Why is 1 not equal to 0 in this proof?

[mark2142]

This is as simple as I can make it:
$$1 \times 0 = 2 \times 0$$But, we cannot cancel the zero to give $1 = 2$.

You can explain that to yourself however you like, but zero cannot be cancelled from both sides of an equation.

Yeah! Because $0/0 \neq 1$. That’s what I think.

 

[fresh_42]

Isn’t that what I said. Since $\frac {(x-1)}{(x-1)} \neq 1$ but is undefined we are stuck. We can’t write the final equation $x=0$.

Yes, sorry.

 

[mark2142]

thanks for the reminder that the goal is to be helpful, not smart alecky. here is a try:

to the OP: this question is a trick, using disguised symbols to hide from you what is being said.

when x= 1, then x-1 = 0, so writing x(x-1)/(x-1) is a disguised way to write (1.0)/0. what do you think that means? if 1.0 = 0, then (1.0)/0 = 0/0, and what does that mean?

here one needs to know that a/b = a.(1/b), so 0/0 = 0.(1/0), but what does 1/0 mean? In fact, as several people have tried to say, it does not mean anything in the real number system.

I.e. although I myself did not learn this until college!, in fact 1/b means: "the number which multiplies b into 1".

Now that means that 1/0 must equal the (real) number that multiplies 0 into 1, but sadly, or actually happily, there is no such real number.

this is a little technical, but i claim 0 multiplies every real number into 0. Again I did not learn this either until college, but there is a rule for all real numbers, called associativity, that says (a+b)c = ac + bc, for all real numbers a,b,c.

Applying this to a=b = 0, using 0 = 0+0, we get 0.c = (0+0)c = 0.c + 0.c. Then subtracting 0.c from both sides, gives us 0 = 0.c, for all c.

now, since 0 multiplies every number into 0, it cannot multiply any number into 1.
Wait! For this proof I need to know that 1≠0!

So you cannot prove there is anything wrong with this proof unless you already know that 1 ≠ 0!

I.e. close analysis of this proof shows that in fact 1 = 0 if and only if 1/0 makes sense, i.e if and only if there is a multiplicative inverse of 0.

So this takes me back to my point, namely one cannot prove anything unless one says what one is assuming. Here we are apparently assuming that we are working with real numbers. But that takes for granted that we know some facts about real numbers. usually people just take for granted that real numbers are (infinite) decimals, and that two such decimals are different except basically that 1.0000... = .99999.....

hence also 3.141599999999.... = 3.141600000...., etc...

in which case we do NOT have 1.0000.... = 0.000000....

So since 1 ≠0 is an assumption about real numbers, it follows that 1/0 makes no sense within the realm of real numbers. So the original question involves trickery, by writing x(x-1)/(x-1), where x=1, and where thus the factor 1/(x-1) = 1/0, which makes no sense.

sorry to be so lengthy, but the point is, you cannot prove anything, ever, unless you agree on what you are assuming. and here we are basically assuming that 1≠0 in the real numbers. hence 1/0 makes no sense, hence the argument originally given involves some symbols that do not represent actual numbers. hence the argument is gobbledy gook.

another point of view is that the argument correctly shows that if 1/0 makes sense, then 1=0. but equivalently, this shows that if 1≠0, then 1/0 does not make sense. so the point is, to understand whether 1=0, try to understand whether 1/0 makes sense.

so apparently this problem is taking advantage of the fact that many of us know that 1≠0, but do not realize that this implies that 1/0 does not make sense. or more likely, they count on us forgetting that if x=1 then 1/(x-1) = 1/0, hence does not make sense, as pointed out several times here above.

and don't feel too naive, i am an old hand, and I just learned something by going through this in detail.
peace.

Yeah! I feel that but sometimes even a person of low experience can be right. I said this before $0/0 =$ undefined. So we cannot replace it with 1.
You wrote a big post for me. Thank you.

 

[mark2142]

Yes, sorry.

Can you again explain the less simple part but in detail in post #26?
I don’t get it.

 

[fresh_42]

It simply says you cannot divide by $0$, which you did by setting $x=1$ so $x-1=0$ and you divided by $x-1$. Division is not defined for $0.$

--------------

You started from $x=1$. Then you multiplied by $x$ to get $x^2=x.$ But this adds another solution that we didn't have at $x=1$ namely the possibility $x=0.$ So you made an equation with two solutions out of an equation with one solution. That would be ok if it could be reversed. In this case, however, it can't because the additional solution is $0$ and we cannot divide it.
______________

The long version would mean to explain what "division" actually is. For mathematicians, division by $x-1$ is simply multiplying by $\cdot \dfrac{1}{x-1}.$ However, multiplication is only defined for numbers unequal zero. Therefore, the construction $\cdot \dfrac{1}{x-1}$ is not available for multiplication.

