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This is a trivial task:

Given $a \in \mathbb{N}$ and $$a=4$$

Show $$a > 1$$

Part of the challenge for newcomers like me is that "easy" tasks actually make it harder to think about the real reasons something is true.

Question: What is the mathematically correct reason we can conclude $a>1$?

Discussion

I am self-teaching so forgive my ignorance if this is very easy.

The five common Peano axioms don't define order as such (is this correct?) They define successor, which we chose to name/label with "numbers".

However, we define order when we define the integers. We say that $a$ is greater than $b$ if $a$ is a successor of $a$, or recursively, if $a$ is greater than another integer $b'$ which is greater than $b$. (Is this correct?)

So the actual reason we can conclude $a>1$ is by the ordering we have defined on the integers, of which the naturals are a subset. (Is this correct?)

Penelope
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  • well i guess the question is how do you define $4$? do you mean it is the successor function applied $3$ times to $1$? then this should follow from the idea that a successor is greater than the original number and by the transitivity of the $>$ operator. – AnCar Jun 16 '24 at 12:35
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    I think most mathematicians would struggle to tell you how to prove that $4 > 1$, because most of them just don't think about it, its too basic of a question, this is something that should be obvious. But it isn't, we are faced with a challenging task of choosing our foundations, what do we already know and what do we need to deduce. – Jakobian Jun 16 '24 at 13:10
  • $4=s(s(s(s(0))))=s(s(s(0)))+s(0)$ using definition of $4$ and axioms for +. And thus $\exists n (4=s(n)+1)$. – Mauro ALLEGRANZA Jun 16 '24 at 14:02
  • I think this https://math.stackexchange.com/questions/3432423/in-peano-arithmetic-can-we-define-inequality-using-successor answers your question. Actually, more than answers your question. I think one of the answers suggests $1<4$ doesn’t make sense, if you are really strict. – Steen82 Jun 16 '24 at 15:46

2 Answers2

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It suffices to clarify the definition of $ \le $ over natural number $ \mathbb{N} $. One way is as follows:

Definition. Let $ m, n \in \mathbb{N} $, $ m $ is said to be greater than or equal to $ n $, denoted by $ m \ge n $, if there is some $ a \in \mathbb{N} $, such that $$ m = n + a. $$

I think that you will be able to show that $ 4 > 1 $ by yourself with this definition, as long as you know how to define the number $ 4 $ and how to define $ + $. For example, notice that $$ 4 = 3++ = 3 + 1, $$ and $ 3 \in \mathbb{N} $, thus $ 4 \ge 3 $. But evidently $ 4 \ne 3 $, because for all $ a \in \mathbb{N} $, one has $ n \ne n++ $. Hence we complete the proof.

Long-Ping Li
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The $>$ is not part of the language used in the Peano Axioms, so if we want to prove this statement, we'll need to add a definitional axiom. For example, we can use:

$\forall x \forall x (x > y \leftrightarrow \exists z \ x = s(y + z))$

And now the proof is easy:

enter image description here

Bram28
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