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I want to prove that $1 > 0$ using the field and order axioms. So far I am trying to use the Peano axiom which states that if two numbers $n$ and $m$ have the same successor, then $n = m$.

Specifically, if we have $1$ to both sides, we obtain:

$$1 > 0 \\ 1 + 1 > 0 + 1 \\ 1 + 1 > 1$$

From this, we can say that $1$ and $0$ are not the same number, because they have different successors.

If we had now the additive inverse of $1$ to both sides:

$$1 + 1 > 1 \\ 1 + 1 + (-1)> 1 + (-1) $$

I have just add the same number to both sides, this is possible according to the order axioms.

$$1 + (1 + (-1))> (1 + (-1))$$

By the field axioms, we know that $-1$ is the additive inverse of $1$, so their sum is $0$. Also, I am using the associativity of field axioms.

$$ 1 + 0> 0 \\ 1 > 0$$

Is this proof correct using the order and field axioms?

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    It is impossible to prove this from the field axioms because some fields have no ordering (i.e., no "$>$"). – Zev Chonoles Apr 14 '15 at 14:50
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    What do Peano axioms have to do with field axioms? Please, clarify the setting. – egreg Apr 14 '15 at 14:51
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    Also, you started from the assumption that $1>0$ and concluded that $1>0$, which is a clear case of circular reasoning. – Zev Chonoles Apr 14 '15 at 14:51
  • I seem to recall that there is an ordered field axiom that says that the product of two negative numbers is positive, and so is the product of two positive numbers. The result should follow easily. – Arthur Apr 14 '15 at 15:15
  • The axioms you mention are redundant. It is enough to know the product ot two positive numbers is positive. – Bernard Apr 14 '15 at 15:22

3 Answers3

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Based on an order axiom we know that if $ a$ and $b $ have the same sign then $a*b>0$ let $a=b=1$ , clearly a and be have the same sign therefore $a*b=1*1=1>0$

Maths Survivor
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This is not a proof unfortunately.

In your analysis you were doing fine until you started with the inverse. At this point your equation:

1+1 > 1

is based on your starting assumption that 1>0. As such, what you have shown is that so long as 1>0, you know that 1>0, which is technically true, but trivial.

Sam Smith-ferrier
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  • If $1 > 0 $ it means that $(1 \geq 0) \wedge (1 \neq 0)$.
  • Further $1 \geq 0$ means that $1 = 0 + m$ for some $m \in \mathbb{N}$.
  • Now $1 \neq 0$ because $ 1 = 0\text{++}$ by definition and by peano's axioms we know that zero is not the successor of any natural number hence $0\text{++} \neq 0$.
  • By definition of addition $(0\text{++}) + 0 = (0+0)\text{++}$ ; $0 + 1 = 0\text{++} = 1.$
  • Hence if we choose $m = 1$ we get $1 = 0 + 1$

Here $\text{++}$ is the increment operator as defined in Terence Tao Analysis I.

AdityaRamesh
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