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## Algebra 1

### Course: Algebra 1 > Unit 15

Lesson 3: Proofs concerning irrational numbers# Proof: square roots of prime numbers are irrational

CCSS.Math:

Sal proves that the square root of any prime number must be an irrational number. For example, because of this proof we can quickly determine that √3, √5, √7, or √11 are irrational numbers. Created by Sal Khan.

## Want to join the conversation?

- Didn't he prove even more than he set out to prove? He didn't state it explicitly, but at3:55,
`p`

wouldn't even need to be prime to continue with the proof. Sal chose`p`

to be one single factor, but really`p`

could be a series of factors, where at least one of the factors is**not a duplicate**. Ie, in the list of factors of`a^2`

he has there,`p`

could be`f1 * f1 * f3`

. This new`p`

isn't prime since it has three factors, and in fact even 2 of the factors are duplicated (f1 and f1), but as long as one factor is unique (f3) then we can say`a`

must be a multiple of`p`

and the proof continues on. Doesn't this show that the square root of a number is irrational if that number is made of anything other than pairs of duplicated factors?(17 votes)- No, I believe the proof would break down in this case, and here's why.

You wrote "in the list of factors of a^2 he has there, p could be f1 * f1 * f3."

Obviously, this would mean that p is a factor of a^2.

But in the case of your p, which isn't a prime number anymore, we wouldn't automatically know whether p is a factor of a as well.

In your example, we'd need to know that the factors of a^2 contained f1 * f1 * f1 * f1 * f3 * f3

i.e. we'd need to know that p^2 is a factor of a^2 to know that p is a factor of a

Otherwise it could happen for example that the factors of a^2 are p * f3 = (f1 * f1 * f3) * f3, making a = f1 * f3

a is still rational, but p is not among the factors of a, ergo a is not a multiple of p, and the proof doesn't work from this point on**EDIT**the proof works as long as you set your new p as only the unpaired factor (in this example, the "f3") Then we can say a is a multiple of f3, which is a prime number, so the same proof applies...(19 votes)

- is the square root of a non prime number rational?(6 votes)
- That is a great question and one that is not at all difficult to decide whether it is true or false.

In math, if you can find just one example that is counter to the claim, then the claim is invalid.

CLAIM: the square root of a non prime number is rational

Take 8 for example. 8 is not prime, correct. But, √8 = √4·√2 = 2·√2.

Now the 2 in √2**is prime**and therefore the square root of it**IS**irrational, and an irrational number times a rational number is**ALWAYS**irrational. Yikes! we have found one non-prime number whose square root is irrational. Therefore we MUST conclude that the claim "the square root of a non prime number is rational" is not true for all non-prime numbers.

Nowfind examples where it is true, for example √4=2. Obviously 4 is NOT a prime and its square root, 2 IS rational - this is an example of a non-prime whose square root is rational. But since we found one non prime whose square root is not rational we cannot say for all non-prime numbers that "the square root of a non prime number is always rational."**we can**

PS: there are infinitely many other non prime numbers whose square roots are irrational - keep studying and one day you will be able to prove it to yourself!

Have Fun & Keep Asking Questions!(19 votes)

- How is p squared definitely irrational when all he did was assume that this equals that and that equals this and a over b can't be reduced?(2 votes)
- That is a great question whose answer is at the same time both profound and subtle.

First - The only assumption, that is, something we do not know if it is true or not, made in this video is the proposition that square roots of prime numbers are irrational. Below I will get to the a/b thing and perhaps some of the other what-you-think-are-assumptive statements you ask about, but you didn’t say what they were. In order to get there, I need to go here first:

Math is like a game in that it has rules. If you follow those rules you win.

What do you win? You win a truth, which up till now, you have been calling “a correct answer” – but there is more. . .

You already know many math rules, like keeping an equation balanced, or a negative number multiplied by another negative number results in a positive number etc. etc. . . . . There are lots of these kinds of procedural rules for playing with numbers.

But there are other types of rules that will become more and more important as you get further into math and they have to do with the properties of numbers and consequences of results. The most amazing thing about these rules is that they, combined with other rules can lead you to a truth than no one has discovered yet, in other words, this rule based game of exploration can take you where no one has gone before – and it is**that**aspect which keeps us pushing the boundaries of mathematics. Many new discoveries are beautiful (fractal geometry), or useful (chaos theory), or weird (quantum mechanics) and some are even a bit disturbing (the n-body problem).

Now, the path that leads to a truth in mathematics is called a proof. Guess what? You have been doing proofs all this time, right since you first started to add numbers up until now. For example, if you have a problem like 7x - 10 = 5x + 6, you can prove, using the rules of the game of math, that x can only be equal to 8 in order that 7x - 10 = 5x + 6 becomes a true statement. This is an example of a proof using just the procedural rules. If at each step of the way, you obey the rules, you prove (arrive at the truth) that x=8.

The proof you are asking about in this video is a proof that uses some properties of numbers and some concepts and their consequences. This takes a bit more considered thinking – so let’s get to it.

We are being asked to prove:*when you take the*. To start the proof, what do we already know in terms of rules, definitions, and properties?**square root**of any**prime number**, the result is an**irrational number**

1)*What is a prime number*? A number, let’s call it p, is prime if it can be evenly divided (no remainder) by only 1 or itself.

