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I am doing problems in Induction. What i know is:

First we need to test Base case, that is $P(n_0)$ should be verified.

Second we need to assume that $P(k)$ is True which is the inductive Hypothesis

Third we need to prove $P(k+1)$ is True using the above Hypothesis which gives the conclusion.

But the below problem made me to fall in a confusion.

When we assume $P(k)$ is True, does it mean all of $P(1),P(2),...P(k-1)$ are also True?

Here is the problem:

Prove using induction that $10^n-(5+\sqrt{17})^n-(5-\sqrt{17})^n$ is divisible by $2^{n+1}, \:\forall n \in N$

My try:

Obviously $P(1)$ is True.

I assumed that $P(k)$ is True: So we have:

$$10^{k}-(5+\sqrt{17})^{k}-(5-\sqrt{17})^{k}=2^{k+1} Q \to (1)$$ for some $Q \in N$

Now let $$P(k+1)=10^{k+1}-(5+\sqrt{17})^{k+1}-(5-\sqrt{17})^{k+1}$$ Using $(1)$ we get

$$\begin{array}{l} P(k+1)=10\left(10^{k}\right)-(5+\sqrt{17})^{k+1}-(5-\sqrt{17})^{k+1} \\ =10\left(2^{k+1} Q+(5+\sqrt{17})^{k}+(5-\sqrt{17})^{k}\right)-(5+\sqrt{17})^{k+1}-(5-\sqrt{17})^{k+1} \\ =5\left(2^{k+2}\right) Q+(5+\sqrt{17})^{k}(5-\sqrt{17})+(5-\sqrt{17})^{k}(5+\sqrt{17})\\ =5(2^{k+2})Q+(5+\sqrt{17})^{k-1}(8)+(5-\sqrt{17})^{k-1}(8)\\ =5(2^{k+2})Q+8P(k-1) \end{array}$$

Now according to my book, the author has taken $P(k-1)$ as True and proved the result. So my question is: When we assume $P(k)$ is True, does it mean all of $P(1),P(2),...P(k-1)$ are also True?

Umesh shankar
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  • Yep. That's called "strong induction" which is the same as "regular" induction. They are logically equivalent. – morrowmh Mar 06 '21 at 03:37
  • Instead of assuming only that $P(k)$ is true, you may assume that $P(1),P(2),\dots,P(k)$ are true. It doesn't matter. – saulspatz Mar 06 '21 at 03:40

1 Answers1

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It is not true that $P(k)$ being true implies $P(i)$ is true for $i<k$. A usual induction proof does the following after the base case: $$P(k) \implies P(k+1)$$

But, the technique the author has used is often referred to as strong induction. It works according to the following: $$P(i) \space (\forall \space i \leqslant k) \implies P(k+1)$$ It can be used when you need more than just the previous case to solve a problem. The reason why it works is similar to the working of induction.

Essentially, you assume that all the previous cases are true, and use some (or all) of them to show that the next case is true. Now, the case you have proved becomes part of the list of initial cases you have proved and so on (similar to the domino effect of induction).

Haran
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  • Is it possible to prove the above problem using weak induction? – Umesh shankar Mar 06 '21 at 03:43
  • I doubt that you can use $P(k)$ alone, it looks like when you use $P(k)$, you naturally have to use $P(k-1)$. Strong induction provides an elegant way of doing the same. Although, I might be wrong. Anyways, strong induction is just another type of induction (just stronger, as the name suggests :)), so it is pretty useful to use techniques like this as they apply to many other problems. – Haran Mar 06 '21 at 03:50
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    Every problem provable by strong induction can be proven with regular induction! If you have a predicate $P(n)$ which is provable by strong induction, you make a new predicate:$$Q(n) = P(1) \land P(2) \land \ldots \land P(n).$$Then $Q(1)$ is true if and only if $P(1)$ is true, and $Q(n) \implies Q(n+1)$ is true if and only if $P(1) \land P(2) \land \ldots \land P(n) \implies P(n+1)$, i.e. the strong induction step. – Theo Bendit Mar 06 '21 at 04:03
  • @TheoBendit Yes, strong induction is a corollary of regular induction. However, I think OP was referring to weak induction specifically on $P(n)$ being the statement $$2^{n+1} \mid (10^n-(5 + \sqrt{17})^n-(5 - \sqrt{17})^n)$$ – Haran Mar 06 '21 at 06:00