Can $\frac {100-100}{100-100}=2$?

Question

\begin{align*} \frac{0}{0} &= \frac{100-100}{100-100} \\ &= \frac{10^2-10^2}{10(10-10)} \\ &= \frac{(10+10)(10-10)}{10(10-10)} \\ &= \frac{10+10}{10} \\ &= \frac{20}{10} \\ &= 2 \end{align*}

I know that $0/0$ isn't equal to $2$, the what is wrong in this proof? I wasn't satisfied after searching it on Google, any help would be appreciated.

You are never allowed to divide by zero from elementary school... — TTY, Mar 13 '14 at 07:50
Thanks for your feedback. :) I know that but thenIf I assume this proof is correct => 0/0=2 or 0=0 (on cross multiplying), does this mean 0/0 can be equated to anything we want? — Harshal Gajjar, Mar 13 '14 at 07:55
no, "equality" is a relation which hold only for numbers in this case, the expression $0/0$ is not a real number, you cannot equate it to anything. — TTY, Mar 13 '14 at 07:58
@Awesome. For me, you made the best and shortest answer to a question which is not a question -> +1 — Claude Leibovici, Mar 13 '14 at 08:11
@Awesome. Yes, do it ! Be sure I shall upvote (in fact, if I could do it $10^{99}$ times, I should do it). Cheers. — Claude Leibovici, Mar 13 '14 at 08:23
@ClaudeLeibovici Any good way of using the minimum of 30 characters other than dots? OH. $. Done. — evil999man, Mar 13 '14 at 08:26
I adjusted the tags; the ones that were there are not intended for this sort of question. — Carl Mummert, Mar 13 '14 at 11:07
Possible duplicate of Explanation of method for showing that $\frac{0}{0}$ is undefined — Kenta S, Oct 07 '19 at 11:06

score 14 · Accepted Answer · answered Mar 13 '14 at 07:59

The problem of this proof is your first proposition, that the number $\frac00$ exists, or, in other words, that the operation of division is defined for the value of $0$. In fact, the operation $(x,y)\mapsto \frac{x}{y}$ is defined on $$\mathbb R\times (\mathbb R\setminus \{0\}).$$

The operation is defined like so: the number $a=\frac{x}{y}$ is the unique number for which $a\times y = x$. This uniqueness allows you to do all the algebraic operations on these numbers, so it is quite necesary.

In the case of $\frac00$, this would mean that $\frac00$ is the unique number $a$ for which $0\cdot a = 0$. However, since $0\cdot a = 0$ for any value of $a$, such a unique number does not exist, therefore we cannot sensibly define $\frac00$.

"This uniqueness allows you to do all the algebraic operations on these numbers, so it is quite necessary."is very important, not sure it is every taught clearly in school though! +1 indeed. — NoChance, Dec 25 '15 at 22:59

score 9 · Answer 2 · answered Mar 13 '14 at 07:56

9

I think the best way to see the folly here is to work backwards. The problem occurs when we encounter the "equality" $$ \frac{10+10}{10}=\frac{(10+10)(10-10)}{10(10-10)} $$ But to obtain this equality, we must multiply the rational number $\displaystyle\frac{10+10}{10}$ by the undefined expression $\displaystyle\frac{0}{0}$.

answered Mar 13 '14 at 07:56

Brian Fitzpatrick

26,559

You don't need to work backwards at all, the proof literally starts with $0/0$. – Najib Idrissi Mar 13 '14 at 11:17
1

@nik Nowhere did I indicate that one needs to work backwards. I simply point out that if OP is confused by this "equation", then he or she might consider working backwards to see why it doesn't make sense. Obviously there are many reasons why this equation is invalid (as indicated by the seven answers here). – Brian Fitzpatrick Mar 13 '14 at 18:16

score 6 · Answer 3 · edited May 14 '16 at 14:10

6

$1.$ You cannot start with $\frac00$. This is not a number. $2.$ You are trying to simplify the fraction by dividing numerator and denominator with $(10-10)$. This cannot be done.

Assume beginning from the end going towards the start. The wrong part is when you multiply with $(10-10)$. Because then denominator would be $0.$

With your logic $\frac00=\frac{0*a}{0*b}=\frac{a}b$.

In general: In almost all these kind of problems where someone proves that $1=2$ (or something similar), most of the times, somewhere the denominator will be $0.$

edited May 14 '16 at 14:10

Roby5

4,287

answered Mar 13 '14 at 07:54

Thanos Darkadakis

1,869

or a square of a negative number; the logarithm of a negative number, those are the most common but any multivalued function can be used – ratchet freak Mar 13 '14 at 10:45

score 3 · Answer 4 · answered Mar 13 '14 at 07:54

3

You do some thing like this $$\frac{0\cdot 5}{0\cdot 9}=\frac{5}{9}$$you haven't to divide over zero

answered Mar 13 '14 at 07:54

Semsem

7,651

score 2 · Answer 5 · edited May 14 '16 at 14:10

2

The answer is very simple. Look at the denominator of the fraction, especially on $(10−10)$. This is exactly zero! You mustn't divide any number (including zero) by zero, even if in the numerator you have zero in identical expression as in the denominator (I mean of course $(10-10)$). So the problem is that you have above and below the line the same expression, but at first glance you don't notice fact, that it is zero, especially in the denominator and in this way you forget about not dividing by zero. That's very fantastic paradox.

edited May 14 '16 at 14:10

Roby5

4,287

answered Sep 14 '15 at 12:25

Jacob

21

3

There has already been seven answers with essentially the same content as your own answer... Why post an eighth one? – Najib Idrissi Sep 14 '15 at 12:33

score 2 · Answer 6 · answered Dec 25 '15 at 22:41

Your mistake is that you thought, $100-100$ is not a sum. But these $100s$ are still waiting for you to add them. When you thought like that, you began to expand and contract the expression so that $100-100$ takes the shape of a non zero number through which you could carry out the division operation freely. When it became $\frac{20}{10}$, you executed the operation. $100-100$ is already a sum with just having another look but mathematically, it is same as $0$.

