## Why is 0! equals 1?

Many books will tell you that $0!$ equals $1$ is a definition. There are actually a few reasons why this is so – the two of which are shown below.

Explanation 1:

Based on mypost, we can conclude that $n$ taken $n$ at a time is equal to $1$. This means, that there is only one way that you can group $n$ objects from $n$ objects. For example, we can only form one group consisting of $4$ letters from AB, C and D using all the 4 letters.

From above, we know that the $\displaystyle{n \choose n} = 1$. But, $\displaystyle{n \choose n} = \displaystyle\frac{n!}{(n-n)!n!} = \displaystyle\frac{n!}{0!n!} = 1$. To satisfy the equation, $0!$ must be equal to $1$.

Explanation 2:

We can also use the fact that $n! = n(n-1)!$. Dividing both sides by $n$, we have $\displaystyle\frac{n!}{n} = (n-1)!$. If we let $n = 1$, we have $\displaystyle\frac{1!}{1} = (1 - 1)! \Rightarrow 1 = 0!$ which is what we want to show.

Explanation 3

We can also use the following pattern. We know that $n! = n(n-1)(n-2)(n-3) \cdots 3(2)(1)$ which means that

$n! = n(n-1)!$.

Dividing both sides of the equation by $n$, we have

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

Using this fact, we can check the following pattern.

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

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

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

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

Now, we go to $0!$

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

As we can see from the 3 examples, $0! = 1$.

## Derivation of the Quadratic Formula

When we discuss about functions, we usually talk about their roots, or geometrically where their graphs pass through the x-axis. For example, $- 3, 1$ and $3$ are the roots of the graph of the function in Figure 1 because it passes through $x = -3, x =1$ and $x=3$.

Since we are looking for points on the x-axis, it means that all the points that we are looking for have $y$-coordinate $0$. As a consequence, (i) if we want to find the root of a quadratic function we have set $y = 0$ and then solve for the values of $x$.

Figure 1 – The x-axis and the line with equation y = 0 are basically the same line so all points on the x-axis have y-coordinate 0.

With the things above in mind, let us find the roots of two quadratic functions: $(1) y = x^2 + 5x + 6$ and $(2) y = x^2 + 3x + 1$.

Finding the root by Factoring

The roots of the function $y = x^2 + 5x + 6$ are easy to find. As we have said, to get the root of a function, we set $y$ to $0$ and then find the value the value of $x$. Solving by factoring, we have

$x^2 + 5x + 6 = 0 \Rightarrow (x + 2)(x + 3) = 0$

Now, $(x + 2)(x + 3) = 0$ if $x + 2 = 0$ or/and $x + 3 = 0$. Solving for the value of $x$ on both equations, we have $x= -2$ and $x = -3$.

Thus, even though we have not seen the graph of the function yet $y = x^2 + 5x + 6$, we are sure that it will pass through $x = -2$ and $x = -3$. If you want to verify if the graph of the function $y= x^2 + 5x + 6$ indeed passes through the $x$-axis at $x = -2$ and $x = -3$, you can verify its graph using a graphing software or a graphing calculator.

Figure 2 – We are sure that the graph of $y = x^2 + 5x + 6$ will pass through the yellow points.

Factoring: Another Interpretation

Another possible interpretation of the expression $x^2 + 5x + 6$ can be the area of a rectangle with length $x + 3$ and width $x+2$. The distribution of the products of the terms of the expressions are represented by the four rectangles formed shown below.

Figure 3 – Another interpretation of the expression x^2 + 5x + 6.

Let us try another example: Let us find the roots of the quadratic function $y = x^2 + 3x + 1$. You will probably observe that there is no way that we can factor this expression. The last term is $1$, but there are only two factors of 1: $\{1,1\}$ and $\{-1,-1\}$, so this means that that the numerical coefficient of$x$ must be $2$ or $-2$, but it is equal to $3$. Hence, the expression is not factorable.

Since, the expression is not factorable, we cannot find the length and width of a rectangle with area $x^2 + 3x + 1$. The easiest way probably to find its length and width is to assume that it is a square.

Completing the Square

We have a quadratic expression which we assumed a perfect square so its factor must be of the form $(x+b)(x+b)$ where $b$ is a real number. Also, $(x+b)^2 = x^2 + 2bx + b^2$. If we consider $x + b$ as a side of the square, then the product of the expressions will form two squares namely $x^2$ and $b^2$, and 2 congruent rectangles with each having an area of $bx$.

Figure 4 – A square with side x + b, where b is a constant.

If we want to use the product of $x+ b$ above, first, we have to take off $x^2$ as one of the squares. Then we are left with a figure with area $3x$ which we will divide into two congruent rectangles. If we are going to follow the positions of the rectangles in Figure 4, then we will have an $x^2$ and two pieces of $\frac{3}{2}x$ (see Figure 5).To construct a square, we extend one of the sides of each of the congruent rectangles.

Figure 5: Completing the square of x^2 + 3x + 1.

Since we have two small rectangles with area $\frac{3}{2}x$, and the longer side (in the diagram) is $x$, it follows that the other dimension is $\frac{3}{2}$ which gives us a smaller square with area $\frac{9}{4}$.

Figure 6 – The area of the small formed is 9/4 and the side length of the big square formed is x + 3/2.

