is that you can multiply top and bottom by any nonzero quantity and preserve the value of the fraction (for the cognoscenti, I amplify that I’m speaking of fractions over an integral domain). – Lubin Oct 22 '14 at 00:42
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    @cpast: I cannot see how saying it is multiplying by $\frac{1/2}{1/2}$ is any clearer than saying it is dividing by $\frac22$, but in any case it is not about multiplying or dividing a rational number by $1$ (which transforms $4/6$ into $4/6$), it is about changing the representation of a rational number, for which one may cancel, yes cancel, common factors from numerator and denominator. Which is legal because by definition rational numbers are equivalence classes of expressions under a relation that allows such cancellations. – Marc van Leeuwen Oct 22 '14 at 12:50
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    "Cancelling" is one example of a kind of magic, like "moving" a term from one side of an equation to the other by reversing the sign, and requires rote learning of rites and rituals that many students don't follow. By contrast, very few fail to grasp addition, subtraction, multiplication, division, which simple operations these terms obfuscate. The general level of mathematical literacy, even in affluent, well educated parts of the world, is disappointingly low - this widespread practice of teaching tricks like "cancelling" over basic principles must surely be the prime suspect. – DeveloperInDevelopment Oct 22 '14 at 22:36
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    @MarcvanLeeuwen Nope, rational numbers are defined as integers divided by natural numbers. The definition you stated is an abstraction of observational properties rational numbers appear to have. Your definition is an observational one that relied on rational numbers being observed in the first place. Cancellation was an observable phenomena the equivalence class definition captures, yet actually cancellation is an illusion, that results from being able to divide natural numbers by natural numbers too. – Brad Graham Sep 01 '15 at 12:24
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    @BradGraham: I think what you call "your definition" (i.e., mine) is defining rational numbers as equivalence classes of expressions? Not really mine, something like that is unavoidable, see for instance here. Nothing "observational" there either. One needs to have rational numbers availble in order to be able to divide integers by nonzero integers in the case where integer divisibility fails. One cannot define an operation without first having a set in which its values are taken. – Marc van Leeuwen Sep 01 '15 at 13:24
  • @MarcvanLeeuwen one can approach the problem geometrically and define rational numbers without equivalence classes. Actually this is how rational numbers were first observed. Your algebraic construction is an abstraction of observations from this. You are confusing abstractions used to make sense of our observations as the observed themselves. – Brad Graham Sep 01 '15 at 14:28
  • In thinking about rational numbers, it is more realistic to think about dividing lengths into equal parts and noting the ratios of original length to the parts are not dependent on the size of the original length. There is no cancellation. Just the inverse of repeating a length which is turning lengths into repeteated lengths themselves. – Brad Graham Sep 01 '15 at 14:34
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    @BradGraham: Your point of view reflects how the ancient Greeks viewed these matters, who I think viewed geometry as existing without needing any human construction. However modern mathematics does not take this point of view. Rational numbers need to be introduced at some point; doing so as ratios of length word require lengths to be introduced first, and since they are rather more complicated than rational numbers, that would be putting the cart before the horse. But this is no place for philosophical discussion; if you have that point of view, tha's fine with me. But don't attack me for it. – Marc van Leeuwen Sep 01 '15 at 14:54
  • And to add about "abstraction from observation", that is not really what is going on. It is just an instance of the formal definition of a field of fractions of an integral domain, for the case of the domain of integers. That definition is based on logic and algebra, but since integral domains are a pretty abstract notion, there is hardly any observation that could guide the construction. My apologies to your right brain hemisphere. – Marc van Leeuwen Sep 01 '15 at 15:00
  • Actually, again an integral domain is an abstraction of the integers which is an abstraction of the natural number to encompass positional qualities. And i understand what you are saying, for example you could argue that "cancellation" allows computers to do faster mathematics for example. But, when dealing with problems of the real world, it is important to remember that cancellation is based on abstraction of a certain observation. Furthermore, when it comes to words, yeah we are using abstractions to apply geometry to words. – Brad Graham Sep 01 '15 at 15:10
  • @marc From this we can derive geometrical properties of words, but a word is more then geometry. Why do we analyse words? (And a field of fractions is an abstraction of the analytic construction of rational numbers, to apply algebraic analysis to other objects, but these objects may be defined with more then just algebraic properties, hence this approach loses some realism, we are forced to analysis its other properties to fit this model, which is just systematic study building on natural observations again.) So for real world problem solving, we need to understand what really is going on. – Brad Graham Sep 01 '15 at 15:17
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Cancelling out $(x-y)$ is done by dividing through the equation by $(x-y)$ (on both sides), and since we cannot divide by $0$ (I'm assuming you know this), we cannot cancel $(x-y)$ when it is $0$.