By the way, you have created a good explanation that how dangerous may be the "division by zero" for mathematics.

score 2 · Answer 7 · answered Mar 13 '14 at 07:50

2

Well, the first expression in the argument, $0/0$, is meaningless. Everything after that is only connected to the first expression by whimsy, not by mathematics.

What you're seeing here is that, if you try to work with meaningless expressions, then absurd things can happen.

answered Mar 13 '14 at 07:50

G Tony Jacobs

31,218

Thanks for the feedback. :) If I assume this proof is correct => 0/0=2 or 0=0 (on cross multiplying), does this mean 0/0 can be equated to anything we want? – Harshal Gajjar Mar 13 '14 at 07:55
I guess. More sensibly, it can be equated to nothing, because it leads to nonsense, and math stops working. The correct approach is to recognize it as an undefined expression that doesn't equal anything, ever. – G Tony Jacobs Mar 13 '14 at 07:57
1

@HarshalGajjar $\frac00$ cannot be equated to anything. The symbol $\frac00$ makes no sense, as division with $0$ is not defined. Just because you can write it, does not mean it has any meaning. I can write the phraze "what is the color of one mile?", but that does not mean in itself that the question has any meaning. Miles do not have colors, just as the number $0$ has no inverse. – 5xum Mar 13 '14 at 08:03
1

It might help to note that "division" is really just multiplication by reciprocals. The number $0$ simply hasn't got a reciprocal, because the equation $0\cdot x=1$ has no solution. – G Tony Jacobs Mar 13 '14 at 08:05

score 1 · Answer 8 · answered Mar 13 '14 at 07:55

That's a nice one; I haven't seen it before. It would be nicer backwards.

The issue is in the very first step. $0/0$=nothing ever, and for exactly this reason. If $0/0$ were a number, then we could do exactly what you've done here and obtain $0/0=2$. But now multiplying both sides by $1/2$, we get $\frac{1}{2}\times\frac{0}{0}=1$ But now, multiplying together the left side, we end up with $\frac{1\times 0}{2\times 0} = \frac{0}{0}$. So now we have $\frac{0}{0}=1$ and $\frac{0}{0}=2$, which implies $1=2$. That's clearly wrong. If we keep riding this train we can eventually prove that every number equals every other number, up is down, cats are dogs, and Bertrand Russell is the pope. But since division by 0 is not an operation, it is all meaningless.

Actually you don't need to keep riding this train. Let A="any statement". B="1=2". Now we know either A or B is true, (vacuously, since B is true). But we know $1\ne2$. So, either A or B is true, but B is false. So A must be true. Some boolean algebra never hurt anybody. :D — Guy, Mar 16 '14 at 10:44

mvw · Answer 9 · 2015-09-17T19:05:16.597

What you provide is an equality statement $$ e_1 = e_2 $$ of two expressions and the infix equality operator. The expression $e_1$ $$ \frac{0}{0} $$ thus the result of dividing 0 by 0 (which is an instance of dividing a number by zero), is usually considered as undefined, e.g. there is no real number that could be used as result and not cause trouble elsewhere. The second expression $e_2$ is $$ \frac{100-100}{100-100} $$ We could now argue that both expressions are equal for example by one of these arguments:

fractions are equal if they have equal nominators and denominators and this is the case as reduction by partial evaluation of nominator and denominator individually is unproblematic
each expression is an instance of division by zero and thus the same failure
each expression is evaluated to the same special undefined value e.g. div, $\bot$, false (used for partial functions) or undefined (e.g. JavaScript) and thus equality of values occurs

Where the first argument would probably be objected by the argument that these expressions are constructed like fractions (syntactic equality $e_1 = e_2$, the expressions are the same) but are not proper fractions, because zero denominators are not allowed there, thus equality by value ($\sigma(e_1) = \sigma(e_2)$) can not apply because lack of values due to evaluation not happening (semantic $\sigma$ which assigns a value to the expression is not defined).

The second might raise the question if those two are different from other failures like $\infty - \infty$ or $1/-$ or $(1,2,3)+(1,2)$.

Nonetheless we might get convinced that equality holds and then continue boldly with $e_3$ $$ \frac{10^2-10^2}{10(10-10)} $$ and $e_4$ $$ \frac{(10+10)(10-10)}{10(10-10)} $$ and then try it again with $e_5$ $$ \frac{10+10}{10} $$ This is not equality by argument 1, 2 or 3 but by argument

two fractions are equal if nominator and denominator are each the same multiple of the other one $$\frac{n_1}{d_1} = \frac{c \, n_1}{c \, d_1}$$

We note that $e_5$ is now a number while $e_1,\ldots,e_4$ were not (be it $./0$, $\bot$, false but certainly no real number). So inequality because of different type applies (comparing apples with pears). Thus argument 4 was invalid, reduction by common division by $(10-10)=0$ is not valid to derive a fraction with the same value, as it changes the object from being not a number into a number. We rather have $$ e_4 \ne e_5 $$

Summary: Division by zero is the culprit but only at the fourth equality (see above) and not before. And there it is not a single division but a common division to reduce a fraction.

Can $\frac {100-100}{100-100}=2$?

9 Answers9