2009 - Math and Multimedia

Proof Tutorial 1: Introduction to Mathematical Proofs

December 30, 2009 GB College Mathematics, High School Mathematics, Problem Solving and Proofs

Introduction

Routine problems in mathematics usually require one or many answers . If we are asked to find the smallest of the three consecutive integers whose sum is 18, then our answer would be 5. If we are asked to find the equation of a line passing through (2,3), we can have many answers.

Proofs, however, is different. It requires us to think more and to reason with valid arguments. It requires us to be explicit and logical. It requires us to convince our readers and most of all ourselves.

Unless a proof problem is already given, finding mathematical statements to prove requires us to see patterns, generalize, and make conjectures about them. The problem stated above about consecutive integers does not require us to reason much or generalize at all.

The success of proof writing requires intuition, mathematical maturity, and experience. Contrary to mathematical proofs written in books, the ideas behind arriving at a proof are not “cut and dried” and elegant. Mathematicians do not reveal the process they go through, or the ideas behind their proofs. This is also a skill that mathematicians and persons who are good in mathematics possess: they are able to read proofs. The skills of reading proofs may be achieved by learning how to write them.

Proving in higher mathematics, on the other hand, requires formal training. For instance, we have to know how to use logical connectives like and, or, not, and must understand how conditional and biconditional connectives work. Basic set theory concepts are also important. Moreover, we also have to learn proof strategies like direct proof and proof by contradiction to name some.

For now, we will not be discussing these things . Most of the proofs in basic mathematics only require a little intuition and good reasoning.

In the tutorial below, I tried to recreate (amateurishly) the process on how mathematicians see patterns, arrive at a conjecture, and how they prove their conjectures. Of course, in reality, problems mathematicians encounter are a lot harder. In fact, some of the hardest problems take hundreds of years to be solved. For example, no mathematician has proved the Fermat’s Last Theorem for more than 300 years, and the mathematician who proved it solved it for eight years.

The proof that we are about to do below is very elementary. For now, we will highlight the process and not the difficulty. The titles of the processes below are not necessarily in order.

Recognizing Patterns and Making Conjectures

Before mathematicians prove theorems, they usually first see patterns. This happens when they read books, solve problems, or prove other theorems. For example, what do we see when we add two even integers? Let’s add some: 2 + 8 = 10, – 24 + 6 = -18, and – 4 + – 8 = -12. We can easily see that if we add two even integers, then their sum is always even.

From here, we might be tempted to say that if we add two integers, then their sum would always be even. In mathematics, this kind of statement or hypothesis is called a conjecture: an educated and reasonable guess based on patterns observed. Rewriting our guess, we have

Conjecture: The sum of two even integers is always even.

If we want to disprove a conjecture, we only need one counterexample — an example that can make the conjecture false. (Can you think of one?). Note that we only need one counterexample to disprove a conjecture. If we want to prove it, however, we might be tempted to pair a few more integers and say that “oh, their sum is even, so it must be true”. No matter how many integers we pair, if we can’t exhaust all the pairs, then it cannot be considered as a proof. When we say the sum of two even integers above, we mean ALL even integers. Of course, there is no way that we can list all pairs of even integers since there are infinitely many of them.

Generalizing Patterns

Since it is impossible to enumerate all pairs of even integers, we need a representation, algebraic expression in particular, that will represent any even integer. If we can find this expression, then all the even integers would be represented. This process is called generalizing. Like what we have done above, we generalized by representing all members of the set by a single expression. In our case, the members of our set are all even integers.

From the definition, we know that all even integers are divisible by 2. That means that if m is an even integer, then, when we divide m by 2, we can find a quotient which is also an integer. For instance, since 18 is an integer, we are sure that there exists an integer such that 18 divided by 2 is equal to that integer.

In general, suppose that quotient of m/2 is q, then it follows that m/2 = q, for any even integer m. Multiplying both sides of the equation by 2, we have m = 2q. That means that if m is an even integer, then there exists an integer q such that m = 2q. Hence, we may represent any even integer m with 2q for some integer q*. Note that q here is a generalized number, which means that an even integer can also be represented by 2x, 2y, 2z or any variable with the condition that they are integers.

Connecting the ideas

Proving is making logical and relevant statements from definitions, facts, assumptions and other theorems to come to a desired conclusion. Before coming up with an elegant proof, mathematicians usually have scratch work, connecting their ideas to arrive at what they want to prove.

