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The problem I am stuck on is this one: If $n>1$ is a natural number then $n-1$ is also a natural number. I am told to use induction.

Normally I would just do the following:

My statement is $P(n)=n-1$ such that $n-1$ is a natural number.

Base case. Since $n>1$, I use the base case of $n=2$. Thus $P(2)=1$ and this is a natural number. Now I choose some $k \in \mathbb{N}$ such that $k > 2$. So $P(k)=k-1 \in \mathbb{N}$.

Past this point I get speculative.

Now for my inductive step. $P(k+1)=(k+1)-1=k$. I don't know if this is allowed, but I noticed that by rearranging $P(k+1)=(k-1)+1 \Rightarrow P(k+1)=P(k)+1$.

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Dont you mean to say that $P(n)$ is the statement that $n-1$ is a natural number? – fretty Sep 5 '12 at 12:04
Yes, that is correct. I have added that. – emka Sep 5 '12 at 12:13
No you are still a bit confused about it. $P(n)$ has to be a statement and you prove the statement to be true for all $n$ using induction. Your statement $P(n)$ should be something like "the number $n-1$ is a natural number". You then proceed to prove that P(2) is true, i.e. "the number $1$ is a natural number" and that for $k>2$, $P(k+1)$ is true given that $P(k)$ is true, i.e. "the number $k+1$ is a natural number" given that "the number $k$ is a natural number". – fretty Sep 5 '12 at 12:43
up vote 1 down vote accepted

We want to show that if the result is true for $n=k$, then it is true for $n=k+1$. Of course it is true for $n=k+1$, we do not even have to use the induction hypothesis.

To put the point in logical form, we want to show that $A(k)\implies A(k+1)$. If we can prove $A(k+1)$, then we have shown that $A(k)\implies A(k+1)$.

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With regards to this problem, I'm still unsure how to proceed with induction. – emka Sep 5 '12 at 12:14
In my answer above, I gave an induction proof (apart from not verifying the base case, since you had done that). – André Nicolas Sep 5 '12 at 12:24
I guess I'm being tortured because the previous problem in this set, which involved proving the identity $1^2+2^2+...n^2=\frac{n(n+1)(2n+1)}{6}$ involved some computation. – emka Sep 5 '12 at 12:29
@EMKA: Yes, we are dealing here with a trivial case of induction, and sometimes when a problem is too easy, it can cause trouble. Induction basically says that if we start at $1$ and keep taking unit steps forward, we travel through all the positive integers. – André Nicolas Sep 5 '12 at 12:34

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