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

CCSS.Math:

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 set up by assuming the opposite. So 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 is rational, if we get to a contradiction 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 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. And 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 it is we could write this as a
ratio of two integers where this is irreducible,
where these no longer share any factors. If you can write anything as
the ratio of two integers, then 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
setting up our contradiction. So I'm going to assume that this
right over here is irreducible. a and b have no
factors in common. Let me write that
down because that's so important for this proof. a and-- want to do
that same color-- a and b have no factors in
common, other than, I guess, 1. So this is irreducible. 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 2, you're going to get 2. And that's going to be equal
to 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 2 times 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 2. So anything times 2 is going--
this is going to be an integer. We assumed b is an integer, so
b squared must be an integer, and so you have an
integer times 2. 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
that a squared must be even. Now, why is that interesting? Well, a squared is the
product of two numbers or is the product
of 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. We just have to remind
ourselves if we multiply an even times an even,
we get an even number. If we multiply an odd times
an odd, 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 that a is even. And another way of
saying that a is even is to say that a
can be represented as the product of 2
times some integer. So let's say some integer k. 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 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 can rewrite this
expression right over here as 2-- I'll do it
over here-- 2 times b squared is equal
to 2k squared. Instead 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
that 2 times b squared is equal to 4k squared. And then you divide
both sides by 2. You get b squared is
equal to 2k squared. And this tells us
that, well, 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. 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. So here's our contradiction. 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/b, is irreducible. But from that and
the fact that a/b must be equal to the
square root of 2, we were able to deduce that
a is even and b is even. Well, if a is even
and b is even, and they both have
2 as a factor, and then this isn't irreducible. You could divide the numerator
and the denominator by 2. a and b have a common factor of 2. So let me write this down. So this is just
to make it clear. So from this and this, we have
a and b have common factor of 2, which means that a
over b is reducible. And so that's the contradiction. So you assume that
square root of 2 can be represented as an
irreducible fraction a/b, irreducible because you can
say 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.