I should add a small example. Consider the following equation, and we want to find the real solutions $x$: $$x^2=3x $$ Now it would be smart to cancel $x$ from each side, but by dividing through with $x$ assumes that $x\neq 0$, so we have to check that possibility separately. Assuming that $x\neq0$ the equation simplifies to \begin{align} \frac{x^2}{x} &=\frac{3x}{x}\\ x &= 3 \end{align} So one solution is $x=3$, but until now we have assumed that $x\neq 0$, so let us check the possibility that $x=0$: \begin{align} 0^2 &= 3\cdot 0\\ 0 &= 0 \end{align} which is true, so the solutions to this equation are $x=3$ and $x=0$.

Eff
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It sounds as if the 'proof' you were asked to work on was a variation of the following:

$$ \begin{align} a^2 - a^2 & = a^2 - a^2 \tag{1} \\ a \cdot (a-a) & = (a+a) \cdot (a-a) \tag{2} \\ a & = a+a \tag{3} \\ a & = 2a \tag{4} \\ 1 & \overset{?}{=} 2 \tag{5} \\ \end{align} $$

My grandfather, a maths teacher, told this as a joke. I remember confusing at least one of my maths teachers with it (yeah, I know).

The problem is that calling it 'cancelling' hides what's really going on. You're not cancelling anything, you're multiplying or dividing.

To arrive at $(3)$ from $(2)$, you're supposed to 'cancel' $(a-a)$. But it's not cancelling, it's dividing! And dividing by $(a-a)$ is dividing by $0$, which as we all know is not good. Had $(2)$ read $a\cdot0 = (a+a)\cdot0$, you wouldn't have dreamed of 'cancelling' the $0$.

Incidentally, the step from $(4)$ to $(5)$ isn't correct either, since you're disregarding the possibility that $a$ — which you're dividing by — could be $0$. But by then you're already so far down the rabbit hole you probably won't notice.

As an aside, another fun invalid "cancellation" $$\require{cancel} \frac{16}{64} = \frac{1\cancel{6}}{\cancel{6}4} = \frac{1}{4} $$

SQB
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    I disagree with your "incidentally". We start out with an identity that holds for all a; if not for the original error, (4) would still hold for all a. Then, given a = 2a for all a, we don't have to worry that a could be 0 to divide by a; the identity also holds when a isn't 0. – Rawling Oct 23 '14 at 07:32
  • @Rawling I disagree with you disagreement. If presented with $a=2a$, the conclusion should be $a=0$, not $1\overset{?}{=}2$. – SQB Jun 10 '16 at 13:35
  • If you had $xa = ya$ for all real $a$, would you dismiss a conclusion that $x = y$ "since you're disregarding the possibility that $a$ - which you're dividing by - could be $0$"? – Rawling Jun 10 '16 at 14:59
  • @Rawling $xa=ya \implies xa-ya=0 \implies (x-y)a=0 \implies x-y=0 \vee a=0 \implies x=y \vee a=0$ Likewise, $2a=a \implies (2-1)a=0 \implies a=0$ – SQB Jun 10 '16 at 15:08
  • The way it's presented in the "joke", it looks as if $1\overset{?}{=}2$ is the inevitable conclusion, while in reality, the conclusion should be $a=0$ – SQB Jun 10 '16 at 15:22
  • You're missing the "for all $a$". Yes, $x - y = 0$ or $a = 0$, but since it holds for all $a$ then the other condition must be true. And the conclusion should be $a = 0$ because of step 3, not step 5. – Rawling Jun 12 '16 at 16:20
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"Canceling out" is a shortcut procedure that only works in certain circumstances.

If we have $ac = bc$, we can rearrange to get $(a-b)c = 0$. If we have a product of two numbers that is $0$, then one or the other (or both) must be $0$. Thus, if we know $c \neq 0$, then we must have $a-b = 0$ or $a = b$.

That process is simplified as "canceling out", and it relies on knowing that $c \neq 0$. That is why you cannot cancel out a term that is (or might be) $0$.

Intuitively, we know that $a\times 0 = 0 = b \times 0$ regardless of the values of $a$ and $b$, so it would be inappropriate to conclude that $a\times 0 = b \times 0$ implies $a = b$.

user3294068
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When we divide an equation of the form $ac = bc$ by $c$, we are multiplying by the multiplicative inverse of $c$.

\begin{align*} ac & = bc\\ acc^{-1} & = bcc^{-1}\\ a & = b \end{align*}

A real number $c$ has a multiplicative inverse if there exists a real number $d$ such that $cd = dc = 1$. No such number exists if $c = 0$ since $0 \cdot x = 0 \neq 1$ for each real number $x$.

