Saturday, December 31, 2011

Check out my 2011 Greatest Hits Video and learn a pattern as well.

A couple of posts ago, I presented my greatest hits video for my Mathemagics performance. If you haven’t seen it before, it is what it says, a combination of math and magic. Now that 2011 has just twelve hours left, it might be a good time to present my 2011 greatest hits video.

For some upcoming shows, I will be at Gathering for Gardner early spring in Atlanta, Curious1729 Con in St. Louis, and many other exciting performances, listed on

As you probably noticed in the Mathemagics performance, squaring numbers is a big part of my show. I’ve already shown how to square 2-digit numbers that end in five, as well as the close-together method (which works for squaring as well), but there is a certain type of 2-digit number, or any digit number for that matter, that follows a specific pattern. Let me show you:

3^2 = 09 (or just 9)
33^2 = 1089
333^2 = 110889
3333^2 = 11108889
33333^2 = 1111088889

6^2 = 36
66^2 = 4356
666^2 = 443556
6666^2 = 44435556
66666^2 = 4444355556

9^2 = 81
99^2 = 9801
999^2 = 998001
9999^2 = 99980001
99999^2 = 9999800001

See the pattern?  This only works for numbers that are all 3’s, 6’s, or 9’s, but it is pretty cool! It’s not a very impressive effect to do for someone, but a really cool pattern, and trick.

Saturday, December 24, 2011

Algebra + Geometry + Trigonometry + Arithmetic = A Perfect Cool Math Stuff Post...

A few weeks ago, we saw the second and third cube roots of one, which used those complex numbers, with i in it. While proving that their cubes were one, we learned how to do some operations with them using algebra. However, there is a really cool geometric way to do the math as well.

First off, take the Cartesian Plane. We'll label the x-axis with the integers and the y-axis with the imaginary numbers.

To plot a point, just move the constant to the left/right and the coefficient up/down. So, for 2 + 3i, you would move two to the right and three up. For 4 - 8i, you would go four right and eight down.

Let's say you had to add these two numbers together. First off, you could do it algebraically, which isn't that cool, and you get 6 - 5i. However, you could also add them on the imaginary plane.

Let's plot the points:

How about we just see what happens when we plot 6 - 5i. We'll connect all of the points to make a little quadrilateral.

What do you notice? The points have formed a parallelogram, with the opposite sides equal in length and parallel. At this point, we should go over how to figure out the lengths.

To figure out the lengths, we use the Pythagorean Theorem. Remember back to the problems of the week? We saw that a^2 + b^2 = c^2? We will be proving that in a later post, believe it or not. Anyways, we will be doing exactly that, literally! The form for a complex number is a + bi, so we can just plug those in and solve for the length c.

In the case of 2 + 3i, we will do 2^2 + 3^2 = c^2.

2^2 + 3^2 = c^2
4 + 9 = c^2
13 = c^2
3.606 ≈ c

To keep things more simple, we will use two different points; 8 + 6i and -3 + 4i. In this case, the lengths are 10 and 5.

Of course, we could add them together pretty easily with algebra or geometry to get 5 + 10i. However, let's figure out the angles of the lines. To do it, we use trigonometry, which we also used for the problem of the weeks.

To briefly review/explain, we will take the b term (which is the opposite side) and divide it by the c term, or hypotenuse, to get the sine of the angle. To retrieve the angle, press the sin^-1 button on your calculator.

In this case, here are our angles:

8 + 6i: 37° (approx.)
-3 + 4i: 127° (approx.)

Now, we will learn how to multiply them together. All you have to do is two easy steps: multiply the lengths and add the angles. For this one, we multiply 5 and 10 and add 37 and 127.

5 x 10 = 50
37 + 127 = 164

If you do some trigonometry, you will get an approximate answer of 48 + 14i, which happens to be the correct answer.

You can also easily multiply complex numbers by real numbers; just multiply the length by that number. If you think about it, this corresponds to the original method. These are just a handful of the many cool things you can do with the imaginary plane. To be honest, I really love algebra and arithmetic, but I'm not one of those people who is all over geometry (I'm not Archimedes, the famous Greek mathematician who died doing geometry in the sand), but this little method is just so cool.

Saturday, December 17, 2011

To infinity and beyond!! With numbers I mean...

We have done a lot with infinite series already, we used it to supposedly prove the Communitive Property wrong, to practice systems of equations in the problem of the week, and even prove that the prime numbers fall into this category. However, we haven't really looked at infinity. Is it a number? Can we define it? What is greater than infinity?

Let's look at an infinite series. Take the natural numbers; 1, 2, 3, 4, 5, 6, 7, ... This sequence takes us out to infinity. How about 10, 20, 30, 40, 50, 60, 70, ... This takes us to infinity also. However, which sequence has more terms in it?

One side of you is saying, for every term in the multiple of ten series, you need ten terms from the natural number series to get that high. Therefore, the first series must be ten times bigger. The other side of you is saying, for both series, we are going out forever. This means they must be equal. When I first saw this, I was leaning greatly towards the first side. However, this isn't quite right.

For each number in the first series, we can pair it with a number in the second. For example, we can pair 1 with 10. Then 2 with 20. Then 3 with 30, 4 with 40, 5 with 50, and so on. If the first side is greater, then you must run out of terms on the second side. However, the second side is infinite also. If the second side were greater (which I cannot make an argument for that), then we would run out of terms on the first side. This means that the sides must be equal.

In fact, any series is the same size as the set of natural numbers if you can write it out with no infinite gaps in between. For instance, the integers:

... -3, -2, -1, 0, 1, 2, 3 ...

You can rewrite this to get:

0, 1, -1, 2, -2, 3, -3, 4, -4...

Since there are no infinite gaps here, that must mean that it is equal to the set of natural numbers.

What about the fractions (you could say the rational numbers, but rational numbers are simplified and fractions can or cannot be)? It's a big statement, but we can try it.


This is a table of the fractions. Is there a way to write this table without any infinite gaps? Turns out there is. If we draw lines diagonally, and put it all together, it will have no infinite gaps. It would look like this:

1/1, 2/1, 1/2, 3/1, 2/2, 1/3, 4/1, 3/2, 2/3, 1/4, ...

This means the amount of fractions is the same! This quantity is denoted with the hebrew letter alef, also called alef zero, or alef naught. However, do you think there is another type of infinity, or is every set equal? I will come back to this in a couple months, and we will find the answer and why.

