yismxplusc: June 2018

Wednesday, 20 June 2018

Why is 0! equal to 1?

If you're done with your permutations/combinations lecture, you must be wondering what is 0! And even if you know that it is 1, why is it so?

This can be explained very simply using different approaches to combinations; I'll do it in three very basic ways :

Using the formula: n! = n(n-1)!
Using a pattern
The Basic Concept of Factorials.

Proof 1:

We know that n! = n(n-1)(n-2)(n-3)(n-4).....(3)(2)(1)

Hence this can be written as n!= n(n-1)!

Lets use this fact to prove that 0! = 1 :

n!= n(n-1)!

Dividing both sides with n , we get:

$\displaystyle\frac{n!}{n} = (n-1)!$

Now lets suppose that n=1:

we get:

$\displaystyle\frac{1!}{1} = (1 - 1)! \Rightarrow 1 = 0!$

Hence shown that 0! = 1

Proof 2:

Lets simply workout some factorials:

5! = (5)(4)(3)(2)(1)= 120

4!= (4)(3)(2)(1) = 24

3!= (3)(2)(1) = 6

2!= (2)(1) = 2

1!= 1

Do you notice a pattern?

Note that 4! can be written as:

$4! = \displaystyle \frac{5!}{5} = \frac{(5)(4)(3)(2)(1)}{5} = 24$

Similarly,

$3! = \displaystyle \frac{4!}{4} = \frac{(4)(3)(2)(1)}{4} = 6$

$2! = \displaystyle \frac{3!}{3} = \frac{(3)(2)(1)}{(3)} = 2$

$1! = \displaystyle \frac{2!}{2} = \frac{(2)(1)}{(2)}$

Following the same pattern, lets workout 0! :

$0! = \displaystyle \frac{1!}{1} = 1$

Hence both the explanations show that 0! = 1

Proof 3:

Lets just come back to the concept of factorials. What are factorials used for?

Factorials help mathematicians calculate the possible combinations or permutations of something.

Lets explain it with an example:

If there are 3 parking spaces available, and there are also 3 cars to park, in how many different ways can the three cars be parked? The answer is 3! = 6 ways

Remove one parking space and one car from the scenario, if there are 2 parking spaces and 2 cars, how many different ways can they be parked? The answer is 2! = 2 ways

Remove another car, and parking space, how many different ways are available to park 1 car in the only parking space? There is just one way. Hence 1! = 1 way

NOW this is where it gets interesting. How many different ways are there to park no car in no parking space? Just one way. You don't. Yes, the ONE and only possibility is that you don't. This means that 0! = 1 way

HENCE PROVED THAT 0! = 1

Friday, 15 June 2018

THE PASCAL'S TRIANGLE AND BINOMIAL THEOREM

BINOMIAL THEOREM AND THE PASCAL'S TRIANGLE

We have all learned to expand binomials with different powers using the factorial method:

Here we have to calculate the coefficients by calculating the combinations :

using the calculator or the formula given below :

Using this method, we could keep on going but eventually, the exponents would get so large that it would become difficult to simplify them.

Thus, we need a better way, and luckily a 17th-century French mathematician has already found one.

Blaise Pascal found a numerical pattern, called Pascal's Triangle, for quickly expanding a binomial.

Here's what the Pascal Triangle looks like :

The magic that it does is that the rows of this triangle show the coefficients of different powers of binomials :

For example:

The 1st row shows the coefficient of the expansion (a + b)0 = 1
The 2nd row shows the coefficients of the expansion (a + b)1 = 1a + 1b
The 3rd row shows the coefficents of the expansion (a + b)2 = 1a2 + 2ab + 1b2
The 4th row shows the coefficients of the expansion (a + b)3 = 1a3 + 3a2b + 3ab2 + 1b³

And it goes on.

The interesting thing is that you do not need to learn the Pascal's triangle. If you notice, it is a sequence and you can simply go on expanding the triangle for as many rows as you want :

The first row has 1.
For the next row you simply add the above number with its left and right number; as in the first row its zero on both sides, the second row gets 1 and 1
In the third row we write 1, then add the second row's 1 and 2 to get 3 twice, and again a 1.

