Look at that beauty. Oh but what about a number like 0.8

Yikes! Problem here is we have a fractional piece. We have this .8

but don't worry! .8 may not be an integer but it is

rational which means it can be expressed as a ratio between two other integers. In

this case 0.8 is equal to 8 over 10. 8 divided by 10. 8 and 10 are clearly both

integers which means that since point 8 is equal to them it is at least rational

It can be expressed as a ratio between two integers. Now integers and rational

numbers are beautiful. Ancient mathematicians love them. There's a

problem. There were some things we could think of that didn't appear to be either

For instance think of a square with side length 1. Every side of this shape has a

length of 1. What is the length of this diagonal. I'll call this line

C. Well the Pythagorean theorem tells us that the length of C squared is equal to

the length of this side squared plus the length of this other side squared.

1 squared plus 1 squared is just 2 so C squared equals 2 which means that C

equals the square root of 2. Perfect alright so the length of this diagonal is

square root of 2. What's the square root of 2? It could be an integer.

It could be rational. Well one thing you could do is you could measure the

diagonal of a perfectly drawn square of side 1 and measure it better and better

and better so that you've got more and more precision. As you did that you would

accumulate new digits for your answer to the square root of 2 and at each step of

the way you could find a ratio that equals that number but here's

the problem will you ever be done? Will you ever reach a point where you've

reached the last digit in the decimal expansion of the square root of two? If

you do then it's a rational number. The numerator and denominator might be

really big numbers but who cares at least it's not irrational or is it?

If it is how do you prove it? It is seems like the only way you could do it is by

calculating for some unknown possible infinite amount of time or making

completely infinitely precise measurements. Yikes.

But here's what is so fantastic about our universe. We have been able to prove

that the square root of 2 is not an integer and is not rational and today

we're gonna do just that but we need to cover four preliminaries so that this

proof is nice and complete. The first thing I want to do is define an even

number. What is an even number? We all are very familiar with even numbers 2 4 6 8

negative 2 negative 4 negative 68 these are all even numbers. What do they have

in common? Well they are divisible by two. That is a definition of evenness which

by the way means that 0 is even because 0 divided by 2 is just 0

there's no fractional piece left over so 0 is even but 1 and negative 1 on either

side are odd and that's how numbers go even odd even odd in that kind of a

pattern. Let's look at this definition: an even number like 8 is even because it

can be evenly divided by 2 8 divided by 2 equals 4 4 as a nice

whole number there's nothing left over that's perfect but what this also means

is that eight is equal to some number in this case four times two so here we have

a nice generalized definition of an even number. A number is even if it can be

expressed as two times some integer. I'll just call that integer C and because the

pattern of even odd even odd is what it is we can also define an odd number as

being equal to two times any integer plus one so negative 12 is even because

negative 12 can be expressed as two times negative six which is an integer

but negative 13 well negative 13 is odd because it can be expressed as two times

negative seven plus one. These are literally the definitions of even and

odd. The next thing we need to do is show that if you take an even number and

square it the result will also be even and if you take an odd number and square

it the result will always be odd. Now here's how we do that. Let's take an even

number which as we know is expressed by two times some integer and let's square

it. Now 2C squared equals two times C times two times C it's just two C times

itself. This is equal to four C squared. But what does 4c squared look like? We

can pull a 2 out of there and wind up with two times two C squared. Alright.

uh-oh look what we've got. This is 2 times some integer we know that two

times an integer is even so this is an even number. An even number squared is

even. Now let's square an odd number. An odd number looks like this. It's two

times any integer you like plus one. Now if we square this we wind up with my oh my

favorite thing, binomial multiplication. Let's take a look at this. We've got two

C plus one and it's squared so we're multiplying it by itself two

c plus 1 times 2c plus 1. Let's use foil to work this out. This means f we will take

we will find the product of the first two terms here. 2c times 2c is 4c

squared we actually already knew that. Then we're going to add to that the

product of the outer terms 2C times 1. Well that's just 2C. We add to that the

product of the inner terms 1 times 2 C which is just 2 C and then finally we

add the product of the last terms 1 times 1 which is 1. This simplifies into

4 C squared 2 C plus 2 C is just 4 C and we've got this one on the end. Ohh alright.

Now how about this? Let's pull a 2 out of this thing

2 times 2 C squared + 2 C and we've got this plus 1 at the end

Yowza! Look at this result. We have this thing right here in the parenthesis

which is some integer and we're multiplying it by 2 but then we're

adding 1. This is the form of an odd number so this is odd. An odd number

squared is odd and even numbers squared is even. How beautiful. Next I want to

talk about squaring rational numbers. Now this is something that we've all learned

before but I want to prove it. When we have some fractions like let's say a

over B and we want to multiply it by itself so we have A over B times a over

B. This is pretty easy to do. You literally just find the product of the

numerators, A squared, and divide them by the product of the denominators B

squared. Boom. Pretty simple but how can we be sure that is true because after

all fractions can be a little bit weird right I mean if I want to add A over B to

