1. ## Prove by induction

1. Prove by induction:
(a) $1 \le n\ \forall n \in N;$
(b) $\forall n \in N,$ either $n = 1$ or $n-1 \in N$; and
(c) $\forall n \in N,$ there is no element $m$ of $N$ in the range $n < m < n+1.$

2. Prove by induction on $n$ that $\forall m, n \in N\ \exists q \in N$ such that $qm > n.$
Also prove that for every $n \in N$, there is an $m \in N$ such that $2^m > n$.

2. Originally Posted by mathwizard
1. Prove by induction:
(a) $1 \le n\ \forall n \in N;$
(b) $\forall n \in N,$ either $n = 1$ or $n-1 \in N$; and
(c) $\forall n \in N,$ there is no element $m$ of $N$ in the range $n < m < n+1.$

2. Prove by induction on $n$ that $\forall m, n \in N\ \exists q \in N$ such that $qm > n.$
Also prove that for every $n \in N$, there is an $m \in N$ such that $2^m > n$.
You can prove some of these without induction, if that is okay with you.
Like Problem 1(c), you can do it by definition of what it means being a natural number.

3. Originally Posted by ThePerfectHacker
You can prove some of these without induction, if that is okay with you.
Like Problem 1(c), you can do it by definition of what it means being a natural number.
No. The question specifies to prove the statements by induction. I think the proof relies on the four axioms that a field $F$ must satisfy. This is from my advanced calculus textbook (first chapter & first section, btw).