Saturday, December 10, 2011

A Few Awesome Properties of Quadratics

In the problem of the week, I would occasionally throw in some quadratics, or problems with squaring involved. If you’ve seen my performances, you might have noticed I like things that have to do with squaring, and quadratics is definitely one of them. Quadratics is normally done in the form f(x) = ax^2 + bx + c, which looks nice, but isn’t so useful. However, there is a form that is pretty commonly used that is almost magical in a way.
This form is the form f(x) = a(x - h)^2 + k. First, I will show you how to get into this form, then we will look at its properties.
The technique we will use is called “completing the square.” The first thing you do is factor the a out of the equation. Let’s use 1/2x^2 + 3x + 5 as an example.
1/2x^2 + 3x + 5
1/2(x^2 + 6x + 10)
Next, plug that new b term into b^2/4. What I find easier is to divide the b term by two and then square it, so the division doesn’t get too messy.
6/2 = 3
3^2 = 9
What this means is that the equation (x^2 + 6x + 9) is square, or a “perfect square trinomial.” It is in fact (x + 3)(x + 3). To figure out that three, all you have to do is divide that b term by two.
There is only one problem though. We have x^2 + 6x + 10, not x^2 + 6x + 9. However, we can put the (x + 3)^2 there, but we must add one to the (x + 3)^2, which ends up getting multiplied by the 1/2 to get a constant of one-half at the end. So, we have:
1/2(x + 3)^2 + 1/2
And that is vertex form. Let’s look at what is cool about it. 
First off, both a’s happen to be equal. This is not coincidence, as we factored out the a in order to switch forms.
What I find really cool is those two terms we couldn’t control, h and k. In this case, they are -3 and 1/2 (the equation is a(x - h)^2 + k, not a(x + h)^2 + k, so h is -3). What does that have to do with anything? Take a look at the graph of this equation.
If you’ll notice, the vertex of this parabola (graph of a quadratic function) is in fact (-3, 1/2). In fact, you can actually graph a quadratic equation using only vertex form, just with this fact and a. I think that is cool that there is a format out there that can do that.
Bonus: Say that instead of doing a(x - h)^2 + k, you did (bx - h)^2 + k. b is unfortunately not the same as the other b, but much more useful. This form is messier, but there is a cool thing about it. In this case, the vertex is actually (bh, k), which is interesting.
What the a factor did was it told us the factor for the “vertical stretch/compression” of the  parabola, or the number you multiply every single y value by. In the case before, every y value of plain (x + 3)^2 + 1 was multiplied by 1/2 to create the graph.
For this form, there is actually a “horizontal stretch/compression,” which isn’t common in quadratics. This form gives you a stretch/compression of 1/b. You might see it as just b, but it is usually written like this:
((1/b)x - h)^2 + k)
Either way, it is pretty cool.

Saturday, December 3, 2011

Fibonacci Day: Fibonacci Magic Trick

I don’t know if you noticed, but today is a Fibonacci Day! It is December third, and three is a Fibonacci number. We’ve looked at some cool patterns in Fibonacci numbers, but it’s time we learn how to do some magic with them! Let’s look at the Fibonacci numbers, but this time in lines.

Line 1: 1
Line 2: 1
Line 3: 2
Line 4: 3
Line 5: 5
Line 6: 8
Line 7: 13
Line 8: 21
Line 9: 34
Line 10: 55
Line 11: 89
Line 12: 144
Line 13: 233
Line 14: 377
Line 15: 610
Line 16: 987
Line 17: 1597
Line 18: 2584
Line 19: 4181
Line 20: 6765

What do you think the sum is of all of the numbers up until line, say thirteen? I can tell you immediately that it is 609. How?

Wait, we know this! Remember when we were adding Fibonacci numbers? The answer was always two ahead minus one! But we can take it one step further.

Let’s make our own Fibonacci sequence this time, starting with any two numbers we want. If you are doing it on pencil and paper, you might want to stick 1 – 10, or you can make an excel or numbers spreadsheet and do it as high or little as you want. In this case, we’ll start with 4 and 7.

Line 1: 4
Line 2: 7
Line 3: 11
Line 4: 18
Line 5: 29
Line 6: 47
Line 7: 76
Line 8: 123
Line 9: 199
Line 10: 322
Line 11: 521
Line 12: 843
Line 13: 1364
Line 14: 2207
Line 15: 3571
Line 16: 5778
Line 17: 9349
Line 18: 15127
Line 19: 24476
Line 20: 39603

What is the grand total up to line eight, you can do any line. The answer is 315. It’s still the same exact principle, except for one little thing.

We move up two lines, and then subtract line two. It is the easiest of all things to do! No matter how gigantic the numbers are, you can still pull it off. What’s even cooler is that there is no specific line you are adding up to, unlike other methods that only go up to line ten.

Bonus Trick: Make one of these sequences yourself, and make sure you have at least ten lines. Now, divide the last line by the one before it. In this case, we would be doing 39603 ÷ 24476. You should have 1.61, right?

This is the same thing as the golden ratio appearing in the Fibonacci sequence. To prove it, we will actually do something a little different than usual. We will add fractions “badly.” If you were a young kid, how would you guess adding fractions works?

I’d say add the numerators, then add the denominators. Like ½ + ¼ should be 2/6, or 1/3. This doesn’t give you the right answer, but it does assure that the answer is in between the two fractions. In this case, we are using line ten and line nine.

Line nine has its own formula: 13x + 21y, assuming that line 1 is x and line 2 is y. Line ten has formula 21x + 34y. So, we have:

21x + 34y/13x + 21y

This is the same as adding fractions badly. This says that this ratio is between 21x/13x and 34y/21y.

21x/13x = 21/13 = 1.61538…
34y/21y = 34/21 = 1.61904…

Both of these numbers begin with 1.61, meaning any number in between them will begin with 1.61. This proves that line ten over line nine is always 1.61, a great bonus prediction effect to the trick.

Saturday, November 26, 2011

Why √2 is not Rational. Or is it...

Rational numbers are numbers that are the quotient of two integers. In other words, a number is rational if it is p/q, and p and q are integers (and q ≠ 0). An irrational number is a number that is not rational. For instance, numbers like pi, or e, the golden ratio, all of these are irrational. However, these numbers are hard to work with, so let's use √2.

What we will do is prove it is irrational. To do this, we will use a technique called proof by contradiction, by assuming the opposite and later getting into confusion. In this case, we will assume √2 is rational.

