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# Proof: square roots of prime numbers are irrational

CCSS Math: HSN.RN.B.3

## 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. And we're going to set this up to be a proof by contradiction. So we're going to assume that the square root of p is rational and see if this leads us to any contradiction. So 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 the 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. So I'm assuming that this fraction that I'm writing right over here, a/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 fraction. So I'm assuming that's where we are right here. So 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 1. 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/b, the whole thing squared, that's the same thing as 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. 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. Let me write this down. So 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-- and any number-- can be rewritten as a product of primes. Or any integer, I should say. So 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. I don't know 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 the prime factorization of a squared going to be? Well, a squared is just a times a. Its prime factorization is going to be f1 times f2, all the way to fn. And then that times f1 times f2 times, all the way to fn. Or I could rearrange them if I want. f1 times f1 times f2 times f2, all the way to fn times fn. Now, we know that a squared is a multiple of p. p is a prime number, so p must be one of these numbers in the prime factorization. p could be f2, or p could be f1, but p needs to be one of these numbers in the prime factorization. So 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 f2. If p is f2, then that means that p is also a factor of a. So this allows us to deduce that a is a multiple of p. Or another way of saying that is that we can represent a as being some integer times p. Now, why is that interesting? And actually, let me box this off, because 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 2 being irrational, let's now substitute this back into this equation right over here. So we get b squared times p. We have b squared 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 some integer k 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 times p squared. We can divide both sides by p, and we get b squared is equal to p times k squared. Or k squared times p. Well, the same argument that we used, 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 b squared is a multiple of p. And by the logic that we applied right over here, that lets us know that b is a 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 1. We assumed that this cannot be reduced. But we've just established, just from this, we have deduced that is a multiple of p and b is a multiple of p. Which means that this fraction can be reduced. We can divide the numerator and the denominator by p. So that is our contradiction. We started assuming it cannot be reduced, but then we showed 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. Square root of p cannot be rational. Square root of p is irrational. Let me just write it down. The square root of p is irrational because of the contradiction.