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• Integers are numbers that have no fractional component like the number 8.

• 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.

• 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