√2 = p/q

Assume p/q is in lowest terms, as we can write any rational number in lowest terms. In order to get rid of that square root sign and deal with easier numbers, let's square both sides.

(√2)^2 = (p/q)^2
2 = p^2/q^2

Now, let's multiply both sides by q^2 to get rid of the fraction.

q^2(2) = q^2(p^2/q^2)
2q^2 = p^2
p^2 = 2q^2

This means that p^2 must be even. In that case, p is even because an even squared is always even, an odd squared is always odd. So, we know p is even, or of the form 2a.

Let's plug 2a in for p.

(2a)^2 = 2q^2
4a^2 = 2q^2
2a^2 = q^2
q^2 = 2a^2

In this case, we are running into the same thing. Here, q^2 is even, meaning that q is even. So, we have p as even and q as even. However, we said that p/q is in lowest terms. This leads us to say that √2 is irrational.

I would never have thought that you could actually prove a number to be irrational. I'd assume you can with other numbers, such as π, e, the golden ratio, all of those guys. That is definitely one of my favorite proofs.

Saturday, November 19, 2011

First, Geometry fails us. Next, Arithmetic. Now, it's Algebra's turn...

Last week, we took a universal property of mathematics and watched it fail to work. This week, we will do it again, but with clear, simple algebraic proofs.

A little over a month ago, we proved that 64 = 65. Now, let's take that down a notch, and prove that 1 = 2. We will prove it two ways, one will be a little easier to understand, and one will involve complex numbers, which we worked with around four weeks ago.

Easy Proof: Let's say that a = b. Pretty simple. Now, we'll multiply both sides by a. This will be step one.

a = b
a(a) = a(b)
a^2 = ab

How about we add a^2 to both sides. This is step two.

a^2 = ab
a^2 + a^2 = a^2 + ab
2a^2 = a^2 + ab

For step three, we will subtract 2ab from both sides.

2a^2 = a^2 + ab
2a^2 - 2ab = a^2 + ab - 2ab
2a^2 - 2ab = a^2 - ab

For step four, we will factor out a two from the left hand side.

2a^2 - 2ab = a^2 - ab
2(a^2 - ab) = a^2 - ab

For step five, we will divide both sides of the equation by a^2 - ab to give us 2 = 1.

2(a^2 - ab) = a^2 - ab
(2(a^2 - ab))/(a^2 - ab) = (a^2 - ab)/(a^2 - ab)
2 = 1

Complex Proof (literally!): This one does involve complex numbers, so it might become a little bit challenging. However, it is pretty easy to understand, as long as you realize that √(-1) = i.

Let's remind ourselves that -1/1 = 1/-1. For step one, let's take the square root of both sides.

-1/1 = 1/-1
√(-1/1) = √(1/-1)

We can now simplify that to give us:

√(-1)/√(1) = √(-1)/√(1)

For step three, we can eliminate all of the square roots and replace them with 1s and is.

i/1 = 1/i

For step four, let's divide each side by two.

i/2 = 1/2i

For step five, how about we add 3/2i to both sides. Strange attempt, but we can go ahead and do it.

i/2 + 3/2i = 1/2i + 3/2i

Let's multiply through by i. That will be our step six.

i(i/2 + 3/2i = 1/2i + 3/2i)
i^2/2 + 3i/2i = i/2i + 3i/2i

For step seven, we can simplify this whole mess and see what we get.

i^2/2 + 3i/2i = i/2i + 3i/2i
-1/2 + 3/2 = 1/2 + 3/2
2/2 = 4/2
1 = 2

Again, we are ending up with the strange solution of 1 = 2.

Why on earth could this be? Maybe arithmetic and geometry can make mistakes, but algebra! Turns out, algebra is fine. These proofs are fallacies. See if you can figure out which step is incorrect, and then read the below.

Easy Proof's Fallacy: Turns out, step five was a fallacy. Where were we there?

2(a^2 - ab) = a^2 - ab

We divided both sides by a^2 - ab to get 2 = 1. The problem lies in the a^2 - ab division. Let's look at it. We know that a = b, so let's plug a in for the b.

a^2 - ab
a^2 - a(a)
a^2 - a^2

We have ended up dividing by zero. Since this is not allowed in mathematics, we cannot do this step. That means that 2 ≠ 1, or at least this does not prove it.

Complex Proof's Fallacy: Again, our simplification was where the fallacy lied. In this problem, it was in step two, when we separated the square roots. Let's look at it:

√(-1/1) = √(1/-1)
√(-1)/√(1) = √(-1)/√(1)

We have jumped one step too far. This property we just used is only true for positive square roots. Remember, a square root is a number when multiplied itself gives you the radicand, or number inside the square root symbol. If the square root is negative, then it does not hold true all the time.

If that was hard to understand, let me lay it out in more simple terms. Say we have:


One way we could do it is multiply together the -1s and get 1.


But if you used this property, you would have:

i • i

This gave us two different answers, so this property just doesn't hold in these circumstances. There are tons and tons of proofs like this, even some that involve calculus. I think that these false statements are really cool to look at and try to figure out what the mistake is. So fortunately, algebra has not went under yet.

Saturday, November 12, 2011

Watch the Communitive Laws fail right before our eyes!!!

A couple of months ago, we were working with infinite series, and even proved that the primes are an infinite series. Today, we will work with another infinite series, known as the harmonic series. It goes like this:

1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + 1/7 - 1/8...

Let's add this up. Because of its nature, we can tell it won't go to infinity. If you can picture it, it kind of moves over a certain amount, then goes back in between, then a little forward again, and so on, zeroing in on a specific number. In fact, it is zoning in on a number called ln2, which is somewhere around .683. I don't really know the proof, but it involves a little bit of calculus.

However, let's try something else. How about we rearrange the numbers. Let's go for every odd denominator, we do two even ones.

1 - 1/2 - 1/4 + 1/3 - 1/6 - 1/8 + 1/5 - 1/10 - 1/12 + 1/7 - 1/14 - 1/16...

It is still the same series because we are adding every odd denominator once and every even denominator once. Let's tackle this in chunks. Let's just group together some terms every so often.

