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# Proof: square roots of prime numbers are irrational

CCSS.Math:

## Video transcript

in a previous video we used a proof by contradiction to show that the square root of 2 is irrational what I want to do in this video is essentially use the same argument but do it in a more general way to show that the square root of any prime number is irrational so let's assume that P is prime P is prime and we're going to set this up to be a proof by contradiction so we're going to assume that the square root square root of P is rational and see if this leads us to any contradiction so if something is rational if something is rational that means that we can represent it as the ratio of two integers and if we can represent something as the ratio of two integers that means that we can also represent it as a ratio of two co-prime integers or two integers that have no factors in common or that we can represent it as a fraction that is irreducible as a fraction that is irreducible so I'm assuming that a this fraction that I'm writing right over here a over B that this right over here is an irreducible fraction you say well how can I do that well this being rational says I can represent the square root of P as some fraction as some ratio of two integers and if I can represent anything as a ratio of two integers I can keep dividing both the numerator and the denominator by the common factors until I eventually get to an irreducible an irreducible fraction so I'm assuming that's where we are right here so this this cannot be reduced and this is important for our proof cannot be reduced which is another way of saying that a and B are co-prime which is another way of saying that a and B share no common factors other than one so let's see if we can manipulate this a little bit let's take the square of both sides we get P is equal to well a over B the whole thing squared that's the same thing as a squared over B squared a squared over B squared we can multiply both sides by B squared and we get B squared times P is equal to a squared is equal to a squared well what does this tell us about a squared well B is an integer so B squared must be an integer so an integer times P is equal to a squared well that means that P must be a factor of a squared so let me write this down so a squared a squared is a multiple multiple of a squared is a multiple of P now what does that tell us about a does that tell us that a must also be a multiple of P well to think about that let's think about the prime factorization of a let's say that a can be read and any number can be rewritten as a product of primes or any integer I should say so let's let's write this out as a product of primes right over here so let's say that I have my first prime factor times my second prime factor all the way to my nth prime factor and I don't know whether how many prime factors a actually has I'm just saying that a is some integer right over here so that's the prime factorization of a what is a prime factorization of a squared going to be well a squared is just a times a its prime factorization is going to be F 1 times F 2 all the way to F N and then that times F 1 times F 2 times all the way to F N or I could rearrange them if I want F 1 times F 1 times F 2 times F 2 all the way to F n times F n now we know that a squared is a multiple of P P is a prime number so P must be one of these one of these numbers in the prime factorization P could be F 2 or P could be F 1 but P needs to be one of these numbers in the prime factorization so P P needs to be one of these factors well if it's let's say and I'll just pick one of these arbitrarily let's say that P is f 2 if P is F 2 then that means that P is also a factor of a so this allows us to do this allows us to deduce that a this allows us to deduce that a is a multiple a is a multiple of pee or another way of saying that is that we can represent a we can represent a as being some integer times P times some integer some integer times P now why is that interesting and actually I'm going to let me box this off because this we're going to reuse this part later but how can we use this well just like we did in the proof of the square root of the square root of two being irrational let's now substitute this back into this this equation right over here so we get B squared times P we have B squared times P times P is equal to a squared well a we're now saying we can represent that as some integer K times P so we can rewrite that as we could rewrite that as some integer right with this some integer K times P times P and so let's see if we were to multiply this out so we get B squared times P and you probably see where this is going is equal to K squared K squared times P squared times P squared we could divide both sides by P and we get we get B squared is equal to P times K squared is the or K squared times P K squared times P well the same argument that we used if B squared times P is if a squared is equal to B squared times P that let us know that a squared is a multiple of P so now we have it the other way around B squared is equal to some integer squared which is still going to be an integer times P so B squared must be a multiple of P so this lets us know that V squared is multiple is a multiple of P of P and by the logic that we apply it right over here that lets us know that B is a multiple of P B is multiple of P and that's our contradiction or this establishes our contradiction that we assumed at the beginning we assumed that a and B are co-prime that they share no factors in common other than one we assumed that this cannot be reduced but we've just established just from this we have deduced that a is a multiple of P and B is a multiple of P which means that this fraction can be reduced we can divide the denominator and the denominator by P so that is our contradiction we started assuming it cannot be reduced but then we show that no it must be able to be reduced the numerator and the denominator have a common factor of P so our contradiction is established P square root of P cannot be rational square root of P is irrational let me just write it down square root of P is irrational is irrational because of the contradiction