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I attempted proving this but came across many circular arguments. How would one prove such a claim without circular arguments?

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    $\begingroup$ Using what axioms? $\endgroup$
    – Hayden
    Sep 15, 2015 at 22:34
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    $\begingroup$ 1 equals 0 in modulo 1 arithmetic (which is trivial). What system are you working in? $\endgroup$
    – user76284
    Sep 15, 2015 at 22:37
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    $\begingroup$ If $1=0$ and $a$ is any element, then $a=a\cdot 1 = a \cdot 0 =0$ so there's only one element in your set. [Uses some defiinitions from what a unitary ring is, and some consequences of axioms] Note that many collections of ring axioms do not rule out $1=0$ but call such a thing the "zero ring" ... $\endgroup$
    – coffeemath
    Sep 15, 2015 at 22:42
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    $\begingroup$ Axiom 1. $1\neq0$. Proposition 1. $1\neq0$. Proof. From axiom 1, we have that $1\neq0$. $\blacksquare$ $\endgroup$ Sep 15, 2015 at 22:44
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    $\begingroup$ Before we answer the question we need to know what you mean by $0$ and $1$. $\endgroup$ Sep 15, 2015 at 23:45

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If you start with Peano axioms for the natural numbers, then $0$ is part of the language, but $1$ is not. We use $1$ as a shorthand for the term $s0$.

Now we can use the axiom $\forall x(sx\neq0)$, and infer that in particular for $x=0$ it is true that $s0\neq0$. Congratulations, we proved that $0\neq1$ axiomatically.

You can choose different contexts, like set theory, field theory, ring theory or other contexts in which we can interpret $0$ and $1$. You can also find contexts in which $0=1$ is a provable statement. For example the theory whose single axiom states $0=1$. True this theory describes very little of what we expect from the natural numbers, or the symbols $0,1$ to mean. But it is a mathematically valid thing to do.

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1) Prove $a \times 0 = 0$ for all $a$.

$a \times 0 = a(1 - 1) = a - a = 0$.

2) If $0 = 1$ then $a = a \times 1 = a \times 0 = 0$

So all terms equal 0.

Which isn't actually a contradiction. It just means we are working with a trivial field. If the field isn't trivial (say the Reals) than $0 \ne 1$.

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  • $\begingroup$ Assuming the axioms are field axioms. $\endgroup$
    – fleablood
    Oct 8, 2015 at 21:35
  • $\begingroup$ It is actually the zero or trivial ring. Good sets field axioms avoid it by explicitly assuming $1 \not=0$ as an axiom. That then makes an axiomatic proof rather easy $\endgroup$
    – Henry
    Apr 7, 2017 at 19:15
  • $\begingroup$ Yikes. I said that 2 years ago! I've seen $0\ne 1$ explicitly omitted from field axiom. I''m sort of reluctant because defining things by axiom seems weak when we can prove something instead. And $1=0\implies $ trivial ring seems worth discovering Still, defining a field from a ring by defining it to have multiplicative identity and non-trivial inverses makes a lot of sense. $\endgroup$
    – fleablood
    Apr 7, 2017 at 20:29
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Maybe we have an axiom that says $0$ is the additive identity and another axiom that says $1$ is the multiplicative identity, that is to say, $x + 0 = x$ and $x \times 1 = x$. Then, in order for $0 = 1$, we'd need to have $1 + 1 = 1$ and $0 \times 0 = 0$. But this implies that $1 \times 2 = 1$, contradicting the multiplicative identity axiom which requires $2 \times 1 = 2$. (Maybe we also need an axiom saying addition and multiplication are commutative).

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    $\begingroup$ $1\times 2=1$ only contradicts $2\times1=2$ if $1\ne 2$. $\endgroup$
    – celtschk
    Oct 6, 2015 at 21:22
  • $\begingroup$ So maybe I have run into one of the same circularities as Neel... $\endgroup$
    – James47
    Oct 6, 2015 at 21:25
  • $\begingroup$ If I understand correctly, $1 \times 2 = 1$ contradicts the axiom that $1 \times x = x$. In this case, $x = 2$, so $1 \times 2$ should be 2, not 1. $\endgroup$ Oct 7, 2015 at 3:07
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    $\begingroup$ @RobertSoupe: If $1=2$ then $1\times 2=1$ does not contradict $1\times 2=2$; indeed, you can derive $1\times 2=1$ from $1\times 2=2$ and $1=2$ (Just use the Euclidean relation property of equality). $\endgroup$
    – celtschk
    Oct 7, 2015 at 18:04
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    $\begingroup$ @celtschk I think I see your point. But at the same time I also feel like maybe this is shifting Neel Shah's question up a notch. $\endgroup$ Oct 8, 2015 at 0:55
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This is one of those deceptively simple questions that is actually rather tough. Declaring $1 \neq 0$ seems rather cheap, but I can't think of a better way. In the hopes that it helps someone come to be a better answer, here's my attempt:

If $x = y$, then $x - y = 0$ and $x \div y = 1$. Obviously $1 - 0 = 1$ but that doesn't help us since maybe $1 = 0$. Nor does $0 \div 1 = 0$ help either. However, $1 \div 0$ is normally considered undefined, and attempts to define usually go for infinity.

Of course the big flaw in this whole argument is that $0 \div 0 \neq 1$ unless we explicitly defined it that way for the purpose of proving $0 = 1$.

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Some good ideas so far, but trying to bring in multiplication is problematic. I think you have to focus on addition here.

Axiom $0$. The additive identity is $0$. Thus $n + 0 = n$.

