Prove, axiomatically that $1$ does not equal $0$.

Question

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

Answer 1

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.

Answer 2

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 $0 \ne 1$.

Answer 3

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).

Answer 4

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 $0 = 1$.

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

Adding on to other answers here. From a Real Number system approach, instead of the ordering property (There exists a non-empty subset 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.