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## Algebra 1

### Course: Algebra 1 > Unit 15

Lesson 2: Sums and products of rational and irrational numbers- Proof: sum & product of two rationals is rational
- Proof: product of rational & irrational is irrational
- Proof: sum of rational & irrational is irrational
- Sums and products of irrational numbers
- Worked example: rational vs. irrational expressions
- Worked example: rational vs. irrational expressions (unknowns)
- Rational vs. irrational expressions

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# Proof: product of rational & irrational is irrational

The product of any rational number and any irrational number will always be an irrational number. This allows us to quickly conclude that 3π is irrational. Created by Sal Khan.

## Want to join the conversation?

- At1:44it is said that you should multiply both sides by the reciprocal. Reciprocals only exist for rational numbers that are nonzero. What would happen if the rational number happens to be zero?

A very subtle counterexample.

Can anyone see how to reword the statement to make it true?(29 votes)- There's now a correction at the beginning of this video indicating that it only works for non-zero rationals. (At1:55the division by
`a`

would be invalid if`a`

were zero.) Zero times an irrational is of course the rational zero again.(10 votes)

- So what is an irrational number times another irrational number?(7 votes)
- A irrational number times another irrational number can be irrational or rational. For example, √2 is irrational. But:

√2 • √2 = 2

Which is rational. Likewise, π and 1/π are both irrational but:

π • (1/π) = 1

Which is rational.

However, an irrational number times another irrational number can also be irrational:

√2 • √3 = √6

Which is irrational.

Comment if you have questions.(28 votes)

- an irratinal number can also be expresed as irrational number/1 . so i am confused. is my question valid? because it becomes a ratio of 2 numbers.(4 votes)
- You do not have to stop there, you could divide an irrational by any whole number, √/2/2 and √3/3 are common ones you will see in Math. However, the division of a irrational by a rational will still result in an irrational number. The question is valid, but the answer is not the one you thought. You can divide an irrational by itself to get a rational number (5π/π) because anything divided by itself (except 0) is 1 including irrational numbers.

The issue is that a rational number is one that can be expressed as the ratio of two integers, and an irrational number is not an integer.(7 votes)

- What about irrational times irrational?(4 votes)
- √2 and √3 are both irrational.

√2•√2=2, which is rational.

√2•√3=√6, which is irrational.

So a product of two irrationals can be either rational or irrational.(6 votes)

- if pi = 22/7 then why pi is considered an irrational numaber?(2 votes)
- 22/7 is a close approximation of pi which can be useful for some calculations, but it does not equal pi.(9 votes)

- 1:28in this what is that dot for?(3 votes)
- The dot is the same as a multiplication symbol. Check this out.

https://www.khanacademy.org/math/algebra/introduction-to-algebra/modal/v/why-aren-t-we-using-the-multiplication-sign(4 votes)

- Couldn't m/n divided by a/b equal a rational number, x?(4 votes)
- But 22/7 is irrational and is Pi, but Pi times 7 gives rational so irrational time rational can be rational?(2 votes)
- 22/7 is not irrational, because you just wrote it down as 22/7. An irrational number cannot be written as a fraction of two integers.

Additionally, 7pi is not rational either.(4 votes)

- Why do we have to assume?

Isn't that true?

Why do we only use variables while proving?(3 votes)- We assume because we do not know that it is true. In this case, it is not true. By assuming the contrary, if there is a way to contradict that, then we know it is not true. We use variables when proving so we can generalize the proof.(2 votes)

- What do you mean?

If you meant "1+0!=?" instead of "1+!=?" then your answer would follow:

1+ (0!) = ?

1 + (1) = ?

? = 2(2 votes)

## Video transcript

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.