(1 - 1/2) - 1/4 + (1/3 - 1/6) - 1/8 + (1/5 - 1/10) - 1/12 + (1/7 - 1/14) - 1/16
1/2 - 1/4 + 1/6 - 1/8 + 1/10 - 1/12 + 1/14 - 1/16

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

Just with that little adjustment, we have turned the same series into 1/2ln2. What we've just seen is that the little rule we learned back in second grade with the turn-around facts, fact families, fact triangles, all of that failing right before our eyes. In fact, this Communative Law that we learned can fail when dealing with infinite series involving negative and positive numbers.

What I find odd is that you can rearrange this series to get whatever number you want. If you tried hard, you could rearrange this to get π, e, or whatever else you want!! I haven't really looked into this, but it seems pretty cool.

Bonus Proof: While we are watching the Communative Law fail, we should ask a question. How do we know it is true? Why should 7 bags of 4 apples be the same as 4 bags of 7 apples? This proof is so obvious, yet I would never had thought of it! In fact, one of the things I've wondered for a while is why the Communative Law is true.

Think of it this way. Take a 4 x 7 rectangle made up of dots. How would we figure out how many dots there were total? Well, we could say, "there are 4 rows made up of 7 dots in each row," or, "there are 7 columns made up of 4 dots in each column." Both ways, we are finding the amount of dots in the rectangle. Which one is right? They both are, which proves why the Communative Law must be true.

Saturday, November 5, 2011

Fibonacci Day: Even Fibonacci Numbers

Today is a Fibonacci Day. It is the fifth of November and five is a Fibonacci number. Let's move on from the squares of Fibonacci numbers and just deal with plain old Fibonacci numbers. Here they are:

1  2  3  4  5  6   7    8    9   10  11   12
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144

Which Fibonacci numbers are even? Well, we've got the 2, 8, 34, 144... Which numbers are those? They are F3, F6, F9, and F12. This time, the pattern is right in our faces!! They are all multiples of three! Why on earth is that?

Notice that the first three Fibonacci numbers follow the pattern of odd-odd-even. The next one is made up by adding the previous two. So, an odd plus an even is an odd. Then, the even plus the odd makes another odd. Then, the odd plus the odd makes an even. And we are back to where we started, odd-odd-even.

What's interesting is that every fourth Fibonacci number is a multiple of three. Every fifth Fibonacci number is a multiple of five. Every sixth Fibonacci number is a multiple of eight. Every seventh Fibonacci number is a multiple of thirteen, and the only multiples of thirteen! I think that is pretty cool!!

I'm not sure why that pattern continues, but if you know, please tell us. It would be pretty cool to see.

Bonus Proof: Since that was pretty short and simple, I'd like to show you one more little thing. You've probably seen that x^0 = 1, and wondered why. Why isn't it like zero or something? Well, let's take the number x/x. How do we simplify that?

Some people might say, it's a number over the same number, and everything over itself is one. That is absolutely correct. However, you might tackle it a little more algebraically, and realize that you have the same base raised to an exponent. You have x^1/x^1. We can use the law of exponents to subtract the 1 from the 1 to get 0, giving us x^0.

One way, we got one. The other way gave us x^0. Either way, we have that x^0 = 1. I thought that was pretty cool.

Saturday, October 29, 2011

Cool Divisibility Stuff (There's a lot!!)

This week, we are going to talk about divisibility, a pretty broad area of mathematics that we were lucky to be introduced to in fifth grade. I’ve looked into it more, and even use it for a lot of tricks in math.

There are a bunch of tricks for it, and I will explain them all and prove them to you. This post will almost be a lesson, but all of the rules are very cool! Let’s go through them all.

Divisibility by one: If the number has no decimal, it is divisible by one. For these rules, we are only working with decimals. To prove it, we look at the Closure Property, which states that any number multiplied by one will remain the same number. This means every integer is a multiple of one.

Divisibility by two: If the number ends in two, four, six, eight, or zero, it is divisible by two. The definition of even number is, “A natural number that is divisible by 2.” It is also defined as, “A whole number that has 0, 2, 4, 6, or 8 in the ones place.” Since these two things are the same, this means that this rule is valid.

Divisibility by three: This rule is probably one of the coolest, and most magical. If you add up the digits in the number and this gives you a multiple of three, this means that the number is a multiple of three. If the sum is too big for you to know if it has a three factor, go ahead and add up the digits again until you do know. I will prove this one along with the divisibility by nine rule, as it is very similar, and a number that is divisible by nine is also divisible by three.

Divisibility by four: Look at the number’s last two digits, and completely ignore the rest. If the number you are looking at is a multiple of four, then the whole number is a multiple of four. If you think about it, every number is 100x + y with y being the last two digits. Since we know 100x is a multiple of 4 (25x • 4 = 100x), all that leaves us with is y. This means that if y is a multiple of 4 also, then adding 100x won’t change this.

Divisibility by five: If the number ends in five or zero, it is a multiple of five. If you think about it, the digits from 0 to 9 end in either five or zero when multiplied by five. Since the last digit of a multiplication problem is the last digit of the last two digits multiplied together, we are left with all of the integral numbers ending in five or zero.

Divisibility by six: Since six is the product of three and two, we can just put both of these rules to use since they share nothing in common. If the number follows through with the divisibility by three rule (just adding to a multiple of three, no need for a multiple of six, 1 + 2 = 3 and 12 is a multiple of six), and is even, it is a multiple of six. Check the divisibility by two and three rules for more.

Divisibility by seven: This is also a really cool one, one my teacher didn’t even know. Take the last digit of the number and double it. Then, subtract that from the rest of the number. Keep repeating this until you have a one or two digit number (negatives are okay, you can just turn them positive). If the number you have is a multiple of seven, the original number is a multiple of seven. Let’s do one real quick, the number 224.

First, double the 4 to get 8. Subtract 8 from 22 to get 14. Since 14 is definitely a multiple of seven, the 224 is as well. 224 happens to be 32 • 7, so we did our math correctly. To prove this, we will use a little bit of Algebra. Our formula states that if 10r + x is a multiple of seven, then r – 2x is a multiple of seven. Let’s say the number is a multiple of seven. This means that r – 2x is a multiple of seven.

r – 2x = multiple of seven

Let’s try multiplying both sides by ten.

r – 2x = multiple of seven
10(r – 2x) = multiple of seven (and seventy)
10r – 20x = multiple of seven (and seventy)

Forget it is a multiple of seventy. We are not looking for that. However, let’s go in a different direction and add 21x to both sides. Since 21 is a multiple of seven, we still have a multiple of seven on the right-hand side of the equation. However, the left becomes:

10r – 20x = multiple of seven
10r – 20x + 21x = multiple of seven
10r + x = multiple of seven

And this brings us back to where we started. It is a little complicated, but I think it is pretty cool.