$0$ is a child of addition. The number that doesn't add something.
Division by $a$ is defined as multiplication, namely by $\dfrac{1}{a}.$
Addition and multiplication meet in the distributive law: $a\cdot (b+c)=a\cdot b+a\cdot c.$
This is all you can use to get formulas.

 

[malawi_glenn]

0 is the additive identity element, 1 is the multiplicative identity element. Pretty neat

 

[mark2142]

The long version would mean to explain what "division" actually is. For mathematicians, division by x−1 is simply multiplying by ⋅1x−1. However, multiplication is only defined for numbers unequal zero. Therefore, the construction ⋅1x−1 is not available for multiplication.

Multiplication is defined for 0. You mean to say multiplication is not defined for $1/0$ ?

 

[fresh_42]

Multiplication is defined for 0.

Not really defined. It is derived from the distributive law ($a\cdot (b+c)=a\cdot b+a\cdot c$) which is given for integers, and multiplication by $0$ is a consequence of it:
$$
4\cdot 0= 4 \cdot (1-1)=4\cdot (1+ (-1))=4\cdot 1 + 4\cdot (-1)=4+(-4)=4-4=0
$$
So $x\cdot 0=0$ is actually a theorem, derived from the distributive law.

The distributive law is the only allowed contact between multiplication and addition, and therefore between $1$ that belongs to multiplication and $0$ that belongs to addition.

You mean to say multiplication is not defined for $1/0$ ?

Yes. I meant that any division is actually a multiplication. E.g., what does $\dfrac{4}{2}$ mean? I mean, how is it defined? Truth is, it is an abbreviation: $\dfrac{4}{2}=4\cdot \dfrac{1}{2}.$ That's its definition. But what is $\dfrac{1}{2}$ then? This is again an abbreviation defined as the solution to the equation $x\cdot 2 =1.$ See? Everything is in the end reduced to multiplication.

We do not really divide in mathematics. We multiply with the inverse.

However, your problem can also be seen as a logical problem, not a numerical one.
\begin{align*}
x=1 &\Longrightarrow x\cdot (x-1)=0 \not\Rightarrow x=0 \\
x=1 &\Longrightarrow x\cdot (x-1)=0 \Longrightarrow x=0 \text{ or }x=1
\end{align*}

The situation is

and you started in the $x=1$ bubble. Then you multiplied by $x$ and got a formula with suddenly two solutions: both pink bubbles. This is all still a logical statement with truth value TRUE. The error occurs when you try to get rid of the $x=1$ bubble. There is no way. Your primary condition is still $x=1$:
$$
x=1 \Longrightarrow x(x-1)=0 \text{ AND }x=1\Longrightarrow x\in \{0,1\} \text{ AND }x=1 \Longrightarrow x=1
$$

The problem with this question is how deep you want to climb into the rabbit hole of mathematics. Possible answers reach from "You cannot divide by $0$" to what was called integral domain above. I tried to answer your question on the basic level and simultaneously provide a glimpse into the theory behind this prohibition.

 

[mathwonk]

This question is more interesting than I realized, thanks to all these nice answers.
Here is one more remark:

There are many number systems possible in the world of mathematics, with different properties. However there are very few in which 1/0 makes sense. In fact the only number system having the usual rules for products and sums and differences, i.e. associativity and subtraction, and in which 1/0 makes sense, is the extremely simple system in which the only number is 0.

I.e. if 1/0 makes sense in your number system, by which I mean 0 has a multiplicative inverse, then I claim your number system only has one number in it, namely zero. To see this recall that 1 is the multiplicative identity, i.e. if x is any number then 1.x = x. And we have shown that 0 multiplies every number into 0. Moreover, the original "proof" does show correctly that if 1/0 makes sense in a number system, then also 1=0 in that system.

But I claim that if 1=0, then every number equals zero. I.e. let x be any number in the system. Then x = 1.x = 0.x = 0, where we have used in the second equality that 1=0. So 1=0 implies x=0 for every x.

