Proof: product of rational & irrational is irrational
About this video
The video uses proof by contradiction to show that a rational number times an irrational number is irrational. It assumes (a/b) times x = m/n with x irrational, solves to get x = mb/na, and notes that this is a ratio of integers, which contradicts x being irrational.
You'll be able to
- describe how a proof by contradiction works.
- solve (a/b) x = m/n for x by multiplying by the reciprocal.
- explain why a rational times an irrational must be irrational.
Key words
- Proof by contradiction
- Assume the opposite of what you want to prove, then show this leads to something impossible.
- Irrational number
- A number that cannot be written as a ratio of two integers.
- Reciprocal
- A fraction flipped upside down; the reciprocal of a/b is b/a.
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.
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