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.

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
0

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:

√((-1)(-1))

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

√((-1)(-1))

√(1)

1

But if you used this property, you would have:

√(-1)√(-1)

i • i

i^2

-1

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.

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.

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.