Thus assuming 1/0 makes sense, forces you to accept that in your number system, all numbers equal zero, i.e. you are dealing with only one number. Now that is not a very useful number system, so this explains why one does not want to divide by zero in any useful number system. I.e. if you want to have a number system with more than one number in it, then you cannot allow division by zero.

In particular since there is more than one real number, the real numbers 1 and 0 must be different, and division by 0 does not make sense in the real numbers. Or to answer directly the OP's title question, we must have 1≠0 because we want to have more than one number to work with.to quote forrest gump: "and that's all i have to say about that".

 

[mark2142]

The problem with this question is how deep you want to climb into the rabbit hole of mathematics. Possible answers reach from "You cannot divide by 0" to what was called integral domain above. I tried to answer your question on the basic level and simultaneously provide a glimpse into the theory behind this prohibition.

Yeah! And I don’t understand very much. I am trying to learn precalulus right now. But I think it’s clear to me division by 0 is not allowed. It’s undefined.
$\frac {(x-1)}{(x-1)} \neq 1$.
Thank you.

 

[mark2142]

The following is a quadratic equation: $x^2-x=0$

It has 2 solutions: $x= 1$ and $x= 0$. It doesn't mean $1=0$, it just means those two values give the same result in the [arbitrary] equation you stated. Graphic solution:

View attachment 324358​

I think I am getting what you mean. It’s called the language of maths. We have to be careful in what we say and understand.
$$x=1$$
Then after some manipulation we write
$$x=0$$
And I think it’s clear $x=0$ is not the solution of $x=1$ because $0 \neq 1$. So $x$ does not have a value $0$ and so $1 \neq 0$. But this last line is absurd. Even if $x=1$ and $x=0$ are solutions ,doesn’t mean $1=0$. Yes?

 

[jack action]

I think I am getting what you mean. It’s called the language of maths. We have to be careful in what we say and understand.
$$x=1$$
Then after some manipulation we write
$$x=0$$
And I think it’s clear $x=0$ is not the solution of $x=1$ because $0 \neq 1$. So $x$ does not have a value $0$ and so $1 \neq 0$. But this last line is absurd. Even if $x=1$ and $x=0$ are solutions ,doesn’t mean $1=0$. Yes?

Yes. Understanding that, reread the posts from @PeroK to understand what you did wrong:

• post #12
• post #20

 

[jbriggs444]

Yes. Understanding that, reread the posts from @PeroK to understand what you did wrong:

• post #12
• post #20

Be careful with #20. It is backward. We started with $x=1$, casually added the possibility that $x=0$ and then incautiously eliminated the case where $x=1$.

 

[mark2142]

Yes. Understanding that, reread the posts from @PeroK to understand what you did wrong:

• post #12
• post #20

We cannot multiply by or divide by $(x-1)=0$.
Well upon close inspection I think the way they are treating x is wrong. It’s not like there are two quantities equal to x and so both will also be equal namely 1=0. Instead if we assume it’s not a trick, x=1 is our original equation not that value of x is 1 ,x=1 and x=0 are solutions to the original equation. But we cannot say two numbers are equal to each other (1=x=0).
(My last post is not so clear)

 

[jbriggs444]

We cannot multiply by or divide by $(x-1)=0$.
Well I think the way they are treating x is wrong. It’s not like there are two quantities equal to x and so both will also be equal namely 1=0. Instead if we assume it’s not a trick, x=1 is our original equation. x=1 and x=0 are solutions to the original equation. But we cannot say two numbers are equal to each other.

Apart from the "backwards" problem, @PeroK delivered an apt summary in #20. Let me repair that problem...

In fact, you can characterise what's happening as:

Start with x=1. It follows that x=0 or x=1. Sneakily get rid of the case x=1. You are left with x=0.

Let us take the original sequence of statements one at a time and explore them in [painful] detail.

1: $x=1$

In some contexts we would write this as "Let $x = 1$". This in turn is mathematical shorthand for "Let x be a variable such that the predicate $x=1$ is true".

Before we are allowed to make such a statement, we must have a guarantee of existence. We do indeed have such a guarantee. We are implicitly using a number system that is rich enough to support the notation used in the argument. This could be the ring of integers or the field of real numbers. It matters little to the argument. What matters is that both algebras include a guarantee of existence and uniqueness for an element denoted by "1".

In any case, we end up with a predicate in one free variable ("$x=1$") which is true for the variable "$x$" that we have created.

