Video transcript
What I want to dowith this video is do a quick proof that ifwe take a rational number, and we multiply it timesan irrational number, that this is going to giveus an irrational number. And I encourage you toactually pause the video and try to think if youcan prove this on your own. And I'll give you a hint. You can prove it by a proofthrough contradiction. Assume that a rationaltimes an irrational gets you a rational number, andthen see by manipulating it, whether you can establish thatall of a sudden this irrational number must somehow be rational. So I'm assuming you'vegiven a go at it. So let's think aboutit a little bit. I said we will do it througha proof by contradiction. So let's just assume that arational times an irrational gives us a rational number. So let's say that this-- torepresent this rational right over here, let's representit as the ratio of two integers, a over b. And then this irrationalnumber, I'll just call that x. So we're saying a/b times x canget us some rational number. So let's call that m/n. Let's call this equaling m/n. So I'm assuming that arational number, which can be expressed as theratio of two integers, times an irrational number canget me another rational number. So let's see if we can setup some form of contradiction here. Let's solve for theirrational number. The best way tosolve is to multiply both sides times the reciprocalof this number right over here. So this, let's multiplytimes b/a, times b/a. And what are we left with? We get our irrational numberx being equal to m times b. Or we could justwrite that as mb/na. So why is this interesting? Well, m is an integer,b is an integer, so this whole numeratoris an integer. And then this wholedenominator is some integer. So right over here, I havea ratio of two integers. So I've just expressedwhat we assumed to be an irrational number,I've just it expressed it as the ratio of two integers. So now we have xmust be rational. And that is ourcontradiction, because we assumed that x is irrational. And so therefore,since this assumption leads to this contradictionright over here, this assumption must be false. It must be that a rational timesan irrational is irrational.
