1

For example, in proving the following lemma

If $n\in\mathbb{N}$ then $n+1=1+n.$

Little gave in his book entitled THE NUMBER SYSTEMS OF ANALYSIS the following proof: Clearly the lemma holds if $n=1.$ Assume as an inductive hypothesis that $n+1=1+n$ for some $n\in\mathbb{N}.$ Then \begin{align*} (n+1)+1&=(1+n)+1\\ &=1+(n+1) \end{align*} by associativity. Thus the lemma follows by induction. $\Box$

My question is: Is it true that the italicized "some" equals "any"? That is, the sentence
\begin{gather*} \text{Assume as an inductive hypothesis that $n+1=1+n$ for some $n\in\mathbb{N}$} \end{gather*} says the same thing as the sentence \begin{gather*} \text{Assume as an inductive hypothesis that $n+1=1+n$ for any $n\in\mathbb{N}$?} \end{gather*} Many thanks.

azc
  • 1,259
  • 1
    No "some $n$" in this context means "a particular $n$." In this context, "any $n$" would mean the same thing as "every $n$" and once you assume that, there's nothing left to prove! – saulspatz Aug 27 '18 at 04:26
  • We know it is true for $n=1$ since we checked before hand in our base step. So, we know that there is at least one $n$ for which it works. Now... we say "suppose it is true for some $n$", in this case initially meaning just the case where $n=1$, but eventually meaning any particular values for which we know it to be true for. We go on to show that it follows that it must be true for $n+1$ as well. That is, originally we know it is true for $1$ since we checked that, this and the induction step implies it must also be true for $n=2$, which in turn implies its true for $n=3$ and so on... – JMoravitz Aug 27 '18 at 04:47

1 Answers1

1

No. If you use the second statement, then you implied that the proposition is right, which is the thing that you are proving. This is cycle reasoning.

What i am saying is, the 2nd statement is just the proposition above. You cannot assume that.

xbh
  • 8,835