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 $x=0$.