Are you an EPFL student looking for a semester project?
Work with us on data science and visualisation projects, and deploy your project as an app on top of Graph Search.
Concept
Proof by exhaustion
Summary
Proof by exhaustion, also known as proof by cases, proof by case analysis, complete induction or the brute force method, is a method of mathematical proof in which the statement to be proved is split into a finite number of cases or sets of equivalent cases, and where each type of case is checked to see if the proposition in question holds. This is a method of direct proof. A proof by exhaustion typically contains two stages:
A proof that the set of cases is exhaustive; i.e., that each instance of the statement to be proved matches the conditions of (at least) one of the cases.
A proof of each of the cases.
The prevalence of digital computers has greatly increased the convenience of using the method of exhaustion (e.g., the first computer-assisted proof of four color theorem in 1976), though such approaches can also be challenged on the basis of mathematical elegance. Expert systems can be used to arrive at answers to many of the questions posed to them. In theory, the proof by exhaustion method can be used whenever the number of cases is finite. However, because most mathematical sets are infinite, this method is rarely used to derive general mathematical results.
In the Curry–Howard isomorphism, proof by exhaustion and case analysis are related to ML-style pattern matching.
Proof by exhaustion can be used to prove that if an integer is a perfect cube, then it must be either a multiple of 9, 1 more than a multiple of 9, or 1 less than a multiple of 9.
Proof:
Each perfect cube is the cube of some integer n, where n is either a multiple of 3, 1 more than a multiple of 3, or 1 less than a multiple of 3. So these three cases are exhaustive:
Case 1: If n = 3p, then n3 = 27p3, which is a multiple of 9.
Case 2: If n = 3p + 1, then n3 = 27p3 + 27p2 + 9p + 1, which is 1 more than a multiple of 9. For instance, if n = 4 then n3 = 64 = 9×7 + 1.
Case 3: If n = 3p − 1, then n3 = 27p3 − 27p2 + 9p − 1, which is 1 less than a multiple of 9. For instance, if n = 5 then n3 = 125 = 9×14 − 1. Q.E.D.
Official source
About this result
This page is automatically generated and may contain information that is not correct, complete, up-to-date, or relevant to your search query. The same applies to every other page on this website. Please make sure to verify the information with EPFL's official sources.
Related courses (32)
CS-101: Advanced information, computation, communication I
Discrete mathematics is a discipline with applications to almost all areas of study. It provides a set of indispensable tools to computer science in particular. This course reviews (familiar) topics a
COM-406: Foundations of Data Science
We discuss a set of topics that are important for the understanding of modern data science but that are typically not taught in an introductory ML course. In particular we discuss fundamental ideas an
MATH-432: Probability theory
The course is based on Durrett's text book
Probability: Theory and Examples.
It takes the measure theory approach to probability theory, wherein expectations are simply abstract integrals.
It takes the measure theory approach to probability theory, wherein expectations are simply abstract integrals.