This can be better understood by the diagram below:

Hence, it shows how the Pascal's triangle can be used to work out Binomial expansions.

The Pascal's triangle has many applications as well as many hidden wonders in itself (some of which have been discovered while some are yet to be explored), and they'll be discussed in detail in later articles.

Wednesday, 13 June 2018

Geometric proof of the CUBIC FORMULA (a+b)^3

CUBIC FORMULA PROOF

Just like we expressed the geometric proof of expansion of quadratic formula - we prove the expansion of (a + b)3

the expansion is: (a+b)³ = a³ + 3a²b + 3ab² + b³

PROOF:
Lets draw a cube with side length (a+b) , hence we know that the volume of this cube would be equal to (a + b)3

Note that in the above diagram, the red part itself is a cube with volume a3 and the blue part is a cube with volume b3
Further more, note that the yellow part is a cuboid with height a, width a, and length b ; thus the cuboid has volume a x a x b = a²b, and as the diagram has 3 yellow cuboids, the total volume of the yellow part would be 3a²b
Similarly, the green part is a cuboid with height b, length b, and width a ; thus a cuboid with volume ab², and as the diagram has 3 green cuboids, the total volume of the green part is 3ab²

HENCE, THE VOLUME OF THE WHOLE CUBE CAN BE DISSECTED AS FOLLOWS :

We first remove the primary a3 cube and b3 from the image as shown below:

Further evaluating the diagram allows us to remove other inscribed combination of shapes :

THEREFORE,

the volume of the cube = a³ + 3a²b + 3ab² + b³

(a+b)³ = a³ + 3a²b + 3ab² + b³

Monday, 11 June 2018

PROOF of the formula for SUM OF INTERIOR ANGLES OF A POLYGON

SUM OF INTERIOR ANGLES OF A POLYGON

A polygon is any 2-dimensional shape formed with straight lines.

Triangles, quadrilaterals, pentagons, and hexagons are all examples of polygons.

The name tells you how many sides the shape has. For example, a triangle has three sides, and a quadrilateral has four sides.

We all learnt in our math class that the sum of the interior angles of a polygon can be calculated by using the formula : (n-2) x 180 , where n is the number of sides of the polygon.

But How did this formula come to be ?
Lets learn it step by step:

We know that the sum of all the angles of a triangle is equal to 180:

Now lets draw a square(a polygon)

Choose a point on one of its sides and draw lines to join with all other sides (only one in case of a square)

Hence, two triangles are formed as shown below. Therefore, as one triangle has sum of angles of 180, the sum of angles of the square would be : 2 x 180 = 360

Now, lets draw a five sided polygon (a pentagon):

Select a point on one of its sides.

Draw lines joining the other points on the other sides

Note that three triangles are formed inside the pentagon. Hence the sum of the interior angles would be :

3 x 180= 540

Lets look at another example:

This an 8 sided polygon (an octagon)

Starting from a point and drawing lines joining it with the other points, we get 6 triangles in the polygon.

So, the sum of the angles would be:

6 x 180 = 1080

Have you noticed a relation between the polygons and the number of triangles that can be formed inside them?

Basically, the number of triangles that can be formed is 2 less than the number of sides of a polygon!

We saw that 2 triangles were formed in a 4-sided polygon i.e square, 3 triangles were formed in a 5 sided polygon i.e a pentagon, 6 triangles were formed in an 8 sided polygon i.e an octagon.

Similarly, n-2 triangles would be formed in an n sided polygon.

Therefore, the sum of the interior angles of a polygon would be:

Sum of int angles = No. of triangles x 180

Sum of int angles = (n-2) x 180

Hence proved!

Saturday, 9 June 2018

GEOMETRIC PROOF OF (a+b)^2

PROOF OF THE FORMULA (a+b)²

We all know that (a+b)² = a²+2ab+b²
EVER WONDERED WHY? Lets prove this formula geometrically!

Area of a square with side L is L² , and we want to find (a+b)² .

Do you find any similarity between these two above?

