[MUSIC PLAYING]
 Can you produce
a rational number
by exchanging infinitely
many digits of pi and e?
The discipline of math has a
wonderful sense of community.
Mathematicians love to share
their ideas and questions
with each other, to
problem solve together,
just like you all do in the
comment section of this show.
There are several online
forums where mathematicians
post questions and responses.
Recently on one of them,
mathematician Erin Carmody
asked, if I exchange infinitely
many digits of e and pi,
are the resulting
numbers transcendental.
She received some
curious answers,
ones which provide an
awesome preview into the way
mathematicians approach
research-- with intuition,
collaboration, and a strong
desire to explore the unknown.
Transcendental numbers
are a very specific kind
of irrational
number, and certainly
the subject of some
future episodes.
But we'll focus on
irrational numbers
in general, which is what most
mathematicians in the post did.
Pi and e, also known
as Euler's number,
are two famous
irrational numbers.
So we'll ask, can you exchange
infinitely many digits of pi
and e to get a rational number.
All the real numbers,
any number that
appears on the
number line, can be
divided into two categories--
rational and irrational.
Rational numbers
are those that can
be written as n divided
by m, where n and m are
two integers, like 73 over 5,
or 1 over 7, or 133 over 300.
An equivalent definition
is that rational numbers
are those whose decimal
expansions terminate
or eventually repeat, like
14.6, or 0.142857 repeating,
or 0.44333333333.
Even rational numbers that we
think of as having terminating
decimal expansions, like 14.6,
actually have a repeating 0
if we write it as 14.60000000.
Irrational numbers are
all the real numbers
that aren't rational, meaning
their decimal expansions don't
repeat.
Some well-known examples include
pi, e, the square root of 2,
and the golden ratio.
Actually, the proof that the
square root of 2 is irrational
is a fantastic classic.
You can try it for
yourself or check out
the link in the description.
Before we return to
the original question,
let's start with a
warm-up question.
Do there exist two
irrational numbers such
that we can exchange
infinitely many digits of them
and one of the resulting
numbers is rational?
Pause here if you want
to think about it.
The answer is yes.
Maybe you came up with
a different example,
but here's mine.
The number 0.010110111
is irrational.
It's made up of strings of 1's
of increasing length with a 0
between each string of 1's.
Since the string of ones
becomes longer and longer,
it never repeats.
Therefore, it's irrational.
Similarly, the number
0.101001000 and so on is
irrational.
But if we exchange infinitely
many of their digits,
like this, we get two
rational numbers, 0 and 1/9.
Or, we could trade these
digits of these two
irrational numbers.
In this case, one of the
resulting numbers is rational
and one is irrational.
OK, back to the
original question
where Carmody is asking about
two very specific numbers.
If I exchange infinitely
many digits of pi and e,
are the resulting
numbers irrational?
Here's pi and here's e.
There's a lot of different
ways we could exchange
infinitely many digits.
We could exchange
the odd digits to get
this pair, or every fourth
digit to get this pair.
Or the second, eighth,
and ninth digits,
and all the digits after the
16th digit, to get these.
With infinite time
and patience, we
could just write down
a big endless list
of all the possible
pairs of numbers
that arise by switching
infinitely many digits of pi
and e.
What kind of numbers
appear on that list?
Are there rational numbers?
Are there irrational numbers?
Mathematician Joel
David Hamkins,
whose work we featured in
the infinite chess episode,
began his response
to Carmody's question
with a question of his own.
How often do the digits
of pi and e differ--
as in, the first digit after
the decimal place is different.
It's one for pi and seven for e.
The second digit is different,
it's four for pi and one for e.
But the 12th digit is the
same, it's nine for both.
If we exchange the 12th
digit, it doesn't do anything.
Same thing for the
16th and 17th digits.
What if the digits of pi
and e were mostly the same?
If there were only finitely
many places they differed,
what would happen?
Well, then there would only
be finitely many new numbers
produced by the digit swaps.
And these new numbers would
only differ from pi or e
at finitely many places.
This means that all
the numbers on the list
would be irrational.
Why?
Let's just look at pi.
Assume we change finitely
many digits of pi.
Let's look at the last
digit that we changed.
Everything after
it is unchanged.
Since pi is irrational,
these digits
don't have a repeating pattern,
so the modified version of pi
also doesn't have
a repeating pattern
and is therefore irrational.
Notice that if pi and
e are only different
at finitely many digits,
then pi minus e is rational.
OK, so the answer is
pretty easy if pi and e
happen to only differ
at finitely many digits.
The numbers on the list
are all irrational.
But it seems more likely
that the digits of e and p
differ at infinitely
many places.
In this case, we can produce
infinitely many new numbers
by swapping their digits.
