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:3:15

CCSS.Math:

What I want to do
with this video is do a quick proof that if
we take a rational number, and we multiply it times
an irrational number, that this is going to give
us an irrational number. And I encourage you to
actually pause the video and try to think if you
can prove this on your own. And I'll give you a hint. You can prove it by a proof
through contradiction. Assume that a rational
times an irrational gets you a rational number, and
then see by manipulating it, whether you can establish that
all of a sudden this irrational number must somehow be rational. So I'm assuming you've
given a go at it. So let's think about
it a little bit. I said we will do it through
a proof by contradiction. So let's just assume that a
rational times an irrational gives us a rational number. So let's say that this-- to
represent this rational right over here, let's represent
it as the ratio of two integers, a over b. And then this irrational
number, I'll just call that x. So we're saying a/b times x can
get us some rational number. So let's call that m/n. Let's call this equaling m/n. So I'm assuming that a
rational number, which can be expressed as the
ratio of two integers, times an irrational number can
get me another rational number. So let's see if we can set
up some form of contradiction here. Let's solve for the
irrational number. The best way to
solve is to multiply both sides times the reciprocal
of this number right over here. So this, let's multiply
times b/a, times b/a. And what are we left with? We get our irrational number
x being equal to m times b. Or we could just
write that as mb/na. So why is this interesting? Well, m is an integer,
b is an integer, so this whole numerator
is an integer. And then this whole
denominator is some integer. So right over here, I have
a ratio of two integers. So I've just expressed
what we assumed to be an irrational number,
I've just it expressed it as the ratio of two integers. So now we have x
must be rational. And that is our
contradiction, because we assumed that x is irrational. And so therefore,
since this assumption leads to this contradiction
right over here, this assumption must be false. It must be that a rational times
an irrational is irrational.