High Performance Tutoring was a proud sponsor of the Utah Education Association (UEA) convention this week. We had a great time handing out candy, speaking to students and parents, and discussing with teachers the major issues facing the education system in Utah.

We spoke with many interested people about the merits of our company including the quality of our tutors and customer service. We also spread the word about our first Free Tutoring Night scheduled for Wednesday, November 15th from 3pm – 7pm at the East Millcreek Library. However, the main attraction was our *Stump The Professor* content in which students could win a $25 Amazon gift card by asking Victor Camacho, our founder and owner, a math question he could not solve or explain how to solve. Many people tried, but in the end, nobody could stump him. He received many excellent questions, but we found three of them to be particularly intriguing. These questions along with a summary Victor’s explanations are presented below. Take a look!

### What is the surface area of a donut?

Mathematicians normally refer to a donut-shaped object as a *torus*. Many 3D shapes can be described as the space occupied by a 2D shape when it is moved along a path. For example, a *box* can be described as the space occupied by moving a rectangle through a straight-line path. A *cylinder* can be described as the space occupied by moving a circle through a straight-line path. A *torus* (donut) can be described as the space occupied by moving a circle around a circular path as seen in the animation below.

Provided that the 2D shape always faces in the direction of motion, there is a simple way to calculate the surface area and volume of the 3D shape it generates. To calculate the volume, simply multiply the area of the 2D shape by the length of the path through which the center of the shape moved. For example, the volume of a cylinder is the area of its base times its height. In other words, it is the area of the circular base times the distance it moved through space to create the cylinder. Therefore, to find the volume of a *torus*, take the area of the circular cross-section (see the figure above) and multiply it by the length of the circular path through which the center of the cross-section moved to create the *torus*.

It is just as simple to calculate the surface area of these types of 3D shapes. In the case of a *torus*, simply multiply the circumference of the circular cross-section by the distance its center moved through space to create the *torus*.

For the donut in the above figure, *r* is the radius of the circular cross-section, and *R* is the radius of the circular path through which the center of the cross-section is moved through space to create the *torus*. Let us use the above reasoning to calculate the surface area of this torus using the variables *r* and *R*. The circumference of the circular cross-section is 2π*r, *and the distance its center moves through space to form the *torus* is 2π*R. *Multiplying these two quantities gives the surface area of the *torus* which is

**S = 4π ^{2}rR**

### Why is 1/0 undefined?

Zero does not abide by the same rules as the other real numbers. However, let us assume for the sake of argument that it does. Let’s let the variable x represent the value of 1/0, whatever that may be. In other words,

1/0 = *x*

Next, cancel the 0 in the denominator by multiplying both sides of the equation by 0. After simplifying we get

1 = 0*x*

Consider the possibilities for *x*. What value could x be to make this equation work? Any number times 0 is 0. So what value could you multiply by 0 so that the product is 1? There is no real number that could do this. Therefore, we say that *x* is *undefined*. By extension, the quantity 1/0 is *undefined*.

It is interesting to point out a related property of 0. It turns out that the quantity 0/0 is NOT considered *undefined. *Instead, 0/0 is considered to be *indeterminable*. To see this, let’s assume once again that 0 follows the same rules as the other real numbers. Let’s let the variable *y* represent the value of 0/0. In other words,

0/0 = *y*

Next, cancel the 0 on in the denominator by multiplying both sides of the equation by 0. After simplifying we have

0 = 0*y*

Consider the possibilities for *y*. What value could *y* be to make this equation work? Any number times 0 is 0, so that means *y* could be ANY number. So the problem here is not that *y* is *undefined*, it is that *y* could be literally any number. Therefore, we say that *y* is *indeterminable*. By extension, the quantity 0/0 is *indeterminable*.

### Why is the square root of 2 irrational?

A rational number is any number that is the ratio of two whole numbers. That is, it is any fraction in which both the numerator and denominator are whole numbers. Some examples include 3/5, -7/29, and 8/1. The one exception, of course, is that the denominator may not equal 0. All whole numbers are rational numbers because any rational number with a denominator of 1 is a whole number. Examples include 8/1, 27/1, and -15/1.

Let’s let the variable *x* represent the square root of 2. This means that if we square *x* (multiply it by itself) we must get 2. In other words,

*x*^{2} = 2

Our goal is to show that *x* must be an irrational number. We can do a quick check to verify that *x* cannot be a whole number. We know that 1^{2} = 1 meaning *x* must be larger than 1, and 2^{2} = 4 meaning *x* must be smaller than 2. So, the value of *x* is some number between 1 and 2.

Let’s assume that *x* is a rational number. This means its numerator and denominator are both whole numbers. Further, since *x* cannot be a whole number, as we established, the denominator of *x* cannot be 1. By the *Fundamental Theorem of Arithmetic, *states that any whole number can be decomposed into the product of one or more prime numbers, both the numerator and denominator can be broken into the product of one or more prime numbers. That is,

*x* = (*p*_{1}*p*_{2}*p*_{3}…*p*_{n}) / (*q*_{1}*q*_{2}*q*_{3}…*q*_{m})

where all of the *p*‘s are prime numbers and all of the *q*‘s are prime numbers. Provided that *x* has been reduced to its simplest form, then none of the *p*‘s can be the same as any of the *q*‘s. If any of them were equal, then we could cancel them out to further reduce *x*, but we assumed *x* is already in its most reduced form. Now let’s square *x*. This yields

*x ^{2}* = (

*p*

_{1}

*p*

_{1}

*p*

_{2}

*p*

_{2}

*p*

_{3}

*p*

_{3}…

*p*

_{n}

*p*

_{n}) / (

*q*

_{1}

*q*

_{1}

*q*

_{2}

*q*

_{2}

*q*

_{3}

*q*

_{3}…

*q*

_{m}

*q*

_{m})

There are still only *p*‘s in the numerator and only *q*‘s in the denominator, but twice as many of each. It is not possible to simplify this fraction because it is still the case that none of the prime numbers in the numerator match any of the prime numbers in the denominator. Therefore, this expression cannot possibly be a whole number. That is, the denominator cannot be reduced to 1.

What we have learned from this little experiment is that squaring a simplified fraction will produce another simplified fraction–never a whole number. Therefore, *x* cannot possibly be a simplified fraction because it would mean that *x ^{2}* is a simplified fraction too, but we want

*x*= 2. We have established that

^{2 }*x*is not a whole number and

*x*is not a rational fraction. Therefore, the only possibility left is that

*x*must be an

*irrational*number!

There are other ways to prove this, but with this reasoning, it is easy to show that the square root of any number that is not a perfect square is also irrational. For example, suppose we wanted to find the square root of 3. Let’s call the number y. Then *y ^{2}* = 3. Clearly, no whole number can be squared to give us 3. Further, if

*y*were any simplified fraction, then its square would produce another simplified fraction, not 3. Therefore,

*y*cannot be a whole number or a fraction, meaning it must be an

*irrational number*.

In fact, this idea can be extended to cube roots, fourth roots, and *n*th roots in general. The *n*th root of any number that is not a perfect power of *n* is irrational. See if you can work through the proof given what we established above.

### Free Tutoring Nights!

The *Stump the Professor* contest was a lot of fun, and we plan on doing it again many times in the future. In fact, High Performance Tutoring has begun offering *Free Tutoring Nights* each month in various cities throughout Utah. Check our events page for the details about the next one. We hope you can make it! Victor Camacho will be in attendance and you will win a prize if you are able to stump him with a math question.