Why is 1 not equal to 0 in this proof?

[mark2142]

Homework Statement: I am having trouble in understanding the proof. Which step is wrong and why?
Relevant Equations: Why $1 \neq 0$ ?
Mentor note: Moved from Homework Section
$x=1$
$x^2=x$
$x^2-x=0$
$x(x-1)=0$
$\frac {x(x-1)}{(x-1)}=\frac 0{(x-1)}$
$x=0$
$1=0$

 

[fresh_42]

You divided by $0$. If $x=1$ then $x-1=0$ so does not have a multiplicative inverse, i.e. cannot be divided by. If you include a FALSE statement in a proof, then you can prove any statement.

You basically used $1=0$ to prove $1=0.$

 

[malawi_glenn]

No own effort made?

 

[Mark44]

No own effort made?

More of a general question than a specific homework-type question. I've moved it to the Math technical section.

 

[fresh_42]

No own effort made?

Isn't the whole thing obviously own effort?

 

[malawi_glenn]

Isn't the whole thing obviously own effort?

No. This is a problem in a math book I use for teaching and the pupil is to figure out where in the "proof" things went wrong.

 

[fresh_42]

No. This is a problem in a math book I use for teaching and the pupil is to figure out where in the "proof" things went wrong.

Sneaky.

 

[malawi_glenn]

Sneaky.

It seems to me when I read OP that the question is "which step is wrong and why". Thus there is no own effort made in trying to figure out what was wrong in that "proof".

 

[pinball1970]

Can I ask why the mentors have ChapGPT as names? Are they using it to answer?

 

[fresh_42]

Are they using it to answer?

No. Not good enough. It is because of the 54th day before Towel Day.

 

[pinball1970]

No. Not good enough. It is because of the 54th day before Towel Day.

Right. Douglas Adams. Towel day....

Erm. Right.

 

[PeroK]

Homework Statement: I am having trouble in understanding the proof. Which step is wrong and why?
Relevant Equations: Why $1 \neq 0$ ?
Mentor note: Moved from Homework Section
$x=1$
$x^2=x$
$x^2-x=0$
$x(x-1)=0$
$\frac {x(x-1)}{(x-1)}=\frac 0{(x-1)}$
$x=0$
$1=0$

Note that in the first line we have $x =1$, but by the second line we have $x =0$ or $x =1$.

 

[mathman]

Start with x=1, then $x^2-(a+1)x+a=0$ is true. So roots are x=1 and x=a. Your example has a=0.

 

[DaveC426913]

Can I ask why the mentors have ChatGPT as names? Are they using it to answer?

Man! I came this close...

I actually went to the Feedback forum to start a thread complaining about the name of this new user.
I just happened to stumble across the thread there and went down the Douglas Adams rabbit hole, which distracted me just long enough for my hindbrain to catch up.
How nearly embarrassing...

 

[jack action]

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:

 

[Vanadium 50]

Can I ask why the mentors have ChapGPT as names?

Are you suggesting that's why the OP asked what he did when he did?

 

[pinball1970]

Are you suggesting that's why the OP asked what he did when he did?

No.

 

[jbriggs444]

Note that in the first line we have $x =1$, but by the second line we have $x =0$ or $x =1$.

Well yes. There is no problem with such a deduction. If $x=1$ then it does indeed follow that either $x=0$ or $x=1$.

The structure of the argument is an assertion that $x=1$ followed by a series of equalities, each of which is a valid consequence of the one immediately above. Except, of course, for the one that divides both sides of the previous equality by zero. Which yields $0/0 = 0/0$.

The use of $x$ is valid but pointless. The same proof could proceed using the numeric literal "$1$" instead. The utility of using a variable named $x$ is to obfuscate the eventual attempt to divide by zero.

 

[PeroK]

Well yes. There is no problem with such a deduction. If $x=1$ then it does indeed follow that either $x=0$ or $x=1$.

It's still worth observing, IMO.

The use of $x$ is valid but pointless.

The point is to create an apparent paradox.

The same proof could proceed using the numeric literal "$1$" instead. The utility of using a variable named $x$ is to obfuscate the eventual attempt to divide by zero.

Precisely!

 

[PeroK]

If $x=1$ then it does indeed follow that either $x=0$ or $x=1$.

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

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

 

[Drakkith]

Can I ask why the mentors have ChapGPT as names? Are they using it to answer?

April Fools joke on the forum I expect.

 

[mark2142]

f you include a FALSE statement in a proof, then you can prove any statement.

Yes. That’s true.

x−1=0 so does not have a multiplicative inverse, i.e.

Ok. x-1=0 and we are dividing by 0. But I do not understand “so does not have a multiplicative inverse” part.
Are you trying to say since $\frac{x-1}{x-1} \neq 1$ we cannot use that ? $\frac 00= undefined$. Any number can be multiplied by $0$ to get $0$.
We write $x(x-1)*1=0$ but $\frac{x-1}{x-1} \neq1$ so we can’t replace 1 by that fraction. Yes?