Scratch work

From the statement above, we have shown that any even integer m, there exists an integer q, such that m = 2q. That means, that if we can show that the sum of two even integers is in the form 2q (or that the sum is divisible by 2), then we can be sure that it is always an even integer.

Since we need two integers, we let m and n are the two integers that we will add. Since both of them are even integers, then we can represent them as 2q and 2r respectively for some integers q and r. Adding both of them, we have m + n = 2q + 2r = 2(q + r). Now, q + r is an integer since q is an integer and r is an integer from our definition above. This means that 2(q + r) isof the form 2x for some integer x. This means that m + n is of the form 2x for some integer x. Therefore, m + n is even.

Writing (elegantly) the final proof

Here, we write our proof in a shorter and more elegant way. Conjectures that are proven are called theorems. So let us write the proof of our first theorem.

Theorem 1: The sum of two even integers is always even.

Proof. Let m, n be even integers. Then m = 2q for some integer q and n = 2r for some integer r. Now, m + n = 2q + 2r = 2(q + r). Since q + r is an integer, clearly, 2(r + s) = m + n is divisible by 2. Therefore, the sum of two even integers is even.

Most proofs are written in a concise way, leaving some details for the reader to fill in. For example, the statement “Since q + r is an integer” did not really state the reason why this is so. This is stated in our scratch work, but not in the proof.

Going Further

If m is an even integer, then m – 1 and m + 1 are odd integers. Since m = 2r, then 2r – 1 and 2r + 1 are also odd integers. In our example below, we will use 2r + 1, to prove that the sum of two odd integers is always even. As an exercise, use 2r – 1 in your proof.

Theorem 2: The sum of two odd integers is always even.

Let p, q be odd integers. Then p = 2a + 1 and q = 2b + 1 for some integers a and b. Now, adding we have p + q = 2r + 1 + 2s + 1 = 2r + 2s + 2 = 2(r + s + 1). Since r + s + 1 is an integer, then 2(r + s + 1) is divisible by 2. Hence, p + q is divisible by 2. Therefore, the sum of two odd integers is even.

Math and Hardwork

Being good in math requires hard work. Andrew Wiles worked on the Fermat’s Last Theorem for seven years, have given up several times thinking that it was impossible. In 1995, he finally thought he had proved it, and presented it in a conference. A month later, his reviewer thought that there is a part of the proof which was vague (or wrong), so he had to review his work and found out that there was a part which was actually wrong. He almost gave up. He worked more than a year to correct the error.

Now, he has carved his place in history.

*In technical language, the word “for some” is equivalent to the word “there exists”.

Exercises:

Prove that the sum of an even number and an odd number is always odd.
Prove that the difference of two odd integers is always even.
Prove that the product of two even integers is always even.
Prove that the product of two odd integers is always odd.
Prove that the product of an even number and an odd number is always even.

Leave a comment Andrew Wiles, Fermat's Last Theorem, how to write proofs, introduction to proofs, mathematical proofs, proof basics, proof tutorial, sum of even numbers

Plotting with Ivan Johansen’s Graph Software

December 29, 2009 GB Software Tutorials

Graph, a graphing software created by Ivan Johansen, was the first graphing software I have learned to use. Although the graphics quality is not that impressive, it has some features that are not available in other graphing software, the most notable of which is the polynomial of best-fit.

Figure 1 – The Graph Window.

The tutorial below teaches the basics of Graph, and most examples are related to elementary and high school mathematics.

The basic capabilities of Graph are enumerated below.

I. Graph Functions

As an example, we will plot the graph of the function f(x) = x² + 3x from x = -3 to x = 2, the graph of which is shown in Figure 2.

To graph a our function, just click the Function menu from the menu bar, then click Insert function, then type the equation of your function. The enumerated steps below are associated with the numbers shown in Figure 2.

Choose Standard from the Function type drop-down list box. The other options are Parametric and Polar.
Type the equation of our function f(x) = = x^2 + 3x in the Function equation text box. Like other software, Graph uses ^ to denote exponential notation.
Specify the domain of the function which is from -3 to 2.
Choose the type of start and endpoint at the Endpoints drop-down list boxes. We choose circle for our left end point and arrow for our right endpoint.
Change the color, line style, draw type, and width of the graph. Click the OK button when you are done.

Figure 2 – The Function dialog box and the graph of the function f(x) = x2 + 3x.

