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!