If we look at the solution set for that predicate, $\{x : x = 1\}$, that solution set is $\{1\}$. Obviously.

2. $x^2 = x$

This is a new predicate that we can get from the previous one by squaring both sides, using the mathematical fact that $1 \times 1 = 1$ and substituting equal values $x$ for $1$.

This new predicate has an enlarged solution set, $\{x : x^2 = x\}$. That solution set is $\{0, 1\}$.

The new solution set is a superset of the old solution set. But that is fine. The new predicate holds whenever the old predicate does. It "follows" from the first predicate, even though it is a strictly weaker predicate.

3. $x^2 - x = 0$

Again, we derive a new predicate from the old. We have subtracted $x$ from both sides of an equality and we've used the property that $x - x = 0$. That is a valid manipulation. And a reversible one. We could add $x$ to both sides to recover the previous predicate. So the two predicates are equivalent. The new solution set is the same as the old.

The solution set is still $\{0, 1\}$.

4. $x(x-1) = 0$

Again, we derive a new predicate from the old. This is an application of the distributive law. The manipulation is reversible. We could apply the distributive law to reverse the manipulation.

The solution set is still $\{0, 1\}$.

5. $\frac {x(x-1)}{(x-1)}=\frac 0{(x-1)}$

Now we've divided both sides of the equation by $x-1$. This is a valid manipulation, provided that $x-1 \neq 0$.

If, contrariwise, $x-1 = 0$ then the manipulation is invalid.

If we were obeying the proper rules of mathematical discourse, we would have to split the analysis at this point into two cases:

Case 1: $x - 1 \neq 0$

And the rest of the proof goes through as written. Without running through that, we properly conclude that $x=0$ and we eventually end up with a solution set under this case of $\{x : x = 0\}$. i.e. $\{0\}$.

Case 2: $x - 1 = 0$

The rest of the proof does not go through. We properly conclude that $x = 1$. The solution set under this case is $\{x : x = 1\}$. i.e. $\{1\}$.

The solution set we end up after considering both cases is the union of the solution set under case 1: $\{0\}$ and the solution set under case 2: $\{1\}$. That union is, of course, $\{0,1\}$.

[So proper case management for this step has neither erroneously strengthened nor accidentally weakened anything. We got to the same solution set]

What we can conclude from the completed and corrected manipulation sequence is that "if $x = 1$ then either $x = 1$ or $x = 0$". This is true but pointless.

What we cannot conclude from the completed and corrected manipulation sequence is that "if $x = 1$ then $x = 0$"

The version of events in #20 from @PeroK is far more pithy: "Start with x=1. It follows that x=0 or x=1. Sneakily get rid of the case x=1. You are left with x=0." I like that description a lot.

 

[jack action]

Case 1:

Statement:
$$x(x-1)=0$$
Solution:
$$x= 0\ \ \ \ \text{or}\ \ \ \ x=1$$
Case 2:
Statements:

$$x(x-1)=0$$
and
$$x=0$$​

Solution:

$$x=0$$​

Case 3:

Statements:

$$x(x-1)=0$$
and
$$x=1$$​

Solution:

$$x=1$$​

Case 4:

Statements:

$$x(x-1)=0$$
and
$$x=1$$
and
$$x=0$$​

Solution:
There are no solutions because $x$ cannot be equal to $0$ and $1$ at the same time.

Case 5:

Statements:

$$x(x-1)=0$$
and either
$$x=1$$
or
$$x=0$$​

Solution:
$$x= 0\ \ \ \ \text{or}\ \ \ \ x=1$$
You started with case 3 and somehow introduced $x=0$ along the way transforming it into case 4.

Note also that rewritting $x(x-1) = 0$ into $x^2=x$ doesn't change the fact that both $(0)^2=(0)$ and $(1)^2=(1)$ satisfy the equation.

 

[mark2142]

Can you add something to what I said instead of pouring extra knowledge? I am a beginner. You can start by telling if I am right or where have I went wrong.

 

[fresh_42]

We cannot multiply by or divide by $(x-1)=0$.

You can multiply by $0$ but that will result in $0=0$ and every bit of information is lost because we cannot retrieve anything from $0=0.$

However, you are right, division is prohibited. You simply cannot define it in any meaningful way. Every attempt to do so is doomed and will ultimately result in contradictions. And contradictions are one of the few things that are banned in mathematics. Whenever we end up in a contradiction, then we can be sure that something earlier on was definitely false.

