# Proof that 0=1

Here is a proof that $0=1$:

\begin{aligned} &\mbox{Let} & & 1=x \quad &(1) \\ &\mbox{We know that} & & 1=x^2 \quad &(2) \\ &\mbox{Taking the difference, we get} && 0 = x^2 - x = x(x-1) \quad & (3)\\ &\mbox{Hence, } && x=0, 1 & (4)\\ &\mbox{From (4), we have} & & x=0 & \\ &\mbox{From (1), we have} & & x=1 & \\ &\mbox{Hence, we have} && 0=x=1. & \\ \end{aligned}

At which step did we go wrong?

We did not divide by 0. We may have introduced an extra solution $x=-1$, but that didn't affect anything.

Note: The common objection is that I can't claim that the roots of a quadratic are equal. I agree with that, and that they in fact are (often) not equal. However, that wasn't how the proof was obtained. We obtained $x=0$ from (4) and have the original condition that $x=1$ from (1).

This post arose from a discussion that we had on a problem. The original problem that led to this discussion is:

[Nov 11 Algebra and Number Theory] The point $(x,y)$ lies on both conics $x^2+xy+x=81$ and $y^2+xy+y=51$. Given that $x+y$ is positive, determine the value of $x+y$. Note by Calvin Lin
5 years, 7 months ago

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At step 4. We get two solutions. One of them is our number.

Eg:

a.Let's say $x={x}_{1}$ is root for a quadratic equation. Does that mean that ${x}_{1}={x}_{2}$?

b. Let $f=0$ a polynomial. We have $x=0$ a root but also $x=12$ is a root. Does that mean that 0=12?

That other problem is easy. If you add the equations and use S and P for the sum and product you get: ${S}^{2}-2P+2P+S=132$ and S will be 11.

- 5 years, 6 months ago

I agree. We can toss out the "x=0" solution we obtain in step 4, after we substitute the value of "0" for x in steps 1 and 2. We see that it creates an equation that is obviously incorrect ("0=1"), so we know "x=0" is an extraneous solution.

- 5 years, 6 months ago

In step (4) you get ${x} _ {1} = 0, {x} _ {2} = 1$. And yes 1 is the root of this equation. Some numbers have some common properties that doesn't mean that the numbers are equal.

- 5 years, 6 months ago

The "first" problem in the proof lies in step (1) -- you "solve" an equation you already know the answer to (i.e. you overdetermine it's value prior to computation). However, I believe the answer you're looking for is in step (3), where you introduce another variable rather than "taking the difference". Therefore, throughout the rest of the proof, you continue to conflate variables for known values.

Specifically, in step (2), you perform a valid operation on the full equation (i.e. square both sides); in step (3), however, you plug in previous value(s) instead of performing the operation stated [ i.e. $0 = x^2 - x$ should be $0 = x^2 - 1$ ].

[By the way, nice hint on the (-1) root solution--I was originally only going to point out that it couldn't be a solution to $x^2 - x$, then realised after I made the point what you were saying...sneaky!]

- 2 years, 9 months ago

Close, and that might be what you're saying. In a certain sense, it could be conveyed by what you say about "conflate variables for known values", but the misconception is much broader.

The issue is that the common solutions to $f(x) = 0$ and $g(x) = 0$ is a subset of the solutions to $f(x) - g(x) = 0$, but not vice versa. So, the solution of 0 to "taking the difference" need not be the solution to the first 2 equations.

Staff - 2 years, 9 months ago

If I understand you correctly, you are stating step (3) is a completely different equation than the preceding steps?

The solution set for the equation in step (3) is, as you state, $x = {0, 1}$.

The solution set for the equation in step (2) is, however, $x = {1, -1}$.

[And, of course, the solution set for the equation in step (1) is $x = {1}$.]

If we view these as indeterminate functions, we would (I assume) have the following:

$[ (1) \: f(x) = x ]; \; [ (2) \: g(x) = x^2 \, \to \, g \circ f(x) = x^2 ]; \; [ (3) \: h(x) = g(x) - f(x) ]$

The second equation introduces (unnecessary) ambiguity--it allows $x$ to be either negative of positive. The third equation, however, introduces an entirely new implicit assumption:

$[ \text{If:} \: h(x) = 0 \: \implies \: g(x) = f(x) ]$

Of course, if you view steps (2) & (3) as a system of equations (both derived from the first), the solution set of the variable becomes the union of these two solution sets--this, of course, requires they first be established (read: noted) as separate equations.

[In other words, if I understand you properly, we needn't read the proof steps as subsequently connected--we only need to read them as valid in their own right and both following from the first?]

- 2 years, 9 months ago

Right. When we write up our steps, it's important to know if we indeed have "if and only if" statements. In this case, $(1) \Rightarrow (2), (1) + (2) \Rightarrow (3)$. However, we do not have $(2) \rightarrow (1)$ or $(3) \Rightarrow (2)$ or $(3) \Rightarrow (1)$. As such, we could have solutions to (3), which are not solutions to (1), or (2).

Staff - 2 years, 9 months ago

I think that the second step is invalid because you squared without accounting for the second root that is produced. :P

- 5 years, 5 months ago

a^0=1 how give proof

- 5 years, 7 months ago

simple proof

1= $a^n/a^n$

1=$a^{n-n}$

hence 1=$a^0$

- 5 years, 5 months ago

It's easy to prove that ${a}^{0}=1$ for any $a>0$.

We know that ${a}^{m}:{a}^{n}={a}^{m-n}$.

For $m=n$ we get $1={a}^{m}:{a}^{m}={a}^{0}$.

- 5 years, 6 months ago

this happens because whenever we square any equation after arriving at the answer we must substitute and into our original condition as squaring always stands and gives possibilities of +/- value of roots

- 5 years, 1 month ago

Squaring gives us an extraneous root, after the final solution we check that the solutions satify the original equation or not.

- 5 years, 1 month ago

It should be either x=0 or x=1, the two values of x can't be equal at the same time..

- 5 years, 1 month ago

what if we take, 1/∞=0, then, 1=0×∞ since (z×0=0) Therefore , 1=0 Sir, please tell me if i am wrong

- 5 years ago

The thing is that the two solutions don't work at the same time. Either $x = 1$ or $x = 0$. This is the very essence of a quadratic.

- 1 year, 10 months ago