Alright this maybe really funny but I want to know why is this wrong. We often come across identities which we prove by multiplying both the sides of the identity by a certain entity but why don't we multiply it by $0$. That way every identity will be proved in one single line. That is so stupid. I mean, by that way we may also say that $1=2=3$. I know it is wrong. But why? I mean if we can multiply both the sides by $2$ then why not by $0$. For example, consider the following trigonometric identity :

Prove the identity : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta$

Usual way

To Prove : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta $

$\displaystyle \implies {\sin^2 \theta \over \cos^2 \theta } = {\tan^2 \theta \cos ^2 \theta \over \cos^2 \theta}$ (multiplying both the sides by $\displaystyle 1 \over \cos^2 \theta$)

$\implies \tan ^2 \theta = \tan^2\theta$

$\implies LHS=RHS$

$\therefore proved$

Funny way

To Prove : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta $

$\displaystyle \implies {\sin^2 \times 0} = {\tan^2 \theta \cos ^2 \theta \times 0}$ (multiplying both the sides by $0$)

$\implies 0 = 0$

$\therefore proved$

Please explain why is this wrong.

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    $\begingroup$ Try reversing the steps... $\endgroup$ – David Mitra Jan 28 '13 at 14:47
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    $\begingroup$ Because all you've proved is that if the first statement is true, then $0=0$. You haven't proved the converse $\endgroup$ – Thomas Andrews Jan 28 '13 at 14:48
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    $\begingroup$ References to World Takewondo Foundation does not agree with this forum's style. Please reword. $\endgroup$ – Maesumi Jan 28 '13 at 15:24
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    $\begingroup$ The "usual way" is an unfortunate high-school process that misrepresents the logic of the real argument. Later, students have to be trained to abandon it. $\endgroup$ – André Nicolas Jan 28 '13 at 16:33
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    $\begingroup$ With the same trick, you could also prove that 1 = 2 because 1 × 0 = 2 × 0. This might make you understand better why it's wrong. $\endgroup$ – vsz Jan 28 '13 at 18:14

14 Answers 14


$a = b$ implies $ac = bc$, but $ac = bc$ doesn't imply $a = b$. (Not immediately. Read below.)

The way you usually get $a = b$ from $ac=bc$ is by multiplying both sides with $1/c$, which is only available when $c \ne 0$.

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    $\begingroup$ One can also prove in abstract algebra that if $1/0 = 0^{-1}$ is permitted to exist in a field, then $1=0$ in that field: $\exists \, 0^{-1} \implies 1 = 00^{-1} = (0+0)0^{-1} = 00^{-1} + 00^{-1} = 1 + 1 \implies 0 = 1$. One can in turn show that this implies $a=b$ for all $a$, $b$ in that field. $\endgroup$ – Emil Lundberg Jan 29 '13 at 0:00
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    $\begingroup$ @EmilLundberg: Maybe an easier way to say it is that the only field which in which 0 has an inverse is {0}. $\endgroup$ – Mark Jan 29 '13 at 2:29
  • $\begingroup$ @Mark: Indeed that is simpler. $\endgroup$ – Emil Lundberg Jan 29 '13 at 10:33

You are confusing derivation with proving.

If you want to prove that some $X = Y$ statement is true, you have to show that that statement can be derived from some other statement which is already known to be true. You're doing it backwards: you're deriving from $X = Y$ some statement which is true, namely $0 = 0$.

To generalize this, let us observe that in a typical mathematical proof-by-algebraic-derivation, you start with some questionable statement $S_0$ and then you go through some derivations to show that it is equivalent to some truthful statement $S_T$: $S_0\Leftrightarrow S_1\Leftrightarrow S_2\cdots\Leftrightarrow S_T$. This proof method works only because the arrows go both ways, and so the opposite derivation is possible: although you are proceeding from $S_0$ toward $S_T$, you are in fact at the same time showing that $S_0$ can be derived from $S_T$. For this method to work, however, none of those arrows must be a one way implication (denoted by $\implies$). If you have such a one-way "trap door" in the logic, then the crucial reversal of implication cannot happen, and so the proof does not hold. Even though you arrive at a truthful statement $S_T$, that statement does not imply the truth of your starting proposition $S_0$.