The biggest square formed in Figure 6 has area $x^2 + \frac{3}{2}x +\frac{9}{4}$, which is $\frac{5}{4}$ more than our original quadratic expression, so we will deduct $\frac{5}{4}$ to preserve the original expression. So our final expression is $x^2 + \frac{3}{2}x +\frac{9}{4} - \frac{5}{4}$

Algebraically, if we have the expression $x^2 + bx + c$, and we want to “compete its square”, we want to transform it to an expression of the form $(x-h)^2 + k$. For example, $x^2 + 6x + 12$ can be expressed as $(x + 3)^2 + 3$. Another example is that $x^2 + 8x + 12$ can be written as $(x + 4)^2 - 4$. Note that the coefficient of $x^2$ should be $1$ so that we are sure that a square $x$ by $x$ is formed as shown in figures 4,5 and 6. In general, the possible steps that we are going to create using the general equation $y = ax^2 + bx + c$ is to set $y$ to $0$ and then find the value of $x$.

In constructing the square in Figure 6, we went through the following processes:

(ii) We made sure that the numerical coefficient of $x^2$ is $1$ to ensure that we have a square with factors (side length) $x$.

(iii) We isolated the constant term $c = 1$, and we just used the first $x^2$ and the second term $3x$.

(iv) To get the area of the smaller square, we divided the numerical coefficient of the $3x$ by $2$ then squared it to get $\frac{9}{4}$.

Shown in Figure 7 is the summary of the steps we did to get the roots of the quadratic function $y = x^2 + 3x + 1$. The rightmost column of the table shows the generalization of our steps, which is getting the roots of the quadratic function $y = ax^2 + bx + c$.

Figure 7 – A step-by-step derivation of the quadratic formula.

The formula $x = \displaystyle\frac{-b \pm \sqrt{b^2-4ac}}{2a}$ located at the bottom part of the rightmost column of the table in Figure 7 is called the quadratic formula. We have derived the quadratic formula from completing the square of a quadratic equation. From the formula, the roots o the quadratic function $y = ax^2 + bx + c$ are $\displaystyle\frac{-b + \sqrt{b^2-4ac}}{2a}$ and $\displaystyle\frac{-b - \sqrt{b^2-4ac}}{2a}$.

## Parallel Lines and Transversals

Given a triangle and a line segment, let us place the triangle such that one of its side and the line segment coincide. Next, we replicate the triangle and slide to the right hand side  such that two of their vertices meet as shown in Figure 1.

Figure 1 – Translated triangles have parallel corresponing sides.

The angles with the same color are basically the same angle, so we will call them corresponding angles. The sides which are parallel are also the same side, so we will call them corresponding sides.

What can you say about their corresponding angles? Their corresponding sides?

Figure 2 – The sum of the angles in point T is 180 degrees.

Next, we replicate one of the triangles, rotate it and place it between the two triangles as shown in Figure 2. Click here to rotate the triangles using GeoGebra. In the figure, we can observe and deduce the following:

1. The measure of the three adjacent angles in T is half the circle and is therefore 180 degrees. This implies that the sum of the interior angles of triangle PQT is also 180 degrees, since they are the sum of the three adjacent angles in T.
2. PQRT and QRST are parallelograms. (Explain why the opposite sides are parallel).
3. The opposite angles of parallelogram are congruent, and their opposite sides are also congruent. (Explain why).
4. We also observe that the sum of the consecutive angles of a parallelogram is 180 degrees. For example, the sum of angle PQR and angle QRT is 180 degrees.
5. The interior angles of a parallelogram consists of 2 green angles, 2 yellow angles and 2 red angles, which means that the angle sum of the interior angles of a parallelogram is twice 180 or 360 degrees.
6. PQRS is a trapezoid and the sum of its interior angles is also 360 degrees. (Explain why QR and PS are parallel).

If we extend QR and QT, then we will two parallel line segments and a transversal. In Figure 3, OR and PS are parallel lines a NU is a transversal. From Figure 1 and Figure 2, the yellow angles are corresponding angles, so RQT and PTQ are congruent.  We can also see that  the blue angle and yellow angle are supplementary (the sum of their measure is 180 degrees).

Figure 3 – NU is a transversal to the parallel lines OR and PS.

With the knowledge that the blue angle and the yellow angle are supplementary, we can deduce the following:

1. Interior angles on the same side of a transversal of parallel lines (RQT and STQ) are supplementary.
2. Angles OQN and OQT are supplementary so NQT must have the same measure as that of the yellow angle.
3. Angles RQT and RQN are supplementary so RQN must have have the same measure as that of the blue angle.
4. From 2 and 3, it follows that PTU is a blue angle and STU is a yellow angle.

With all the observations and deductions above, we came up with the following diagram. Note that angles symbols with the same color are congruent angles.

Figure 4 – Angle signs with the same color are congruent.

Pairs of these angles have special names  for distinction purposes. Most of the names are descriptive, so it is easy to determine other pairs. Some pairs of angles and their names are listed below.

1.) TQR and QTS are same side interior angles.

2.) NQR and STU are same side exterior angles.

3.) RQT and PTQ are alternate interior angles.

4.) STU and NQO are alternate exterior angles.

5.) NQO and QTP are corresponding angles.

Parallel lines and its relationship with its interval was mentioned in Euclid’s book The Elements. It is called the Fifth postulate or the Parallel Postulate. The parallel postulate states that

If a line segment intersects two straight lines forming two interior angles on the same side that sum to less than two right angles, then the two lines, if extended indefinitely, meet on that side on which the angles sum to less than two right angles.

All the relationships we have deduced above can be derived from this single statement.

I have also written why the angle sum of the interior angles of a triangle is 180 degrees. C

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