 

[malawi_glenn]

A multiplicative inverse to $a$ is $a^{-1}$ and is defiened as $a^{-1}a =aa^{-1} = 1$. All real numbers except 0 have an unique multiplicative inverse.

 

[fresh_42]

Yes. That’s true.

Ok. x-1=0 and we are dividing by 0. But I do not understand “so does not have a multiplicative inverse” part.
Are you trying to say since $\frac{x-1}{x-1} \neq 1$ we cannot use that ? $\frac 00= undefined$. Any number can be multiplied by $0$ to get $0$.
We write $x(x-1)*1=0$ but $\frac{x-1}{x-1} \neq1$ so we can’t replace 1 by that fraction. Yes?

You cannot even write $\dfrac{1}{x-1}.$ You pretend it equals $1$, by going from $\dfrac{a}{x-1}=\dfrac{b}{x-1}$ to $a=b$ whereas it actually equals $0.$ In this sense, you put $1=0$ into your proof and showed $1=0 \Longrightarrow 1=0$ which is a TRUE statement, but $1=0$ is still FALSE.

 

[mark2142]

You cannot even write $\dfrac{1}{x-1}.$ You pretend it equals $1$, by going from $\dfrac{a}{x-1}=\dfrac{b}{x-1}$ to $a=b$ whereas it actually equals $0.$ In this sense, you put $1=0$ into your proof and showed $1=0 \Longrightarrow 1=0$ which is a TRUE statement, but $1=0$ is still FALSE.

Can you be a bit simple in your language?

And am I right in post #22 ?

 

[fresh_42]

Can you be a bit simple in your language?

Sure. You set $x=1.$ Then you wrote $\dfrac{x(x-1)}{x-1}$ which equals $\dfrac{x(x-1)}{0}$ which you are not allowed to do. Even if you say $\dfrac{x(x-1)}{x-1} =\dfrac{1\cdot 0}{0}$ you will be stuck here because you cannot cancel $0$ in a quotient. The zero in the denominator makes it automatically forbidden.

And am I right in post #22 ?

No. See above.

And one last more in a less simple language:

We do not even have $0\cdot x =0$ if we take a closer look. $0$ and $1$ are only, and I mean exclusively connected by the distributive law. All we are allowed to do is
$$
x\cdot (y+z) = x\cdot y +x\cdot z
$$
That's it. In any other case, $0$ has to stay away from multiplication. They are not defined on the same set of numbers! From the distributive law, we get $x\cdot (1+(-1))= x\cdot 1 +x\cdot(-1)=x+(-x)=0$ and thus $x\cdot 0 =0$ but it was the distributive law that gave us this equation. You cannot do something similar for division.

 

[PeroK]

Yes. That’s true.

Ok. x-1=0 and we are dividing by 0. But I do not understand “so does not have a multiplicative inverse” part.
Are you trying to say since $\frac{x-1}{x-1} \neq 1$ we cannot use that ? $\frac 00= undefined$. Any number can be multiplied by $0$ to get $0$.
We write $x(x-1)*1=0$ but $\frac{x-1}{x-1} \neq1$ so we can’t replace 1 by that fraction. Yes?

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.

 

[mathwonk]

I was tempted to ask how we are supposed to know we are working in some integral domain here, (such as the reals as assumed in post #23), where indeed 1 and 0 are distinct, and not possibly the integers modulo 1, but then I realized this would most likely be unhelpful to the OP. But I do confess to a habit of asking for hypotheses before trying to make a valid proof, (the importance of which is signaled in post #2).

 

[fresh_42]

I was tempted to ask how we are supposed to know we are working in some integral domain here, (such as the reals as assumed in post #23), where indeed 1 and 0 are distinct, and not possibly the integers modulo 1, but then I realized this would most likely be unhelpful to the OP. But I do confess to a habit of asking for hypotheses before trying to make a valid proof.

I had the same thought but decided that this would have led deep into the rabbit hole.

 

[mathwonk]

of course in post 2 you remarked on the most sweeping consequence of faulty hypotheses!

 

[PeroK]

How is it helpful to a struggling student to bring up integral domains?

 

[WWGD]

@mark2142 : Some operations on your equations will preserve roots. Others, like multiplying terms may introduce new roots.

 

[mcastillo356]

Hi, @mark2142, to check if transmissions between electronic devices are right we make use of the parity bit. $0$ is even, and $1$ is odd. It's how it works.
Greetings!

 

[mathwonk]

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.

 

[mark2142]

Sure. You set x=1. Then you wrote x(x−1)x−1 which equals x(x−1)0 which you are not allowed to do. Even if you say x(x−1)x−1=1⋅00 you will be stuck here because you cannot cancel 0 in a quotient. The zero in the denominator makes it automatically forbidden.

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 $x=0$.

 

[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.