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Current time:0:00Total duration:7:28

CC Math: HSN.RN.B.3

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.