YES YOU DID! Basically, (a+b)² is the area of a square with side length = (a+b)

We know that the area of this square is (a+b)²

Now, lets divide this square into following parts:

Now, our square contains two squares and two rectangles.

So, the area of the square with side (a+b) = area of two squares + area of rectangles

Now, you can observe that the area of the two squares is a² and b², and the area of the two rectangles is ab and ab.

So, lets add area of all parts of the square of length (a+b) :

= a² + ab + ab + b²

= a² + 2ab + b²

As the Area of square of length (a+b) = a² + 2ab + b²

Hence, (a+b)² = a²+2ab+b²

Friday, 8 June 2018

Geometric proof of DIFFERENCE OF TWO SQUARES

GEOMETRIC PROOF OF THE DIFFERENCE OF TWO SQUARES

The difference of two squares states that for all numbers $a$ and $b$ , $a^2 - b^2 = (a + b)(a-b)$ .
The visual representation below, however, only covers the condition that $a^2 - b^2 \geq 0$

To proceed with the visual proof, we create a square with side length $a$ as shown in (1). Then, we cut a square with side length $b$ from its corner as shown in (2). Since the area of the larger square is $a^2$ and the area of the smaller square is $b^2$ , the area of the remaining figure is $a^2 - b^2$ .
Next, we draw a horizontal line segment cutting the remaining figure into two rectangles as shown in (3). We change the color of the smaller rectangle for clearer representation, and move it to the right hand side as shown in (4).

The resulting rectangle in (4) has length $a + b$ and width $(a - b)$ . Its area is $(a+b)(a-b)$ .

The area of the figure in (2) is equal to the area of the rectangle in (4).

So, $a^2 - b^2 = (a + b)(a-b)$ .

Thursday, 7 June 2018

Geometric Proof of the Quadratic formula!

Proof of the quadratic formula:

Where did it come from?

The quadratic formula allows us to easily obtain the roots of any quadratic equation

a x 2 + b x + c = 0

. There is a very nice proof of this formula that uses geometry to give an intuitive understanding of this result. We show this proof below alongside the typical purely algebraic proof.

Theorem 1 (The Quadratic Formula): Let

a

b

, and

c

be real numbers with

a \neq 0

. Then the solutions to the quadratic equation

a x 2 + b x + c = 0

are

x = - b \pm b 2 - 4 a c - - - - - - - \sqrt 2 a

We require that $a \neq 0$ to ensure that $a x 2 + b x + c = 0$ is indeed a quadratic equation. If $a = 0$ then $a x 2 + b x + c = 0$ is the same as $b x + c = 0$ , which is a linear equation.

Proof (Geometric): Consider the quadratic equation $a x 2 + b x + c = 0$ with $a$ , $b$ , and $c$ as real numbers with $a \neq 0$ . Then:

(1)

a x 2 + b x + c a x 2 + b x x 2 + b a x = 0 = - c = - c a

We illustrate the equation above as follows. We denote $x 2$ to be the area of square. Therefore, this square has side length $x$ .

We denote $b a x$ to be the area of a rectangle whose side lengths are $x$ and $b a$ .

Lastly, we denote $- c a$ to denote the area of a rectangle whose side lengths are unimportant.
Consider the rectangle $b a x$ . We separate this rectangle into two equal rectangles whose side lengths are $b 2 a$ and $x$ :
We take the first of these rectangles and attach the side with side length $x$ to the top of the square $x 2$ . WE take the second of these rectangles and attach the side with side length $x$ to the right side of the square $x 2$ . We add a smaller square with side length $b 2 a$ (and area $b 2 4 a 2$ to the lefthand side of the equation and we add the same amount of area to the righthand side of the equation to balance the equality:

From above we get the following equation:

(2)

(x + b 2 a) 2 = - c a + b 2 4 a 2 = - - 4 a c 4 a 2 + b 2 4 a 2 = b 2 - 4 a c 4 a 2

Therefore:

(3)

x + b 2 a x x = \pm b 2 - 4 a c 4 a 2 - - - - - - - \sqrt = - b 2 a \pm b 2 - 4 a c - - - - - - - \sqrt 2 a = - b \pm b 2 - 4 a c - - - - - - - \sqrt 2 a ■