The Magic of Algebra

Think of a number from one to ten.
Multiply that number by two.
Add thirteen.
Subtract seven.
Divide by two.
Subtract your original number.
You are thinking of the number three.

This is a classic mind reading trick that many people will learn to do when they are little. The premise is simple; the answer is always three.

Why does this work? As a young child, you probably think back to the arithmetic skills you learned in school and realized that you can't prove this trick because you don't know what number they'll choose.

This is a good time to note that when you learned this trick, the first thing you thought was "why?" Many people see proofs as a tedious process that could be avoided just by accepting things to be true. However, that why spark that went off when you learned the trick is where proofs can be applied. I use proofs so often in my posts primarily for that reason: to foster that why spark.

Anyways, this situation is where you need to define that unknown quantity of the number the spectator is thinking of. How do you define an unknown quantity? Well, this is when you pull out a little tool called algebra. You might not expect to apply algebra to a practical situation, but this is when you can. With algebra, we can stop saying "the number that they are thinking of" and start saying "n."

n

Now, the first step was to multiply the number by 2. n times 2 is pretty simple; it is 2n.

n • 2
2n

Next, we had to add thirteen. That can't be simplified much.

2n + 13

Now, we subtract seven. 13 - 7 is 6, so we can simplify it to 2n + 6.

2n + 13 - 7
2n + 6

Now, we have to divide by two. 2 and 6 both have a factor of two in it, so we will not need any fractions.

(2n + 6)/2
2(n + 3)/2
n + 3

The last step was to subtract the original number. But, we already defined that number to be n, so we just subtract n.

n + 3 - n
3

As you see, the n's cancelled, and you ended up with three. Basically, you undid all of the previous operations to end up with one number.

I think that this is a perfect way to show young children magic, but also introduce them to concepts like algebra and proofs by lighting that spark in such a fun and interesting way.

History of Math: Leonhard Euler

Today, I am giving a talk about Mending Mathematics Education in America at TEDxBushnellPark in Hartford, Connecticut. I did a post on this issue (click here to see it), as well as a huge school project on the issue (click here to see the research paper).

In the talk, I will be bringing up a famous math problem that originated with Leonhard Euler. I won't post about that problem today, but I will give a little story from the life of Euler.

Leonhard Euler was born on April 15, 1707 in Basel, Switzerland. He chose the path of a mathematician (if he didn't, I probably wouldn't be talking about him right now), and ended up as one of the most prolific mathematics writers of all time.

Some of his biggest contributions include standardizing the notation for the number e and π, coming up with Euler paths and circuits, and revamping most of the branches of mathematics that were known in his time. He was said to be able to entertain a child, scratch a cat, and calculate math problems simultaneously.

On this blog, I have mentioned sequences a lot. For instance, the triangular numbers are a sequence. The negative-first powers are a sequence. The powers of 1/2 are a sequence. Let's look at these three sequences:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55...
1, 1/2, 1/3, 1/4, 1/5, 1/6, 1/7, 1/8, 1/9, 1/10...
1, 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, 1/128...

We haven't done it before, but a common thing to do is add up all of the numbers in these sequences. In some sequences, this seems odd to be doing, but in other ones, it seems completely understandable. Let's try it with those three.

1 + 3 + 6 + 10 + 15 + 21 + 28 + 36 + 45 + 55 +...

The individual numbers themselves are going towards infinity, so the sum of them is probably infinity as well. Generally, if the numbers in a sequence are growing, then the sum will be infinite. This type of sequence is called a divergent sequence, and questions like their sum are generally not asked.

What about the next series. This one is a little harder to tell, but if we group some things together, it will be easier to see.

1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ...

Compare this to the following sequence, which is clearly less than this one.

1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + ...
1 + 1/2 + (1/4 + 1/4) + (1/8 + 1/8 + 1/8 + 1/8) + ...

Since all of the terms in the first sequence seem to be bigger than the second one, then the first one has to sum to a greater total. But, the second sequence simplifies to 1 + 1/2 + 1/2 + 1/2 +..., which will end up as infinity. So, a sequence that has a bigger sum must also be infinite as well. This makes that a diverging sequence.

What about the third one? Well, combining terms together won't do much, but let's try figuring out what the sum is by setting the sequence equal to x.

x = 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 +...

Now, what if you were to multiply that by two? You would get:

2x = 2(1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + 1/128 +...)
2x = 2 + 1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 +...
2x = 2 + (1 + 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 +...)
2x = 2 + x
x = 2

So, the sum of this sequence is not infinity, but rather the natural number two. Since the sequence does not diverge to infinity, it is called a convergent sequence. These are the type that are generally summed to their total, since that number can then be applied.

