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There is no formal Induction assertion given. But there is an assertion $P(n)$ for $n\ge2$.

(a)We know that $P(2)$ is true.

(b) we know for any $a\ge2$, truth of $P(a)$ implies truth of $P(2a)$.

(c) we know for any $b\ge3$, truth of $P(b)$ implies truth of $P(b-1)$.

We have to use induction to prove that $P(n)$ is true for all $n\ge2$.

My Solution: I use the Well-ordering property (WOP) of the natural numbers.

T = { k>= 2; p(k) is True}

F= {l>= 2, p(l) is False}

so T U F = N(all natural Numbers), and T∩F = ∅

Case 1: (F = ∅ Then T = N so P(n) is always True)

Case 2: (F ≠ ∅ Then ∅ ≠ F ⊆ N s.t there is an element m = min F (WOP))

now, m ≠ 2 because 2 ∈ T (by Hypothesis)

I cant figure out how to use p(a) and p(b) to prove the statement p(n).

General Grievance
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Harry Solomon
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  • What have you tried? Where are you stuck? How much background do you have? Please edit the question to include this information. Questions that don't demonstrate some independent effort to solve the problem usually are not well received here. Are you familiar with the well-ordering principle? If there's some $n \geq 2$ such that $P(n)$ is false, then there must be a least such $n$. Use what you know to prove that there must be a smaller such $n$, contradicting the assumption that you already chose the least such $n$. – Robert Shore Nov 05 '20 at 02:50
  • What Robert Shore commented. His approach is one of descent. If you are going to explore his idea, experiment by assuming that P(10) (for example) is false and try to work backwards to obtain a contradiction. The alternative is to take the opposite approach. Try to prove, based on your premises, that P(10) is true. Then try to formalize that into a proof where P(n) is true, for any n. – user2661923 Nov 05 '20 at 03:12
  • I am sorry Mr Shore, I am new to this page, I will edit right away – Harry Solomon Nov 05 '20 at 06:40

2 Answers2

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I think it is easiest to prove using strong induction.

Induction hypothesis: Suppose $P(2), P(3), ..., P(n-1)$ are true.
From this we need to deduce that $P(n)$ is true.

If $n$ is even, then $n=2a$ for some $a<n$. By the induction hypothesis, $P(a)$ is true. From statement (b) it follows that $P(2a)=P(n)$ is true.

If $n$ is odd, then $n+1=2a$ for some $a<n$. As before it follows that $P(2a)=P(n+1)$ is true. From statement (c) with $b=n+1$ it follows that $P(b-1)=P(n)$ is true.

In either case, $P(n)$ is true. Therefore by induction $P(n)$ is true for all values of $n\ge2$.

Jaap Scherphuis
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Using (a), (b) and induction you can derive that $P(2^n)$ is true for any $n\in\mathbb{N}.$
Now suppose you want to show the validity of $P(n)$ for some random positive integer $n\ge 2.$ Notice that $2^n\gt n,$ and $P(2^n)$ is true. Then use (c) to verify the result for the (decreasing) sequence $P(2^n-1), P(2^n-2),\cdots, P(n+1), P(n).$

This method is known as (Cauchy's) forward backward induction. See here for an example.

Bumblebee
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