That is, the list would
be infinitely long.
Remember back to the hierarchy
of infinities episode
where we talked about
different sizes of infinity.
There was a smallest infinity,
which is called countable.
And then there were
bigger infinities.
Anything bigger than
countable infinity
is called uncountable infinity.
Well, if pi and e differ
at infinitely many digits,
then the list of
numbers produced
by all possible exchanges of
infinitely many digits of e
and pi is uncountably
long, which
actually means it's not even
possible to write it in a list.
So this is now a
metaphorical list.
I'll leave it as a
challenge problem
for you to show that
it's uncountable.
Let us know your
solutions in the comments.
Here's why that's
important for us.
There are only countably
many rational numbers.
That means there are more
numbers on the list than there
are rational numbers.
So only countably many pairs of
numbers have a rational number.
That means, for most of
the infinitely many pairs
of numbers on the list,
both numbers are irrational.
So if pi and e differ
infinitely often,
most of the numbers on
our list are irrational.
But are any of them rational?
Mathematician Terry Tao had an
answer, sort of, for that part.
There is a conjecture, which is
basically an unproven educated
guess, that both e and
pi are normal numbers.
A normal number is one where
any digit or sequence of digits
is equally likely to occur.
So as we look through
pi, the number 3
occurs just as often as 4.
And the sequence 1, 2, 3 occurs
just as often is the sequence
8, 8, 8.
In other words, the digits of
a normal number are balanced.
A crazy fact about
normal numbers
is that we know almost every
number is normal, but have
very few concrete examples.
In particular, no one has proved
pi and e are normal numbers,
even though it's
believed to be the case.
Tao pointed out that if pi
and e are normal numbers,
as we suspect, then it's
impossible to produce
a rational number by
exchanging their digits.
If you want the
technical argument,
I recommend reading the original
post linked in the description.
But here's the intuition--
if you did produce
a rational number
by exchanging the digits
of pi and e, that number
would have a repeating part
of its decimal expansion,
like this.
But normal numbers don't
like to repeat themselves.
And blending two
of them together
doesn't change that fact.
So on the list of all
possible numbers generated
by exchanging infinitely
many digits of pi and e,
are there rational numbers?
Probably not.
There's three things I
want to highlight about why
this question is so cool.
1, it's about e and pi.
2, we just saw two
very different answers,
one coming from set theory and
one coming from number theory.
3, and this is my
favorite part, this post
is a small and
accessible preview
into the way math
research really happens.
It's a curious question
with different perspectives
and no known correct answer.
We've reduced a
complicated problem
to other complicated
problems, ones
about whether pi and
e are normal numbers,
or whether their digits
differ infinitely often.
And we're playing with
mathematical intuition.
I recommend checking
out the original post
in the description for
some of the subtle details
that I omitted.
And don't forget to answer
the challenge in the comments.
Also, if anyone can prove
that e and pi differ
at infinitely many
digits, that'd
be a great fact to
know for certain.
See you next time on
"Infinite Series."
Hello, I just want
to respond to some
of the comments
about our episode
on the Wolverine
problem and solving
it using graph colorings.
So first, let's get to
some important stuff--
my T-shirt.
A lot of you had a lot of
opinions about my shirt.
It said, let epsilon
be less than 0.
So people usually use
the variable epsilon
to symbolize a really
small positive number.
It's most commonly
used in calculus,
but also shows up in
other areas of math.
So let epsilon be less than 0
is just kind of a goofy joke,
maybe funny.
All right, on to more
important things.
[INAUDIBLE] has asked, for
the Hadwiger-Nelson problem,
the colored regions don't have
to be the same shape, right?
Correct.
They can be any shape.
And in fact, if you
place restrictions
on the kind of shapes that
the colored regions can be,
or what color each of
those regions can be,
it changes the problem.
It changes the bound on
the chromatic number.
And if you check out
the Wikipedia article
for Hadwiger-Nelson
problem, it actually
tells you a lot
of those results,
and they're pretty cool.
So go check it out.
Finally, Brian Willett
gave a great answer
to our challenge question.
Why is the lower bound on the
chromatic number of the plane
4?
Here's his answer.
You can bend the graph,
we showed, in the plane
such that all edges had
length one centimeter.
Any coloring of the plane must
have a coloring number bigger
than the one at the graph.
So to show that the
chromatic number of the graph
must be bigger than
3, we need to note
that it takes at
least two colors
to color the outer hexagon.
And then this center vertex
must be a third color.
The three vertices of
the central triangle
must be three different
colors, but they are all
adjacent to the vertices of
the hexagon of the same color.
So these can't be the same
three colors from before.
And so we need at
least four colors.
An example of a
four-coloring is shown.
Good job, Brian.
We'll see you next week.
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