A Proof That The Square Root of Two Is Irrational

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. 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 that 2 goes away and this 4 becomes a 2. Now oh my goodness gracious look what we have. Now we have 2 times C squared which is some integer well this is divisible by 2 so this is even but if this is even and it's equal to B squared then B squared must also be even and since an even number squared creates an even number B must be even and here we have our result. A must be even and B must be even but if both of them are even they cannot be co-prime because they both share 2 as a common factor. This could go on forever because every ratio of integers must be reducible to the ratio between 2 co-prime integers and this one can't. The square root of 2 is not rational. This result is beautiful because what we're able to do in this proof is learn discover something about our universe using just mathematics and logic inside our own minds without looking at the universe itself. Stay curious and as always thanks for watching

Channel: D!NG

Views: 3,820,763

Rating: 4.917017 out of 5

Keywords: vsauce, michael stevens, math, square root, proof, mathematical proof, integer, rational number, learn, fractions, irrational number, stem, ding, d!ng, dingsauce

Id: LmpAntNjPj0

Length: 17min 21sec (1041 seconds)

Published: Sat Oct 13 2018

Reddit Comments

[u/Babu679]

Gotta Love discreet math

[u/throwaway12343234321]

While he never said anything technically incorrect, it feels like he might be a bit misleading when he's talking about attempting to prove irrationality by taking closer and closer estimations.

He says that if you ever finish by finding all the digits, then you know it's rational. There are rational numbers with an infinite number of digits, like 1/9. The difference is that rational numbers have no repeating pattern to their infinite digits. Again, he actually said anything incorrect, just thought I'd mention it.

The reason that some rational numbers have an infinite number of decimal places has more to do with what base you're writing the number in. If you were to start testing different fractions/different denominators, what you would find is that if the denominator and the base you are working in are coprime, then the fraction will have an infinite number of decimal places. In typical mathematics, we've worked in base 10. Fractions with denominators like 3,7,9,11... Will have infinite decimal places, since each of those numbers are coprime with 10. For example, 1/3 in base 10 is 0.333333.... infinitely repeating. 1/3 in base 3, however, is just 0.1.

[u/[deleted]]

The square root of any integer that is not a perfect square is irrational. So not just the square root of 2, but also of 3, 5, 6, 7, 8, 10, etc. All irrational.

[u/namkash]

Beautiful and very entertaining.

[u/conalfisher]

^Oh

My

Goodness

Gracious