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I kinda understand the logic and motivation behind the proof, but what bothers me is the fact, why is the base case (the first statement that we write) is always $P(1)$ when we are proving a proposition is true for all $\mathbb{N} $? Why can't it be anything other than $1$? What about $P(5)$ or something? Does it change anything?

William
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  • If you take $P(5)$, then the conclusion will be proved for integers $n\ge 5$ only. That may or may not be good enough for your purposes. – Angina Seng Aug 09 '18 at 11:18
  • @LordSharktheUnknown Yes, my teacher did say something like that, can you please elaborate on that? I don't understand why is that the case? Thank you. – William Aug 09 '18 at 11:20
  • Is not the paradox that all horses have the same color the striking example, that $P(1)$ does not always work? – dEmigOd Aug 09 '18 at 11:22

3 Answers3

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A typical induction hypothesis (not always) is saying that if $k$-th statement is true, then $(k+1)$-th statement is true. Imagine that you have some dominoes, the $(k+1)$-th dominoes fall because $k$-th domino hits it. Now, if you started proven that the first is fallen, now you can use the induction hypothesis to conclude that the $2$nd domino falls as well and so on. Then your conclusion is valid for $n \ge 1$.

If you only started pushing from the $5$-th dominos, the first $4$ dominos could still be standing. Your conclusion can only be made for $n \ge 5$.

Also, try proving $n! \ge 2^n, \forall n \ge 4$ by induction, it is likely that you will use a different base case.

Siong Thye Goh
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  • If my base case uses a number, say $ 7$. Does that mean $k≥7, , \forall , k \in \mathbb{N} $ if so, why can't $k$ represent a number less than $7$? Sorry for the trouble. – William Aug 09 '18 at 17:19
  • Suppose your induction hypothesis is $P(k) \implies P(k+1)$. Imagine you have a rows of dominoes that are labeled from $1$ onwards, you start pushing the dominoes from the $7$-th domino, it hit the $8$, which in turn hit the $9$ and so on, the first $6$ could still be standing right (they were untouched). It's like I learn mathematical induction today, so from today onwards, you know what is mathematical induction, but what about yesterday? – Siong Thye Goh Aug 10 '18 at 00:43
  • Thank you, I finally understand it :) – William Aug 10 '18 at 09:29
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The short answer is that we don't always start with $P(1).$ It depends on what we're trying to prove. Sometimes we'll have multiple base cases, or start with a greater/lesser base case.

The idea of mathematical induction is more that we build up on the base case(s) with the induction step(s), thereby handling infinitely-many cases without having to tackle them one by one.

Added: For an example where we start with a larger base case, see here. For an example where we want multiple base cases, see here. Here is an example of a proof with multiple induction steps. Here's an outline of a proof that induces in both directions from the base case $P(0),$ proving a claim for all integers.

Cameron Buie
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We need to construct proof for a statement $P(n)$ for all $n\ge 1$

As basis case, we need to show $P(1)$ is true.

Then in induction step, we prove $P(k) \implies P(k+1)$.

Hence $P(1) \implies P(2), P(2) \implies P(3) \cdots \cdots$

If you take $P(5)$ as basis case, then

$P(5) \implies P(6), P(6) \implies P(7) \cdots \cdots$

Thus in case you take $P(5)$ as basis case, You need to prove your statement explicitly for $n=1,2,3,4$

hanugm
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  • I think the OP should see some example, than $P(1)$ is just not true, while $P(2)$ for example is. – dEmigOd Aug 09 '18 at 11:27