Under most derivation steps in algebra, you don't have to worry about the direction because the derivations establish equivalence: this means that there is a two-way implication between the statements. For instance $3X = 3Y$ can be derived from $X = Y$, but also $X = Y$ can be derived from $3X = 3Y$. These statements are equivalent, and so we can connect them with a double arrow: $X = Y \Leftrightarrow 3X = 3Y$.

Some derivation steps, however, only go one way, because they involve some "trapdoor" function: an operation which cannot be reversed, because it erases information. One example of a trapdoor function is multiplication by zero, around which your question revolves. Another example is taking a remainder in a division.

For instance, suppose $X$ and $Y$ are integers. Then have $$X = Y \implies (X\mod 3) = (Y\mod 3)$$

(If X equals Y, then the remainder left when dividing X by 3 is the same as the remainder left when dividing Y by 3). However, the converse isn't true. Just because two numbers have the same remainder when divided by three doesn't mean that they are equal.

More about "trap doors"

More formally, we can define "trapdoor function" as any function which fails to be one-to-one (or injective), because such functions have inverse functions. If if $g$ is an injective function covering the entire domain of $X$ and $Y$, then we have $X = Y \Leftrightarrow g(X) = g(Y)$. If $h$ fails to be injective (is not invertible) then we have $X = Y \implies h(X) = g(Y)$. The function $g$ does not have to be onto, only one-to-one.

An example of an injective function is $e^x$, over the real numbers. It is a one to one function in that it maps each domain value to a unique value in its range. (But it is not onto: it does not map a domain value to every real number: its output is only positive real numbers. That doesn't matter.)

Therefore, we know that $X = Y \Leftrightarrow e^X = e^Y$. In the real domain only!

In the domain of complex numbers, $e^x$ is not one to one. More than one value of $x$ will map to the same range value. The inverse function, $\ln x$, is not actually a function in the complex plane, because it is multi-valued. (When the complex logarithm is used as a function anyway, it has to be restricted to a particular "branch".) Therefore if our proof involves complex numbers, $X = Y \Leftrightarrow e^X = e^Y$ does not hold. Complex exponentiation is a "trap door" and so the implication only goes one way: $X = Y \implies e^X = e^Y$.

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    $\begingroup$ This is the most conceptually thorough answer I've read so far. $\endgroup$ – snapfractalpop Jan 29 '13 at 7:31

Because to prove $a=b$ you don't have to prove that $$a=b \implies c=c$$You have to prove that some trivial statement, such as $$c=c$$or another axiom, or logical tautology, or proved statement, and derive from that that $a=b$. So in your example,

$$\begin{array}{lcl}&& \sin^2 \theta & = & tan^2 \theta \cos ^2 \theta \\ & \implies & {\sin^2 \theta \over \cos^2 \theta } & = & {tan^2 \theta \cos ^2 \theta \over \cos^2 \theta} \\ & \implies & \tan ^2 \theta & = & \tan^2\theta \\ & \implies & LHS & = &RHS \\ & \therefore & proved\end{array}$$

Doesn't prove the initial $\sin^2 \theta = tan^2 \theta \cos ^2 \theta$, it's simply an easy way to "work backwards" in reviersible way to arrive at the actual proof: $$\tan^2\theta=\tan^2\theta\implies \sin^2 \theta = tan^2 \theta \cos ^2 \theta $$ by reversing multiplication and using division. This very much doesn't work for $0$, because, well. You're doing arithmetic and dividing by $0$.