Divisibility by eight: Similar to divisibility by four, we will be looking at a block of digits on the right. However, note the fact that eight isn’t a multiple of one hundred. However, it’s a multiple of one thousand. So, we will look at the last three digits and see if it is a multiple of eight. Since this might be a little hard for you, as I don’t have my eights memorized that far, you can look at the hundreds place and see if it is odd or even. If it is even, this means that you can check the last two digits for divisibility by eight because 200 is a multiple of eight. If it is odd, check the last two digits for divisibility by four, but then make sure it is not a multiple of eight, or a multiple of eight plus four. Since 104 is a multiple of eight, this should work as well.

You can even take these principles to divisibility by any power of two. The power you are raising two to is the amount of digits you need to test at the end. So, for divisibility by 64, just check to see if the last six digits are divisible by 64 because 2^6 = 64.

Divisibility by nine: This is unquestionably the most magical of all of the proofs, and is used for so many things. It is the answer to numerous magic tricks, math tricks, and even provides loads of ways to check answers, including divisibility, digital roots, and mod sums. In fact, you can even cube root numbers based on this rule. To find divisibility by nine, you simply add up the digits of the number, and if that is a multiple of nine, the whole number is a multiple of nine. To prove it, let’s test the number 234. 2 + 3 + 4 = 9, so it is definitely a multiple. But why? Let’s write the number 234 in expanded notation, something you learn around third or fourth grade.

2 x 100 + 3 x 10 + 4 x 1

We can rewrite these to get:

2 x (99 + 1) + 3 x (9 + 1) + 4 x (0 + 1)

Let’s distribute the 2, 3, and 4 into the groups of numbers beside them.

(2 x 99) + 2 + (3 x 9) + 3 + (4 x 0) + 4

We can rearrange this to get:

[(2 x 99) + (3 x 9) + (4 x 0)] + 2 + 3 + 4

You might notice, but the sum inside of the brackets is definitely a multiple of nine, as it is being created by all multiples of nine. So, we can eliminate it to get the sum of all of the digits. Here, I don’t think the proof lives up to the magic in the actual trick, but it is pretty cool!

Divisibility by ten: To test divisibility by ten, all we need to do is see what the last digit is. If it is a zero, then the number is a multiple of ten. To prove it, we look at the Power of Zeros rule, which says that to multiply a number by a power of ten, just tack on that power amount of zeros. This means that a number multiplied by 10 has one zero at the end. You can elevate this to say that a multiple of 100 has two zeros at the end, a multiple of 1000 has three and so on.

Divisibility by eleven: To test divisibility by eleven, alternately subtract and add the digits and if you end with a multiple of eleven (it may be zero or negative), then your original number is a multiple of eleven. So for the number 1353, you would go 1 – 3 + 5 – 3 = -2 + 5 – 3 = 3 – 3 = 0. Since zero is a multiple of eleven, the full number is a multiple of eleven.

To prove it, we know that if you subtract eleven from a number constantly, the number still keeps its status as a multiple of eleven. Therefore, let’s put ten to its powers (creating the place values) and see what happens when you constantly subtract multiples of eleven.

100 – (0 • 11) = 1
101 – (1 • 11) = -1
102 – (9 • 11) = 1
103 – (91 • 11) = -1
104 – (909 • 11) = 1
105 – (9091 • 11) = -1
106 – (90909 • 11) = 1
and so on…

This convinces me enough, but I’m not sure of where to take it from there. If you want to add anything, please comment. It would be interesting for all of us.

Divisibility by twelve: Let’s take two distinct ones, divisibility by four and three. If the number is divisible by four and its digits add to a multiple of three, it is a multiple of twelve. Like the divisibility by six rule, this works as well. You can even take this principle to other numbers like this, like 14, 15, 18, 21, 22, 24, 28, and so on.

Divisibility by thirteen or greater (the “create a zero, kill a zero method”): Since this is a little more complicated, let’s learn it through an example. Is 2756 divisible by thirteen? First, we must look at the last digit of 2756, six. We need to find a multiple of thirteen that ends in six. If there is one, that means the number is not divisible, and you have learned a new property about that number. However, 13 x 2 = 26, so we have found one. We must subtract the 26 from 2756 to create a zero. So, 2756 - 26 = 2730. Now, we kill the zero, or just ignore it. Now, we create a zero by subtracting 13 (273 – 13 = 260). After killing it, we have 26. Since 26 is a multiple of 13, 273 and more importantly, 2756 are multiples of 13. You could even have subtracted 156 from the original number if you knew that it was 12 x 13. This would make it only take two steps instead of three.

Even though it is a lot in one dose, it is definitely worth practicing. These are really cool principles, and you could probably figure out a few tricks out of them. If you watch my Mathemagics show closely, I will use divisibility to perform one of the tricks, so you can investigate that as well.

Saturday, October 22, 2011

How many cube roots of one are there?

When getting into algebra, you will hear the term "real number" a lot, without an explanation as to why you can't just put "number." Since no one asks the question, it just slides by, and when the question is asked, the teacher responds, "You'll learn that in Algebra II," or something along those lines.

The answer to that question is that there is such thing as "imaginary numbers," or complex numbers, which are made up of a constant, a coefficient, and the letter i, which symbolizes the square root of negative one.

What is the square root of negative one? Some say negative one. Well, (-1) x (-1) = 1, so that is incorrect. People then turn around and say one. Well, 1 x 1 = 1, so that is incorrect. Then, they might try 1/2. Well, 1/2 x 1/2 = 1/4, so that is wrong. They will keep trying things until they give up.

What is the answer? If you think about it, a negative times a negative, or a negative squared, is a positive. A positive times a positive, or a positive squared, is a positive. So, you cannot square a real number and get a negative. So, mathematician Heron of Alexandria came up with the letter i, and began using that as the square root of -1. So, then, by the Multiplication Property of Square Roots, you can conclude that √(-9) would be 3i because you can break that into √(9)√(-1) = 3√(-1) = 3i. Rafael Bombelli built on his works, and make this concept a regular part of Algebra.

Now, let me introduce you to one more thing about imaginary numbers before I show you about the cube rooting. When you write these terms out, you write like 5 + 3i, which means 5 + √(-9), just like how you'd write a real square root in Algebra. The 5 + 3i would be known as a complex number, which is a number involving i. The conjugate of a complex number is to keep the same expression, but switch the operation separating them. For instance, the conjugate of 5 + 3i = 5 - 3i because we kept the same term, but switched the operation.