Exercise: Refer to the steps above in graphing the following functions:

Standard Functions

f(x) = x^3 + 3x – 1
f(x) = sin(x)
f(x) = sqrt(x)
x(t) = cos(t), y(t) = tan(t)
e^(sin(t)) – 2cos(4t) + sin((t – pi/2) /12)^5

Note: a, b and c are standard functions; d is parametric and e is polar. Choose their appropriate function type the Function type box before typing the equations.

II. Graph Inequalities

To get the intersection of the graph y < x^3 + 3 and y > 2x, we first transform the inequality to equation, and choose the shaded portion later. This is the part of Graph that I do not quite like. It’s more like manual drawing rather than graphing.

Click the Function menu from the menu bar, the click Insert Function.
Type x^3 + 1 in Function equation box.
To graph y > 2x, repeat step a and type y = 2x in the Function equation box.
To shade the graph below x³ + 3, be sure that the equation of the function is selected in the equation window (left pane). Click Function from the menu bar and then click Insert shading… from the list.
Choose Below function icon (see Figure 3). Take note of the other options. Click the OK button.
As an exercise, shade y > 2x.

Figure 3 – The Insert Shading dialog box.

III. Plot Points Series and Determine Line (or Polynomial) of Best Fit

To plot a point series, click the Function menu from the menu bar, click Insert point series. Type the ordered pairs on the Insert point series window as shown below.

Figure 4 – The Insert Points series dialog box.

To insert a line of best fit, click the Function menu from the menu bar, then click Insert trendline…. In the Insert trendline window, choose Linear. Notice that you can also choose polynomial of a chosen order fit.

Figure 5 – The Trendline dialog box.

Change the Line width to 3 and click the OK button. The line of best fit of our point series is shown below. The line of best fit graph is shown in Figure 6.

Figure 6 – The Line of Best Fit of the given data in Figure 4.

IV. Find the area under a curve

Graph is capable of finding the area under a curve or technically, perform definite integration. In the example below, we will find the area under the curve of y = sin(x) from -2 to 3.

To plot y = sin(x), click the Function menu from the menu bar and click Insert function from the drop-down list box.
Type sin(x) in the Function equation box.
To get the area of the curve under -2 through 3, click the Calc menu from the menu bar, then click Area from the list.
A dialog box will appear located at the bottom-left of the Graph window. Type -2 in the From text box and type 3 in the To text box. Notice that the area the curve is displayed on the Area box below the To text box.

V. Generate a table from a graph

Graph is capable of generating table from a graph. If we want to generate table of values of the graph y = sin(x) in (4), be sure that the graph is selected in the left pane of the Graph window, then do the following steps:

Click the Calc menu and then click Table.
Type the minimum value, say -10, in the From text box and the maximum value, say 10, in the To text box.
In the $\delta x$ text box, type the interval, say 2, of your table, the click the Calc button.

Notice that not only the x and f(x) are displayed but also the value of the first and second derivatives.

Leave a comment graph math functions, graph visualization software, graphing, graphing calculator, graphing inequalities, mathematical function graph, plot graph freeware

The Exterior Angle Theorem

December 26, 2009 GB High School Calculus, High School Geometry, High School Mathematics

In the angle sum of a triangle post, we have discussed that the angle sum of a triangle is $180$ degrees. In the angle sum of a polygon post, we also have discussed that and that the angle sum of a polygon with $n$ sides is $180(n-2)$ . For example, a pentagon has $5$ sides, so the sum of its interior angle is $180(5-2) = 180(3) = 540$ degrees.

Figure 1 – The interior and exterior angles a triangle and a quadrilateral.

The angle sums that we have discussed in both blogs refer to the sum of the interior angles. What about the exterior angles?

The exterior angle is formed when we extend a side of a polygon. In the triangle above, $\alpha$ is an exterior angle. The sum of the interior angle and the exterior angle adjacent to it is always $180$ degrees (Why?). Angles whose sum is $180$ degrees are called supplementary angles. If two angles are supplementary, we call them a linear pair. For example, angles $\alpha$ and $a_1$ are supplementary angles and at the same time a linear pair, so $\alpha + a_1 = 180$ degrees. Now this means, that $\alpha = 180 - a_1$ .Therefore, if we want to compute the measure of an exterior angle adjacent to an interior angle, we can always subtract the measure of the interior angle from $180$ as shown in Figure 1.

