HOHOHO :) Welcome to the 2018 Christmas Mathologer video. In this edition I'd like
to give you a present, a beautiful and easy to use precision tool for spotting
irrational numbers. Interested? Let's start with something really simple: Is
root 2 a rational number, that is, is it possible to write root 2 as a ratio of
integers, as a fraction. Yes, yes I can hear your thoughts:
Is he kidding? Of course even preschoolers know that root 2 is irrational. True, but
now try this rooty number on your local preschooler. Rational or irrational?
And how about this one? Or this one? Well all these numbers look pretty damn
irrational, right? But one of them isn't. This number here actually turns out to
be equal to 1, which is very much not irrational. Surprised? Hmm so, given a root
expression, how can we tell if it's truly ruly irrational? My Christmas present to
you is a Mathologerisation of the famous root theorems. These are the
ultimate irrationality zippers, the master tools for deciding whether
rooty numbers like the ones over there are rational or irrational. Many of you
will have glimpsed these irrationality zappers in school though probably with
different packaging. A standard middle school maths game is to find the
solutions to equations like this one there. Okay, so say you forgot your
hi-tech calculator and your life, hmm or at least your test score depends upon
finding the solutions, what do you do? Well since this is a school game chances
are the equation has been rigged to ensure there's a nice simple integer
solution and if that is really the case then there's a simple trick to narrow
the hunt for such a solution. It turns out that if an equation like this has an
integer solution then that solution must divide the constant term 6. And so what
are the factors of 6. Well there is 1, 2, 3 what else?
6. Is that it? Not quite since there are all the negatives as well. So if this
equation has an integer solution it's gotta be one of these eight numbers. So
you cross your fingers, hope that your teacher or your Mathologer has cooked up
things correctly and you start going through them. Well let's start with 1,
let's see okay, ... okay so that didn't work out and you can check that
2 also doesn't work. But we're lucky 3 does. You can also check that none
of the other integer candidates work which means that 3 is the only
integer solution to this equation.
Great now who among you learned this trick in school and in what form? Let me
know in the comments including where in the world you're from. It will be
interesting to see who gets served what at school. Okay the official name for
this trick is the Integral Root Theorem. What does any of this have to do with
irrational numbers? Well, if you actually state this trick as a proper theorem and
not as the contrived guessing game presented in schools, then the integral
root theorem runs like this: Given a polynomial equation with integer
coefficients and, very important, leading coefficient 1, and, also very important,
nonzero constant term. Then any real solution of this equation is either an
integer that divides the constant term, like that, or, and here it comes, or the
solution is an irrational number. So the extra cool bit that is missing from the
school trick is that any real solution that is not an integer must be an
irrational number, no proper rational numbers like 1/2 or 17/13 can sneak
in as a solution. Again only integer or irrational solutions are possible. Let's
now use the integral root theorem to pin down some irrational numbers, starting
with our good old friend root 2. Root 2 is a solution of
this quadratic equation here. Being approximately equal to 1.4
root 2 is definitely not an integer. Therefore by the integral root theorem
it has to be irrational. How slick is that? A different and important way to
come to the same conclusion runs like this: the factors of the constant term 2
are what. Well they are 1, - 1, 2 and - 2. And so, by the integral root
theorem, these four factors are the only possible integer solutions. But clearly
none of them is a solution, right? And that means that there are no rational
solutions at all and so every solution including root 2 must be irrational. Exactly the same argument shows that the cube root of 2 is irrational and in fact
the same argument works for any root of 2 and for the nth root of any positive
integer that is not itself an nth power of an integer. Of course, I'm
sure that for quite a few of you the fact that these roots are irrational is
old hat and not terribly exciting. So let's have a look at some more elaborate
rooty combos like this one here. Rational or irrational? To find out let's call
this number, well let's see, let's call it something mysterious, something exotic
like .... x :) squaring on both sides gets rid of the outer square root. Shuffling 2 to
the right side gets us this and squaring again get gets rid of the second root sign.
Okay now some quick algebra autopilot.
Okay what we've shown is that our rooty combo number is a solution of this
polynomial equation and now the integral root theorem is ready to pounce again.
The only factors of the constant term are well 1 and -1, both of which are
definitely not solutions of the equation and so all real solutions of this
equation, including our rooty combo number must be irrational. How nifty is
that? Next, how about this curious number here? I'll leave it as another challenge
for you figure it out for yourself and let the
rest of us know in the comments. And, yes, it really does equal 3.14 dot dot dot :)
Now for the next step. Remember that the integral root theorem only applies to
equations with leading coefficient 1. But what if the leading coefficient is different
from 1. Well there's a super nifty generalization of the integral root
theorem that takes care of this. It is called the Rational Root Theorem. Okay to
turn the statement of the integral root theorem into the rational root theorem
we simply have to replace the integer dividing the constant term by a
fraction u/v with u dividing the constant term, like that, and v
dividing the leading coefficient, like that. This means that in practice and,
just as before, we can easily determine the very few candidates for rational
solutions of an equation like this, right? And by doing so we will have
automatically proved that any other real solution must be irrational.
