If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:7:47

Video transcript

what I want to do in this video is prove to you that the square root of 2 is irrational and I'm going to do this through a proof by contradiction and the proof by contradiction is is set up by assuming the opposite let's assume so this is what this is our goal but for the sake of our proof let's assume the opposite let's assume that square root of 2 is rational and then we'll see if we lead to a contradiction that this actually cannot be the case and if it cannot be the case that it is rational if we get to a contradiction by setting by assuming the square root of 2 is rational then we have to deduce that the square root of 2 must be irrational so let's let's assume the opposite square root of 2 is rational well if the square root of 2 is rational that means that we can write the square root of 2 as the ratio of two integers a and B a and B and we can also assume we can also assume that these have no factors in common let's say that they did have some factors in common if we divided the numerator and the denominator by those same factors then you're getting into a situation where they have no factors in common or another way of saying is that a and B are co-prime or another way of saying is we can write this as a ratio of two integers where this is irreducible where they no longer share any factors if you can write anything as entered as the ratio of two integers and you could obviously simplify it further factor out any common factors to get it to a point where it is irreducible so I'm going to assume that my a and B that this fraction right over here is irreducible and this is important for our for setting up our contradiction so I'm going to assume that this right over here is irreducible they have no factors a and B have no factors in common let me write that down just because that's so important for this proof a and that same color a and B a and B have no factors in common no no factors in common other than I guess one common no factors in common other than one so this is irreducible these are these two numbers are co-prime so what does that do for us well let's just try to manipulate this a little bit let's square both sides of this equation so if you square the principal root of two you're going to get two and that's going to be equal to a squared over B squared a squared over B squared and that just comes from a over B squared is the same thing as a squared over B squared and now we can multiply both sides of this by B squared and so we get we get two times B squared B squared is equal to a squared now what does this tell us about a squared well a squared is some number B squared times two so anything times two is going if you this is going to be an integer we assumed eight we assumed B is an integer so B squared must be an integer and so you have an integer times two well that must give you an even number that must give you an even integer so this right over here a squared must be so this tells us this tells us that a squared must be a squared is even now why is that interesting well a squared is the product of two numbers or is its product of two the same number it's a times a so this is another way of saying that a times a is even so what does that tell us about a let's just remind ourselves a is either going to be we're assuming a is an integer a is either going to be even or odd and we just have to remind ourselves if we multiply an even times an even and even times an even we get an even number if we multiply an odd times an odd we get an odd number we get an odd number so we have a number times itself we got an even number well the only way to get that is if that number is even so this tells us this tells us that a a is even and another way of saying that a is even is to say that a can be represented that a can be represented as the product of two times some integer so let's say some integer some integer K so I know so where is all of this going well as you'll see we can then use this to show that B must also be even so let's let's think about that a little bit so let's go back to this step right over here if we say that a can be represented as two times the product of some integer and that comes out of the fact that a is even then we could write rewrite this expression right over here as - a bit over here - two times B squared 2 times B squared is equal to 2k squared is a set of a squared I could write 2k squared we're claiming or we're deducing that assuming everything we've just assumed that a is even so if a is even it can be represented as a product of 2 and some integer and then we can write we can write that 2 times B squared 2 times B squared is equal to is equal to 4 K squared 4 K squared and then you divide both sides by 2 you get B squared B squared is equal to 2 K squared is equal to 2 times K squared and this tells us that well you know K squared is going to be an integer you take any integer times 2 you're going to get an even value so this tells us that B squared is even V squared B squared is even so that tells us that B squared is even well if B squared is even by the same logic we just used that tells us that B is even B is even so here's our contradiction we we assumed in the beginning that a and B have no common factors other than 1 we assumed that this fraction right over here a over B is irreducible but from the that and the fact that a over B must be equal to the square root of two we were able to deduce that a is even and B is even well if a is even and B is even then they both have to as a factor then this isn't irreducible you could divide the numerator and the denominator by two a and B have a common factor of two so that's so let me write this down so that this is this make it clear so from from this and this we have a and B have common factor common factor of two which means that a over B is reducible and so that's the contradiction that's the contradiction so you assume that square root of two can be represented by as an irreducible fraction a over B irreducible you can see it's a ratio of two integers right over here that leads you to the contradiction that no it actually can be reducible so therefore you cannot make this assumption it leads to a contradiction square root of 2 must be irrational