Today we will use induction to solve some more problems.
Problem 4. Prove that $$1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + n (n+1)(n+2) = \frac{1}{4} n(n+1)(n+2)(n+3).$$
Math for the curious
Mathematical induction
Today, we will learn about mathematical induction. We usually use induction to prove a certain statement to be correct for all natural numbers.
Let us use $P(n)$ to denote a statement that involves a natural number $n$. To prove that $P(n)$ is correct for all natural number $n$, an induction proof will have the following steps
Step 1: is called the initial step. We will prove that the statement $P(n)$ is correct for the case $n=0$.
Step 2: is called the induction step. This is the most important step. In this step,
- we assume that for any $0 \leq n \leq k$, the statement $P(n)$ is correct;
- with this assumption, we will prove that the statement $P(n)$ is also correct for the case $n=k+1$.
With these two steps, by the mathematical induction principle, we conclude that the statement $P(n)$ must be correct for all natural number $n$.
Pascal's triangle
Today, we will look at a famous number pattern, the Pascal's number triangle.
 |
| Pascal's triangle |
Pythagorean triples
In geometry, there is a well known theorem, called the Pythagorean Theorem, which says that in a right triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.
 |
| Pythagorean Theorem: $BC^2 = AB^2 + AC^2$ |
That is why we call the equation $$x^2 + y^2 = z^2$$ the Pythagorean equation and its solution $(x,y,z)$ is called a Pythagorean triple. Of course, we only consider integer solutions.
Today, we will solve the Pythagorean equation and show that this equation has an infinite number of solutions.
Wilson's Theorem
Today, we will study Wilson's Theorem - a theorem concerning prime numbers. Wilson's theorem says that if $p$ is a prime number then the number $(p-1)! + 1$ will be divisible by $p$.
Here, the notation $n!$ denotes $$n! = 1 \times 2 \times 3 \times \dots \times n.$$
For example,
- $1! + 1 = 2$ is divisible by $2$
- $2! + 1 = 3$ is divisible by $3$
- $4! + 1 = 25$ is divisible by $5$
- $6! + 1 = 721$ is divisible by $7$
Some problems on prime numbers
In previous posts, we have learned about prime numbers, and we know from Euclid's Theorem that there exists an infinite number of primes. In this post, we continue to look at prime numbers. Leading mathematicians like Fermat, Euler, Gauss were all fascinated by prime numbers. There are many problems about primes that even the statements are simple, but they still remain unsolved even to this day.
Euclid's theorem on prime numbers
Continuing with our story about prime numbers, today we will prove that there exists an infinite number of primes. This is called the Euclid's theorem on prime numbers. This theorem has a very simple proof but it is probably one of the most beautiful proofs ever in mathematics.
Prime numbers
Today we will learn about prime numbers - a basic building block of arithmetic.
A prime number is a natural number greater than 1 and has no divisors other than 1 and itself. For example, the numbers 2, 3, 5, 7, 11, 13 are prime numbers. The number 9 is not a prime number because it is divisible by 3. The number 2012 is not a prime number because it is divisible by 2.
The Fermat's point of a triangle II
In the previous post, we have analyzed the Fermat's problem, the problem of finding a point $M$ for a given triangle $ABC$ such that $MA + MB + MC$ is the minimum.
 |
| the Fermat's problem: find $M$ so that $MA + MB + MC$ is minimum |
The Fermat's point of a triangle
In previous posts about modulo, we learn about the mathematician Fermat and his famous problem $$x^n + y^n = z^n.$$
Today, we will look at a geometry problem that bears his name. As we already know, Fermat was not a professional mathematician, but was a lawyer. He was doing math probably just for fun and most of his achievements that we know of today originated from his letters to his friends and also from his occasional writings on the margin of books that he read. The most famous is, of course, the problem $x^n + y^n = z^n$ and his note "I have found a beautiful proof but there is not enough space" that he wrote on the margin of the book by Diophantus.
The problem that we investigate today was raised in a letter that Fermat sent to an Italian mathematician, Torricelli. In his letter, Fermat challenged Torricelli to find a point such that the total distance from this point to the three vertices of a triangle is the minimum possible. Well, this problem was not hard for Torricelli. Since Torricelli knew how to find such a point, today some people refer to this point as the Fermat's point, and others refer it as the Torricelli's point of the triangle.
|
| the Fermat's problem: find a point $M$ so that $MA + MB + MC$ is minimum |
A problem about finding shortest path and a property of the ellipse
Today we will look at two problems that seem to be unrelated. The first one is a beautiful geometry problem about finding shortest path and the other one is about a property of an ellipse.
But first, let us introduce the ellipse. An ellipse is drawn below.
Solve for special cases first!
I would like to share with you a lesson that I have learnt. That is when facing a problem and we do not know what to do, the first thing we can do is to look at special cases of that problem. Investigating special cases can help us gain a greater understanding of the problem. To illustrate the point, let us solve some problems.
Subscribe to:
Posts (Atom)