I was reading about proof by infinite descent, and proof by minimal counterexample. My understanding of it is that we assume the existance of some smallest counterexample $A$ that disproves some proposition $P$, then go onto show that there is some smaller counterexample to this which to me seems like a mix of infinite descent and 'reverse proof by contradiction'.
My question is, how do we know that there might be some counterexample? Furthermore, are there any examples of this?
Consider, for instance, the statment
Every $n\in\mathbb{N}\setminus\{1\}$ can be written as a product of prime numbers (including the case in which there's a single prime number appearing only once).
Suppose otherwise. Then there would be a smallest $n\in\mathbb{N}\setminus\{1\}$ that would not be possible to express as a product of prime numbers. In particular, this implies that $n$ cannot be a prime number. Since $n$ is also different from $1$, it can be written as $a\times b$, where $a,b\in\{2,3,\ldots,n-1\}$. Since $n$ is the smallest counterexample, neither $a$ nor $b$ are counterexamples and therefore both of them can be written as a product of prime numbers. But then $n(=a\times b)$ can be written in such a way too.
Such a proof will often go as follows.
Assume the $\sqrt{2}$ is rational. Then there are whole numbers $a$ and $b$ such that $\sqrt{2}=a/b$ and $a$ is the smallest number with this property but then we find that $a/2, b/2$ are also whole numbers with this property. So there is a smaller example.
I think this must be the example people are the most familiar with and it's a kind of infinite descent even though we often don't refer to it as such.
A fundamental example in number theory is descent by (Euclidean) division with remainder (or, equivalently, by repeated subtraction), as in the following basic result.
Lemma $\ \ $ Let $\,S\,$ be a nonempty set of positive integers that is closed under subtraction $> 0,\,$ i.e. for all $ \,n,m\in S, \,$ $ \ n > m\ \Rightarrow\ n-m\, \in\, S.\,$ Then the least $ \:\ell\in S\,$ divides every element of $\, S.$
Proof ${\bf\ 1}\,\ $ If not there is a least nonmultiple $ \,n\in S,\,$ contra $ \,n-\ell \in S\,$ is a nonmultiple of $ \,\ell.$
Proof ${\bf\ 2}\, \,\ \ S\,$ closed under subtraction $ \,\Rightarrow\,S\,$ closed under remainder (mod), when it is $\ne 0,$ since mod is simply repeated subtraction, i.e. $ \ a\bmod\ b\, =\, a - k b\, =\, a\!-\!b\!-\!b\!-\cdots\! -\!b.\,$ Therefore $ \,n\in S\,$ $\Rightarrow$ $ \, (n\bmod \ell) = 0,\,$ else it is in $\, S\,$ and smaller than $ \,\ell,\,$ contra minimality of $ \,\ell.$
Remark $\ $ In a nutshell, two applications of induction yield the following inferences
$\begin{eqnarray}\rm S\ closed\ under\ {\bf subtraction} &\:\Rightarrow\:&\rm S\ closed\ under\ {\bf mod} = remainder = repeated\ subtraction \\ &\:\Rightarrow\:&\rm S\ closed\ under\ {\bf gcd} = repeated\ mod\ (Euclid's\ algorithm) \end{eqnarray}$
This yields Bezout's GCD identity: the set $ \,S\,$ of integers of form $ \,a_1\,x_1 + \cdots + a_n x_n,\ x_i\in \mathbb Z,\,$ is closed under subtraction so Lemma $\Rightarrow$ every positive $ \,k\in S\,$ is divisible by $ \,d = $ least positive $ \in S.\,$ Therefore $ \,a_i\in S$ $\,\Rightarrow\,$ $ d\mid a_i,\,$ i.e. $ \,d\,$ is a common divisor of all $ \,a_i,\,$ necessarily the greatest such because $ \ c\mid a_i$ $\Rightarrow$ $ \,c\mid d = a_!\,x_1\!+\!\cdots\!+\!a_nx_n$ $\Rightarrow$ $ \,c\le d.\,$ When interpreted constructively, this yields the extended Euclidean algorithm for the gcd.
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