Axiom $1$. The successor interval is $1$. Thus the successor of $n \in \textbf{Z}$ is $n + 1$. If $n'$ is the successor of $n$, then $n < n'$ and $n' > n$.

Axiom $2$. Addition is commutative. Thus $a + b = b + a$.

Therefore, $0' = 1$, because $1 + 0 = 0 + 1 = 1$. This sits well with Axiom $0$ and Axiom $1$.

Maybe $0 = 1$. But by Axiom $1$, we have $1 > 0$ and $0 < 1$, flatly contradicting $0 = 1$. We conclude that $0 \neq 1$ after all.

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  • $\begingroup$ This doesn't work - you need some axiom asserting that $a=b$ implies $\neg(a<b)$. $\endgroup$ Oct 10, 2015 at 20:16
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    $\begingroup$ Sometimes it's hard to tell the difference between rigor and pettiness. A comment along the lines of "you haven't defined the comparison operators" would lead me to think it was the former rather than the latter. $\endgroup$
    – Bob Happ
    Oct 12, 2015 at 20:42
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Proving $0\neq 1$ was shown above (@Asaf Karagila) in the context of a Peano system. If, however, you are trying to prove $0\neq 1$ in the context of the Real number system, the order axioms are required as shown below.

(Note: here the Reals are based directly from the field axioms, order axioms, and the completeness axiom, as opposed to being considered an extension(s) of the Rationals, Integers, or Natural number systems.)

It is sufficient to use the field and order axioms to prove that $0\neq 1$ (so this result also applies in the context of the Rational numbers). The order axioms can be presented as follows:

  • (O0): There exists a non-empty subset $P$ of $\mathbb{R}$, called the set of "strictly positive real numbers."
  • (O1): If $a,b\in P$, then $a+b\in P$.
  • (O2): IF $a,b\in P$, then $a\cdot b\in P$.
  • (O3) "Trichotomy Property": If $a\in\mathbb{R}$, then exactly one of the following holds: $a\in P,\ a=0,\ -a\in P.$

First, it was shown above (@fleablood) using the field axioms, that if one assumes $0=1$, then the result is a trivial situation in which all elements of $\mathbb{R}$ equal $0$.

However, a contradiction arises in the trivial solution of $a=0$ for all $a\in\mathbb{R}$. This is because Order Axiom (O0) requires $P$ to be non-empty, but if $a=0$ for all Real numbers, then the Trichotomy Property (axiom (O3)) indicates that $-a,a\notin P$ for any $a$. Hence, the assumption that $0=1$ must be false.

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    $\begingroup$ This depends on your axiomatizations. In many places, fields are given a nontriviality axiom, namely $0\neq 1$. So proving $0\neq 1$ there is even shorter than the proof I suggested in PA. $\endgroup$
    – Asaf Karagila
    Feb 19, 2016 at 21:26
  • $\begingroup$ @AsafKaragila, that is something good to know, as I was not aware of it (the Real Analysis textbook I used in college does not assume a nontiviality axiom). In any event, the above proof at least shows that the nontriviality axiom is redundant (and therefore consistent) with the field/order axioms. $\endgroup$
    – Stephen K.
    Feb 19, 2016 at 21:44
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    $\begingroup$ Alternatively, given the nontriviality axiom, one could weaken the Order Axioms by not requiring $P$ to be non-empty. Then the property $0\neq 1$ can be used to prove that $P$ is non-empty. $\endgroup$
    – Stephen K.
    Feb 19, 2016 at 23:22
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I attempted proving [$1 \neq 0$] but came across many circular arguments. How would one prove such a claim without circular arguments?

Any proof of that is circular in substance.

There is a non-circular proof only in a thin definitional sense particular to a given theory, such as valid derivation from axioms in Peano Arithmetic or Zermelo-Fraenkel set theory. In almost any other meaningful sense, a proof that $1 \neq 0$ assumes more than its conclusion. You need a certain minimal amount of mathematics to get an axiomatic proof system up and running, such as the ability to parse strings of characters and tell them apart. Distinguishing $1$ from $0$ is far less than what is needed for that.

A proof of the inequality is also circular in the meta-theoretical sense that the theories accepted as foundational schemes for arithmetic have been pre-selected for their apparent inability to prove $0=1$, as well as their demonstrated ability to prove $0 \neq 1$. A theory that proves $0=1$ is (for most purposes) deemed to be inconsistent, and one that does not prove $0 \neq 1$ is too weak for serious use. One is "deriving" 0 < 1 from axioms chosen for their ability to reach that conclusion, and not only their a priori feeling of correctness as a basis for mathematical reasoning.

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Adding on to other answers here. From a Real Number system approach, instead of the ordering property (There exists a non-empty subset 𝑃 of ℝ, called the set of "strictly positive real numbers."), you can also prove the same progression with "There exists at least 2 distinct real numbers".

This is to say, 1 = 0 iff ∀ℝ = 0. This feels like the weakest axiom possible here, as it makes no claim to order, just being distinct. It also gives more information than just assuming 0≠1.

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Based on the recursive definition of Natural Numbers from the Naive Set Theory:

We start by defining $0 := \emptyset$. Then given any natural $n$ define next as $n^+ := n \cup \{n\}$.


Thus we get:

$0 = \emptyset$ and $1 = 0^{+} = 0 \cup \{0\} = \emptyset \cup \{\emptyset\} = \{\emptyset\}$.

So $0 = \emptyset$ while $1 = \{\emptyset\}$.

Hence $0 \neq 1$ because $1 \not \subset 0$ due to $\emptyset \not \in \emptyset$.

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