The trick is: From FALSE we can derive false or true statements, whatever we like. But from TRUE, we can only derive true statements. At least if we make no mistakes.

Well upon close inspection I think the way they are treating x is wrong. It’s not like there are two quantities equal to x and so both will also be equal namely 1=0.

Yes and no. The step $x=1$ to $x^2=x$ adds another solution to the equation since there are two numbers that obey that equation: $0^2=0$ and $1^2=1.$ Hence, $x=1$ has a unique solution, namely $x=1,$ and $x^2=x$ has two different possible solutions. We added one.

However, our original statement was $x=1.$ This keeps true. So whatever we derive from there, $x=1$ doesn't get false out of thin air. Therefore we always have $x=1,$ then $x^2=x$ AND $x=1,$ then $x^2-x=0$ AND $x=1,$ then $x(x-1)=0$ AND $x=1,$ then ($x=0$ or $x=1$) AND $x=1.$

$x=1$ doesn't get lost. We started with it and there is no reason that it gets lost. And the logical formula $\{\;[\;x=0 \text{ OR }x=1\;] \text{ AND }x=1\;\}$ has only the solution $x=1.$

Instead if we assume it’s not a trick, x=1 is our original equation not that value of x is 1 ,x=1 and x=0 are solutions to the original equation. But we cannot say two numbers are equal to each other (1=x=0).
(My last post is not so clear)

Yes. More or less.

If we have an equation with an $x$, then $x^2-x=0$ means $\{x\in \mathbb{R}\,|\,x^2-x=0\},$ in words: the set of all real numbers $x$ which satisfy the equation $x^2-x=0.$ In this sense we have
$$
\{x\in \mathbb{R}\,|\,x^2-x=0\}=\{0,1\}\; , \;\{x\in \mathbb{R}\,|\,x=1\}=\{1\}
$$

 

[jack action]

In the attachment of post #49, you do 2 transformations, namely:
$$\frac{x^2}{x} = \frac{x}{x}$$
$$x\frac{x-1}{x-1} = \frac{0}{x-1}$$
The first one is valid for every value of $x$ except $0$ where $x=\frac{0}{0}$;
The second one is valid for every value of $x$ except $1$ where $x=\frac{0}{0}\frac{0}{0}$.

So neither can guarantee to give the complete picture. But both of them together do give the complete picture, i.e. for every value of $x$.

The best way to find the values that make the equation $x^2-x=0$ true is by the following transformation:
$$(x-0)(x-1)=0$$
We can easily see that if the result of either set of parentheses is $0$ then the equation is true. There are two sets of parentheses, thus two possible answers, namely, $0$ and $1$.

 

[mathwonk]

jack makes a good point that underlies why this is so confusing. I.e. is x a variable or a number? In making any argument about elements of a number system you have to say and understand which system you are working in. E.g. in post #45 it is asserted that (x-1)/(x-1) ≠ 1, but on first glimpse I would say this is wrong. I.e. if we are working, as it appears from the notation, in the quotient field of the ring of polynomials, then in fact 1/(x-1) is the inverse of x-1, so as a "rational function", in fact (x-1)/(x-1) does equal 1. The problem comes when we try to evaluate these rational expressions at specific values of x. I.e. there is an "evaluation" mapping from polynomials to numbers, defined by any number a, that sets x=a in that polynomial and evaluates it.

However it is not possible to evaluate a "rational function" or "rational expression" at all numbers a; indeed that can only be done when the denominator does not vanish at x=a. Thus the expression (x-1)/(x-1) does equal 1 as a rational function, but that does not mean it gives the value 1 when evaluated at every number. Namely that rational "function" cannot be evaluated at the number x=1.

So, in the field R(x) of rational functions over the reals R, (x-1)/(x-1) = 1 is true, where the 1 on the right hand side is the multiplkicative identity of the field k(x). But that does not mean that the VALUE of (x-1)/(x-1) at x=1 is equal to 1, because there is no such value.

Thus this problem is more confusing because it uses notation involving x's which suggest the objects being manipulated are polynomials, but it has been assumed that actually "x=1", i.e. that we are trying to evaluate the polynomials at x=1, and hence the objects involved are numbers, but this evaluation process is not always valid. To repeat, it matters what number system or "ring" the elements belong to, and confusion is purposely introduced by using symbols like x that imply we are working in the ring of rational functions, while the words imply we are working with real numbers.