If you think about it, a complex number's conjugate is equal to the number because any number has two square roots, a positive one and a negative one. So by making the i term negative, we are just looking at the other root.

Now that we've gotten that out of the way, let's get to the good part! Let's take the complex number -1/2 + i/2√(3). It is the same thing, with a constant of -1/2 and coefficient of 1/2√(3). How about we cube it.

(-1/2 + i/2√(3))(-1/2 + i/2√(3))(-1/2 + i/2√(3))

First, we'll square it. We can use FOIL for that. If you don't know, it stands for "First, outer, inner last." It's basically the distributive property made simpler for multiplying binomials.

(-1/2 + i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) - i/4√(3) - 3/4
-1/2 - i/2√(3)

We ended up with the conjugate of before. That's interesting. Let's finish off by multiplying by the -1/2 + i/2√(3).

(-1/2 - i/2√(3))(-1/2 + i/2√(3))
1/4 - i/4√(3) + i/4√(3) + 3/4
1 - i/4√(3) + i/4√(3)

What do we do with that? Well, there are i's in both terms, so we can combine them. However, look closer. They are opposites of each other, or the additive inverse of each other. What does that mean? The definition of additive inverses are two numbers in which when added together give you zero. So, these two confusing numbers simplify to zero!

1 - i/4√(3) + i/4√(3)
1 + 0

So, we are left with 1 as our answer! We did nothing wrong there. -1/2 + i/2√(3) is in fact the cube root of one, as well as its conjugate and of course, the integer one. Was this random? No! Mathematics is never random!

If you take the Cartesian Plane, and make the numbers going up the y-axis i, 2i, 3i, 4i, 5i, etc. and -i, -2i, -3i going down, you have the Imaginary Cartesian Plane. If you make a circle going through the points (1,0), (0, i), (-1, 0), and (0, -i), then you will have a unit circle. To find the 1st root of one, we of course start at (1, 0) and that is it. For the square root, or the second root, we would split the 360° of the circle in half to get 180°. So, we have the 1, and then we travel 180° to get -1, the other square root. For the fourth root, we could split 360 in fourths to get 90°, and at ninety degrees, all of the roots are found, 1, -1, i, and -i.

What about if we split in thirds, or 120°. Then, we end up at the points 1, -1/2 + i/2√(3), and -1/2 - i/2√(3). You can check that if you'd like. At the 72 degree marks, you will find the fifth roots, and the 60 degree marks give you the sixth roots.

If you know anything else about this, please tell us! Also, we will probably be taking more about imaginary numbers, so if you want me to show anything in particular, let me know.

Saturday, October 15, 2011

Why Does 64 = 65? Or Does It...

Why does 64 = 65? What kind of a question is that? Any pre-schooler probably knows that 64 doesn't equal 65. Algebra clearly shows that they are not equal. However, geometry might throw us off track.

Let's take a chessboard. It's an 8 x 8 grid, with a total area of 64 square units. I'd like you to make the following cuts in the chessboard, as well as along the 5th row from the top (3rd row from the bottom).

Now, arrange these shapes into a rectangle. Let's check out its dimensions. We have a side that is 5 units long, and a side that is 13 units long. To figure out the area of the rectangle, we would do 5 x 13 = 65.

Not good enough? Make the shapes into a triangle. We have 10 units for the base, and 13 units for the height. To find area, we do (bh)/2. So, 10 x 13 = 130 ÷ 2 = 65.

How is this possible? We have taken a grid with area 64 and just by rearranging the shapes, end up with a grid with area 65. No, there was no human error involved, your cuts probably were very accurate, and even a perfectly straight cut would still give you 65 as your area. So, how is this possible?

Even I completely understand that this is completely invalid. However, I still can't wrap my head around why it is wrong. I've been told that the squares along the cut after you assemble the shape are not valid squares, which is the only thing that seems accurate. However, with this, I just find it fascinating to take knowledge you learned in kindergarten, or even pre-school, is being challenged with this very contradictory proof.

I also saw this as a proof of last week's maneuver with Fibonacci numbers. I am not quite sure of why, but it does make some sense, that you are turning eight squared into thirteen times five. It does work again with a 5x5 or 13x13 grid, as long as you make the correct cuts. 

Region Revenge: In August, I gave you guys a problem called region revenge. The goal was to find its explicit formula. The answer 2^n-1 is incorrect, as the sixth cut can only make 31 regions, the seventh makes 57, and so on. Here is the correct answer:

An = (n^4 - 6n^3 + 23n^2 - 18n + 24)/24

You could have solved this with the techniques we used for the other problems, just by creating a five way system.

Saturday, October 8, 2011

Fibonacci Day: More Patterns in the Squares

Today yet again is a Fibonacci Day! It is October 8th, and 8 is the sixth Fibonacci number. To keep the theme of last week, let's use the square Fibonacci numbers again. Here they are:

1     1     2     3     5     8     13     21     34
1     1     4     9   25   64   169   441 1156

Let's take each Fibonacci number and move one away from it. Now, we'll multiply those numbers and see how close we get to the square.

1) 0 x 1 = 0 = 1^2 - 1
2) 1 x 2 = 2 = 1^2 + 1
3) 1 x 3 = 3 = 2^2 - 1
4) 2 x 5 = 10 = 3^2 + 1
5) 3 x 8 = 24 = 5^2 - 1
6) 5 x 13 = 65 = 8^2 + 1
7) 8 x 21 = 168 = 13^2 - 1

And so on and so forth. Basically, Fn-1 x Fn+1 = Fn^2 ± 1, or even more accurate, Fn-1 x Fn+1 = Fn^2 + (-1)^n. Let's try it again, this time looking two away from the number. Keep in mind that 1 is the negative-first Fibonacci number.

1) 1 x 2 = 2 = 1^2 + 1
2) 0 x 5 = 0 = 1^2 - 1
3) 1 x 5 = 5 = 2^2 + 1
4) 1 x 8 = 8 = 3^2 - 1
5) 2 x 13 = 26 = 5^2 + 1
6) 3 x 21 = 63 = 8^2 - 1
7) 5 x 34 = 170 = 13^2 + 1

This time, we have pretty much the same pattern. Fn-2 x Fn+2 = Fn^2 ± 1, or  Fn-2 x Fn+2 = Fn^2 - (-1)^n. How about we move three away. It's the same type of pattern, but a little different. Keep in mind that -1 is the negative-second Fibonacci number (since a Fibonacci number is the two numbers before it added together, than the zero comes from x + 1, or -1 + 1).

