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I need to prove 0<1, I'm not able to give any of what I tried because I'm not sure how to prove it, maybe starting by some axioms I would be able to prove it.

Steve
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1 Answers1

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It depends which axioms you are starting from. One extremely basic definition of an order on ${\mathbb{N}}$ is $$ a\leq b\Leftrightarrow \exists\ c \in \mathbb{N}\ |\ a + c = b $$ we say ${a<b}$ iff ${c\neq 0}$.

From the definition of ${0,1}$ and addition, we have ${0 + 1 = 0 + S(0) = S(0 + 0)}$, but ${0 + 0 = 0}$ by definition, so this is ${S(0)=1}$. So ${0+1=1}$. From the definition of order, this proves ${0\leq 1}$. To show strict inequality, i.e. ${0<1}$, note ${1\neq 0}$ since if it did, ${S(0)=0}$ which directly contradicts one of the Peano axioms.

This is a very specific proof to a very specific set of axioms, though. There are slight variations all over the place. You should really specify what axioms you want to start from.