Here's another mini-challenge for you to sort out in the comments. Remember
we also require that the constant be different from zero. What happens and
what do you do if the constant term IS equal to zero. That's pretty easy. Anyway
the real numbers solutions to polynomial equations like this are super important
and have a special name. They're called the algebraic numbers.
Numbers that are not algebraic such as e and pi are called transcendental numbers.
So while the root theorems are fantastic tools for determining the irrationality
of an algebraic number they are of no use for transcendental numbers. But we also
have an algebraic confession to make :) It's definitely true that the root
theorems are super powerful at showing which solutions to polynomial equations
are irrational but there's a Grinchy problem. Algebraic numbers are often
presented in ways that don't automatically come with one of those
special equations and of course we can't apply our root theorems unless we
somehow produce the right polynomial equation. For example, it's true but quite
tricky to prove that the rooty combo numbers like the ones I mentioned earlier
are algebraic and for complicated rooty numbers it's not at all obvious
how you come up with such an equation. Just pick one of these guys over there
and have a go at finding an equation for it.
Take your time I'll wait. No I won't :)
However, all is not lost even for very complicated rooty expressions there are
complex algorithms that given the input of any root expression will output a
suitable polynomial equation. These algorithms have also been implemented as
part of computer algebra systems. For example, in Mathematica or Wolfram Alpha
the relevant command is Minimal Polynomial. In fact, unleashing this
command on a rooty number will spit out a special polynomial, an extra
special polynomial, the so called minimal polynomial, that is, a polynomial of
lowest possible degree. But wait a minute, what is a polynomial equation with
integer coefficients of least degree for a rational number. Think about it for a
second. The answer is, the simplest possible such equation is a
linear equation, this one here but that means that you can tell at a glance from
the minimal polynomial whether a rooty expression is a rational number and if
so what rational number. For example, on input of this expression here
Mathematica spits out this linear equation. And what does that mean? Well
obviously that our number is rational and equal to 1. Surprising but true. Here
had two final very challenging challenges for you: First algebraically
massage this rooty number into 1 by hand. Second, again by hand, try to find a
polynomial equation that has this number as a root. Well, let's see what you come up with.
That's all great but if you are like Marty and me, you won't be very satisfied with
this Mathematica black box solving the problem for us. Well promise eventually
there will be another Mathologer video dedicated to proving that every real
rooty expression is algebraic by showing how to construct its minimal
polynomial. Very very tricky stuff but at the same time extremely beautiful
mathematics. Also on my to-do list is a video on the super surprising fact that
even though all real root expressions are algebraic not all algebraic numbers
are rooty. For example, the one real root of this equation there is not rooty.
Time for the grand finale. Let me animate a super simple proof of
the rational root theorem for you. Enjoy :) As usual I'll focus on a proof for a
sufficiently general example. What must a fractional solution of this equation
look like? Let's say u/v is a solution of the equation and let's
assume we've canceled out such that u and v have no more common factors. Then
what i'll do is show that u is a factor of the constant 3 and V is a factor of the
leading coefficient 2 and that's all we need to do, right? Ok 2, 3 play a special
role so let's highlight them just a little bit. Ok so u/v is a
solution, there. Autopilot, right, let's shuffle the
constant 3 to the right side of the equation, like that.
To get rid of the fractions on the left we'll multiply through by the largest
denominator v^6 like that. Again just autopilot, right, that's
what you all would do, cancel, cancel cancel and now we have integers
throughout and all the terms on the left have the common factor u. Pull out the
u in front. Now u divides the left side and so the right side as well. Think
about it for a moment. Okay? However we began by making sure that u
and v have no common factor and so u and v^6 also have no
common factor, and so u must be a factor of 3. Easy, isn't it? And with a
slight rearrangement we can pin down v in exactly the same way. Let's back up
one step. We'll now move everything to the right side except for the u^6 term like that. Now all the terms on the right have the common factor v.
Pull the v out in the back. Now v divides the right side and therefore also the
left side and just like before we can conclude that v divides 2, easy-peasy
What a fantastically simple proof for such a powerful theorem :) Just a final
footnote. The integral root theorem is a special case of the rational root
theorem. In turn the rational root theorem is a special case of a wonderful
result called Gauss's lemma named after the mathematical superstar Carl
Friedrich Gauss. Further googling is definitely recommended. I hope you
enjoyed this mathematical Christmas present. Fröhliche Weihnachten :)(Merry Christmas in German)
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