Observe the computation in the two diagrams. If we let $S_t$ be the angle sum of the exterior angles of a triangle, then $S_t = (180 - a_1) + (180 - a_2) + (180 - a_3) = 540$ . Rearranging the terms, we have $S_t = 540 - (a_1 + a_2 + a_3)$ . But $a_1 + a_2 + a_3$ is the sum of the interior angles of a triangle which is $180$ degrees, so $540 - (a_1 + a_2 + a_3) = 540 - 180 = 360$ degrees.

Now, try calculating for the sum of the exterior angles of the quadrilateral above. What is your answer?

To verify our hunch, we will try to compute for the sum of the exterior angles of a pentagon.

Let $S_p$ be the sum of the exterior angles of the pentagon in Figure 2. Then

$S_p =(180 - c_1)+ (180 - c_2) + (180 - c_3) +(180 - c_4) +(180 - c_5)$ . Simplifying, we have $S_p = 900 - (c_1 + c_2 + c_3 + c_4 + c_5)$ . But according to the angle sum theorem for polygons, $c_1 + c_2 + c_3 + c_4 + c_5 = 540$ . Therefore, $900 - (c_1 + c_2 + c_3 + c_4 + c_5) = 900 - 540 = 360$ degrees.

We have three polygons – triangle, quadrilateral, pentagon – whose angle sums of exterior angles are always $360$ degrees. Now, is this true for all polygons? Try to compute polygons up to $10$ sides and see if the sum is $360$ degrees.

Delving Deeper

We know that in a polygon, the number of exterior angles is equal to the number of interior angles. Furthermore, we know that the angle sum of an interior angle and the exterior angle adjacent to each is always latex 180 degrees. If we have a polygon with 5 sides, then

interior angle sum + exterior angle sum = 180(5)

In general, this means that in a polygon with n sides

interior angle sum* + exterior angle sum = 180n

But the interior angle sum = 180(n – 2). So, substituting in the preceding equation, we have

180(n – 2) + exterior angle sum = 180n

which means that the exterior angle sum = 180n – 180(n – 2) = 360 degrees. More formal proofs using these arguments are shown below.

Theorem: The sum of the measure of the exterior angles of a polygon with n sides is 360 degrees.

Proof 1:

Let $a_1, a_2, \cdots a_n$ be measures of the interior angles of a polygon with n sides. Let $b_1, b_2, \cdots b_n$ be measures of the exterior angles of the same polygon where all angle names with the same subscripts are adjacent angles from $a_1$ and $b_1$ all the way up through $a_n$ and $b_n$ . We know that adjacent interior and exterior angles are supplementary angles, so this implies that their measures add up to 180 degrees. Hence,

(a₁ + b₁) + (a₂ + b₂) + … + (a_n + b_n) = 180 + 180 + … +180 (n of them) = 180n

Regrouping the terms of the preceding equation, we have

(a₁ + a₂ + … + a_n) + (b₁ + b₂ + … + b_n) = 180n

But the sum of the interior angles is a₁ + a₂ + … + a_n = 180(n – 2)

So,

180(n – 2) + (b₁ + b₂ + … + b_n) = 180n

b₁ + b₂ + … + b_n = 180n – 180(n – 2) = 360

Therefore, the sum of the exterior angles of any polygon is equal to 360 degrees.

Proof 2:

Let a₁, a₂, …, a_n be measures of the interior angles of the polygon with n sides. Since each adjacent interior and exterior angle is a linear pair, it follows that the measure of the exterior angles adjacent to them respectively are 180 – a₁, 180 – a₂, …, 180 – a_n.

If we let S, be the sum of the measure of the exterior angles, we have

S = (180 – a₁) + (180 – a₂) + (180 – a₃) + … + (180 – a_n)

= (180 + 180 + 180 + … +180 (n of them)) – a– a₂ – a₃– … – a_n

S = 180n – (a₁+ a₂ + a₃+ … + a_n)

But a₁+ a₂ + a₃+ … + a_nis the sum of the measures of the interior angles of a polygon with n sides which equals

180(n – 2), so, S = 180n – 180(n – 2) = 360, which is want we want to show.

Therefore, the sum of the exterior angles of any polygon is equal to 360 degrees.

Leave a comment angle sum, angle sum of polygons, exterior angle math, exterior angle theorem, exterior angles convex polygon, exterior angles definition, interior angles, linear pair, polygon exterior angle sum theorem, sum of exterior angles of a polygon, sum of interior angles, triangle angle sum

1 2 3 … 12 »