Which is smaller? which is larger?

Today let us look at the following two tricky questions:

Question 1: Among these two numbers 4444 x 55555 and 5555 x 44444, which one is smaller? which one is larger?


Question 2: How about  5000 x 5000 and 4999 x 5001, which one is smaller? which one is larger?

Pythagorean Identity

This year we will start a blog series about mathematical identities. Let us begin today with the Pythagorean Identity $$ a^2 + b^2 = c^2 $$

Happy New Year!!!

Fibonacci sequence and Pascal triangle

In previous post, we used the result of a tile matching puzzle to prove an identity for Fibonacci sequence. Today, we will continue on and prove another identity. We will show that $${2011 \choose 0} + {2010 \choose 1} + {2009 \choose 2}+ {2008 \choose 3}+ \dots + {1007 \choose 1004}+ {1006 \choose 1005} = F_{2012},$$ $${2012 \choose 0} + {2011 \choose 1} + {2010 \choose 2}+ {2009 \choose 3}+ \dots + {1007 \choose 1005}+ {1006 \choose 1006} = F_{2013}.$$
In general, we have the following identity $$\sum_{v+u=n}{v \choose u} = F_{n+1}.$$ Through this identity, we will see an interesting connection between the Fibonacci sequence and the Pascal triangle.

Fibonacci identities

In previous post, we have learnt about Fibonacci sequence and a tile matching puzzle. Today, we will use the result of the tile matching puzzle to prove the following identity on Fibonacci sequence $$F_{n+m+2} = F_{n+1} F_{m+1} + F_{n+1} F_m + F_n F_{m+1}.$$

Like an art painting, a mathematical problem reflects different beauties if we view it at different angles. Take, for example, the Fibonacci identity that we are considering, we can prove this identity by different methods. One method is to use the formula $$F_n = \frac{1}{\sqrt{5}} \left[ \left( \frac{1 + \sqrt{5}}{2} \right)^n - \left( \frac{1 - \sqrt{5}}{2} \right)^n \right]$$ to calculate the two sides of the identity and make them equal by straightforward algebraic manipulations. Another method is to make use of the tile matching puzzle that we learned in the previous post. I hope that you will find this "combinatorial" method more interesting.

Fibonacci numbers and a tile matching puzzle

Today we will learn about Fibonacci sequence and a tile matching puzzle. We will soon see that the tile matching puzzle seems to be unrelated to the Fibonacci sequence, but it turns out that the solution to the puzzle is, in fact, the Fibonacci sequence!

Let us first introduce the Fibonacci sequence. The Fibonacci sequence $\{F_n\}$ is determined by the following rule: $$F_0 = 0, ~F_1 = 1, ~F_{n+1} = F_n + F_{n-1}.$$

Thus, $$F_0 = 0, ~F_1 = 1, ~F_2 = 1, ~F_3 = 2, ~F_4 = 3, ~F_5 = 5, ~F_6 = 8, ~F_7 = 13, ~F_8 = 21, \dots$$

de Moivre's formula

In previous post we have learned about complex numbers. Today, we will learn about the trigonometric form of a complex number and the famous de Moivre's formula.

Complex number

Today we will learn about the complex numbers. The crucial point about complex numbers is that we accept a very special number that we will denote it by $i$. This number $i$ is very special because it satisfies the following identity $$i^2 = -1.$$

So a complex number will have the form $$a + ib$$ where $a$ and $b$ are two real numbers. When $b=0$ then $a + ib = a$ is just a normal real number, and when $a=0$ then $a + ib = ib$ is called a pure imaginary number. Here are some examples of complex numbers: $$1+ i, ~~ 2 - 3i, ~~ -\sqrt{3} + 4i, ~~5i - 4, ~~6, ~~i, ~~-3i, ~~4 + 2i, \dots$$

Power sum and Wolstenholme's theorem

In the last post, we have learned how to derive formulas for the power summation $$S_k(n) = 1^k + 2^k + 3^k + \dots + n^k.$$

Today we will investigate some divisibility properties of the power summation $S_k(n)$. We will prove that if $p$ is a prime number and $k$ is not divisible by $p-1$ then $$S_k(p-1) = 1^k + 2^k + 3^k + \dots + (p-1)^k = 0 \pmod{p}.$$

We will also investigate the following summation $$S_{-k}(n) = \frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + \dots + \frac{1}{n^k}.$$

There is a theorem in number theory concerning this summation, it is called Wolstenholme's theorem.

Wolstenholme's Theorem. If $p$ is a prime number $>3$ then $$S_{-1}(p-1) = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{p-1} ~=_{Q} ~0 \pmod{p^2} $$ and $$S_{-2}(p-1) = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \dots + \frac{1}{(p-1)^2} ~=_{Q} ~0 \pmod{p}.$$

We will prove a general result, that is, if $p$ is a prime number and $k$ is not divisible by $p-1$ then $$S_{-k}(p-1) = \frac{1}{1^k} + \frac{1}{2^k} + \frac{1}{3^k} + \dots + \frac{1}{(p-1)^k} ~=_{Q} ~0 \pmod{p}.$$

Power sum

Today we will learn about how to derive the formula for power summation $$S_k(n) = 1^k + 2^k + 3^k + \dots + n^k .$$


Our main tool is the following binomial identity
$$(x+y)^k = x^k + {k \choose 1} x^{k-1} y + {k \choose 2} x^{k-2} y^2 + {k \choose 3} x^{k-3} y^3 + \dots + {k \choose {k-1}} x y^{k-1} + y^k.$$

Binomial identity

As we are learning about mathematical induction, in this post, we are going to use induction to prove the following:

• formula for the Pascal's triangle $$p_{n,k} = {n \choose k} = \frac{n!}{k! (n-k)!}$$
• the binomial identity $$(x+y)^n = x^n + {n \choose 1} x^{n-1} y + {n \choose 2} x^{n-2} y^2 + \dots + {n \choose {n-2}} x^{2} y^{n-2} + {n \choose {n-1}} x y^{n-1} + y^n$$

Mathematical induction II

Today we will use induction to solve some more problems. 

Problem 4. Prove that $$1 \times 2 \times 3 + 2 \times 3 \times 4 + \dots + n (n+1)(n+2) = \frac{1}{4} n(n+1)(n+2)(n+3).$$

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

What does a 4-dimensional space look like?

We often hear people in physics and mathematics talking about 4-dimensional, 5-dimensional spaces, etc... so what are those spaces? how do we picture these high dimensional spaces?