False proof $m = 1$ for any $m$

Question

Is it possible to find a fake proof that results in $$ 1 = m, $$ where $m$ is any number you chose?

The idea is to find a generalization to this fallacy:

\begin{align}a &= b\\ &\Downarrow\\ a^2 - b^2 &= ab - b^2\\ (a+b) (a-b) &= b (a-b)\\ a+b &= b\\ 2b &= b \\ 2 &= 1 \end{align}

You can use division by zero or anything a little bit elaborated you can think of.

Answer 1

Using $2=1$, you can prove using induction that $$\forall n\in\mathbb N,\,n=1.$$

Indeed, it is true for $1$ and also for $2$ by assumption. Assume that it's true for a certain $n>2$.

\begin{align} n+1&=n-1+2\\ &=n-1+1\\ &=n\\ &=1. \end{align}

Edit:

Another method which looks like a real generalisation of your given proof as follows: let $n\in\mathbb N$. We have:

\begin{align} a&=b\\ a^n-b^n&=ab^{n-1}-b^n\\ (a-b)\sum_{k=0}^{n-1}a^kb^{n-1-k}&=b^{n-1}(a-b)\\ \sum_{k=0}^{n-1}a^kb^{n-1-k}&=b^{n-1}\\ nb^{n-1}&=b^{n-1}\\ n&=1. \end{align}

Answer 2

Let's prove that, for every positive integers $m$ and $n$, we have $m=n$; in particular, for every positive integer $m$, we have $m=1$.

We do it by induction on $\max(m,n)$ in the form “if $\max(m,n)=k$, then $m=n$”.

The base case, $\max(m,n)=1$ is trivial.

Suppose the statement holds when $\max(m,n)=k$ and take $m,n$ such that $\max(m,n)=k+1$. Then $\max(m-1,n-1)=k$, so by the induction hypothesis we get that $m-1=n-1$, hence $m=n$.

(Taken from “What is Mathematics?” by R. Courant and H. Robbins.)

Answer 3

Let $a=b$.

Then \begin{align}a^m-b^m&=a^m-b^m \\ a^m-b^m&=ab^{m-1}-b^m \\ (a-b)(a^{m-1}+a^{m-2}b+...+b^{m-1})&=(a-b)b^{m-1} \\ a^{m-1}+a^{m-2}b+...+b^{m-1}&=b^{m-1}\\ mb^{m-1}&=b^{m-1}\\ m&=1\end{align}

Answer 4

Let $m$ be an arbitrary number and $x= \frac{m+1}{2}$.

Then we have \begin{align} x&=\frac{m+1}2\\ 2x&=m+1\\ 2x(m-1)&=(m+1)(m-1)\\ 2xm-2x&=m^2-1\\ x^2+2xm-2x&=x^2+m^2-1\\ x^2-2x+1&=x^2-2xm+m^2\\ (x-1)^2&=(x-m)^2\\ x-1&=x-m\\ -1&=-m\\ m&=1 \end{align}