Hence, in your example, we cannot multiply both sides of the equation

$$x(x - y) = (x + y)(x - y)$$

by $(x - y)^{-1}$ since $x - y = 0$ and $0$ does not have a multiplicative inverse.

N. F. Taussig
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A very basic explanation: $$3x=6\Longrightarrow \dfrac{1}{3}\times3x=\dfrac{1}{3}\times6$$ Also, $$x=2\Longrightarrow 3\times x=3\times2$$ so overall: $$3x=6\Longleftrightarrow x=2$$

The two are equivalent, that is both contains exactly the same information. This is because $\dfrac{1}{3}$ and $3$ are both $\neq 0$.

Now, $$3x=6 \Longrightarrow 0\times 3x=0\times 6$$ but $$0=0\quad\textbf{ does not imply}\quad 3x=6$$

So $0=0$, which contains in fact no information, does not imply $3x=6$.

In your example, dividing by $x-y$ if $x=y$ leads to something similar to saying that $0=0$ implies $3x=6$ by diving by 0 (duh, I wrote it :/).

anderstood
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Generally you should never cancel but rather exclude terms from equation. Is such situation you make use of main properties of arithmetic operations and transform equation like $$a\cdot c=b\cdot c$$ into $$ac-bc=0$$ (because two expressions are equal iff their difference is zero), then distributivity of multiplication over addition allows you to convert the left hand side to $$(a-b)c=0$$

Now from properties of multiplication you get that at least one of the two terms is zero: $$(a-b=0)\lor (c=0)$$ which finally results in an alternative: $a$ equals $b$ or $c$ equals zero: $$a=b \lor c=0$$

The right term of the alternative can be easily lost when 'cancelling', so you should always keep in mind, that 'cancelling' the term like $(x-y)$ introduces additional equation $(x-y)=0$ as an alternative solution.

CiaPan
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There is no such thing as "cancelling out". In this case you want to divide both sides of the equation by (x-y). But always remember before dividing: you can't divide by zero. But if x=y then in fact (x-y) = 0 and so you would in fact divide by zero which you are not allowed to. But only professors are allowed to divide by zero. :-) Just kidding.

moonring
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  • Yes. First answer to explain (to a certain extent) the (x - y) = 0 part, which is assumed to be obvious to the OP. Further explanation: if x = y, you can, in (x - y), substitute in x for y, yielding (x - x), which equals 0. Combine that knowledge with the you're-not-cancelling-you're-dividing-both-sides-of-the-equation-by-x-minus-y reasoning, and tada! – Mathieu K. Nov 08 '15 at 08:05
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If $x=y$ then $x-y=0$ and the equation is equivalent to $0\cdot x =0\cdot (x+y)$
which means that $0=0$.
So you may have $x=300, y=500$ or $x=1 ,y =3.14$ or anything at all.
This means you cannot find the actual values of $x,y$ because the equation is always $0=0$.

Konstantinos Gaitanas
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  • No, in this case you cannot have $x=300$ and $y=500$, as you've already established $x=y$. For OPs equation $x(x−y)=(x+y)(x−y)$, the "canceled" version $x=x+y$ happens to be a perfectly valid statement (which however just isn't derivable via cancellation). – Nikolaj-K Oct 22 '14 at 11:01
  • @NikolajKI did not say that $x=300$ AND $y=500$ – Konstantinos Gaitanas Oct 22 '14 at 11:36
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$x=0 \implies 1\cdot x=0 \cdot x \implies \dfrac{1 \cdot x}{x}= \dfrac{0 \cdot x}{x}\implies 1=0$ $!!$

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Remember that if you divide both sides of an equation by a certain quantity and assert that the result is still true, you are also tacitly asserting that the quantity you divided by was not zero. If it was zero, the equality might hold, but you can't be absolutely sure anymore. You now have to split the problem into two cases and see what happens if y = x. (But this whole thing is kind of silly because your instructor essentially gave you two equations and two unknowns, where they are obviously not parallel lines or skew, so you could just solve for x and y directly... what was the point of all of this?)

So, you are dividing by (y-x), which is to assert that y does not equal x (because you kept the equal sign at full strength), but at the outset, you asserted that y did equal x. You can't have it both ways.

Donald Dade
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