How to find the error in a proof? (that $1=0$)

Question

So I devised this proof that $1=0$. Of course it is false, but I don't know why. Why?

$$\begin{align*} x+1&=y\\ \frac{x+1}{y}&=1\\ \frac{x+1}{y}-1&=0\\ \frac{x+1}{y}-\frac{y}{y}&=0\\ \frac{x-y+1}{y}&=0\\ x-y+1&=0\\ x-y+1&=\frac{x-y+1}{y}\\ y(x-y+1)&=x-y+1\\ y&=1\\ x+1&=1\\ x&=0\qquad * * * *\\ y-1&=x\\ \frac{y-1}{x}&=1\\ \frac{y-1}{x}-1&=0\\ \frac{y-1}{x}-\frac{x}{x}&=0\\ \frac{y-x-1}{x}&=0\\ y-x-1&=0\\ y-x-1&=\frac{y-x-1}{x}\\ x(y-x-1)&=y-x-1\\ x&=1\qquad * * * *\\ 1&=0\\ \end{align*}$$

Answer 1

You have a problem already by going from $$x-y+1=0= y(x-y+1)$$ to $$y=1$$ because you are dividing by zero. The fact that you are deducing $y=1$ from a starting place, namely $$x+1=y$$ that does not assume anything about $x$ or $y$, should have been a signal that something was wrong.

Answer 2

Here is the problem. You claim x=0. Then you say $\frac{y-1}{x}=1$. Can't divide by 0.

Answer 3

When debugging proofs on abstract objects, the error may become simpler to locate after specializing to more concrete objects. Your proof begins with the equation $\rm\:y = x\!+\!1.\:$ So you are working with a general point $\rm\:(x,y)\:$ on the line $\rm\:y = x\!+\!1.\:$ It is easy to find simple special points on the line, e.g. the integer points $\rm\:(x,y) = (n,n\!+\!1).\:$ In particular, it is easy to choose such special points that do not satisfy your inference that $\rm\:y = 1,\:$ e.g. the point $\rm\:(x,y) = (1,2).\:$ Now substitute these values into your proof, and find the first place where it yields an $\rm\color{#c00}{incorrect\ equality}$ between integers. Then the inference yielding that incorrect equation must be invalid. Let's do that, successively evaluating all equations in the proof at $\rm\,(x,y) = (1,2).\,$ Omitting some steps we get $$\begin{align*} \rm x+1 &\rm = y & 1+1 &= 2 & 2 = 2\,\ \color{#0a0}\checkmark\\ \rm x-y+1&=0 & 1-2+1 &= 0 & 0 = 0\,\ \color{#0a0}\checkmark\\ \rm y\:\!(x-y+1)&\rm =x-y+1 & 2\,(1-2+1) &= 1-2+1 & 0 = 0\,\ \color{#0a0}\checkmark\\ \rm y&=1 & \color{#c00}2\ & \color{#c00}{= 1} & \color{#c00}{ 2 = 1}\phantom{\,\ \color{#0a0}\checkmark} \end{align*}\qquad\qquad$$

Thus the final inference is invalid. Indeed, it was erroneously derived by dividing by (or cancelling) the expression $\rm\:x-y+1\ = 0.\:$ Note how this method allowed us to quickly pinpoint the location of the error using only knowledge of simpler objects (arithmetic of integers versus polynomials). For some similar examples see here.

Analogous methods prove helpful generally: when studying abstract objects and something is not clear, look at concrete specializations to gain further insight on the general case. It is only by such back-and-forth journeys between the abstract and the concrete that we can ever hope to develop intuition on such abstract objects. Once we do, the abstract objects become more concrete, more intuitive. Then, with such mastery, we can understand these objects better by considering further abstractions, taking one step higher on the ladder (web) of abstraction.

For example, consider various abstractions of the notions of "number", integers, rationals, algebraics, (hyper) reals, (hyper) complexes, quaternions, octonions, surreals, polynomials, etc., all of which are abstracted in the algebraic structure known as a ring. When studying general rings, it proves quite helpful to construct a catalog of concrete prototypical (counter)examples exhibiting various properties, to help one develop better intuition of the abstract case from experience with these prototypical examples.