Why is any number raised to 0 equals 1?

Why is any number raised to 0 equals 1?

If we raise a number to an exponent, we are multiplying it by itself a certain number of times.  For example, 3^4 means you have to multiply 3 by itself 4 times.   In exponential notation, we call 3 the base and 4 the exponent.

Shown below are examples of exponential expressions and their expansion.

2^3 = 2 \cdot 2 \cdot 2

4^7 = 4 \cdot 4 \cdot 4 \cdot 4 \cdot 4\cdot 4\cdot4

x^4 = x \cdot x \cdot x \cdot x

(a + b)^3 = (a+ b)(a+b)(a+b)

x^m = x \cdot x \cdot x \cdot \ldots \cdot x \cdot x (Multiply x by itself, m times)

The\ldots symbol means “and so on.” It represents x’s that are missing.  It is convenient to use the said symbol for large values of m.

Multiplying Expressions with Exponents

If we want to multiply expressions with the same base, let us see what happens. For example, what will happen if we multiply 2^3 and 2^5?

From above, 2^3 = 2 \cdot 2 \cdot 2 and 2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2. Multiplying the two expressions, we have

2^3 \cdot 2^5 = (2 \cdot 2 \cdot 2)(2 \cdot 2 \cdot 2 \cdot 2 \cdot 2) = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 = 2^8

From our computation, we can conclude that if we multiply to expressions with the same base, we have just have to add their exponents (Can you see why?). That is, for expressions x^m and x^n,

x^m \cdot x^n = x^{m+n} (*)

Q1: What if the base of the two expressions are not the same? Will our formula above still apply?

Dividing Expressions with Exponents

What about dividing expressions with exponents? Suppose, we want to divide 2^5  by 2^3.

We know that 2^5 = 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 and 2^3 = 2 \cdot 2 \cdot 2. Dividing the two expressions, we have

\displaystyle \frac{2^5}{2^3}= \displaystyle\frac{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2}{2 \cdot 2 \cdot 2} = 2^2

Since, three 2‘s are canceled out, we can therefore conclude that in dividing two expressions with the same base, we just have to subtract their exponents. That is, for expressions x^m and x^n,

\displaystyle\frac{x^m}{x^n} = x^{m-n}. (**)

What happens if the exponent of the denominator is larger? For example, \displaystyle \frac{2^2}{2^7}?

From (**), \displaystyle\frac{2^2}{2^7} = 2^{2-7} = 2^{-5}.

Now, let us compare this result when we expand our expression:

\displaystyle \frac{2^2}{2^7}= \displaystyle\frac{2 \cdot 2}{2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2 \cdot 2} = \frac{1}{2^5}

Our observation tells us that, 2^{-5} = \displaystyle\frac{1}{2^5}. Therefore, x^{-m}= \displaystyle\frac{1}{x^m}.

Q2: In general, what is the value of \frac{x^m}{x^n} if m < n?

What if the exponents of the numerator are equal? For instance, \frac{2^5}{2^5}. This is practically dividing the same number, so obviously the answer is 1. However, we can also use our conclusion above.

From (**), \displaystyle\frac{2^5}{2^5} = 2^{5-5} = 2^0 = 1

Conclusion

Here we observe that raisingx (or any expression) to 0  means that the number of factors in the numerator and the number of factors in the denominator is the same. Therefore, x^0 can be expressed as x^{m-m} for any value of m. But from (**), x^{m-m} is equivalent to \displaystyle\frac{x^m}{x^m} = 1

Therefore, any number raised to 0 (with the exception of 0) equals 1.

GeoGebra Tutorial 6 – Parameterization of Length and Area

This is the sixth tutorial of the GeoGebra Intermediate Tutorial Series. If this is your first time to use GeoGebra, please read the GeoGebra Essentials Series.

In this tutorial, we are going to learn the following:

  1. use variables in GeoGebra
  2. compute using these variables
  3. use variables as parameters of objects

Problem: Given a rectangle with perimeter 10 units, find the dimension of the rectangle that can be formed that has the largest area.

This problem can be easily solved algebraically, but we are going to use GeoGebra to parameterize the length and the area of the rectangle to find its maximum area. The output of this tutorial is shown above.

Figure 1 – Rectangle ABCD with two of its sides on the x and y-axis

Before doing the tutorial, let us first solve the problem. We know that the rectangle’s perimeter is constant, so we choose the width w.  It follows that the height h will depend on the width. For instance,  if  w=4 units, then h = (10 – 2*4)/2 which is equal to 1. Hence, h = (10 – 2w)/2. Using this information, we plan the GeoGebra construction.

  1. First we make our maximum width 5 (Why?). We will create segment AL with length 5 with A at the origin and L at (5,0)
  2. Next, we create point D on AL. With D = (w,0), AD will be the width of the rectangle.
  3. We compute for h = (10 – 2w)/2, then, and take the value as the height of the rectangle. Then, we create point B  with coordinates (0,h).
  4. We create the fourth vertex of the rectangle by getting the intersection of the horizontal line passing through B and a vertical line passing through D.
  5. Lastly, create ABCD using the polygon tool, and then produce point P (w, A_r) where A_r is the area of the rectangle.