Anyways, some of Euler's students were analyzing a much more complicated convergent sequence to figure out its sum. They had been adding numbers, and after they got to the seventeenth number in the sequence, they found their answers to be different at the fiftieth decimal place. Since they didn't want to go back through that tedious process, they argued and argued over who got it right.

Eventually, they went to their professor, Euler, to ask him for assistance. After asking for the problem, he mentally calculated out the answer to the fiftieth decimal, and was able to determine who was correct.

I found that story kind of funny, but also really impressive. To be able to calculate that difficult of a problem to that high of a degree shows some real mathematical talent.

Problem of the Week Day 5: Week of 6/17/13 - 6/21/13

Today is the final day of the first 2013 problem of the week! The goal of the problem is to find x, so I have saved the x variable for the last part. In the hard problem, we used x with a subscript, but the plain x will represent the solution to today's problem. We have already covered trigonometry, algebra, number theory, and geometry, so I thought it would be good to devote Friday to probability. I might incorporate some game theory as well on Fridays in later problems.

Easy:

Take the following set of data:

59, 38, 45, p, 29, g, 27, y + 1.24, q, 41, z + 5.92, 37, b, 21

What is the median of this sequence?

x = _____

Medium:
First, solve for the following six variables, which will come up in the data set you will be analyzing.

d = a ÷ 1000
g = h ÷ 7
j = u ÷ 2
k = u ÷ 4
r = p ÷ 10
s = q ÷ 3

The following data represents the correlation between the age of test subjects (x) and the average dollar amount that they spend at a nearby shopping mall per visit (y). Some of the data was rounded by the data collectors, and some of it is as accurate as possible, meaning that there will be some longer decimals.

x	y
s	q + r + g
t	u
q - h	q
h - u	h - t
r	h + t
r - t	h - r
d + t	p - h
g	h - g
d	h
2t	q + t
k	q + t
k	u + t
j	4s

Find the function that describes the line of best fit for this data set. Use the traditional y = mx + b to make the equation of the line, using m and b as your variables. Then, determine the average spending of a 65-year-old woman using the equation. The average spending will be the x value that we always end the problem with.

m = ______
b = ______
x = ______

Hard:
First, use the variables throughout the week to solve for m.

        t – 0.77 – p(100q₁₀ + q₁₂)(q₉ – q₁₂)
m = –––––––––––––––––––––––––––––––
       q₀ + 2c – a + b + q₆ + q₁₀ + q₁₁ + q₁₄

Now, pretend that there is a room with m people inside of it. Determine what the odds are that in this room, two people share the same birthday (just the month and date, not the year). In other words, solve the birthday paradox for a room of m people. x will represent the odds of this happening.

x = _____%

Problem of the Week Day 4: Week of 6/17/13 - 6/21/13

Today is day four of the problem of the week! We already covered trigonometry, algebra, and number theory, so I will devote today to geometry. Good luck!

Easy:
Take the following rectangle:

What is the area and perimeter of this rectangle? Since a and p were already used on the triangle, use b for the area and q for the perimeter (just move up a letter).

b = _____
q = _____

Medium:
First, determine the values of u and q.

u = n ÷ 1000
q = p ÷ 2

Now, take the following trapezoid, using these values:

What is the area of this trapezoid? Use a for area.

u = _____
q = _____
a = _____

Hard:
Find the area t of a regular polygon with x₃(b - a - c) sides and a perimeter of q. Round to the nearest hundredth.

t = _____

Good luck!

Problem of the Week Day 3: Week of 6/17/13 - 6/21/13

Today is day three of the problem of the week! Monday was trigonometry, Tuesday was algebra, so we will make Wednesday be number theory. This branch of math might sound foreign to you, but the concepts within it that I plan to use in the problems have all been taught to you at some point in your life.

Easy:
Find the Greatest Common Factor of 100y and 100z. We will call this number g.

g = _____

Medium:
What is the number of dots in a regular t-gon array whose sides are of length h? In other words, look at the figurative families, and find the tth family. Then, find the hth number in that family. Call that number n.

n = _____

Hard:
Take the following sequence:

x₁, x₂, p + y₁, (q + 110x₃ + 4)/(t), q0 – q9 – q10 – q11 – q12 – q13

Now, find the explicit formula for this sequence. For the simplicity of the variables, I will tell you that the formula should be a quadratic function. Use the ax² + bx + c model, with a, b, and c being the answers to this problem. They are also the coefficients of the equation.

a = _____
b = _____
c = _____

Good luck!