1) -1 x 3 = -3 = 1^2 - 4
2) 1 x 5 = 5 = 1^2 + 4
3) 0 x 8 = 0 = 2^2 - 4
4) 1 x 13 = 13 = 3^2 + 4
5) 1 x 21 = 21 = 5^2 - 4
6) 2 x 34 = 68 = 8^2 + 4
7) 3 x 55 = 165 = 13^2 - 4

We have the same idea. We are stuck with a four, giving us the pattern of  Fn-3 x Fn+3 = Fn^2 + 4(-1)^n. Let's look at our neighbors four away and see if we can see the pattern better. What do you think the negative-third Fibonacci number is? If you got two, then good job.

1) 2 x 5 = 10 = 1^2 + 9
2) -1 x 8 = -8 = 1^2 - 9
3) 1 x 13 = 13 = 2^2 + 9
4) 0 x 21 = 0 = 3^2 - 9
5) 1 x 34 = 34 = 5^2 + 9
6) 1 x 55 = 55 = 8^2 - 9
7) 2 x 89 = 178 = 13^2 + 9

Same idea again. We have Fn-4 x Fn+4 = Fn^2 - 9(-1)^n. However, the four and nine aren't there randomly. Let's look these differences closer.

1, 1, 4, 9

Recognize them? They are the squares of the Fibonacci numbers again! If you go five away, it is the square of the fifth Fibonacci number, six away is the square of the sixth Fibonacci number, one hundred away is the square of the hundredth Fibonacci number. Basically, a general formula is Fn-a x Fn+a = Fn^2 ± (Fa^2)(-1)^n. Or, you can use the below formula to be even more accurate:

Fn-a x Fn+a = Fn^2 - ((-1)^a)(Fa^2)((-1)^n)

I have no clue why this works, but please put up a proof if you know it. This is one of the coolest things about Fibonacci numbers!

Saturday, October 1, 2011

Fibonacci Day: Adding the Squares

Today is a Fibonacci Day! It is the first, and one is a Fibonacci number. One is in fact two Fibonacci numbers. In the last post, you learned how to square numbers that end in five. To keep this squaring theme, how about we square the Fibonacci numbers.

1, 1, 2, 3, 5, 8, 13, 21, 34...
1, 1, 4, 9, 25, 64, 169, 441, 1156...

We've been doing lots of adding Fibonacci numbers. Let's finish it off by adding the square Fibonacci numbers.

1 = 1
1 + 1 = 2
1 + 1 + 4 = 6
1 + 1 + 4 + 9 = 15
1 + 1 + 4 + 9 + 25 = 40
1 + 1 + 4 + 9 + 25 + 64 = 104

Do you see a pattern? It is a little hard to find, but definitely present. Look at this:

1 = 1 =                                       1 x 1
1 + 1 = 2 =                                 1 x 2
1 + 1 + 4 = 6 =                           2 x 3
1 + 1 + 4 + 9 = 15 =                   3 x 5
1 + 1 + 4 + 9 + 25 = 40 =           5 x 8
1 + 1 + 4 + 9 + 25 + 64 = 104 = 8 x 13

They are the product of the two consecutive Fibonacci numbers! Why on earth would that be? I had recently looked for one on the internet, and found an amazing geometric proof for it.

Have you ever heard of the golden rectangle, or the golden ratio? We touched on the golden ratio when I gave you the explicit formula for Fibonacci numbers (the golden ratio is the same as the greek letter fi). The golden rectangle is a rectangle of which the ratio of the length and width is the golden ratio. Something else whose ratio is the golden ratio is Fibonacci numbers! So, the side lengths of the rectangle are consecutive Fibonacci numbers!

Since we are dealing with squares of Fibonacci numbers, let's make some squares.

We've just taken these squares and organized them in a fashion that makes the side lengths two Fibonacci numbers. We went up to 34 squared, so let's see what the side lengths are.

They are 34 and 55. So, to figure out the area of the whole thing, you can add up the areas of all the squares, or just multiply the 34 by 55. And because of the way it is laid out, you can do it with any Fibonacci numbers! I think that is really cool!

Bonus Pattern: How about we add the squares of consecutive Fibonacci numbers.

1 + 1 = 2
1 + 4 = 5
4 + 9 = 13
9 + 25 = 34
25 + 64 = 89
64 + 169 = 233

The sums of the consecutive square Fibonacci numbers is in fact a Fibonacci number. I don't know a proof for this, but please tell me if you find one!

Saturday, September 24, 2011

Check out my Mathemagics Performance, and even learn a trick!!!

I don't know if you have seen my performances before, so today is your opportunity to see it. Basically, I spend the show doing very big multiplication and squaring problems, create magic squares, tell people the day of the week they were born on, and more. Here is a compilation of some of my performances.

If you liked it, you can see me live at the Chicago Toy and Game Fair in November, and many other possible shows. These are listed on my website,

In this video, you saw me square numbers. This is a method like the close-together method, but even easier than squaring numbers, and even the things I've taught you all along is squaring a two-digit number ending in five. A crowd of fifth graders could understand it, so hopefully, you guys are smarter than those fifth graders!!

Let's take the number 35. In order to square it, you need to remember two things:

1) The last two digits are always 25.
2) The start of the number is the first digit times the first digit plus one.

So, first, we take the first digit, or 3. Three times four (3 + 1) is equal to 12. Since the last two digits are always 25, 35^2 is 1225.

Let's look at, say 85^2. 8 x 9 = 72, so the answer is 7225. Very simple, right.

Let me make it a tad harder on you. Try 145^2. All we need to do is think of it as a two-digit number, with fourteen as the first digit and five as the last one. So, 14 x 15 isn't too bad, as you remember from the first post. If you try it yourself, you will get the number 210. Since it always ends in 25, the answer is 21025.

With very little practice, you will be able to square two-digit numbers ending in five. If you practice the close-together method I taught on August 20, you will be able to square three-digit numbers ending in five. Have fun!!

August Problem of the Week Answers:


b = 35
Explicit Formula: n - 1
Recursive Formula: an-1 + 1
g = 34
a = 908
p = 5%


z = 42.5
a = 1/2
b = 1/2
c = 1
y = 55
x = 10

Saturday, September 17, 2011

What does .9999999999999... really mean?

