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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.