# When does one equal two [duplicate]

When does 1=2 I once saw a proof that 2=1. Of course there was an error in the proof. (some where in the proof they were dividing by 0 - a/a) Does anybody remember seeing this proof. It was a very simple thing but it has been years and I don't remember how it went.

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## marked as duplicate by Zev Chonoles, Old John, Thomas Andrews, T. Bongers, Ayman HouriehSep 5 '13 at 23:26

you mean this? getdigital.de/images/produkte/t4/t4_drinkandderive.jpg –  Kofi Sep 5 '13 at 22:52
These sorts of fake proofs are everywhere. If you just search the web you're bound to find a ton. –  Jeremy Sep 5 '13 at 22:52
math.stackexchange.com/questions/446040/why-does-2-equal-1 and math.stackexchange.com/questions/117998/… are a couple of examples on this site. –  JB King Sep 5 '13 at 22:55

Here is a valid proof that $1=2$ in a particular ring.

First, a ring is a set $R$ together with two binary operations $+$ and $\cdot$ satisfying some very straightforward axioms that follow from our intuition of working with real numbers. One of these axioms is that the ring has a special element, denoted by the symbol $0$, such that $0+r=r=r+0$ for all $r\in R$.

Not a part of the axioms is another condition that some rings satisfy: the ring has an identity if there is a special element in $R$, denoted by the symbol $1$, such that $r\cdot 1=r=1\cdot r$ for all $r\in R$.

In any ring with an identity, denote by the symbol $2$ the value of $1+1$.

It turns out that the one element set $R=\{\star\}$ is a ring with identity, using the only possible definitions of the binary operations: $\star+\star=\star$ and $\star\cdot \star=\star$. Notice that in this case, we have $\star=0$, $\star=1$, and $\star=2$. By the transitivity property of equality, it follows that $0=1=2$.

So to address the question in the title, with these definitions of the symbols $1$ and $2$, it can be shown that $1=2$ if and only if $R=\{\star\}$.

Your question most likely addresses the symbols $1$ and $2$ in the ring of integers, denoted $\mathbb{Z}$. Since $\mathbb{Z}$ is not 'the same' as the ring $\{\star\}$, it follows that the symbols $1$ and $2$ are different elements in $\mathbb{Z}$. Hence, as you've noted, any proof that claims they are the same element must be faulty. There are plenty of such proofs, many of which you can find in the links provided in the comments.

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Taking the title question literally: Structures in which $1=2$ include the circle group $\mathbb R/\mathbb Z$, the group of rationals modulo integers $\mathbb Q/\mathbb Z$, and the trivial ring $\mathbb Z/\mathbb Z$. However, these structures are not necessarily related to the fake proofs you have in mind.

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