# Proposition involving natural numbers [closed]

How can I prove the following proposition: A certain property $P(n)$ is true for every natural number $n$.

Thank you very much.

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## closed as not a real question by Grigory M, anon, Qiaochu YuanSep 10 '11 at 18:56

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

I found this question in Terence Tao's text Analysis(volume I).I am a beginner.But when Professor Tao asks the reader to prove it in the form as above,do you consider it 'rhetorical'?I was under the impression that a fields medalist wouldn't write trash. –  Eisen Sep 10 '11 at 19:27
There is nothing to prove unless you specify what the property $P$ is. –  Zev Chonoles Sep 10 '11 at 19:33
But then why has Terence Tao asked this of the reader?I am a novice though.In fact, he hasn't specified what it is.I now believe that someone should teach the Master the rules of his own game. –  Eisen Sep 10 '11 at 19:34
Sabyasachi, this rhetorical question is not "trash"; it has pedagogical value and naturally motivates discussion of general approaches to a broad spectrum of very typical questions arising in analysis. I believe Tao knows perfectly well what he's doing. The only issue here is that it is (arguably) overly broad for this Q&A site, and more specifically, as it was posed it suggests an actual proof is desired to a rhetorical question, as opposed to (more sensibly) a list of potential techniques for a variety of circumstances of this form (which is ultimately the point of the question). –  anon Sep 10 '11 at 19:51
HINT $\$ The extensional view of property $P$ of $\mathbb N$ is simply that subset $P\subset \mathbb N$ of all naturals satisfying the property. So the problem amounts to proving that there exists a subset of $\mathbb N$ that includes every natural, which is quite easy. –  Bill Dubuque Sep 10 '11 at 20:07

Without specifying what the property $P\,\,\,$ is, this is an extremely broad question. But some of the main strategies for proving a statement about every natural number are:
• Direct proof - explicitly demonstrate that any natural number $n$ will have property $P$.
• Induction (see here) - prove that $1$ has property $P$, and prove that for any natural number $n$, $$n\text{ has property }P\implies n+1\text{ has property }P.$$ Intuitively, this lets us say \begin{align} (\text{base case}) \qquad\qquad\qquad\qquad\qquad\qquad\qquad&1\text{ has property }P\qquad\checkmark\\ {1\text{ has property }P,\text{ and }\atop (n\text{ has property }P\implies n+1\text{ has property }P)}\bigg\}\implies&2\text{ has property }P\qquad\checkmark\\ {2\text{ has property }P,\text{ and }\atop (n\text{ has property }P\implies n+1\text{ has property }P)}\bigg\}\implies&3\text{ has property }P\qquad\checkmark\\ \vdots\end{align}
• Contradiction (see here) - assume that there were some natural number $n$ that didn't have property $P$, and show that a contradiction would follow from this assumption.