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Current time:0:00Total duration:3:24

CCSS.Math:

So I'm curious as
to what happens if I were to take
a rational number and I were to add it to
an irrational number. Is the resulting number going
to be rational or irrational? Well, to think about
this, let's just assume it's going to be
rational and then see if this leads to any
form of contradiction. So let's assume
that this is going to give us a rational number. So let's say that this
first rational number we can represent as the ratio
of two integers, a and b. Let's call this irrational
number, let's just call this x. And their sum gives us
another rational number. Well, let's express
that as the ratio of two other integers, m and n. So we're saying that a/b
plus x is equal to m/n. Well, another way of
thinking about it-- we could subtract
a/b from both sides and we would get our
irrational number x is equal to m/n
minus a/b, which is the same thing as n
times b in the denominator. And then let's see. m/n is the same
thing as mb over nb. So this would be mb. I'm just adding
these two fractions. mb minus-- Let's see. a/b is the same thing
as n times a over n times b. So minus n times a. All I did is I added
these two fractions. I found a common denominator. So to make it clear, I
multiplied this one b and b, and then I multiplied
this one n and n then I just added these two things,
and I got this expression right over here. So this denominator
is clearly an integer. I have the product
of two integers. That's going to be an integer. That's going to be an integer. And then this numerator,
mb, is an integer. na is an integer. The difference of two integers. This whole thing is
going to be an integer. So it looks like, assuming
that the sum is rational, that all of a sudden we
have this contradiction. We assumed that x is irrational,
we're assuming x is irrational, but, all of a sudden, because
we made that assumption, we're able to assume
that we can represent it as this ratio of two integers. So this tells us that
x must be rational. And that is the contradiction. That is a very large
contradiction right over there. The assumption was
that x is irrational. Now we got that x
must be rational. So, therefore, this
cannot be the case. A rational plus an
irrational must-- so this is not right--
a rational plus an irrational must
be irrational. Let me write that down. So a rational plus an
irrational must be irrational.