0

First Principle of Mathematical Induction

Let S be a set of integers containing a. Suppose S has the property that whenever some integer $n \ge a$ belongs to S, then the integer $n+1$ also belongs to S. Then, S contains every integer greater than or equal to a.

Proof: Let $S$ be an arbitrary set containing $a$. Suppose $S$ has the property that whenever some integer $n\ge a$belongs to $S$, then the integer $n+1$ also belongs to $S$. Since $n+1 \ge a$, it follows $n+2 \ge a$, and similarly for all $n+k$ where $k$ is a positive integer. Therefore, S contains every integer greater than or equal to a.

EDIT: Is this going to work?

Since $a \ge a$, by the property, $a+1 \in S$. Then, since $a+1 \ge a$, $a+2 \in S$. We can continue this as long as we want, so we can get a set of all integers greater than $a$ in the set $S$.

This makes perfect sense to me but I think it's wrong because it seems way too easy and simple for being a proof. This is my problem, how can I be sure if my proofs are correct or not? What questions should I ask to myself to make sure my proof is solid?

user3000482
  • 1,516
  • The idea is correct, but definitely it is not the way to write a formal proof of this statement. What background can you assume for the proof? Do you have the usual principle of mathematical induction? – Darío G Sep 01 '16 at 15:34
  • 1
    I think the question is ill-posed, since the validity of a proof can be checked only if the assumptions are made clear. In ZFC, the problem is not formulated in these terms (and there is a specific axiom that can arguably be interpreted as "the definition of $\Bbb N$ as the set where mathematical induction makes sense"). Typically, for an exercise of this leve, a professor is very likely to assume that every time you use the word "similarly" you are trying to mask your incapacity to properly prove the next statement in finitely many words (as it should be in a mathematical proof). –  Sep 01 '16 at 15:38

0 Answers0