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    $\begingroup$ Please check the trig identity, as mentioned in comment by Michael Joyce to OP. $\endgroup$ – Maesumi Jan 28 '13 at 15:20
  • $\begingroup$ @Maesumi Thanks. I hadn't actually payed attention. $\endgroup$ – guest196883 Jan 28 '13 at 15:21

One can look at the issue from the angle of information.

When we multiply $b$ and $c$ by a non-zero number $a$, no information is loss:

$$b = c \to a b = a c\\ b \ne c \to a b \ne a c$$ Since no information is lost, we can reverse the "logic" and cancel $a$ in both side of equation. In contrast, when we multiply $b$ and $c$ by $0$, one lose the information of "equality": $$b = c \to 0 b = 0 c\\b \ne c \to 0 b = 0 c$$ This means one can no longer reverse the "logic" and deduce $b = c$ from $0 b = 0 c$.

  • $\begingroup$ clear and concise. $\endgroup$ – Lucas Jan 29 '13 at 16:16


$a \neq b$

$a\cdot0 \neq b\cdot0$

But, since $a\cdot0=0$ and $b\cdot0=0$

We get, by substitution


Combine this with your discovery, and I expect that this site will self-desctruct in 3,...2,...1,... (Wait, don't do it, it's just a jo........


When you go from $ax=b$ to $x=\frac ba$ you have multiplied both sides by $\frac 1a$ and it would be good to remark that this can only be done if $a \ne 0$ (though people often forget this). What you are wanting to do under Funny Way is to start with $0=0$, then divide both sides by zero (which is not allowed) and get whatever you want on each side. As you say in your introduction, this would allow us to conclude that all things are equal.


The "usual way" is often used with inequalities, but we take care to make sure that we can reverse our steps. For instance, suppose we want to prove $$a+b \geq 2\sqrt{ab}.$$ Since both sides are positive, this statement is true iff $$(a+b)^2 \geq 4ab,$$ which is true iff $$(a+b)^2-4ab \geq 0 \Leftrightarrow (a-b)^2 \geq 0.$$

Since this last inequality is true, and we can reverse all of our steps, the original inequality is true. Notice that at each step, we carefully ensure that we only do reversible operations (each step we want if and only if statements).

When you try to multiply by zero, however, that's a non-reversible step (you can't divide by zero). In fact, the "usual way" that you explain is not valid either, because you divide by $\cos^2{\theta}$, which is reversible only if $\cos \theta \neq 0$. Thus, neither really constitute a proof of the identity $\sin^2 \theta = \tan^2 \theta \cos^2 \theta$.

Why does it matter that the steps are reversible? Otherwise, you CAN'T go back from a true statement (0 = 0) to prove the original statement ($\sin^2 \theta = \tan^2 \theta \cos^2 \theta$), and so you have not really proved what you wanted to prove.

  • $\begingroup$ The "usual way" is valid for all nonzero values of $\cos \theta$, and the zero value you can simply verify by yourself. $\endgroup$ – Joe Z. Jan 29 '13 at 15:00
  • $\begingroup$ Yes, I know, so I'm saying that the "usual way" doesn't really work in that you supposedly address the zero value of $\cos \theta = 0$ in the same step as you address non-zero values of $\cos\theta$, i.e. the "usual way" that the OP talks about overlooks a case. Sure, it might be trivial, but it still is overlooked. $\endgroup$ – notes Jan 30 '13 at 2:11

A function actually consists of two things, a domain (in first courses in calculus the domain is assumed to be the natural domain) and a rule for assigning a unique value to any real number in that domain. The domain of $\sin^2(\theta)$ is all of $\mathbb R$ and the domain of $\tan(\theta)$ is all of the reals that are not odd integer multiples of $\pi/2$. Hence this is not really an identity because the functions on each side of the $=$ have different (natural) domains. It is an identity if you restrict the domains of $\sin$ and $\cos$ to the reals which are not of the form ${(2n+1)\pi}\over2$.

