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It may sound stupid but why is $1=1$ or $n=n$ if thats the case does $1/0 = 1/0.$

Kortlek
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Sudheej
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  • Keep in mind that we use equality to capture the idea of "sameness". Something is the same as itself pretty much by the definition of "same". There's not much going on here that's more mysterious than the use of equality/sameness in normal language. – Robert Mastragostino Jun 13 '13 at 17:32

6 Answers6

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Because we define equals as an equivalence relation which needs to fulfill three properties

  1. reflexive meaning $x\sim x$
  2. symmetric $x\sim y \implies y\sim x$
  3. transitive $x\sim y $ and $y\sim z$ imply $x\sim z$ where $\sim$ means it fulfills the relation.
Joshua Meyers
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Dominic Michaelis
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One of the defining properties for equality is that for any $x$ we have $x=x$. Therefore $1=1$ and $n=n$ by definition.

Now we would also have $1/0=1/0$ if $1/0$ existed. However, there's no such number as $1/0$, and for something that does not exist, obviously no equality can be defined.

Note that for equality we additionally demand that there's no other $y$ besides $x$ so that $y=x$. However this is in some sort a "soft" demand because we can always enforce it from an equivalence relation by simply considering equivalence classes. Also, to make that demand, you'll have to have an independent notion of what it means for two things to be "the same".

One example of the "soft" sameness is fractions: Formally, we define fractions as pairs of numbers with certain calculation rules (and a special notation of the pair to signify that we mean the fraction, i.e. those special rules for calculation). Now $\frac12$ and $\frac24$ are clearly different pairs, yet we consider them the same number. Technically, this means that our fractions are actually the equivalence classes of all $(a,b)$ under the equivalence relation $(a,b)\equiv(c,d)\Leftrightarrow ad=bc$.

celtschk
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  • maths showed non-Euclidean geometries can exist.. if you say something "exists", what you mean is that it doesn't introduce a logical contradiction given a particular set of rules. Is there any rule that defines 1/0 – Sudheej Jun 13 '13 at 07:36
  • @xtechkid: $1/0$ would, by definition, be the solution to the equation $0x=1$. If such an $x$ existed, we would have $1 = 0x = (0+0)x = 0x + 0x = 1 + 1 = 2$, but we know $1\ne2$. Therefore $1/0$ cannot exist. – celtschk Jun 13 '13 at 07:39
  • @xtechkid: There are other arithmetic systems (to which celtschk's analysis wouldn't apply). If you want to ask about other arithmetic systems, then you need to do so explicitly -- otherwise, people (rightfully!) assume you are using words and symbols to mean what those words and symbols usually mean. –  Jun 13 '13 at 07:43
  • @Hurkyl: In which arithmetic system does $0x=1$ have a solution? – celtschk Jun 13 '13 at 07:52
  • @celtschk: I don't know (any commonly used ones). But you make a mistake in assuming that 1/0 would be defined to be a solution to $0x=1$. –  Jun 13 '13 at 08:04
  • @celtschk Wheel theory would be the main one. You modify the meaning of division so that it reduces to the normal one for the numbers you know and love, but can be extended to $1/0$ and the like. – Robert Mastragostino Jun 13 '13 at 17:30
  • That's interesting. Of course, looking at the Wikipedia page, it modifies quite a bit more than just the meaning of division; especially the distributive law is modified (indeed, it has quite a few rules which cover what normally would be covered by the distributive law). – celtschk Jun 13 '13 at 19:30
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I think a non math answer would be that if $2 = 1$, then what is $1 + 1$? It isn't $2$ anymore (or is it? ), but we have $4 = 2$.

We are going to arrive at some inconsistencies.

EDIT: Actually to be more precise, we have to redefine '$+$' don't we? Should we go with your definition.

Lemon
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Sticking to the title of your post, I think it is okay for you to define $1$ to be something else. But in that case, you have to deal with consequences of your definition.

Gil
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Take the number 0 then for axioms we have for all n exist -n then we know n-n=0 then n=n

Gmath
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The symbol $1$ is reserved for neutral element of multiplication. The answer to the topic comes from uniqueness of multiplicative neutral element. Assume that $1$ and $n$ are neutral elements of multiplication. Then \begin{eqnarray} n = 1\cdot n = n\cdot 1 = 1 \ . \end{eqnarray} The first equation comes from the properties of $1$ and the third from the properties of $n$. That's why $1$ cannot be anything else.

The questions "Why is $1=1$?" and "Why it is so?" and "Why can't be $1=$ something else?" are separate and the first holds trivially. The second question refers to the first. Hence it is also trivial. The third question was answered above.

Also $n=n$ holds trivially. The equation $1/0=1/0$ is not defined according to standard definitions.

Juha-Matti Vihtanen
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