Instructions

1.) Open GeoGebra and be sure that the Algebra & Graphics view is selected in the Perspectives panel.
2.) Select the Segment between Two Points tool, click on (0,0) and click on (5,0) to construct segment AB. Show the label of the points, and rename point B to L.
3.) Create a point on AL. You may not see the segment, so before doing this, hide the axes by clicking the Axes icon in the upper left of the Graphics View. If the icons are not displayed, click the arrow.
4.) Rename the recently created point to D. Move the point and notice that it can only move between A and L.  Now, hide point L and display the axes. AD will be the width (lower base) of the rectangle.
5.)  We now determine the width and the height of the rectangle. First, we want to determine the AD which is the width. To do this, we get the x-coordinate of D (Why?). To get the x-coordinate of D, type w = x(D) in the Input bar and press the ENTER key. This means that the value of w, a declared variable, will be the x-coordinate of point D which is the same as the length of AD.
6.) Next, we compute for the height h of the rectangle. Type h = (10 – 2w)/2 in the input bar and press the ENTER key. Notice the values of h and were added to the Algebra view.
7.) Next we create point B with coordinates (0,h). To do this, type B = (0,h) in the Input bar and press the ENTER key. This will be the third point on the rectangle.

8.) Move point D. What do you observe?
9.) Next, we locate the fourth vertex of the rectangle. The fourth vertex C will have the y-coordinate the same as B and x-coordinate the same as D. Therefore, we type C = (x(D), y(B))
10.)  Now, we use the polygon tool to construct rectangle ABCD. Click the Polygon tool and then click the points in following order: point A, point B, point C, point D and, again, point A to close the figure.
11.)  Now, let us display the area of the polygon. Right click the interior of the rectangle, then click Object Properties to open the Preferences window. In the Basic tab of the Preferences window, check the Show Label check box and choose Value from the drop down list box. Close the window.

12.)  Move point D. What do you observe? What length of AD gives the rectangle the largest area?
13.)  Now, we create point P, type P = (w, poly1). Note that poly1 is the name of the rectangle and its value is area of the rectangle (see the Algebra view).
14.)  Right click on point P, then click check Trace On. This will trace the path of point P.

15.)  Move point D. What do you observe? What can you say about the curve formed by the traces of point P? Explain why your observations are such.
16.)  Solve the problem algebraically. What is the relationship between the equation formed from getting the solution of the problem and curve formed by traces of point P?

GeoGebra Tutorial 5 – Discovering the Pythagorean Theorem

This is the fifth tutorial in the GeoGebra Intermediate Tutorial Series. If this is your first time to use GeoGebra, please read the GeoGebra Essentials Series.

In this tutorial, we compare areas of squares formed on the sides of a right triangle. To construct this figure, we first construct a right triangle, and form three squares, each of which contains one of the three sides as shown below. Then we observe the relationship among the areas of the squares.

figure5-1

Figure 1 – Squares formed containing the sides of a right triangle

In this tutorial, we learn how to use the Regular Polygon tool.   Most of the constructions that you will make here are review of the first four tutorials.

Instructions

1.) Open GeoGebra and select Geometry from the Perspective panel.
segmentpng 2.) Select the Segment between Two Points tool and click two distinct places on the Graphics view to construct segment AB.
movepng 3.) If the labels of the points are not displayed, click the Move tool, right click each point and click Show label from the context menu.
perpendicularpng 4.) Next, we construct a line perpendicular to segment AB and passing through point B. To do this, choose the Perpendicular line tool, click on segment AB, then click on point B.
newpointpng 5.) Next, we create point C on the line. To do this, click the New point tool and click on the line. Be sure that the label of the third point is displayed.

figure 5-2

Figure 2 – Point C on the line passing through B

You have to be sure that C is on the line passing through B. Be sure that you cannot drag point C out of the line. Otherwise, delete the point and create a new point C.

6.) Hide everything except the three points by  right clicking tmem and unchecking the Show Object option.
7.) Next, we rename point B to point C and vice versa. To rename point B to C, right click point B, click Rename and then type the new name, in this case point C, in the Rename text box, then click the OK button. Now, rename B (or B1) to C.
regpolygonpng 8.) Next we  construct a square with side AC. Click the Regular polygon tool, then click on point C and click on point A.

regpolygonpng

9.) In the Points text box of the Regular polygon tool, type 4. If the position of the square is displayed the wrong way (right hand side of AC) just undo button and reverse the order of the clicks when creating the polygon.

figure5-3

Figure 3 – Square containing side AC

regpolygonpng

10. ) With the Polygon tool still active, click point B and click point C to create a square with side BC. Similarly, click point A, then click point B to create a square with side AB.  After step 10, your drawing should look like the one shown below.

figure5-4

Figure 4 – Squares containing sides of right triangle ABC

11.)  Hide the label of the sides of the side of the squares.
12.)  Rename the sides of the rectangle as shown below.

figure5-5

Figure 5 – Triangle ABC wth side lengths a, b and c.

13.)  Now, let us reveal the area of the three squares. Right click the interior of the square with side AC, then click Object Properties from the context menu to display the Preferences window.
14.)  In the Basic tab of the Preferences window, check the Show Label check box and choose Value from the drop-down list box. Do this to the other two squares as well.

preferences-window

Figure 6 – Properties of squares shown in the Preferences window.

movepng 15.)  Move the vertices of the triangle. What do you observe about the area of the squares?
16.)  You may have observed that the area of the biggest square is equal to the sum of the areas of the two smaller squares. To verify this, we can put a label in the GeoGebra window displaying the areas of the three squares.
17.)  Suppose the side of the two smaller squares are a and b, and the side of the biggest square is c, what equation can you make to express the relationship of the of the three squares?
18.)  What conjecture can you make based on your observation?

 

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