 

[mark2142]

Yes and no. The step x=1 to x2=x adds another solution to the equation since there are two numbers that obey that equation: 02=0 and 12=1. Hence, x=1 has a unique solution, namely x=1, and x2=x has two different possible solutions. We added one.

However, our original statement was x=1. This keeps true. So whatever we derive from there, x=1 doesn't get false out of thin air. Therefore we always have x=1, then x2=x AND x=1, then x2−x=0 AND x=1, then x(x−1)=0 AND x=1, then (x=0 or x=1) AND x=1.

x=1 doesn't get lost. We started with it and there is no reason that it gets lost. And the logical formula {[x=0 OR x=1] AND x=1} has only the solution

I am not saying x=1 is wrong. It’s our assumption. It’s the equation that we have taken to do the math.
And simply putting the final equation x=0 is the equivalent equation to x=1 . But this means 1=0 which is wrong. So somewhere we have made a mistake. We have done something which is not allowed in maths.

(And somebody said it’s proving 1=0 by assuming 1=0. I didn’t get that.)

 

[mark2142]

You can multiply by 0 but that will result in 0=0 and every bit of information is lost because we cannot retrieve anything from 0=0.

Oh! Zero is a very dangerous quantity then. We should be afraid of 0.

 

[jbriggs444]

Oh! Zero is a very dangerous quantity then. We should be afraid of 0.

That is not the lesson to take away from this.

 

[mark2142]

That is not the lesson to take away from this.

Joking…Let’s focus on my other post.

 

[Mark44]

We cannot multiply by or divide by (x−1)=0.

@fresh_42 already responded to this, but it's worth repeating. You can always multiply both sides of an equation by zero, but this is not a useful step, as you end up with 0 = 0, which is vacuously true.

Can you add something to what I said instead of pouring extra knowledge? I am a beginner.

I.e. if we are working, as it appears from the notation, in the quotient field of the ring of polynomials, then in fact 1/(x-1) is the inverse of x-1, so as a "rational function",

Given the comment above by the OP, talk about quotient fields is not helpful, IMO. The expression "drinking from a firehose" comes to mind.

 

[fresh_42]

(And somebody said it’s proving 1=0 by assuming 1=0. I didn’t get that.)

$25\dfrac{m}{s}$ is a quantity of $25$ meter per second. It says what happens within the timespan of one second, namely a change in location of $25$ meter. It normalizes time. One second becomes the unit, i.e. $1$ second = $1$.

By writing $\dfrac{x}{x-1}$ you introduced the unit $x-1$ which thus is treated like $1$ wheras it still equals $x-1=1-1=0.$ In this sense, you treated $1$ as $0$. You made $1$ second $0$ seconds.

 

[DrClaude]

This is a new predicate that we can get from the previous one by squaring both sides

You mean multiplying by $x$ on both side.

 

[DrJohn]

This divide by 0 was a joke / trick we used to play on the younger kids at school when I was a teenager. And that was a long time ago! We used to set them up with the basics and say look, this proves 1 = 0 and watch as they approached the maths teacher for help.

Why are we still discussing this in a TWO page thread?! Dividing by zero is just not allowed. End of.

And remind yourself of when it was posted here...

 

[DrJohn]

PS people my age may remember that The Who released a single in 1971 called "wont get ****** again"

 

[fresh_42]

This divide by 0 was a joke / trick we used to play on the younger kids at school when I was a teenager. And that was a long time ago! We used to set them up with the basics and say look, this proves 1 = 0 and watch as they approached the maths teacher for help.

Why are we still discussing this in a TWO page thread?! Dividing by zero is just not allowed. End of.

And remind yourself of when it was posted here...

No. This is exactly why math in school does not work. Instead of providing any insights, it's simply forbidden. This is not how education works, at least not on PF. Why is it forbidden? You only offer: "because it is". Truth is, the why question can be answered on many levels, and we did not even touch all of them!

Shut up and calculate might work in quantum physics, but it does not work in mathematics. Not even at school, as can be seen at any school of your choice.

 

[PeroK]

No. This is exactly why math in school does not work. Instead of providing any insights, it's simply forbidden.

When I was at school I could see for myself that division by zero was problematic. I don't see any deficiency in my reasoning in those days that required advanced abstract mathematics. I certainly wasn't guilty of "shut up and calculate". Quite the opposite!

 

[fresh_42]

Since this thread has obviously turned into a general mocking about why we discuss it at all, I will close it by now. The original question has been answered on many levels anyway.