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During my studies, I came across with questions that I need to solve with induction and WOP (Well Ordering Principle).

I solved it with induction but I find it hard to solve it using Well Ordering Principle. This is how I solved it with induction: $\space$

Let's check on $k=1$:

$3|4^k +5 \implies 3|4^1 +5\implies 3|9$

assuming it is true for any $k$. Now, check for $k+1$:

$3|4^{k+1} +5\iff 3|4*4^k +5\iff 3|4*\underbrace{(4^k +5)}_\text{$3|4^k +5$} -\underbrace{15}_\text{$3|15$}$

Therfore, the expression is true.

Now, how should I prove it with WOP?

Bill Dubuque
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John
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  • "I solved it with induction but I find it hard to solve it using Well Ordering Principle." They are the same. Or, you need the WOP to justify why induction works. Or something along those lines. My point is, it's really difficult to use one and not the other. – Arthur Apr 16 '20 at 11:10
  • I've been asked to do it with WOP only. I did it with induction just to see if I know how to do it that way... How are they the same? I don't know how to solve it that way. – John Apr 16 '20 at 11:22
  • It's just a contrapositive form of induction - infinite descent on counterexamples (or existence of a "minimal criminal"), e.g. see here and its links. – Bill Dubuque Apr 16 '20 at 15:08

2 Answers2

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You could argue as follows: suppose $\;3\mid(4^n+5)\;$ isn't true for all natural numbers, and let $\;K:=\left\{\,k\in\Bbb N\;|\; 3\nmid(4^k+5)\,\right\}\;$ . As $\;K\neq\emptyset\;$ by assumption, the WOP tells us it has a first element, say $\;k_0\;$. But then $\;3\mid(4^{k_0-1}+5)\;$ , and then

$$4^{k_0}+5=4\cdot4^{k_0-1}+5=(4^{k_0-1}+5)+3\cdot4^{k_0-1}$$

Buth the first term in the rightmost expressios is divisible by $\;3\;$ by the minimality of $\;k_0\;$ , whereas the second term there is trivially divisible of $\;3\;$ , and thus the whole expression is divisible by $\;3\;$ , which provides a contradiction to $\;K\neq\emptyset\;$ .

As you can see, the above is almost exactly the same as a proof by mathematical induction...which isn't a surprise as they both are equivalent.

DonAntonio
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  • The difference is basically whether you phrase the induction proof directly, or as an argument by contradiction. At least as far as I can see. – Arthur Apr 16 '20 at 11:43
  • @Arthur Yes, I agree. It's more a matter of semantics, and in this case WOP goes, as usual, with contradiction. – DonAntonio Apr 16 '20 at 11:50
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Assume that it is not true for every positive integer $n$.

Then according to WOP a positive integer $m$ exists such that $3\nmid4^m+5$ and $3\mid4^n+5$ for every positive integer $n$ that satisfies $n<m$.

Evidently $m>1$ so $m=k+1$ where $k$ is a positive integer that satisfies $k<m$.

Then $4^k+5=3l$ for some integer $l$.

Then $4^m+5=4^{k+1}+5=4(4^k+5)-15=3(4l-5)$ contradicting that $3\nmid4^m+5$.

So a contradiction is found and we conclude that the assumption is wrong.

drhab
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