  • $\begingroup$ This kind of restriction of domains to avoid dividing by $0$ when using some kind of formula to define a function is an essential tool in defining derivatives and appears in many areas of mathematics. Functions are not just rules, they are rules plus domains (and even the set in which he values of the function are taken to lie as in expressions such as 'real-valued function'). $\endgroup$ – Barbara Osofsky Jan 28 '13 at 17:27

The first proof (unlike the second) immediately leads the reader to the following, more definite style of writing essentially the same thing: $$ \sin^2(x)=\sin^2(x)\frac{\cos^2(x)}{\cos^2(x)}=\frac{\sin^2(x)}{\cos^2(x)}\cos^2(x)=\tan^2(x)\cos^2(x)\text{.} $$ This is a more explicit demonstration of what I believe Andre Nicolas is talking about in his comment and Tunococ in his answer. In terms of reversing steps, notice that we can easily do this backwards: $$ \tan^2(x)\cos^2(x)=\frac{\sin^2(x)}{\cos^2(x)}\cos^2(x)=\sin^2(x)\text{.} $$


Here's how you see that something is wrong with your argument:

  1. To prove: $1 = 2$

  2. Multiply both sides by zero:

    $ \Rightarrow 1 \times 0 = 2 \times 0$ (multiplying both sides by $0$)

  3. $ \Rightarrow 0 = 0 $

Conclusion: This is useless as a proof method. Multiplying anything by zero gives you zero. So, although both sides of your equation become zero when you do this, it doesn't mean that they were equal at the start.

Arithmetic proofs of (in)equality are only valid if each step guarantees that the results could only be equal if the inputs were equal. Essentially, the steps have to work backwards as well as forwards. In this case, the reverse step would amount to dividing by zero.


the reason behind this is is a basic primary school rule. to prove x=y

this condition should also hold true.


so if u divide both sides by '0'

then lhs=rhs
how do u answer this
lhs/rhs = 1

This is an interesting point. I will do this on the fly....

OK. So, false proof.

\begin{align*} 1 & = 2\\ 1 \cdot 0 & =2 \cdot 0\\ 0 & =0 \end{align*}


Um... I can't see anything wrong with this. So $0 = 0$, ok. So $1 \cdot 0 = 2 \cdot 0$... ok. So $1 = 2$... Not ok. What is wrong? The flaw must be in the conclusion, so what did we do from one step to the next? We did $0/0$... Ahhhhh. Ok, as this is an indeterminate value (not undefined. Undefined means no solutions: $0 \cdot$ what $= 1$? Indeterminate is infinite. $0 \cdot$ what $= 0$?) we can really prove anything, by substituting one value for another. See, you really proved a (then, if) statement, which is ridiculous.


There is no problem with multiplying by zero, this is quite legal.

Consider your text:

"To Prove : $\sin^2 \theta = \tan^2 \theta \cos ^2 \theta $

$\displaystyle \implies {\sin^2 \times 0} = {\tan^2 \theta \cos ^2 \theta \times 0}$ (multiplying both the sides by $0$)

$\implies 0 = 0$".

This is perfectly right, you proved that $0=0$. Congrats!

  • $\begingroup$ The real problem is when you try to divide by zero. $\endgroup$ – Yves Daoust Feb 17 '15 at 20:19

Zero differs from other numbers and mathematical entities which represent some sort of value because it is the absence of value, not a value itself. Thus, it would be incorrect to say $a \cdot 0 = b \cdot 0$ means that $a = b$. Similarly, in the case of infinity: let's say that $a$ does equal $b$. However, $a \cdot \infty \neq b \cdot \infty$ because infinity does not have a definite or specific value. Rather, it is a concept of something without limit.

  • $\begingroup$ a/0 is not infinity, rather undefined $\endgroup$ – BЈовић Jan 29 '13 at 7:54
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    $\begingroup$ because it is the absence of value, not a value itself... Wow. What does that mean? Zero is very much a value: it is the number of tigers currently in the room where I am typing this. $\endgroup$ – Did Jan 29 '13 at 8:23

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