2)*What is a square root*? The square root of a number, say n, is a value, say k, that, when multiplied by itself, gives the number, that is, k*k=n, so k is a square root of n.

3)*What is an irrational number*? An irrational number is a real number that cannot be expressed as a ratio of integers, commonly called a fraction. So if x is irrational, there are no integer values, say a and b, such that x=a/b..**This property will be really important in the proof**

4)*What is reduced form*? Any fraction can be converted into reduced form. EG 12/18 = 6/9 = 2/3. 2/3 is the reduced form of both 12/18 and 6/9. Consider 8/16 = 4/8 = 2/4 = 1/2 (reduced form). We know we can do this reducing process for any fraction that is not already in reduced form..**This property will be really important in the proof**

With most proofs we can proceed directly, like the 7x - 10 = 5x + 6 example. Other proofs use a different tactic, and this tactic works because each application of a rule of math will always give you a valid, or correct result – and we can use this to our advantage! Here’s how:

A common method of proof in math and other logic systems is called “proof by contradiction” or formally “reductio ad absurdum” (reduced to absurdity).*How this type of proof works is: suppose we want to prove that something is true, let’s call that something S. If we start the proof by assuming that S is instead false, and then, using the rules of the game of math (also known as mathematically correct and valid arguments) we show that we get a nonsense or contradictory result, well then, that*!! In this case, our statement S is “the square roots of prime numbers are irrational”. So our plan of attack it to assume that S is false, that is, let’s assume that “the square roots of prime numbers ARE rational”. So let’s try to prove that . . . Let’s start with the properties #1 to #4 we know about.**must**mean the assumption we made that S was false can’t be correct, so S must be true

Let’s pick any prime number, and let’s call it p. Now contrary to S, which states “the square roots of prime numbers are irrational”,**we are assuming that S is wrong**and that “the square roots of prime numbers ARE rational”. Now let’s play the game of math see what happens when we try to prove that the square roots of prime numbers are rational.

If the square root of our prime number p is rational, that means we can say √p = a/b, where a and b are integers -**recall rule/property #3**.

From**rule/property #4**we know we can reduce the fraction a/b if it is not already in reduced form. Now you might say “what if a/b is not reduced?” That’s OK – we can reduce it and call it c/d. What? c/d is not reduced? OK reduce it and call it e/f. We can keep repeating this process until a/b**is**in reduced form. Since weit can be done, let’s just say we have done all that and our a/b is now in reduced form.**know**

Given**rule/property #2**if √p is the square root of p, then (√p)( √p) = p, right?

Using the “keep an equation balanced” rule where you must do the same thing to bother sides (in this case, squaring) we have, given √p = a/b, that (√p)( √p) = (a/b)(a/b), so p=a^2/b^2.

Now follow along with the video from2:00. Each step is an application of a rule of math. If some of the outcomes of the rules you do not understand, you may want to review that particular subject.

If I missed a detail you want clarified – just ask.

Keep Studying!(17 votes)

- At4:18, shouldn't
**p=f2 x f2**to be a multuple of a^2?(8 votes) - at4:26, Sal said a^2 is a multiple of p, so a is a multiple of p.

but how about this case?:

6^2 = 36, and 36 is a multiple of 4

6 is not a multiple of 4.

how does this work?(5 votes)- He showed explicitly from prime factorization that it applies when p is prime. For the example you gave, you used 4, which is not prime.(6 votes)

- can you prove that the square root of any non perfect square is irrational(3 votes)
- Not all non perfect square's square roots are irrational(1 vote)

- I have a question regarding irrational numbers. How many irrational numbers (with a "NAME" such as Pi) can be found by finding the square root of imperfect squares starting from the smallest imperfect square to the largest imperfect square who's square root has a name?

PS: This idea occurred to Me while watching one of Sal's earlier videos when He said that the square root of ANY imperfect square is irrational (I suppose that also means a number doesn't have to be prime in order for its square root to be irrational.)

PSS: Does anyone know of an algorithm for finding imperfect squares for example 2,3,5,6,7,8,10,11,12,13 etc? (The reason I asked this question is because an algorithm for finding imperfect squares would be helpful for answering the first question above.)(4 votes)- I cannot think of any special number with a name that equals the square root of an imperfect square. I doubt there is many either. Pi isn't represented by the square root of an imperfect square.

One example that ALMOST goes with what your saying would be the golden ratio, which equals

(1 + sqrt(5))/2

But that doesn't entirely fit your criteria of having a special name for a pure square root of an imperfect number.(1 vote)

- I had this question in the previous video as well. Couldn't he have just done b * √(p)= a( the product of a rational and irrational). which he proved previously is a contradiction.(2 votes)
- do I need this information anywhere?(2 votes)
- Yes. You will need the information but not the problem itself.(1 vote)

- Are any prime numbers irrational?(1 vote)
- No. Prime numbers themselves are all whole numbers. They can be written as a ratio of 2 integers by giving them a denominator of 1. Thus, they fit the definition of rational numbers.

Most irrational numbers you will deal with will be well known constants like "Pi" or "e"; or they will be radicals of imperfect roots like: sqrt(6); cuberoot(9); 4throot(24); etc.(3 votes)

## 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.