A over B I don't just add up the numerators and add up the denominators

instead I add up the numerators, 2A, and then I just keep the denominator the

same. This is very different than this.

what's going on? How can we be sure that we're doing this fraction multiplication

correctly? Well my favorite way to do this since we already kind of have an

idea that this is right unless our teachers have been lying to us our

entire lives is to just take advantage of the fact that multiplication and

division are inverse operations so let's take two fractions. I will call one A

over B and I will multiply it by another fraction C over D. What the heck is their

product going to look like? Well here is how we will make this easy. Let's go

ahead and multiply their product by B times D and divide by B times D because

multiplication and division are inverse operations this won't change anything. If

I multiply by some number and then divide by the same number I haven't

changed the thing I started with so all of this junk is equal to what we're

trying to study; the product of A over B times C over D. Now let's start

associating and commuting all of these little things. We can do that in

multiplication so I can take this B here for instance and multiply it by the

product of AB times CD or I can take this B and multiply it just by AB and

then bring in C over D. So let's do that because if I take A over B and I

multiply it by B and then I multiply C over D by that D. I still have to make

sure I don't forget that I'm also dividing by BD and would you look at

this. Multiplication and division are inverse so if you divide by B and then

multiply by B that's the same as just multiplying by one. So A stays the same.

Same over here. Dividing by D and multiplying by D gives us C so what

we're left with is A times C divided by B times D tada

A over B times A over B equals the product of the numerator and the product

of the denominator. Wonderful we are now ready to really

take a big bite out of rational numbers. Every single ratio of integers can be

reduced to lowest terms. In fact if you can imagine a ratio of integers that

cannot be reduced to lowest terms then it is not a ratio of integers and we do

this all the time when we're working with fractions. Take a look at a fraction

like 4/6. That's beautiful. That is a totally legitimate ratio of integers but

it's not in lowest terms because there are factors shared by 4 and 6.

By a factor I mean a number that evenly divides into them. What numbers evenly

divided into 4? Well 1 2 and 4. What numbers evenly divide into 6? Well 1 does so

does 2 so does 3 and so does 6. Yowzas. There is a common factor of 2. I can

divide both of them by 2. Now dividing by 2 over 2 is the same as dividing by 1 so

this ratio won't change, it'll just be in simpler terms. 4 divided by 2 is 2. 6

divided by 2 is 3 and boom 2/3. This is a very pretty looking fraction. It is equal

to 4/6 but the neat thing about it is that it is in a way complete because

it's a ratio between two integers that are co-prime. Co-prime means that two

numbers do not share any factors except for one. The factors of 2 are 1 and 2 and

the factors of 3 are 1 and 3. They share none in common but one so they are

co-prime. Every single ratio between two integers can be reduced to a ratio

between co-prime integers. There's another example

14/15. This one doesn't feel as pretty but it's done these are lowest terms. The

factors of 14 are 1, 2, 7, and 14. The factors of

15 are 1, 3, 5, and 15. The only factor they share in common is

once they are co-prime. 14/15 is in lowest terms. It is a reduced fraction. I

love it. The key here is that every single ratio

of integers can be reduced to a ratio between co-prime integers. If the square

root of 2 is indeed rational it should be 2. So here we begin our proof that the

square root of 2 is in fact irrational. We do this by contradiction. We just

start off by assuming that the square root of 2 is rational which means it

really does equal the ratio between two integers. We'll call them A and B. We

don't know what they are but we're just assuming they exist. If that's true then

A over B squared should equal 2. That's the definition of a square root. We've

also shown that a fraction squared means of course A over B times A over B and

when you multiply fractions you literally just find the product of the

numerators, A times A is A squared and the product of the denominators B times

B is B squared so A squared over B squared should equal 2 if the square

root of 2 is rational. Now we can rearrange this by multiplying both sides

by B squared. This gives us A squared equals 2B squared oh my goodness

gracious. Look at that. Look at that. Now B squared is some integer we don't know

what it is but it's being multiplied by 2 which means that this term is even.

It's divisible by 2. This is the definition of an even number. Where's my

even page look an even number as we said is 2 times some integer. This is 2 times

some integer so 2 B squared is clearly even but if it is even and it is equal

to A squared then A squared must also be even. Now we know that an even

number squared is even so if A squared is even then A must be even as well. Now

that's pretty interesting. It means that we can represent A as two times some

integer. Let's call it...let's call it C. I think that'll be clear enough and then

let's take this representation of A and plug it right back into this equation so

if A is 2 C then we have A squared so that means 2 C squared equals 2 B

squared. Now this 2 C squared is just 2 times C times 2 times C so 4 C squared

equals 2 B squared. Good. Oh we can divide both sides by 2 so