Haven't we all heard people say that there is a 99.999 "repeating" percent chance that something will happen. They think that this means they are almost sure it will happen. Though I don't want to make things complicated, this is actually equal to 100 percent.

To make things more simple, let's look at if 0.9999... = 1. First off, realize that 1/3 x 3 = 1, but .33333... x 3 = .99999... Same with 1/7, or other repeating fractions.

However, this won't convince people. So, I like to show them my favorite way to look at it, the algebraic proof, the one that proves this fact, or all of algebra and other mathematics incorrect. Say that .9999... is equal to S.

S = .9999...

Try multiplying that all by ten.

10S = 9.9999...
     S =    .9999...

What if we subtract both of these equations from each other.

10S = 9.9999...
   -S =    .9999...
  9S = 9
     S = 1

By dividing both sides by nine, it shows that S is both of these numbers.

This is another thing that is absolutely shocking. Before I saw this, I always thought that it was too close to call, but not exactly one. This is probably one of my favorite proofs in number theory, and all of mathematics.

Saturday, September 10, 2011

Another Probability Paradox: What's your birthday?

As I said about a month and a half ago, probability is a topic that people have difficulty understanding. I showed you the Monty Hall Paradox, where the average person is almost 17% off on the odds! 17% makes a big difference in what you should do in that scenario! Let's look at another thing that completely fools people, the Birthday Paradox.

Say you walk into a room of 23 people. What do you think the odds are that two of the people have the same birthday? Maybe like 23 in 365 because there are 365 days in a year, being about 6.3%. Would you be surprised if the true answer was over 50%?

The mistake people make is they try to determine the odds that someone has the same birthday as them. This is not correct, as there is no specification in the problem as to which two people it is. If you were to calculate those odds, it would be around 0.14%, which is nowhere close to the true odds.

In the Monty Hall Problem, I showed you how we know this. For this problem, we don't need much proof, because we can figure out the odds. Not of this question, but of the opposite question.

First off, what if we want to make a list of how many possibilities there are for 23 people. There are 365 for the first person, 365 for the second, and so on. Basically, there are a total of about 85 octodecillion (8.5 x 10^58), or 365^23. Basically, a list for n people is 365^n.

Now let's cover the rest of it. How many of these different possibilities have all different birthdays? Well, the first person's birthday could be any of 365 days. Since the second must be different, it only can choose between 364 days. The third only has 363, all the way up to the 23rd, who has 343 different possibilities. For n people, this is equal to:

365 x 364 x 363 x ... x (367 - n) x (366 - n)

You can very easily shorten this expression. On your calculator, there may be a button that is an x with an exclamation point after it. All this does is takes the number you type in and multiply it by every single whole number before it. For instance, 6! is 720 because 6 x 5 x 4 x 3 x 2 x 1 = 720. So, this expression is equal to:


And this is out of a list of 365^n possibilities. Therefore, the total possibilities there are for no one having the same birthday is:


If you subtract this decimal from one, you will know the probability that two people have the same birthday. For 23 people, we 365!/(365^n)(365-n)!]

This is about a 50.7% chance. For 30 people, we are already at 70%. 50 people is already at a 97% chance. 100 people is like a 99.99996% chance that two people will have the same birthday.

Saturday, September 3, 2011

Fibonacci Day: Adding the Odds

Today is another Fibonacci day. It is the 3rd, which is a Fibonacci number. We've looked at a bunch of addition patterns in the Fibonacci numbers, like adding them, or the even ones. What about the odds? How about we look.

1 = 1
1 + 2 = 3
1 + 2 + 5 = 8
1 + 2 + 5 + 13 = 21

See the pattern? We are getting the Fibonacci numbers! Why? Let's look at each one as the previous two.

1 + (1 + 1) + (2 + 3) + (5 + 8)

You are basically adding the Fibonacci numbers up, and then adding one! that's a Fibonacci number minus one plus one, or a plain Fibonacci number.

Bonus: We've done quite a bit with explicit formulas lately. I think now is the time to mention the explicit formula for Fibonacci numbers. For the nth Fibonacci number, you do:
[1/(5^.5)](φ^n - φ^n)

If you don't know, the Greek letter fi means the golden ratio, or 1.618... Fi with a bar on top is what you get with a small change in the golden ratio's formula, -0.618...

This is definitely complicated, but I couldn't believe that worked. I checked many numbers just to convince myself it worked!!

Saturday, August 27, 2011

Do the Primes go to Infinity and Beyond?!!

We've been doing a lot of work with patterns, or more formally called sequences or infinite series. Infinite series are a little different though, as they are guaranteed to go on forever without stopping. For instance, the sequence with formula n^2, or 1, 4, 9, 16..., is an infinite series. However, one like √-n + 3 is not infinite, as it will go √2, 1, 0, and then will be hitting complex numbers, which are not valid for these sequences.

Most patterns, it is pretty obvious, and would never be on a test, or even a valid question for a teacher to ask. However, what about the prime numbers. If you don't know, primes are numbers with only two factors, one and itself. So, 5 is prime, because its factors are 1 and 5. However, 6 is not prime, because it has more than two factors, namely 1, 2, 3, and 6. That means that 6 is composite, which is having three or more factors. Numbers such as one are known as universal, as they have only one factor.

Anyways, are prime numbers infinite? This is definitely a valid question, and answerable too! Can you create a list of all prime numbers on it? How about you try to. I'll bet you can't.

Say someone pops up and says,"I have made a list with all of the prime numbers that exist on it." The list would be much longer if someone did say that, but I made a small list below:


Okay. Let's multiply all of these numbers you've found together. 2 x 3 x 5 x 7 x 11 = 2310. Great. According to you, this is the product of all of the prime numbers out there. Try adding one. Now, we have 2311, which is a multiple of no prime numbers. Since every number is composed of primes, or is prime, this is not possible. That means the number is either prime, as 2311 happens to be, or could be a multiple of another prime that is not present on the list.

Is that all of the primes? No! We can do that process forever, and always find a prime that is missing. This is not a formula to generate prime numbers, as the first ten primes multiplied together is 6,469,693,230 which if you add one gives you 6,469,693,231, which you have no clue if it is prime or not! You guys can figure that one out. However, it is a cool proof that answers a question that definitely gets you thinking!