Why is 1 not equal to 0 in this proof?

[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.