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Let $P(n)$ be any collection of $n$ coins that can be obtained using a combination of $3$ cent and $5$ cent coins. Use strong mathematical induction to prove that $P(n)$ is true for all integers $n \ge 14$.

Basis: $P(14), P(15), P(16)$ can be made up of $5$ cents and $3$ cents.

Inductive Step: $P(14)$ through $P(k)$ are true

$P(k + 1): k + 1 = (k + 1 – 3) + 3$ where if $k \ge 16$, then $(k + 1 – 3) \ge 14$ so $k + 1 \ge 17$, so $P(k + 1)$ is true by hypothesis.

If this makes sense, why do we need to show the basis steps explicitly? Aren't we assuming them to be true in the inductive step?

user353270
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  • Not sure of the phrasing. Are you saying "show that we can obtain any desired value $≥14$ using some combination of $3$ and $5$ cent coins"? But Ii's clear you can get $8,9,10$, say, and any integer $≥11$ can be written as one of those plus a multiple of $3$. – lulu Jul 10 '16 at 19:56
  • To stress: I can't work out what your $P(n)$ might be. You say it is 'any collection of $n$ coins' but then you ask if it is true or not. How can a collection of coins be true or false? – lulu Jul 10 '16 at 19:59
  • @lulu, I think my book is more articulate than me: problem statement which is referring to the problem. Maybe I should rewrite the the proof completely. – user353270 Jul 10 '16 at 20:03
  • Ah, that's different. It turns out that I answered the problem completely (the argument in my first comment shows that you can get any value $≥8$ using $3$ and $5$ cent coins). I still don't understand their definition of $P(n)$ but it is different than yours. – lulu Jul 10 '16 at 20:21
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    To stress: I would define $P(n)$ to be the statement "we can get a total value of $n$ cents using only $3$ and $5$ cent coins." Thus $P(7)$ is false, but $P(n)$ is true for $n≥8$. No idea where the $14$ comes in. Also not sure if this is what they intended to write. – lulu Jul 10 '16 at 20:25
  • Here's the given hint – user353270 Jul 10 '16 at 20:50
  • Like I said...you can get $8,9,10$ and any higher number is equal to one of those plus a multiple of $3$. That's all you need. I have absolutely no idea why they are talking about $14$, seems arbitrary and irrelevant. – lulu Jul 10 '16 at 20:55
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    "Aren't we assumingg the basis steps in the induction?". Um, yes. We are assuming. Things aren't true just because we assume them. Consider this. All odd numbers are divisible by 2. Proof: assume True for odd 2k+1. 2k+1 = 2m. So 2(k+1) +1 = 2k +2 +1=2k+1+2 =2m +2 =2 (m+1). So is true for k+1. So by induction all odd numbers are divisible by two. The induction step is valid. We assumed the basis case was valid. The only thing we didn't do was shoe the base case. But we assumed it. – fleablood Jul 10 '16 at 21:03

2 Answers2

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This type of induction works for any property $\,P(n)\,$ that is preserved under a shift, such as here where $\,P(n)\,$ true $\,\Rightarrow\,P(n\!+\!3)\,$ true. Writing $\,P\,$ for the subset of naturals where $\,P\,$ is true, the induction works as follows.

Theorem $\ $ Suppose $\,P\subseteq \Bbb N\,$ satisfies $\,n\in P\,\Rightarrow\, n\!+\!3\in P,\ $ for all $\,n\ge a.\ $ Then

$$\,a,a\!+\!1,a\!+\!\color{#c00}2\in P\,\Rightarrow\,n\in P{\rm\, \ for\ all\,\ } n\ge a$$

Proof $\ $ If not there is a least counterexample $\,\ell\not\in P.\,$ Note $\,\ell \ge a\!+\!\color{#c00}3\,$ so $\,\ell\!-\!3\ge a,\,$ Therefore, by our shift-closure hypothesis, $\,\ell = (\ell\! -\! 3)+ 3\in P,\,$ contradiction.

Remark $\ $ Clearly the proof generalizes from shift increment $\,k=3\,$ to arbitrary $\,k\ge 1,\,$ with $\,k\,$ consecutive integers $\,a,a\!+\!1,\ldots,a\!+\!k\!-\!1\,$ serving as the base cases, i.e. the foundation of the induction. Notice that the case $\,k=1\,$ is simply ordinary induction.

Bill Dubuque
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  • See also here for the ubiquitous special case of modular case analysis, when the base cases are a (consecutive) complete set of residues (remainders) $!\bmod k\ \ $ – Bill Dubuque Oct 13 '23 at 23:09
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Yes, you are assuming them to be true in the later step. (But you can assume anything you want.)

Precisely to make the argument valid you need to establish (or to prove) the base cases somewhere.

How do you know that $P(17)$ is true? You know it because you can reduce it to the truth of $P(14)$ (this is what is done in the induction step), and the truth of this was established explicitly.

If you do not do this, then you can only say $P(17)$ is true if $P(14)$ is true. (This is the induction step.) But you need to show the latter somewhere.

Of course, and this is the charm of induction, you can do longer chains.

Why is $P(20)$ true? It is true if $P(17)$ is true. And $P(17)$ is true if $P(14)$ is true. And $P(14)$ is in fact true. So $P(20)$ is true. Yet, if you do not know the truth of $P(14)$ then you cannot conclude anything.

To look at it differently. You could just as well do the argument for the induction step with $k \ge 3$ and assuming $P(1), P(2), P(3)$ true. But you cannot get $4$ with $3$ and $5$ coins! Why? because you cannot get $1$ either. But under the (false) assumption you can get $1$, you could also get $4$.

So you need to check that your base assumption are not false. This is what the base step is for.

quid
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    Stupendously bad answer. Had to downvote. Look at Bill Dubuques answer if you want to know how a good answer looks. – TROLLHUNTER Jul 14 '16 at 15:36